Introduction
Consider the equation
∂ t u + b ( t , x ) ⋅ ∇ u + L t u = f , (1) \partial_t u+b(t,x)\cdot\nabla u+\mathcal{L}_t u=f, \tag*{(1)} ∂ t u + b ( t , x ) ⋅ ∇ u + L t u = f , (1) where b b b L t \mathcal{L}_t L t α = 2 s ∈ [ 1 , 2 ) \alpha=2s\in[1,2) α = 2 s ∈ [ 1 , 2 ) α = 1 \alpha=1 α = 1
In the absence of drift, Felsinger and Kassmann [10 ] proved a weak Harnack inequality and Hölder estimates under quadratic-form assumptions. They also discussed first-order terms for α > 1 \alpha>1 α > 1 [14 ] obtained Harnack estimates that allow negative values outside the cylinder. Kassmann and Weidner [11 ] proved a full parabolic Harnack inequality using time-integrated tails and distinguished the roles of local coercivity and upper jumping smoothness. In the probabilistic setting, Bass and Levin [1 ] proved a Harnack inequality for jump processes. Chen, Kumagai, and Wang [5 ] identified upper jumping smoothness as a necessary condition for the parabolic Harnack inequality for autonomous symmetric jump forms.
For fractional diffusion with drift, Caffarelli and Vasseur [4 ] proved Hölder regularity at the critical order in their study of the quasi-geostrophic equation. Chen and Zhang [6 ] obtained Harnack estimates for the fractional Laplacian with critical and supercritical divergence-free drifts. Liao [12 ] proved weak and full Harnack inequalities and Hölder estimates for pointwise comparable kernels and divergence-free drifts in scale-invariant classes.
The definition of a local solution there requires a locally square-integrable spatial gradient [12 Definition 1.1] . Ciani and Nakamura [7 ] derived regularity estimates for functions in nonlocal parabolic De Giorgi classes.
For bounded incompressible drift, interpolation gives a continuous transport form on H 1 / 2 × H 1 / 2 H^{1/2} \times H^{1/2} H 1/2 × H 1/2 s = 1 / 2 s = 1/2 s = 1/2 § 2 . The energy identity then gives strong convergence of the viscous approximations, while a fractional Nash inequality gives positive-time smoothing (§ 3 ).
For the local estimates, we add the time integral of the negative exterior tail and the negative source bound to the solution before taking powers. The derivative of this shift absorbs both terms in the energy inequality. The localized energy bounds in § 4 and a differential inequality for the weighted mean of the logarithm give the weak Harnack inequality in § 5 . In § 6 we use it to reduce the oscillation, estimating the tail on successive annuli. For the full Harnack inequality, upper jumping smoothness bounds the earlier positive tail in terms of the later infimum, the source, and an arbitrarily small multiple of a supremum on a larger past cylinder. Iteration on nested cylinders absorbs the supremum (§ 7 ).
The examples in § 8 distinguish the kernel assumptions and illustrate the time dependence of the exterior tail. In Example 8.1 the kernel is supported on a union of separated dyadic annuli. Each missing jump splits into two allowed jumps inside the same ball, giving local coercivity. This kernel also satisfies upper jumping smoothness, even after multiplication by a symmetric measurable factor bounded above and bounded away from zero. In Example 8.3, concentrated jumps preserve the form and jump-intensity bounds but violate upper jumping smoothness.
Kernel assumptions and notation
For a ball D ⊂ R n D \subset\mathbb{R}^{n} D ⊂ R n
[ v ] s , D 2 = ∬ D × D ∣ v ( x ) − v ( y ) ∣ 2 ∣ x − y ∣ n + 2 s d x d y , [ v ] s = [ v ] s , R n . [v]_{s,D}^{2} = \iint_{D \times D} \frac{\lvert v(x) - v(y)\rvert^{2}}{\lvert x-y\rvert^{n+2s}}\,\mathrm{d}x\,\mathrm{d}y, \qquad[v]_{s} = [v]_{s,\mathbb{R}^{n}}. [ v ] s , D 2 = ∬ D × D ∣ x − y ∣ n + 2 s ∣ v ( x ) − v ( y ) ∣ 2 d x d y , [ v ] s = [ v ] s , R n . The norm on H s ( R n ) H^{s}(\mathbb{R}^{n}) H s ( R n ) ( ∥ v ∥ 2 2 + [ v ] s 2 ) 1 / 2 (\lVert v\rVert_{2}^{2} + [v]_{s}^{2})^{1/2} (∥ v ∥ 2 2 + [ v ] s 2 ) 1/2
E t , D ( v , w ) = 1 2 ∬ D × D ( v ( x ) − v ( y ) ) ( w ( x ) − w ( y ) ) K ( t ; x , y ) d x d y , E t = E t , R n . (2) \mathcal{E}_{t,D}(v,w) = \frac{1}{2}\iint_{D \times D} (v(x)-v(y))(w(x)-w(y))K(t;x,y)\,\mathrm{d}x\,\mathrm{d}y, \qquad\mathcal{E}_{t} = \mathcal{E}_{t,\mathbb{R}^{n}}. \tag*{(2)} E t , D ( v , w ) = 2 1 ∬ D × D ( v ( x ) − v ( y )) ( w ( x ) − w ( y )) K ( t ; x , y ) d x d y , E t = E t , R n . (2) The kernel K K K x , y x,y x , y 0 < λ ≤ Λ 0 < \lambda\leq\Lambda 0 < λ ≤ Λ
E t , D ( v , v ) ≥ λ [ v ] s , D 2 for every ball D and v ∈ H s ( D ) , (3) \mathcal{E}_{t,D}(v,v) \geq\lambda[v]_{s,D}^{2} \qquad\text{for every ball } D \text{ and } v \in H^{s}(D), \tag*{(3)} E t , D ( v , v ) ≥ λ [ v ] s , D 2 for every ball D and v ∈ H s ( D ) , (3) E t ( v , v ) ≤ Λ [ v ] s 2 for every v ∈ H s ( R n ) , (4) \mathcal{E}_{t}(v,v) \leq\Lambda[v]_{s}^{2} \qquad\text{for every } v \in H^{s}(\mathbb{R}^{n}), \tag*{(4)} E t ( v , v ) ≤ Λ [ v ] s 2 for every v ∈ H s ( R n ) , (4) e s s s u p x ∫ ∣ x − y ∣ ≥ r K ( t ; x , y ) d y ≤ Λ r − α for every r > 0. (5) \mathop{\mathrm{ess\,sup}}_{x}\int_{\lvert x-y\rvert\geq r} K(t;x,y)\,\mathrm{d}y \leq\Lambda r^{-\alpha} \qquad\text{for every } r > 0. \tag*{(5)} ess sup x ∫ ∣ x − y ∣ ≥ r K ( t ; x , y ) d y ≤ Λ r − α for every r > 0. (5) In (1.5), it suffices to take positive rational radii on a common set of spatial points of full measure. Monotone convergence then gives the estimate at every radius. We assume
b ∈ L ∞ ( ( 0 , T ) × R n ; R n ) , div x b = 0 in distributions . (6) b \in L^{\infty}((0,T) \times\mathbb{R}^{n};\mathbb{R}^{n}), \qquad\operatorname{div}_{x} b = 0 \quad\text{in distributions}. \tag*{(6)} b ∈ L ∞ (( 0 , T ) × R n ; R n ) , div x b = 0 in distributions . (6) All spatial suprema and infima of measurable functions are essential extrema.
For z 0 = ( t 0 , x 0 ) z_{0} = (t_{0},x_{0}) z 0 = ( t 0 , x 0 ) C R C_{R} C R P R P_{R} P R F R F_{R} F R H R H_{R} H R
C R ( z 0 ) = ( t 0 − 4 R α , t 0 + R α ) × B 4 R ( x 0 ) , P R ( z 0 ) = ( t 0 − 3 R α , t 0 − R α ) × B 2 R ( x 0 ) , F R ( z 0 ) = ( t 0 , t 0 + R α ) × B R ( x 0 ) , H R ( z 0 ) = ( t 0 − 5 2 R α , t 0 − 2 R α ) × B R ( x 0 ) . (7) \begin{aligned}
C_{R}(z_{0}) &= (t_{0}-4R^{\alpha},t_{0}+R^{\alpha}) \times B_{4R}(x_{0}),\\
P_{R}(z_{0}) &= (t_{0}-3R^{\alpha},t_{0}-R^{\alpha}) \times B_{2R}(x_{0}),\\
F_{R}(z_{0}) &= (t_{0},t_{0}+R^{\alpha}) \times B_{R}(x_{0}),\\
H_{R}(z_{0}) &= (t_{0}-\tfrac{5}{2}R^{\alpha},t_{0}-2R^{\alpha}) \times B_{R}(x_{0}). \tag*{(7)}
\end{aligned} C R ( z 0 ) P R ( z 0 ) F R ( z 0 ) H R ( z 0 ) = ( t 0 − 4 R α , t 0 + R α ) × B 4 R ( x 0 ) , = ( t 0 − 3 R α , t 0 − R α ) × B 2 R ( x 0 ) , = ( t 0 , t 0 + R α ) × B R ( x 0 ) , = ( t 0 − 2 5 R α , t 0 − 2 R α ) × B R ( x 0 ) . (7) We take cylinders whose closures lie in the solution domain, except possibly for the terminal time face. The rescaled drift has norm
M R = R α − 1 ∥ b ∥ L ∞ ( C R ) . (8) M_R = R^{\alpha-1}\|b\|_{L^\infty(C_R)}. \tag*{(8)} M R = R α − 1 ∥ b ∥ L ∞ ( C R ) . (8) Structural constants depend only on n n n s s s λ \lambda λ Λ \Lambda Λ C R C_R C R M R M_R M R Q R Q_R Q R R α − 1 ∥ b ∥ L ∞ ( Q R ) R^{\alpha-1}\|b\|_{L^\infty(Q_R)} R α − 1 ∥ b ∥ L ∞ ( Q R )
For concentric balls with r < ρ r<\rho r < ρ
Θ ( v ; t , r , ρ ) = ess sup x ∈ B r ( x 0 ) ∫ R n ∖ B ρ ( x 0 ) ∣ v ( t , y ) ∣ K ( t ; x , y ) d y . (9) \Theta(v;t,r,\rho)=\operatorname*{ess\,sup}_{x\in B_r(x_0)}\int_{\mathbb{R}^n\setminus B_\rho(x_0)}|v(t,y)|K(t;x,y)\,dy. \tag*{(9)} Θ ( v ; t , r , ρ ) = x ∈ B r ( x 0 ) ess sup ∫ R n ∖ B ρ ( x 0 ) ∣ v ( t , y ) ∣ K ( t ; x , y ) d y . (9) Here and below we omit the common center when it is clear. The weak Harnack inequality involves the negative tail
T R − ( u ) = ∫ t 0 − 4 R α t 0 + R α Θ ( u − ; t , 3 R , 4 R ) d t . (10) T_R^-(u)=\int_{t_0-4R^\alpha}^{t_0+R^\alpha}\Theta(u_-;t,3R,4R)\,dt. \tag*{(10)} T R − ( u ) = ∫ t 0 − 4 R α t 0 + R α Θ ( u − ; t , 3 R , 4 R ) d t . (10) Under parabolic rescaling, T R − ( u ) T_R^-(u) T R − ( u ) u u u v v v (5) gives
Θ ( v ; t , r , ρ ) ≤ Λ ( ρ − r ) − α ∥ v ( t ) ∥ ∞ . (11) \Theta(v;t,r,\rho)\leq\Lambda(\rho-r)^{-\alpha}\|v(t)\|_\infty. \tag*{(11)} Θ ( v ; t , r , ρ ) ≤ Λ ( ρ − r ) − α ∥ v ( t ) ∥ ∞ . (11) Principal results
Write
X T = L ∞ ( 0 , T ; L 2 ( R n ) ) ∩ L 2 ( 0 , T ; H s ( R n ) ) . \mathcal{X}_T=L^\infty(0,T;L^2(\mathbb{R}^n))\cap L^2(0,T;H^s(\mathbb{R}^n)). X T = L ∞ ( 0 , T ; L 2 ( R n )) ∩ L 2 ( 0 , T ; H s ( R n )) . For u 0 ∈ L 2 u_0\in L^2 u 0 ∈ L 2 f ∈ L 2 ( 0 , T ; H − s ( R n ) ) f\in L^2(0,T;H^{-s}(\mathbb{R}^n)) f ∈ L 2 ( 0 , T ; H − s ( R n )) u ∈ X T u\in\mathcal{X}_T u ∈ X T
− ∫ 0 T ∫ u ∂ t ϕ d x d t − ∫ 0 T ∫ u b ⋅ ∇ ϕ d x d t + ∫ 0 T E t ( u , ϕ ) d t = ∫ 0 T ⟨ f , ϕ ⟩ d t + ∫ u 0 ϕ ( 0 ) d x (12) -\int_0^T\int u\,\partial_t\phi\,dx\,dt-\int_0^T\int u\,b\cdot\nabla\phi\,dx\,dt+\int_0^T\mathcal{E}_t(u,\phi)\,dt
=\int_0^T\langle f,\phi\rangle\,dt+\int u_0\phi(0)\,dx \tag*{(12)} − ∫ 0 T ∫ u ∂ t ϕ d x d t − ∫ 0 T ∫ u b ⋅ ∇ ϕ d x d t + ∫ 0 T E t ( u , ϕ ) d t = ∫ 0 T ⟨ f , ϕ ⟩ d t + ∫ u 0 ϕ ( 0 ) d x (12) for every ϕ ∈ C c ∞ ( [ 0 , T ) × R n ) \phi\in C^\infty_c([0,T)\times\mathbb{R}^n) ϕ ∈ C c ∞ ([ 0 , T ) × R n ) Definition 4.2 specifies the local variational inequalities. For distributional energy solutions the equation implies the local time regularity required there.
Theorem 1.1 (Energy solutions and approximation). Under (1.3)–(1.6), for every u 0 ∈ L 2 u_0\in L^2 u 0 ∈ L 2 f ∈ L 2 ( 0 , T ; H − s ( R n ) ) f\in L^2(0,T;H^{-s}(\mathbb{R}^n)) f ∈ L 2 ( 0 , T ; H − s ( R n )) u ∈ X T u\in\mathcal{X}_T u ∈ X T u 0 u_0 u 0
L 2 ( 0 , T ; H s ( R n ) ) ∩ H 1 ( 0 , T ; H − s ( R n ) ) ∩ C ( [ 0 , T ] ; L 2 ) . L^2(0,T;H^s(\mathbb{R}^n))\cap H^1(0,T;H^{-s}(\mathbb{R}^n))\cap C([0,T];L^2). L 2 ( 0 , T ; H s ( R n )) ∩ H 1 ( 0 , T ; H − s ( R n )) ∩ C ([ 0 , T ] ; L 2 ) . It satisfies the energy identity (2.6) and the renormalized testing formulas of § 2 . If f f f u ε u_\varepsilon u ε
∂ t u ε + b ⋅ ∇ u ε + L t u ε − ε Δ u ε = f , u ε ( 0 ) = u 0 , \partial_tu_\varepsilon+b\cdot\nabla u_\varepsilon+\mathcal{L}_tu_\varepsilon-\varepsilon\Delta u_\varepsilon=f,\qquad u_\varepsilon(0)=u_0, ∂ t u ε + b ⋅ ∇ u ε + L t u ε − ε Δ u ε = f , u ε ( 0 ) = u 0 , satisfy, as ε ↓ 0 \varepsilon\downarrow0 ε ↓ 0
u ε ⟶ u in C ( [ 0 , T ] ; L 2 ) ∩ L 2 ( 0 , T ; H s ( R n ) ) , ε ∫ 0 T ∥ ∇ u ε ∥ 2 d t ⟶ 0. (13) u_\varepsilon\longrightarrow u\quad\text{in }C([0,T];L^2)\cap L^2(0,T;H^s(\mathbb{R}^n)),\qquad\varepsilon\int_0^T\|\nabla u_\varepsilon\|^2\,dt\longrightarrow0. \tag*{(13)} u ε ⟶ u in C ([ 0 , T ] ; L 2 ) ∩ L 2 ( 0 , T ; H s ( R n )) , ε ∫ 0 T ∥∇ u ε ∥ 2 d t ⟶ 0. (13) Theorem 1.2 (Weak Harnack inequality). Let u u u C R ( z 0 ) C_R(z_0) C R ( z 0 ) f ∈ L ∞ ( C R ) f\in L^\infty(C_R) f ∈ L ∞ ( C R ) u ≥ 0 u\geq0 u ≥ 0 T R − ( u ) < ∞ T_R^-(u)<\infty T R − ( u ) < ∞
∫ P R ( z 0 ) u d x d t ≤ C ( ess inf F R ( z 0 ) u + R α ∥ f − ∥ L ∞ ( C R ) + T R − ( u ) ) . (14) \int_{P_R(z_0)}u\,dx\,dt\leq C\left(\operatorname*{ess\,inf}_{F_R(z_0)}u+R^\alpha\|f_-\|_{L^\infty(C_R)}+T_R^-(u)\right). \tag*{(14)} ∫ P R ( z 0 ) u d x d t ≤ C ( F R ( z 0 ) ess inf u + R α ∥ f − ∥ L ∞ ( C R ) + T R − ( u ) ) . (14) Here C C C u u u T R − ( u ) = 0 T_R^-(u)=0 T R − ( u ) = 0
For Hölder estimates, set Q R ( t ∗ , x ∗ ) = ( t ∗ − R α , t ∗ ) × B R ( x ∗ ) Q_R(t_*,x_*)=(t_*-R^\alpha,t_*)\times B_R(x_*) Q R ( t ∗ , x ∗ ) = ( t ∗ − R α , t ∗ ) × B R ( x ∗ )
d α ( ( t , x ) , ( τ , y ) ) = ∣ x − y ∣ + ∣ t − τ ∣ 1 / α . d_\alpha((t,x),(\tau,y))=|x-y|+|t-\tau|^{1/\alpha}. d α (( t , x ) , ( τ , y )) = ∣ x − y ∣ + ∣ t − τ ∣ 1/ α . Theorem 1.3 (Hölder regularity). Let u ∈ X T u\in\mathcal{X}_T u ∈ X T Q R Q_R Q R U = ∥ u ∥ L ∞ ( Q R ) U=\|u\|_{L^\infty(Q_R)} U = ∥ u ∥ L ∞ ( Q R ) f ∈ L ∞ ( Q R ) f\in L^\infty(Q_R) f ∈ L ∞ ( Q R ) q > 1 q>1 q > 1
A q = R α ( ∫ t ∗ − R α t ∗ Θ ( u ; t , R / 2 , R ) q d t ) 1 / q < ∞ . A_q=R^\alpha\left(\int_{t_*-R^\alpha}^{t_*}\Theta(u;t,R/2,R)^q\,dt\right)^{1/q}<\infty. A q = R α ( ∫ t ∗ − R α t ∗ Θ ( u ; t , R /2 , R ) q d t ) 1/ q < ∞. There are γ ∈ ( 0 , 1 ) \gamma\in(0,1) γ ∈ ( 0 , 1 ) C C C n n n s s s λ , Λ \lambda,\Lambda λ , Λ q q q R α − 1 ∥ b ∥ L ∞ ( Q R ) R^{\alpha-1}\|b\|_{L^\infty(Q_R)} R α − 1 ∥ b ∥ L ∞ ( Q R ) u u u
∣ u ( z ) − u ( ζ ) ∣ ≤ C ( U + A q + R α ∥ f ∥ ∞ ) ( d α ( z , ζ ) R ) γ , z , ζ ∈ Q R / 4 . (15) |u(z)-u(\zeta)|\le C\left(U+A_q+R^\alpha\|f\|_\infty\right)\left(\frac{d_\alpha(z,\zeta)}{R}\right)^\gamma,\qquad z,\zeta\in Q_{R/4}. \tag*{(15)} ∣ u ( z ) − u ( ζ ) ∣ ≤ C ( U + A q + R α ∥ f ∥ ∞ ) ( R d α ( z , ζ ) ) γ , z , ζ ∈ Q R /4 . (15) If u u u L ∞ L^\infty L ∞ A q A_q A q C C C f = 0 f=0 f = 0 u 0 ∈ L 2 u_0\in L^2 u 0 ∈ L 2
Definition 1.4 (Upper jumping smoothness). The kernel has upper jumping smoothness with constant C J C_J C J
K(t;x,y)\le C_J\fint_{B_r(x)}K(t;z,y)\,dz,\qquad0<r\le|x-y|/4,
for almost every t , x , y t,x,y t , x , y
Theorem 1.5 (Parabolic Harnack inequality). Assume also (1.16). Let u ∈ X T u\in\mathcal{X}_T u ∈ X T C R ( z 0 ) C_R(z_0) C R ( z 0 ) f ∈ L ∞ ( C R ) f\in L^\infty(C_R) f ∈ L ∞ ( C R ) u ≥ 0 u\ge0 u ≥ 0
ess sup H R ( z 0 ) u ≤ C ( ess inf F R ( z 0 ) u + R α ∥ f ∥ L ∞ ( C R ) ) . (16) \operatorname*{ess\,sup}_{H_R(z_0)}u\le C\left(\operatorname*{ess\,inf}_{F_R(z_0)}u+R^\alpha\|f\|_{L^\infty(C_R)}\right). \tag*{(16)} H R ( z 0 ) ess sup u ≤ C ( F R ( z 0 ) ess inf u + R α ∥ f ∥ L ∞ ( C R ) ) . (16) The constant C C C C J C_J C J
At the critical order α = 1 \alpha=1 α = 1 R α − 1 R^{\alpha-1} R α − 1 α > 1 \alpha>1 α > 1 R ↓ 0 R\downarrow0 R ↓ 0 M R M_R M R
Energy solutions and nonlinear testing
Energy forms and the transport operator
Applying (1.3) on balls increasing to R n \mathbb{R}^n R n
λ [ v ] s 2 ≤ E t ( v , v ) ≤ Λ [ v ] s 2 , ∣ E t ( v , w ) ∣ ≤ Λ [ v ] s [ w ] s . (17) \lambda[v]_s^2\le\mathcal{E}_t(v,v)\le\Lambda[v]_s^2,\qquad|\mathcal{E}_t(v,w)|\le\Lambda[v]_s[w]_s. \tag*{(17)} λ [ v ] s 2 ≤ E t ( v , v ) ≤ Λ [ v ] s 2 , ∣ E t ( v , w ) ∣ ≤ Λ [ v ] s [ w ] s . (17) The last inequality follows by Cauchy–Schwarz in L 2 ( K ( t ; x , y ) d x d y ) L^2(K(t;x,y)\,dx\,dy) L 2 ( K ( t ; x , y ) d x d y ) ⟨ L t v , w ⟩ = E t ( v , w ) \langle L_tv,w\rangle=\mathcal{E}_t(v,w) ⟨ L t v , w ⟩ = E t ( v , w ) H s ( R n ) H^s(\mathbb{R}^n) H s ( R n ) H − s ( R n ) H^{-s}(\mathbb{R}^n) H − s ( R n )
Lemma 2.1 (Half-order transport estimate). For b ∈ L ∞ ( R n ; R n ) b\in L^\infty(\mathbb{R}^n;\mathbb{R}^n) b ∈ L ∞ ( R n ; R n ) div b = 0 \operatorname{div}b=0 div b = 0
B b ( v , w ) = − ∫ v b ⋅ ∇ w d x , v , w ∈ C c ∞ ( R n ) , \mathcal{B}_b(v,w)=-\int v b\cdot\nabla w\,dx,\qquad v,w\in C_c^\infty(\mathbb{R}^n), B b ( v , w ) = − ∫ v b ⋅ ∇ w d x , v , w ∈ C c ∞ ( R n ) , extends continuously to H 1 − θ × H θ H^{1-\theta}\times H^\theta H 1 − θ × H θ 0 ≤ θ ≤ 1 0\le\theta\le1 0 ≤ θ ≤ 1
∣ B b ( v , w ) ∣ ≤ C ∥ b ∥ ∞ ∥ v ∥ H 1 − θ ∥ w ∥ H θ . (18) |\mathcal{B}_b(v,w)|\le C\|b\|_\infty\|v\|_{H^{1-\theta}}\|w\|_{H^\theta}. \tag*{(18)} ∣ B b ( v , w ) ∣ ≤ C ∥ b ∥ ∞ ∥ v ∥ H 1 − θ ∥ w ∥ H θ . (18) In particular, it is bounded and skew-symmetric on H 1 / 2 × H 1 / 2 H^{1/2}\times H^{1/2} H 1/2 × H 1/2
Proof. Integration by parts yields
B b ( v , w ) = ∫ w b ⋅ ∇ v d x = − B b ( w , v ) . \mathcal{B}_{b}(v,w)=\int wb\cdot\nabla v\,dx=-\mathcal{B}_{b}(w,v). B b ( v , w ) = ∫ w b ⋅ ∇ v d x = − B b ( w , v ) . This gives the two estimates
∣ B b ( v , w ) ∣ ≤ ∥ b ∥ ∞ ∥ v ∥ H 1 ∥ w ∥ 2 , ∣ B b ( v , w ) ∣ ≤ ∥ b ∥ ∞ ∥ v ∥ 2 ∥ w ∥ H 1 . |\mathcal{B}_{b}(v,w)|\leq\|b\|_{\infty}\|v\|_{H^{1}}\|w\|_{2},\qquad|\mathcal{B}_{b}(v,w)|\leq\|b\|_{\infty}\|v\|_{2}\|w\|_{H^{1}}. ∣ B b ( v , w ) ∣ ≤ ∥ b ∥ ∞ ∥ v ∥ H 1 ∥ w ∥ 2 , ∣ B b ( v , w ) ∣ ≤ ∥ b ∥ ∞ ∥ v ∥ 2 ∥ w ∥ H 1 . The two realizations of b ⋅ ∇ b\cdot\nabla b ⋅ ∇ H 1 H^{1} H 1 L 2 L^{2} L 2 L 2 L^{2} L 2 H − 1 H^{-1} H − 1 H 1 H^{1} H 1 C ∞ C^{\infty} C ∞ (18) , since [ H a , H c ] θ = H ( 1 − θ ) a + θ c [H^{a},H^{c}]_{\theta}=H^{(1-\theta)a+\theta c} [ H a , H c ] θ = H ( 1 − θ ) a + θ c [2 ] . Skew-symmetry extends to H 1 / 2 × H 1 / 2 H^{1/2}\times H^{1/2} H 1/2 × H 1/2
Proposition 2.2 (The threshold for the energy pairing). Let 0 ≤ σ < 1 / 2 0\leq\sigma<1/2 0 ≤ σ < 1/2 b = e 1 b=e_{1} b = e 1 H σ ( R n ) × H σ ( R n ) H^{\sigma}(\mathbb{R}^{n})\times H^{\sigma}(\mathbb{R}^{n}) H σ ( R n ) × H σ ( R n )
Proof. Choose a nonzero real-valued χ ∈ C c ∞ ( R n ) \chi\in C^{\infty}_{c}(\mathbb{R}^{n}) χ ∈ C c ∞ ( R n )
v N ( x ) = χ ( x ) cos ( N x 1 ) , w N ( x ) = χ ( x ) sin ( N x 1 ) . v_{N}(x)=\chi(x)\cos(Nx_{1}),\qquad w_{N}(x)=\chi(x)\sin(Nx_{1}). v N ( x ) = χ ( x ) cos ( N x 1 ) , w N ( x ) = χ ( x ) sin ( N x 1 ) . Direct differentiation and one integration by parts give
B e 1 ( v N , w N ) = − 1 4 ∫ ∂ 1 ( χ 2 ) sin ( 2 N x 1 ) d x − N 2 ∫ χ 2 ( 1 + cos ( 2 N x 1 ) ) d x = − N 2 ∥ χ ∥ 2 2 . \begin{aligned}
\mathcal{B}_{e_{1}}(v_{N},w_{N})=-\frac{1}{4}\int\partial_{1}(\chi^{2})\sin(2Nx_{1})\,dx-\frac{N}{2}\int\chi^{2}(1+\cos(2Nx_{1}))\,dx \\
&=-\frac{N}{2}\|\chi\|_{2}^{2}.
\end{aligned} B e 1 ( v N , w N ) = − 4 1 ∫ ∂ 1 ( χ 2 ) sin ( 2 N x 1 ) d x − 2 N ∫ χ 2 ( 1 + cos ( 2 N x 1 )) d x = − 2 N ∥ χ ∥ 2 2 . The Fourier transforms of these functions are linear combinations of translates of χ ^ \widehat{\chi} χ ± N e 1 \pm Ne_{1} ± N e 1
( 1 + ∣ ξ ± N e 1 ∣ 2 ) σ ≤ C σ ( 1 + N 2 ) σ ( 1 + ∣ ξ ∣ 2 ) σ (1+|\xi\pm Ne_{1}|^{2})^{\sigma}\leq C_{\sigma}(1+N^{2})^{\sigma}(1+|\xi|^{2})^{\sigma} ( 1 + ∣ ξ ± N e 1 ∣ 2 ) σ ≤ C σ ( 1 + N 2 ) σ ( 1 + ∣ ξ ∣ 2 ) σ therefore implies ∥ v N ∥ H σ + ∥ w N ∥ H σ ≤ C ( 1 + N ) σ \|v_{N}\|_{H^{\sigma}}+\|w_{N}\|_{H^{\sigma}}\leq C(1+N)^{\sigma} ∥ v N ∥ H σ + ∥ w N ∥ H σ ≤ C ( 1 + N ) σ N ≤ C ( 1 + N ) 2 σ N\leq C(1+N)^{2\sigma} N ≤ C ( 1 + N ) 2 σ N N N
Set V = H s ( R n ) V=H^{s}(\mathbb{R}^{n}) V = H s ( R n ) H = L 2 H=L^{2} H = L 2 V ′ = H − s ( R n ) V'=H^{-s}(\mathbb{R}^{n}) V ′ = H − s ( R n ) s ≥ 1 / 2 s\geq1/2 s ≥ 1/2 V ↪ H 1 / 2 V\hookrightarrow H^{1/2} V ↪ H 1/2
∣ B b ( t ) ( v , w ) ∣ ≤ C ∥ b ∥ ∞ ∥ v ∥ V ∥ w ∥ V , B b ( t ) ( v , v ) = 0. (19) |\mathcal{B}_{b(t)}(v,w)|\leq C\|b\|_{\infty}\|v\|_{V}\|w\|_{V},\qquad\mathcal{B}_{b(t)}(v,v)=0. \tag*{(19)} ∣ B b ( t ) ( v , w ) ∣ ≤ C ∥ b ∥ ∞ ∥ v ∥ V ∥ w ∥ V , B b ( t ) ( v , v ) = 0. (19) The time-dependent form a t ( v , w ) = E t ( v , w ) + B b ( t ) ( v , w ) a_{t}(v,w)=\mathcal{E}_{t}(v,w)+\mathcal{B}_{b(t)}(v,w) a t ( v , w ) = E t ( v , w ) + B b ( t ) ( v , w )
a t ( v , v ) + ∥ v ∥ 2 ≥ min ( 1 , λ ) ∥ v ∥ V 2 . (20) a_{t}(v,v)+\|v\|^{2}\geq\min(1,\lambda)\|v\|_{V}^{2}. \tag*{(20)} a t ( v , v ) + ∥ v ∥ 2 ≥ min ( 1 , λ ) ∥ v ∥ V 2 . (20) Construction and uniqueness
Lemma 2.3 (Time traces and the quadratic chain rule). If v ∈ L 2 ( 0 , T ; V ) v\in L^{2}(0,T;V) v ∈ L 2 ( 0 , T ; V ) v ′ ∈ L 2 ( 0 , T ; V ′ ) v'\in L^{2}(0,T;V') v ′ ∈ L 2 ( 0 , T ; V ′ ) v v v C ( [ 0 , T ] ; H ) C([0,T];H) C ([ 0 , T ] ; H )
∥ v ( t ) ∥ 2 2 − ∥ v ( r ) ∥ 2 2 = 2 ∫ r t ⟨ v ′ , v ⟩ d τ . (21) \|v(t)\|_{2}^{2}-\|v(r)\|_{2}^{2}=2\int_{r}^{t}\langle v',v\rangle\,d\tau. \tag*{(21)} ∥ v ( t ) ∥ 2 2 − ∥ v ( r ) ∥ 2 2 = 2 ∫ r t ⟨ v ′ , v ⟩ d τ . (21) The corresponding product formula holds for two such functions.
Proof. Time mollification on a compact subinterval gives the identity for the regularized functions. The limits in the integral follow from convergence in L 2 V L^{2}V L 2 V L 2 V ′ L^{2}V' L 2 V ′ V V V H H H 2 ⟨ v ′ , v ⟩ 2\langle v',v\rangle 2 ⟨ v ′ , v ⟩ [13 ] .
Proposition 2.4 (The Cauchy problem). For the data in Theorem 1.1 , there is exactly one variational solution in L 2 V ∩ H 1 V ′ L^2V \cap H^1V' L 2 V ∩ H 1 V ′ t ∈ [ 0 , T ] t \in[0,T] t ∈ [ 0 , T ]
1 2 ∥ u ( t ) ∥ 2 2 + ∫ 0 t E τ ( u , u ) d τ = 1 2 ∥ u 0 ∥ 2 2 + ∫ 0 t ⟨ f , u ⟩ d τ . (22) \frac{1}{2}\lVert u(t)\rVert_2^2+\int_0^t \mathcal{E}_\tau(u,u)\,\mathrm{d}\tau=\frac{1}{2}\lVert u_0\rVert_2^2+\int_0^t\langle f,u\rangle\,\mathrm{d}\tau. \tag*{(22)} 2 1 ∥ u ( t ) ∥ 2 2 + ∫ 0 t E τ ( u , u ) d τ = 2 1 ∥ u 0 ∥ 2 2 + ∫ 0 t ⟨ f , u ⟩ d τ . (22) Every distributional solution in X T \mathcal{X}_T X T u 0 u_0 u 0
Proof. Choose a linearly independent sequence ( ϕ j ) (\phi_j) ( ϕ j ) C c ∞ ( R n ) C^\infty_c(\mathbb{R}^n) C c ∞ ( R n ) V V V H H H P N P_N P N H H H V N = span ( ϕ 1 , … , ϕ N ) V_N=\operatorname{span}(\phi_1,\ldots,\phi_N) V N = span ( ϕ 1 , … , ϕ N )
( u N ′ , w ) + a t ( u N , w ) = ⟨ f , w ⟩ ( w ∈ V N ) , u N ( 0 ) = P N u 0 , (u_N',w)+a_t(u_N,w)=\langle f,w\rangle\quad(w\in V_N),\qquad u_N(0)=P_Nu_0, ( u N ′ , w ) + a t ( u N , w ) = ⟨ f , w ⟩ ( w ∈ V N ) , u N ( 0 ) = P N u 0 , has an absolutely continuous solution because its coefficients are bounded and measurable and its forcing is square integrable. Testing with u N u_N u N
∥ u N ∥ L ∞ H 2 + ∥ u N ∥ L 2 V 2 ≤ C T ( ∥ u 0 ∥ 2 2 + ∥ f ∥ L 2 V ′ 2 ) . (23) \lVert u_N\rVert_{L^\infty H}^2+\lVert u_N\rVert_{L^2V}^2\le C_T\bigl(\lVert u_0\rVert_2^2+\lVert f\rVert_{L^2V'}^2\bigr). \tag*{(23)} ∥ u N ∥ L ∞ H 2 + ∥ u N ∥ L 2 V 2 ≤ C T ( ∥ u 0 ∥ 2 2 + ∥ f ∥ L 2 V ′ 2 ) . (23) The constant C T C_T C T N N N ∥ b ∥ ∞ \lVert b\rVert_\infty ∥ b ∥ ∞
Along a subsequence, u N ⇀ u u_N\rightharpoonup u u N ⇀ u L 2 V L^2V L 2 V u N ⇀ ∗ u u_N\overset{*}{\rightharpoonup}u u N ⇀ ∗ u L ∞ H L^\infty H L ∞ H χ ( t ) ϕ j \chi(t)\phi_j χ ( t ) ϕ j j j j
∂ t u + A t u = f in D ′ ( 0 , T ; V ′ ) , ⟨ A t v , w ⟩ = a t ( v , w ) . \partial_tu+A_tu=f\quad\text{in }D'(0,T;V'),\qquad\langle A_tv,w\rangle=a_t(v,w). ∂ t u + A t u = f in D ′ ( 0 , T ; V ′ ) , ⟨ A t v , w ⟩ = a t ( v , w ) . Boundedness of a t a_t a t u ′ = f − A t u ∈ L 2 V ′ u'=f-A_tu\in L^2V' u ′ = f − A t u ∈ L 2 V ′ H H H u ( 0 ) = u 0 u(0)=u_0 u ( 0 ) = u 0 u u u
If a function in X T \mathcal{X}_T X T A t A_t A t u ′ = f − A t u ∈ L 2 V ′ u'=f-A_tu\in L^2V' u ′ = f − A t u ∈ L 2 V ′ V V V
Renormalized transport and convex tests
Lemma 2.5 (Composition and localized chain rules). Let 0 < s < 1 0<s<1 0 < s < 1 β \beta β β ( 0 ) = 0 \beta(0)=0 β ( 0 ) = 0 H s H^s H s H s H^s H s H s H^s H s s ≥ 1 / 2 s\ge1/2 s ≥ 1/2 v ∈ H s ( R n ) v\in H^s(\mathbb{R}^n) v ∈ H s ( R n ) η ∈ W 1 , ∞ \eta\in W^{1,\infty} η ∈ W 1 , ∞ β \beta β β ( 0 ) \beta(0) β ( 0 )
B b ( v , β ( v ) η 2 ) = − ∫ Γ ( v ) b ⋅ ∇ ( η 2 ) d x , Γ ′ = β . (24) \mathcal{B}_b(v,\beta(v)\eta^2)=-\int\Gamma(v)b\cdot\nabla(\eta^2)\,\mathrm{d}x,\qquad\Gamma'=\beta. \tag*{(24)} B b ( v , β ( v ) η 2 ) = − ∫ Γ ( v ) b ⋅ ∇ ( η 2 ) d x , Γ ′ = β . (24) For a global test with β ( 0 ) = 0 \beta(0)=0 β ( 0 ) = 0 B b ( v , β ( v ) ) = 0 \mathcal{B}_b(v,\beta(v))=0 B b ( v , β ( v )) = 0
Proof. Let v j → v v_j\to v v j → v H s H^s H s v j → v v_j\to v v j → v ∑ j ∥ v j − v ∥ H s < ∞ \sum_j\lVert v_j-v\rVert_{H^s}<\infty ∑ j ∥ v j − v ∥ H s < ∞ v j ( x ) − v j ( y ) v_j(x)-v_j(y) v j ( x ) − v j ( y ) L 2 ( ∣ x − y ∣ − n − 2 s d x d y ) L^2(|x-y|^{-n-2s}\,\mathrm{d}x\,\mathrm{d}y) L 2 ( ∣ x − y ∣ − n − 2 s d x d y ) β \beta β β ( v j ) \beta(v_j) β ( v j ) ∥ β ( v j ) − β ( v ) ∥ 2 ≤ Lip ( β ) ∥ v j − v ∥ 2 \lVert\beta(v_j)-\beta(v)\rVert_2\le\operatorname{Lip}(\beta)\lVert v_j-v\rVert_2 ∥ β ( v j ) − β ( v ) ∥ 2 ≤ Lip ( β ) ∥ v j − v ∥ 2 H s H^s H s
η ( x ) v ( x ) − η ( y ) v ( y ) = η ( x ) ( v ( x ) − v ( y ) ) + v ( y ) ( η ( x ) − η ( y ) ) \eta(x)v(x)-\eta(y)v(y)=\eta(x)(v(x)-v(y))+v(y)(\eta(x)-\eta(y)) η ( x ) v ( x ) − η ( y ) v ( y ) = η ( x ) ( v ( x ) − v ( y )) + v ( y ) ( η ( x ) − η ( y )) and estimate the second summand using the Lipschitz bound on ∣ x − y ∣ < 1 |x-y|<1 ∣ x − y ∣ < 1 η \eta η ∣ x − y ∣ ≥ 1 |x-y|\ge1 ∣ x − y ∣ ≥ 1
Approximate v v v H s H^s H s v j v_j v j v j v_j v j div b = 0 \operatorname{div}b=0 div b = 0 β ( 0 ) \beta(0) β ( 0 ) L 1 L^1 L 1 Γ \Gamma Γ v j → v v_j\to v v j → v L 2 L^2 L 2
For a global test, the same approximation gives B b ( v j , β ( v j ) ) = 0 \mathcal{B}_b(v_j,\beta(v_j))=0 B b ( v j , β ( v j )) = 0
Lemma 2.6 (Chain rule with an absolutely continuous shift). Let J J J η ∈ W c 1 , ∞ ( D ) \eta\in W_c^{1,\infty}(D) η ∈ W c 1 , ∞ ( D ) u ∈ L 2 ( J ; H l o c s ( D ) ) u \in L^2(J;H^s_{\mathrm{loc}}(D)) u ∈ L 2 ( J ; H loc s ( D )) χ u ∈ L 2 ( J ; V ) ∩ H 1 ( J ; V ′ ) \chi u \in L^2(J;V) \cap H^1(J;V') χ u ∈ L 2 ( J ; V ) ∩ H 1 ( J ; V ′ ) χ \chi χ D D D h ∈ W 1 , 1 ( J ) h \in W^{1,1}(J) h ∈ W 1 , 1 ( J ) Γ ∈ C 1 ( R ) \Gamma\in C^1(\mathbb{R}) Γ ∈ C 1 ( R )
t ↦ ∫ Γ ( u ( t , x ) + h ( t ) ) η ( x ) 2 d x t \mapsto\int\Gamma(u(t,x)+h(t))\eta(x)^2\,dx t ↦ ∫ Γ ( u ( t , x ) + h ( t )) η ( x ) 2 d x is absolutely continuous, and
d d t ∫ Γ ( u + h ) η 2 d x = ⟨ ∂ t u , Γ ′ ( u + h ) η 2 ⟩ + h ′ ( t ) ∫ Γ ′ ( u + h ) η 2 d x (25) \frac{d}{dt}\int\Gamma(u+h)\eta^2\,dx = \langle\partial_t u,\Gamma'(u+h)\eta^2\rangle+ h'(t)\int\Gamma'(u+h)\eta^2\,dx \tag*{(25)} d t d ∫ Γ ( u + h ) η 2 d x = ⟨ ∂ t u , Γ ′ ( u + h ) η 2 ⟩ + h ′ ( t ) ∫ Γ ′ ( u + h ) η 2 d x (25) for almost every t ∈ J t \in J t ∈ J supp η \operatorname{supp}\eta supp η
Proof. Choose χ ∈ C c ∞ ( D ) \chi\in C^\infty_c(D) χ ∈ C c ∞ ( D ) supp η \operatorname{supp}\eta supp η U = χ u U=\chi u U = χ u U ∈ C ( J ; H ) U \in C(J;H) U ∈ C ( J ; H ) U U U J J J U j U_j U j
U j → U in L 2 ( J ; V ) ∩ C ( J ; H ) , U j ′ → U ′ in L 2 ( J ; V ′ ) . U_j \to U \text{ in } L^2(J;V)\cap C(J;H), \qquad U'_j \to U' \text{ in } L^2(J;V'). U j → U in L 2 ( J ; V ) ∩ C ( J ; H ) , U j ′ → U ′ in L 2 ( J ; V ′ ) . Choose smooth h j h_j h j h h h W 1 , 1 ( J ) W^{1,1}(J) W 1 , 1 ( J ) Γ ′ \Gamma' Γ ′ Γ \Gamma Γ
The functions Γ ′ ( U j + h j ) η 2 \Gamma'(U_j+h_j)\eta^2 Γ ′ ( U j + h j ) η 2 Γ ′ ( U + h ) η 2 \Gamma'(U+h)\eta^2 Γ ′ ( U + h ) η 2 L 2 ( J ; V ) L^2(J;V) L 2 ( J ; V ) C ( J ; L 2 ) C(J;L^2) C ( J ; L 2 )
sup t ∈ J ∣ ∫ ( Γ ′ ( U j + h j ) − Γ ′ ( U + h ) ) η 2 d x ∣ ≤ C ( ∥ U j − U ∥ C ( J ; H ) + ∥ h j − h ∥ C ( J ) ) → 0. \sup_{t\in J}\left|\int\bigl(\Gamma'(U_j+h_j)-\Gamma'(U+h)\bigr)\eta^2\,dx\right|
\leq C\left(\lVert U_j-U\rVert_{C(J;H)}+\lVert h_j-h\rVert_{C(J)}\right)\to0. t ∈ J sup ∫ ( Γ ′ ( U j + h j ) − Γ ′ ( U + h ) ) η 2 d x ≤ C ( ∥ U j − U ∥ C ( J ; H ) + ∥ h j − h ∥ C ( J ) ) → 0. The two terms on the right of the regularized identity therefore converge in L 1 ( J ) L^1(J) L 1 ( J ) V ′ , V V',V V ′ , V h j ′ → h ′ h'_j\to h' h j ′ → h ′ L 1 ( J ) L^1(J) L 1 ( J ) Γ ( U j + h j ) η 2 \Gamma(U_j+h_j)\eta^2 Γ ( U j + h j ) η 2 Γ ′ \Gamma' Γ ′ L 2 L^2 L 2
For a global function u ∈ L 2 ( J ; V ) ∩ H 1 ( J ; V ′ ) u \in L^2(J;V)\cap H^1(J;V') u ∈ L 2 ( J ; V ) ∩ H 1 ( J ; V ′ ) h = 0 h=0 h = 0
d d t ∫ R n Γ ( u ) d x = ⟨ u ′ , Γ ′ ( u ) ⟩ (26) \frac{d}{dt}\int_{\mathbb{R}^n}\Gamma(u)\,dx = \langle u',\Gamma'(u)\rangle \tag*{(26)} d t d ∫ R n Γ ( u ) d x = ⟨ u ′ , Γ ′ ( u )⟩ (26) whenever Γ ( 0 ) = Γ ′ ( 0 ) = 0 \Gamma(0)=\Gamma'(0)=0 Γ ( 0 ) = Γ ′ ( 0 ) = 0 Γ ′ \Gamma' Γ ′ ∣ Γ ( u ) ∣ ≤ C ∣ u ∣ 2 |\Gamma(u)|\leq C|u|^2 ∣Γ ( u ) ∣ ≤ C ∣ u ∣ 2 ∣ Γ ′ ( u ) ∣ ≤ C ∣ u ∣ |\Gamma'(u)|\leq C|u| ∣ Γ ′ ( u ) ∣ ≤ C ∣ u ∣ C ( J ; H ) C(J;H) C ( J ; H ) Γ \Gamma Γ
If u + h ≥ ε > 0 u+h\geq\varepsilon>0 u + h ≥ ε > 0 Γ ′ \Gamma' Γ ′ [ ε , ∞ ) [\varepsilon,\infty) [ ε , ∞ ) R \mathbb{R} R Γ ( r ) = log r \Gamma(r)=\log r Γ ( r ) = log r Γ ( r ) = r p / p \Gamma(r)=r^p/p Γ ( r ) = r p / p p < 1 p<1 p < 1 p ≠ 0 p\neq0 p = 0 Γ ( r ) = r + 2 / 2 \Gamma(r)=r_+^2/2 Γ ( r ) = r + 2 /2 h = − k h=-k h = − k
⟨ u ′ , ( u − k ) + η 2 ⟩ = 1 2 d d t ∫ ( u − k ) + 2 η 2 d x + k ′ ( t ) ∫ ( u − k ) + η 2 d x , k ∈ W 1 , 1 ( J ) . (27) \langle u',(u-k)_+\eta^2\rangle= \frac{1}{2}\frac{d}{dt}\int(u-k)_+^2\eta^2\,dx + k'(t)\int(u-k)_+\eta^2\,dx,\qquad k\in W^{1,1}(J). \tag*{(27)} ⟨ u ′ , ( u − k ) + η 2 ⟩ = 2 1 d t d ∫ ( u − k ) + 2 η 2 d x + k ′ ( t ) ∫ ( u − k ) + η 2 d x , k ∈ W 1 , 1 ( J ) . (27) Thus k ′ ∈ L 1 ( J ) k'\in L^1(J) k ′ ∈ L 1 ( J ) u u u div b = 0 \operatorname{div} b=0 div b = 0
Strong approximation and positive-time smoothing
Strong vanishing-viscosity convergence
Proposition 3.1. The convergence in (13) holds for the entire family as ε ↓ 0 \varepsilon\downarrow0 ε ↓ 0
Proof. For fixed ε > 0 \varepsilon> 0 ε > 0
a t ε ( v , w ) = a t ( v , w ) + ε ∫ ∇ v ⋅ ∇ w d x on H 1 ( R n ) . a_t^\varepsilon(v,w)=a_t(v,w)+\varepsilon\int\nabla v\cdot\nabla w\,dx\quad\text{on }H^1(\mathbb{R}^n). a t ε ( v , w ) = a t ( v , w ) + ε ∫ ∇ v ⋅ ∇ w d x on H 1 ( R n ) . The construction of Proposition 2.4 applies on this space. Its energy identity is
1 2 ∥ u ε ( t ) ∥ 2 2 + ∫ 0 t E τ ( u ε , u ε ) d τ + ε ∫ 0 t ∥ ∇ u ε ∥ 2 2 d τ = 1 2 ∥ u 0 ∥ 2 2 + ∫ 0 t ⟨ f , u ε ⟩ d τ . (28) \frac{1}{2}\lVert u_\varepsilon(t)\rVert_2^2+\int_0^t\mathcal{E}_\tau(u_\varepsilon,u_\varepsilon)\,d\tau+\varepsilon\int_0^t\lVert\nabla u_\varepsilon\rVert_2^2\,d\tau=\frac{1}{2}\lVert u_0\rVert_2^2+\int_0^t\langle f,u_\varepsilon\rangle\,d\tau. \tag*{(28)} 2 1 ∥ u ε ( t ) ∥ 2 2 + ∫ 0 t E τ ( u ε , u ε ) d τ + ε ∫ 0 t ∥ ∇ u ε ∥ 2 2 d τ = 2 1 ∥ u 0 ∥ 2 2 + ∫ 0 t ⟨ f , u ε ⟩ d τ . (28) Consequently u ε u_\varepsilon u ε L ∞ H ∩ L 2 V L^\infty H\cap L^2V L ∞ H ∩ L 2 V ε ∇ u ε \sqrt{\varepsilon}\nabla u_\varepsilon ε ∇ u ε L 2 L^2 L 2
∥ ε Δ u ε ∥ L 2 H − 1 ≤ ε ∥ ∇ u ε ∥ L 2 = O ( ε ) . \lVert\varepsilon\Delta u_\varepsilon\rVert_{L^2H^{-1}}\leq\varepsilon\lVert\nabla u_\varepsilon\rVert_{L^2}=O(\sqrt{\varepsilon}). ∥ ε Δ u ε ∥ L 2 H − 1 ≤ ε ∥ ∇ u ε ∥ L 2 = O ( ε ) . The equation gives a uniform bound for u ε ′ u_\varepsilon' u ε ′ L 2 H − 1 L^2H^{-1} L 2 H − 1 H 1 H^1 H 1 C ( [ 0 , T ] ; H w e a k ) C([0,T];H_{\mathrm{weak}}) C ([ 0 , T ] ; H weak ) L 2 V L^2V L 2 V u u u
At t = T t=T t = T L 2 V L^2V L 2 V (28) . The terminal squared norm and the integrated form energy have lower limits at least as large as their values for u u u (28) . Hence
∫ 0 T E t ( u ε , u ε ) d t ⟶ ∫ 0 T E t ( u , u ) d t , (29) \int_0^T\mathcal{E}_t(u_\varepsilon,u_\varepsilon)\,dt\longrightarrow\int_0^T\mathcal{E}_t(u,u)\,dt, \tag*{(29)} ∫ 0 T E t ( u ε , u ε ) d t ⟶ ∫ 0 T E t ( u , u ) d t , (29) ε ∫ 0 T ∥ ∇ u ε ∥ 2 2 d t ⟶ 0. \varepsilon\int_0^T\lVert\nabla u_\varepsilon\rVert_2^2\,dt\longrightarrow0. ε ∫ 0 T ∥ ∇ u ε ∥ 2 2 d t ⟶ 0. Weak convergence in L 2 V L^2V L 2 V ∫ 0 T E t ( u ε , u ) d t → ∫ 0 T E t ( u , u ) d t \int_0^T\mathcal{E}_t(u_\varepsilon,u)\,dt\to\int_0^T\mathcal{E}_t(u,u)\,dt ∫ 0 T E t ( u ε , u ) d t → ∫ 0 T E t ( u , u ) d t u ε − u u_\varepsilon-u u ε − u
∫ 0 T [ u ε − u ] s 2 d t ⟶ 0. (30) \int_0^T [u_\varepsilon-u]_s^2\,dt\longrightarrow0. \tag*{(30)} ∫ 0 T [ u ε − u ] s 2 d t ⟶ 0. (30) Cauchy–Schwarz for the forms now gives convergence of E t ( u ε , u ε ) \mathcal{E}_t(u_\varepsilon,u_\varepsilon) E t ( u ε , u ε ) E t ( u , u ) \mathcal{E}_t(u,u) E t ( u , u ) L 1 ( 0 , T ) L^1(0,T) L 1 ( 0 , T )
The functions t ↦ ∫ 0 t ⟨ f , u ε − u ⟩ d τ t\mapsto\int_0^t\langle f,u_\varepsilon-u\rangle\,d\tau t ↦ ∫ 0 t ⟨ f , u ε − u ⟩ d τ ( r , t ) (r,t) ( r , t )
∥ f ∥ L 2 ( r , t ; V ′ ) ∥ u ε − u ∥ L 2 ( 0 , T ; V ) . \lVert f\rVert_{L^2(r,t;V')}\lVert u_\varepsilon-u\rVert_{L^2(0,T;V)}. ∥ f ∥ L 2 ( r , t ; V ′ ) ∥ u ε − u ∥ L 2 ( 0 , T ; V ) . The convergence is therefore uniform. Since the form energies converge in L 1 ( 0 , T ) L^1(0,T) L 1 ( 0 , T ) ∥ u ε ( t ) ∥ H 2 \lVert u_\varepsilon(t)\rVert_H^2 ∥ u ε ( t ) ∥ H 2 ∥ u ( t ) ∥ H 2 \lVert u(t)\rVert_H^2 ∥ u ( t ) ∥ H 2 u ( [ 0 , T ] ) u([0,T]) u ([ 0 , T ]) H H H C t H w e a k C_tH_{\mathrm{weak}} C t H weak
sup t ∈ [ 0 , T ] ∣ ( u ε ( t ) − u ( t ) , u ( t ) ) ∣ ⟶ 0. \sup_{t\in[0,T]}|(u_\varepsilon(t)-u(t),u(t))|\longrightarrow0. t ∈ [ 0 , T ] sup ∣ ( u ε ( t ) − u ( t ) , u ( t )) ∣ ⟶ 0. The polarization identity proves convergence in C t H C_tH C t H L 2 V L^2V L 2 V □ \square □
Corollary 3.2 (Stability of the viscous approximation under changes in the data). Suppose u 0 ε → u 0 u_0^\varepsilon\to u_0 u 0 ε → u 0 L 2 L^2 L 2 and f ε → f f_\varepsilon\to f f ε → f in L 2 ( 0 , T ; V ′ ) L^2(0,T;V') L 2 ( 0 , T ; V ′ ) With these data, the viscous solutions still satisfy (13) .
Proof. Let u ~ ε \widetilde{u}_{\varepsilon} u ε u 0 u_0 u 0 f f f z ε = u ε − u ~ ε z_{\varepsilon}=u_{\varepsilon}-\widetilde{u}_{\varepsilon} z ε = u ε − u ε z ε z_{\varepsilon} z ε
∥ z ε ∥ C ( [ 0 , T ] ; H ) 2 + ∥ z ε ∥ L 2 ( 0 , T ; V ) 2 + ε ∥ ∇ z ε ∥ L 2 ( ( 0 , T ) × R n ) 2 ≤ C T ( ∥ u 0 ε − u 0 ∥ 2 2 + ∥ f ε − f ∥ L 2 ( 0 , T ; V ′ ) 2 ) . \|z_{\varepsilon}\|_{C([0,T];H)}^2+\|z_{\varepsilon}\|_{L^2(0,T;V)}^2+\varepsilon\|\nabla z_{\varepsilon}\|_{L^2((0,T)\times\mathbb{R}^n)}^2
\leq C_T\left(\|u_0^{\varepsilon}-u_0\|_2^2+\|f_{\varepsilon}-f\|_{L^2(0,T;V')}^2\right). ∥ z ε ∥ C ([ 0 , T ] ; H ) 2 + ∥ z ε ∥ L 2 ( 0 , T ; V ) 2 + ε ∥∇ z ε ∥ L 2 (( 0 , T ) × R n ) 2 ≤ C T ( ∥ u 0 ε − u 0 ∥ 2 2 + ∥ f ε − f ∥ L 2 ( 0 , T ; V ′ ) 2 ) . The constant is independent of ε \varepsilon ε Proposition 3.1 and the triangle inequality give both assertions in (1.13).
Contractions and the evolution family
Proposition 3.3 (Comparison and contraction). The homogeneous evolution preserves nonnegativity and is contractive in L 1 L^1 L 1 L 2 L^2 L 2 L ∞ L^\infty L ∞ L 2 L^2 L 2 u 0 ∈ L 2 ∩ L ∞ u_0\in L^2\cap L^\infty u 0 ∈ L 2 ∩ L ∞ f ∈ L t 2 L x 2 ∩ L t 1 L x ∞ f\in L_t^2L_x^2\cap L_t^1L_x^\infty f ∈ L t 2 L x 2 ∩ L t 1 L x ∞
∥ u ( t ) ∥ ∞ ≤ ∥ u 0 ∥ ∞ + ∫ 0 t ∥ f ( τ ) ∥ ∞ d τ . (31) \|u(t)\|_\infty\leq\|u_0\|_\infty+\int_0^t\|f(\tau)\|_\infty\,\mathrm{d}\tau. \tag*{(31)} ∥ u ( t ) ∥ ∞ ≤ ∥ u 0 ∥ ∞ + ∫ 0 t ∥ f ( τ ) ∥ ∞ d τ . (31) Proof. For every increasing Lipschitz β \beta β β ( 0 ) = 0 \beta(0)=0 β ( 0 ) = 0
E t ( u , β ( u ) ) ≥ 0 , B b ( t ) ( u , β ( u ) ) = 0. \mathcal{E}_t(u,\beta(u))\geq0,\qquad\mathcal{B}_{b(t)}(u,\beta(u))=0. E t ( u , β ( u )) ≥ 0 , B b ( t ) ( u , β ( u )) = 0. Testing the homogeneous equation for a difference v v v − v − -v_- − v − v v v − v -v − v ( v − M ) + (v-M)_+ ( v − M ) + ( − v − M ) + (-v-M)_+ ( − v − M ) + M = ∥ v ( 0 ) ∥ ∞ M=\|v(0)\|_\infty M = ∥ v ( 0 ) ∥ ∞ L ∞ L^\infty L ∞ L 1 L^1 L 1
Γ δ ( r ) = r 2 + δ 2 − δ , β δ ( r ) = r ( r 2 + δ 2 ) − 1 / 2 . \Gamma_\delta(r)=\sqrt{r^2+\delta^2}-\delta,\qquad\beta_\delta(r)=r(r^2+\delta^2)^{-1/2}. Γ δ ( r ) = r 2 + δ 2 − δ , β δ ( r ) = r ( r 2 + δ 2 ) − 1/2 . The integral of Γ δ ( u ) \Gamma_\delta(u) Γ δ ( u ) u ∈ L 2 u\in L^2 u ∈ L 2 Γ δ ( r ) ≤ r 2 / ( 2 δ ) \Gamma_\delta(r)\leq r^2/(2\delta) Γ δ ( r ) ≤ r 2 / ( 2 δ ) (26) . Monotone convergence as δ ↓ 0 \delta\downarrow0 δ ↓ 0 L 1 L^1 L 1 L 2 L^2 L 2 (22) .
For (31) , use the moving level
k δ ( t ) = ∥ u 0 ∥ ∞ + δ + ∫ 0 t ∥ f ( τ ) ∥ ∞ d τ . k_\delta(t)=\|u_0\|_\infty+\delta+\int_0^t\|f(\tau)\|_\infty\,\mathrm{d}\tau. k δ ( t ) = ∥ u 0 ∥ ∞ + δ + ∫ 0 t ∥ f ( τ ) ∥ ∞ d τ . The forcing and level terms combine to ∫ ( f − k δ ′ ) ( u − k δ ) + ≤ 0 \int(f-k_\delta')(u-k_\delta)_+\leq0 ∫ ( f − k δ ′ ) ( u − k δ ) + ≤ 0 k δ ≥ δ k_\delta\geq\delta k δ ≥ δ ∥ ( u − k δ ) + ∥ 1 ≤ δ − 1 ∥ u ∥ 2 \|(u-k_\delta)_+\|_1\leq\delta^{-1}\|u\|_2 ∥ ( u − k δ ) + ∥ 1 ≤ δ − 1 ∥ u ∥ 2 Lemma 2.6 gives the global moving-level identity. Indeed,
∥ ( u − k ) + − ( v − k ) + ∥ 1 ≤ δ − 1 ( ∥ u ∥ 2 + ∥ v ∥ 2 ) ∥ u − v ∥ 2 \|(u-k)_+-(v-k)_+\|_1\leq\delta^{-1}(\|u\|_2+\|v\|_2)\|u-v\|_2 ∥ ( u − k ) + − ( v − k ) + ∥ 1 ≤ δ − 1 ( ∥ u ∥ 2 + ∥ v ∥ 2 ) ∥ u − v ∥ 2 for k ≥ δ k\geq\delta k ≥ δ k , ℓ ≥ δ k,\ell\geq\delta k , ℓ ≥ δ L 1 L^1 L 1 ∣ k − ℓ ∣ δ − 2 ∥ u ∥ 2 2 |k-\ell|\delta^{-2}\|u\|_2^2 ∣ k − ℓ ∣ δ − 2 ∥ u ∥ 2 2 { u ≤ δ } \{u\leq\delta\} { u ≤ δ } k δ ′ ∈ L 1 k_\delta'\in L^1 k δ ′ ∈ L 1 − u -u − u δ ↓ 0 \delta\downarrow0 δ ↓ 0 (31) .
Let U ( t , r ) U(t,r) U ( t , r ) r r r t t t U ( t , r ) U ( r , q ) = U ( t , q ) U(t,r)U(r,q)=U(t,q) U ( t , r ) U ( r , q ) = U ( t , q )
Lemma 3.4 (Fractional Nash inequality). For g ∈ L 1 ∩ H s ( R n ) g\in L^1\cap H^s(\mathbb{R}^n) g ∈ L 1 ∩ H s ( R n )
∥ g ∥ 2 2 + 2 α / n ≤ C [ g ] s 2 ∥ g ∥ 1 2 α / n . (32) \|g\|_2^{2+2\alpha/n}\leq C[g]_s^2\|g\|_1^{2\alpha/n}. \tag*{(32)} ∥ g ∥ 2 2 + 2 α / n ≤ C [ g ] s 2 ∥ g ∥ 1 2 α / n . (32) Proof. By the Fourier characterization of H s H^s H s ρ \rho ρ
∥ g ∥ 2 2 ≤ C ρ n ∥ g ∥ 1 2 + C ρ − α [ g ] s 2 . \|g\|_2^2\leq C\rho^n\|g\|_1^2+C\rho^{-\alpha}[g]_s^2. ∥ g ∥ 2 2 ≤ C ρ n ∥ g ∥ 1 2 + C ρ − α [ g ] s 2 . For g ≠ 0 g\neq0 g = 0 ρ \rho ρ (32) in every dimension, and the inequality is immediate for g = 0 g=0 g = 0
Proposition 3.5 (Positive-time smoothing). For $0 \le r < t \le T$,
∥ U ( t , r ) ∥ L 1 → L 2 ≤ C ( t − r ) − n / ( 2 α ) , ∥ U ( t , r ) ∥ L 2 → L ∞ ≤ C ( t − r ) − n / ( 2 α ) . (33) \|U(t,r)\|_{L^1\to L^2} \le C(t-r)^{-n/(2\alpha)}, \qquad\|U(t,r)\|_{L^2\to L^\infty} \le C(t-r)^{-n/(2\alpha)}. \tag*{(33)} ∥ U ( t , r ) ∥ L 1 → L 2 ≤ C ( t − r ) − n / ( 2 α ) , ∥ U ( t , r ) ∥ L 2 → L ∞ ≤ C ( t − r ) − n / ( 2 α ) . (33) The constants depend on n , s , λ n,s,\lambda n , s , λ ∥ b ∥ ∞ \|b\|_\infty ∥ b ∥ ∞
Proof. For nonzero initial data in L 1 ∩ L 2 L^1 \cap L^2 L 1 ∩ L 2 y ( t ) = ∥ u ( t ) ∥ 2 2 y(t)=\|u(t)\|_2^2 y ( t ) = ∥ u ( t ) ∥ 2 2 P = ∥ u ( r ) ∥ 1 P=\|u(r)\|_1 P = ∥ u ( r ) ∥ 1 L 1 L^1 L 1
y ′ ( t ) ≤ − c P − 2 α / n y ( t ) 1 + α / n . y'(t) \le-cP^{-2\alpha/n}y(t)^{1+\alpha/n}. y ′ ( t ) ≤ − c P − 2 α / n y ( t ) 1 + α / n . Integration gives y ( t ) ≤ C P 2 ( t − r ) − n / α y(t)\le CP^2(t-r)^{-n/\alpha} y ( t ) ≤ C P 2 ( t − r ) − n / α
The adjoint evolution is the backward problem with form E t − B b ( t ) \mathcal{E}_t-\mathcal{B}_{b(t)} E t − B b ( t ) U ( t , r ) ∗ U(t,r)^* U ( t , r ) ∗ g ∈ L 1 ∩ L 2 g\in L^1\cap L^2 g ∈ L 1 ∩ L 2
∣ ( U ( t , r ) v , g ) ∣ = ∣ ( v , U ( t , r ) ∗ g ) ∣ ≤ C ( t − r ) − n / ( 2 α ) ∥ v ∥ 2 ∥ g ∥ 1 . |(U(t,r)v,g)|=|(v,U(t,r)^*g)|\le C(t-r)^{-n/(2\alpha)}\|v\|_2\|g\|_1. ∣ ( U ( t , r ) v , g ) ∣ = ∣ ( v , U ( t , r ) ∗ g ) ∣ ≤ C ( t − r ) − n / ( 2 α ) ∥ v ∥ 2 ∥ g ∥ 1 . Duality proves the second estimate. In particular, ∥ u ( t ) ∥ ∞ ≤ C t − n / ( 4 s ) ∥ u 0 ∥ 2 \|u(t)\|_\infty\le Ct^{-n/(4s)}\|u_0\|_2 ∥ u ( t ) ∥ ∞ ≤ C t − n / ( 4 s ) ∥ u 0 ∥ 2
Localized energy and boundedness
Cutoffs and local solution spaces
Lemma 4.1 (Cutoff bounds). Assumption (1.5) implies
e s s s u p x ∫ ∣ x − y ∣ < r ∣ x − y ∣ 2 K ( t ; x , y ) d y ≤ 2 Λ 2 − α r 2 − α . (34) \mathop{\mathrm{ess\,sup}}_x\int_{|x-y|<r}|x-y|^2K(t;x,y)\,dy\le\frac{2\Lambda}{2-\alpha}r^{2-\alpha}. \tag*{(34)} ess sup x ∫ ∣ x − y ∣ < r ∣ x − y ∣ 2 K ( t ; x , y ) d y ≤ 2 − α 2Λ r 2 − α . (34) If 0 ≤ η ≤ 1 0\le\eta\le1 0 ≤ η ≤ 1 Lip ( η ) ≤ L \operatorname{Lip}(\eta)\le L Lip ( η ) ≤ L
G η ( t , x ) : = ∫ ( η ( x ) − η ( y ) ) 2 K ( t ; x , y ) d y ≤ C Λ L α . (35) G_\eta(t,x):=\int(\eta(x)-\eta(y))^2K(t;x,y)\,dy\le C\Lambda L^\alpha. \tag*{(35)} G η ( t , x ) := ∫ ( η ( x ) − η ( y ) ) 2 K ( t ; x , y ) d y ≤ C Λ L α . (35) Proof. Layer integration and (1.5) give
∫ ∣ x − y ∣ < r ∣ x − y ∣ 2 K d y = ∫ 0 r 2 ρ ∫ ρ < ∣ x − y ∣ < r K d y d ρ ≤ 2 Λ ∫ 0 r ρ 1 − α d ρ . \int_{|x-y|<r}|x-y|^2K\,dy=\int_0^r2\rho\int_{\rho<|x-y|<r}K\,dy\,d\rho\le2\Lambda\int_0^r\rho^{1-\alpha}\,d\rho. ∫ ∣ x − y ∣ < r ∣ x − y ∣ 2 K d y = ∫ 0 r 2 ρ ∫ ρ < ∣ x − y ∣ < r K d y d ρ ≤ 2Λ ∫ 0 r ρ 1 − α d ρ . For (4.2), split at L − 1 L^{-1} L − 1 ∣ η ( x ) − η ( y ) ∣ ≤ 1 |\eta(x)-\eta(y)|\le1 ∣ η ( x ) − η ( y ) ∣ ≤ 1
Definition 4.2 (Local variational inequalities). Let f ∈ L l o c 2 ( I × D ) f\in L^2_{\mathrm{loc}}(I\times D) f ∈ L loc 2 ( I × D ) u = w + c ( t ) u=w+c(t) u = w + c ( t ) w ∈ X T w\in\mathcal{X}_T w ∈ X T c ∈ H l o c 1 ( I ) c\in H^1_{\mathrm{loc}}(I) c ∈ H loc 1 ( I ) I × D I\times D I × D χ ∈ C c ∞ ( D ) \chi\in C^\infty_c(D) χ ∈ C c ∞ ( D ) χ u ∈ H l o c 1 ( I ; H − s ( R n ) ) \chi u\in H^1_{\mathrm{loc}}(I;H^{-s}(\mathbb{R}^n)) χ u ∈ H loc 1 ( I ; H − s ( R n ))
⟨ ∂ t u , ϕ ⟩ + B b ( t ) ( u , ϕ ) + E t ( u , ϕ ) ≥ ∫ f ϕ d x (36) \langle\partial_tu,\phi\rangle+\mathcal{B}_{b(t)}(u,\phi)+\mathcal{E}_t(u,\phi)\ge\int f\phi\,dx \tag*{(36)} ⟨ ∂ t u , ϕ ⟩ + B b ( t ) ( u , ϕ ) + E t ( u , ϕ ) ≥ ∫ f ϕ d x (36) for almost every t t t ϕ ∈ H s ( R n ) \phi\in H^s(\mathbb{R}^n) ϕ ∈ H s ( R n ) D D D E t ( u , ϕ ) = E t ( w , ϕ ) \mathcal{E}_t(u,\phi)=\mathcal{E}_t(w,\phi) E t ( u , ϕ ) = E t ( w , ϕ ) B b ( t ) ( u , ϕ ) = B b ( t ) ( w , ϕ ) \mathcal{B}_{b(t)}(u,\phi)=\mathcal{B}_{b(t)}(w,\phi) B b ( t ) ( u , ϕ ) = B b ( t ) ( w , ϕ ) c ′ ( t ) c'(t) c ′ ( t ) supp ϕ \operatorname{supp}\phi supp ϕ u u u
If u ∈ X T u\in\mathcal{X}_T u ∈ X T I × D I\times D I × D f ∈ L l o c 2 ( I × D ) f\in L^2_{\mathrm{loc}}(I\times D) f ∈ L loc 2 ( I × D ) ∂ t ( χ u ) ∈ L l o c 2 ( I ; H − s ( R n ) ) \partial_t(\chi u)\in L^2_{\mathrm{loc}}(I;H^{-s}(\mathbb{R}^n)) ∂ t ( χ u ) ∈ L loc 2 ( I ; H − s ( R n )) c ∈ H l o c 1 ( I ) c\in H^1_{\mathrm{loc}}(I) c ∈ H loc 1 ( I ) f f f f + c ′ f+c' f + c ′ L 1 L^1 L 1
Lemma 4.3 (Local parabolic Sobolev inequality). Put κ = 1 + α / n \kappa= 1 + \alpha/n κ = 1 + α / n B R B_R B R J J J ∥ ⋅ ∥ 2 \|\cdot\|_2 ∥ ⋅ ∥ 2 L 2 ( B R ) L^2(B_R) L 2 ( B R )
∫ J ∫ B R ∣ w ∣ 2 κ d x d t ≤ C ( ess sup t ∈ J ∥ w ( t ) ∥ 2 2 ) α / n ∫ J ( [ w ( t ) ] s , B R 2 + R − α ∥ w ( t ) ∥ 2 2 ) d t . (37) \int_J \int_{B_R} |w|^{2\kappa}\,dx\,dt \le C\left(\mathop{\operatorname{ess\,sup}}_{t\in J}\|w(t)\|_2^2\right)^{\alpha/n}\int_J\left([w(t)]_{s,B_R}^2+R^{-\alpha}\|w(t)\|_2^2\right)\,dt. \tag*{(37)} ∫ J ∫ B R ∣ w ∣ 2 κ d x d t ≤ C ( ess sup t ∈ J ∥ w ( t ) ∥ 2 2 ) α / n ∫ J ( [ w ( t ) ] s , B R 2 + R − α ∥ w ( t ) ∥ 2 2 ) d t . (37) Proof. On R n \mathbb{R}^n R n σ = s / κ < n / 2 \sigma=s/\kappa<n/2 σ = s / κ < n /2 s s s 2 n / ( n − 2 σ ) = 2 κ 2n/(n-2\sigma)=2\kappa 2 n / ( n − 2 σ ) = 2 κ
∥ w ∥ 2 κ 2 κ ≤ C [ w ] s 2 ∥ w ∥ 2 2 κ − 2 . \|w\|_{2\kappa}^{2\kappa}\le C[w]_s^2\|w\|_2^{2\kappa-2}. ∥ w ∥ 2 κ 2 κ ≤ C [ w ] s 2 ∥ w ∥ 2 2 κ − 2 . On the unit ball, reflection across the sphere in a boundary annulus followed by multiplication by a fixed cutoff gives an extension bounded in both L 2 L^2 L 2 H s H^s H s B R B_R B R [ w ] s 2 [w]_s^2 [ w ] s 2 [ w ] s , B R 2 + R − α ∥ w ∥ L 2 ( B R ) 2 [w]_{s,B_R}^2+R^{-\alpha}\|w\|_{L^2(B_R)}^2 [ w ] s , B R 2 + R − α ∥ w ∥ L 2 ( B R ) 2 [8 ] for the fractional Sobolev and extension theorems. Integrating in time proves (4.4), including the case n ≤ 2 s n\le2s n ≤ 2 s
Truncation inequalities
For a bounded domain D D D η \eta η
Θ D , k ± ( t , x ) = ∫ D c ( u ( t , y ) − k ( t ) ) ± K ( t ; x , y ) d y . \Theta^\pm_{D,k}(t,x)=\int_{D^c}(u(t,y)-k(t))_\pm K(t;x,y)\,dy. Θ D , k ± ( t , x ) = ∫ D c ( u ( t , y ) − k ( t ) ) ± K ( t ; x , y ) d y . Proposition 4.4 (Local truncation inequality). Let u u u w = ( u − k ( t ) ) + w=(u-k(t))_+ w = ( u − k ( t ) ) + k k k 0 ≤ η ≤ 1 0\le\eta\le1 0 ≤ η ≤ 1 D D D
1 2 d d t ∫ D w 2 η 2 d x + E t , D ( w η , w η ) + ∫ D w η 2 Θ D , k − d x ≤ 1 2 ∫ D w 2 G η d x + ∫ D w 2 ∣ b ⋅ ∇ η ∣ d x + ∫ D ( f + − k ′ ) w η 2 d x + ∫ D w η 2 Θ D , k + d x . (38) \begin{aligned}
\frac{1}{2}\frac{d}{dt}\int_D w^2\eta^2\,dx+\mathcal{E}_{t,D}(w\eta,w\eta)+\int_D w\eta^2\Theta^-_{D,k}\,dx \\
&\le\frac{1}{2}\int_D w^2G_\eta\,dx+\int_D w^2|\boldsymbol{b}\cdot\nabla\eta|\,dx+\int_D(f_+-k')w\eta^2\,dx+\int_D w\eta^2\Theta^+_{D,k}\,dx. \tag*{(38)}
\end{aligned} 2 1 d t d ∫ D w 2 η 2 d x + E t , D ( w η , w η ) + ∫ D w η 2 Θ D , k − d x ≤ 2 1 ∫ D w 2 G η d x + ∫ D w 2 ∣ b ⋅ ∇ η ∣ d x + ∫ D ( f + − k ′ ) w η 2 d x + ∫ D w η 2 Θ D , k + d x . (38) The inequality holds after integration against nonnegative time cutoffs. Applying it to − u -u − u
Proof. The admissibility of w η 2 w\eta^2 w η 2
B b ( t ) ( u , w η 2 ) = − ∫ D w 2 η b ⋅ ∇ η d x . \mathcal{B}_{b(t)}(u,w\eta^2)=-\int_D w^2\eta\,\boldsymbol{b}\cdot\nabla\eta\,dx. B b ( t ) ( u , w η 2 ) = − ∫ D w 2 η b ⋅ ∇ η d x . For the interior form, monotonicity of the truncation and the identity
( w ( x ) − w ( y ) ) ( w ( x ) η ( x ) 2 − w ( y ) η ( y ) 2 ) = ( w ( x ) η ( x ) − w ( y ) η ( y ) ) 2 − w ( x ) w ( y ) ( η ( x ) − η ( y ) ) 2 (w(x)-w(y))(w(x)\eta(x)^2-w(y)\eta(y)^2)
=(w(x)\eta(x)-w(y)\eta(y))^2-w(x)w(y)(\eta(x)-\eta(y))^2 ( w ( x ) − w ( y )) ( w ( x ) η ( x ) 2 − w ( y ) η ( y ) 2 ) = ( w ( x ) η ( x ) − w ( y ) η ( y ) ) 2 − w ( x ) w ( y ) ( η ( x ) − η ( y ) ) 2 give a lower bound E t , D ( w η , w η ) − 1 2 ∫ D w 2 G η \mathcal{E}_{t,D}(w\eta,w\eta)-\frac{1}{2}\int_D w^2G_\eta E t , D ( w η , w η ) − 2 1 ∫ D w 2 G η w ( x ) > 0 w(x)>0 w ( x ) > 0
( u ( x ) − u ( y ) ) w ( x ) = w ( x ) 2 + w ( x ) ( k − u ( y ) ) + − w ( x ) ( u ( y ) − k ) + . (u(x)-u(y))w(x)=w(x)^2+w(x)(k-u(y))_+-w(x)(u(y)-k)_+. ( u ( x ) − u ( y )) w ( x ) = w ( x ) 2 + w ( x ) ( k − u ( y ) ) + − w ( x ) ( u ( y ) − k ) + . Discard the nonnegative term w ( x ) 2 w(x)^2 w ( x ) 2 E t \mathcal{E}_t E t 1 / 2 1/2 1/2
All separated cross terms are finite before any tail supremum is taken. Indeed, if A = supp η A=\operatorname{supp}\eta A = supp η d = dist ( A , D c ) > 0 d=\operatorname{dist}(A,D^c)>0 d = dist ( A , D c ) > 0 K d x d y K\,dx\,dy K d x d y
∫ A ∫ D c ∣ w ( x ) u ( y ) ∣ K d y d x ≤ C d − α ∥ w ∥ L 2 ( A ) ∥ u ∥ 2 . \int_A\int_{D^c}|w(x)u(y)|K\,dy\,dx\le Cd^{-\alpha}\|w\|_{L^2(A)}\|u\|_2. ∫ A ∫ D c ∣ w ( x ) u ( y ) ∣ K d y d x ≤ C d − α ∥ w ∥ L 2 ( A ) ∥ u ∥ 2 . For an affine shift u = u ∗ + c ( t ) u=u_*+c(t) u = u ∗ + c ( t ) u ∗ ∈ X T u_*\in X_T u ∗ ∈ X T c c c ∣ k ∣ |k| ∣ k ∣ k k k
A separated-tail local bound
On Q R = ( t ∗ − R α , t ∗ ) × B R ( x ∗ ) Q_R=(t_* - R^\alpha,t_*)\times B_R(x_*) Q R = ( t ∗ − R α , t ∗ ) × B R ( x ∗ )
D R + ( u ) = e s s s u p t ∈ ( t ∗ − R α , t ∗ ) sup R / 2 ≤ r < ρ ≤ R ( ρ − r ) α Θ ( u + ; t , r , ρ ) . (39) \mathcal{D}_R^+(u)=\mathop{\mathrm{ess\,sup}}_{t\in(t_*-R^\alpha,t_*)}\ \sup_{R/2\le r<\rho\le R}(\rho-r)^\alpha\Theta(u_+;t,r,\rho). \tag*{(39)} D R + ( u ) = ess sup t ∈ ( t ∗ − R α , t ∗ ) R /2 ≤ r < ρ ≤ R sup ( ρ − r ) α Θ ( u + ; t , r , ρ ) . (39) By (1.11), D R + ( u ) ≤ Λ ∥ u ∥ L ∞ ( ( t ∗ − R α , t ∗ ) × R n ) \mathcal{D}_R^+(u)\le\Lambda\|u\|_{L^\infty((t_*-R^\alpha,t_*)\times\mathbb{R}^n)} D R + ( u ) ≤ Λ∥ u ∥ L ∞ (( t ∗ − R α , t ∗ ) × R n ) u u u
Proposition 4.5 (Local boundedness with separated tails). *For a subsolution in Q R Q_R Q R
e s s s u p Q R / 2 u + ≤ C [ ( ∫ Q R u + 2 ) 1 / 2 + R α ∥ f + ∥ ∞ + D R + ( u ) ] (40) \mathop{\mathrm{ess\,sup}}_{Q_{R/2}}u_+\le C\left[\left(\int_{Q_R}u_+^2\right)^{1/2}+R^\alpha\|f_+\|_\infty+\mathcal{D}_R^+(u)\right] \tag*{(40)} ess sup Q R /2 u + ≤ C [ ( ∫ Q R u + 2 ) 1/2 + R α ∥ f + ∥ ∞ + D R + ( u ) ] (40) whenever the quantities on the right are finite.*
Proof. Subtract ∥ f + ∥ ∞ ( t − t ∗ + R α ) \|f_+\|_\infty(t-t_*+R^\alpha) ∥ f + ∥ ∞ ( t − t ∗ + R α ) u u u R = 1 R=1 R = 1 t ∗ = 0 t_*=0 t ∗ = 0 u u u D = D 1 + ( u ) D=\mathcal{D}_1^+(u) D = D 1 + ( u ) B = ∥ b ∥ L ∞ ( Q 1 ) B=\|b\|_{L^\infty(Q_1)} B = ∥ b ∥ L ∞ ( Q 1 )
r j = 1 2 + 2 − j − 2 , Q j = ( − r j α , 0 ) × B r j , k j = k ( 1 − 2 − j − 1 ) , w j = ( u − k j ) + . r_j=\frac{1}{2}+2^{-j-2},\qquad Q_j=(-r_j^\alpha,0)\times B_{r_j},\qquad k_j=k(1-2^{-j-1}),\qquad w_j=(u-k_j)_+. r j = 2 1 + 2 − j − 2 , Q j = ( − r j α , 0 ) × B r j , k j = k ( 1 − 2 − j − 1 ) , w j = ( u − k j ) + . Choose a spatial cutoff η j \eta_j η j B r j + 1 B_{r_{j+1}} B r j + 1 B ( r j + r j + 1 ) / 2 B_{(r_j+r_{j+1})/2} B ( r j + r j + 1 ) /2 ∣ ∇ η j ∣ ≤ C 2 j |\nabla\eta_j|\le C2^j ∣∇ η j ∣ ≤ C 2 j B r j B_{r_j} B r j − r j α -r_j^\alpha − r j α ( − r j + 1 α , 0 ) (-r_{j+1}^\alpha,0) ( − r j + 1 α , 0 ) C 2 α j C2^{\alpha j} C 2 α j
A − 1 = ∫ ( − 1 , 0 ) × B 1 u + 2 , A j = ∫ Q j w j 2 ( j ≥ 0 ) . A_{-1}=\int_{(-1,0)\times B_1}u_+^2,\qquad A_j=\int_{Q_j}w_j^2\quad(j\ge0). A − 1 = ∫ ( − 1 , 0 ) × B 1 u + 2 , A j = ∫ Q j w j 2 ( j ≥ 0 ) . For every j ≥ 0 j\ge0 j ≥ 0 j = 0 j=0 j = 0 A − 1 A_{-1} A − 1
∣ { w j > 0 } ∩ Q j ∣ ≤ C 2 2 j k − 2 A j − 1 , ∫ Q j w j ≤ C 2 2 j k − 1 A j − 1 . (41) |\{w_j>0\}\cap Q_j|\le C2^{2j}k^{-2}A_{j-1},\qquad\int_{Q_j}w_j\le C2^{2j}k^{-1}A_{j-1}. \tag*{(41)} ∣ { w j > 0 } ∩ Q j ∣ ≤ C 2 2 j k − 2 A j − 1 , ∫ Q j w j ≤ C 2 2 j k − 1 A j − 1 . (41) The first estimate follows from the gap between consecutive levels (from 0 0 0 k 0 k_0 k 0 j = 0 j=0 j = 0 A j ≤ A j − 1 A_j\le A_{j-1} A j ≤ A j − 1
For x ∈ supp η j x\in\operatorname{supp}\eta_j x ∈ supp η j C 2 α j D C2^{\alpha j}D C 2 α j D α ≥ 1 \alpha\ge1 α ≥ 1
S j : = e s s s u p t ∈ ( − r j + 1 α , 0 ) ∫ w j 2 η j 2 + ∫ − r j + 1 α 0 E t , B r j ( w j η j , w j η j ) d t ≤ C 2 ( α + 1 ) j ( 1 + B + D / k ) A j − 1 . (42) \begin{aligned}
S_j:=\mathop{\mathrm{ess\,sup}}_{t\in(-r_{j+1}^\alpha,0)}\int w_j^2\eta_j^2+\int_{-r_{j+1}^\alpha}^{0}\mathcal{E}_{t,B_{r_j}}(w_j\eta_j,w_j\eta_j)\,\mathrm{d}t \\
&\le C2^{(\alpha+1)j}(1+B+D/k)A_{j-1}. \tag*{(42)}
\end{aligned} S j := ess sup t ∈ ( − r j + 1 α , 0 ) ∫ w j 2 η j 2 + ∫ − r j + 1 α 0 E t , B r j ( w j η j , w j η j ) d t ≤ C 2 ( α + 1 ) j ( 1 + B + D / k ) A j − 1 . (42) The Sobolev estimate (4.4) applied on this inner time interval yields ∫ Q j + 1 w j 2 κ ≤ C S j κ \int_{Q_{j+1}}w_j^{2\kappa}\le CS_j^\kappa ∫ Q j + 1 w j 2 κ ≤ C S j κ w j ≥ k 2 − j − 2 w_j\ge k2^{-j-2} w j ≥ k 2 − j − 2 { w j + 1 > 0 } \{w_{j+1}>0\} { w j + 1 > 0 }
A j + 1 ≤ C 2 N j k − 2 ( κ − 1 ) ( 1 + B + D / k ) κ A j − 1 κ ( j ≥ 0 ) . (43) A_{j+1}\le C2^{Nj}k^{-2(\kappa-1)}(1+B+D/k)^\kappa A_{j-1}^\kappa\qquad(j\ge0). \tag*{(43)} A j + 1 ≤ C 2 N j k − 2 ( κ − 1 ) ( 1 + B + D / k ) κ A j − 1 κ ( j ≥ 0 ) . (43) Here one may take N = ( α + 3 ) κ − 2 N=(\alpha+3)\kappa-2 N = ( α + 3 ) κ − 2 N N N Y j = A j / k 2 Y_j=A_j/k^2 Y j = A j / k 2 Z m = max ( Y 2 m − 1 , Y 2 m ) Z_m=\max(Y_{2m-1},Y_{2m}) Z m = max ( Y 2 m − 1 , Y 2 m ) Z m + 1 ≤ C 1 C 2 m Z m κ Z_{m+1}\le C_1C_2^mZ_m^\kappa Z m + 1 ≤ C 1 C 2 m Z m κ k = C ∗ ( A − 1 1 / 2 + D ) > 0 k=C_*(A_{-1}^{1/2}+D)>0 k = C ∗ ( A − 1 1/2 + D ) > 0 C ∗ C_* C ∗ B B B A j → 0 A_j\to0 A j → 0 A − 1 = 0 A_{-1}=0 A − 1 = 0 Q 1 Q_1 Q 1
Lemma 4.6 (Initial local boundedness under a pointwise upper bound). Assume, in addition,
K ( t ; x , y ) ≤ L ∣ x − y ∣ − n − α . (44) K(t;x,y)\le L|x-y|^{-n-\alpha}. \tag*{(44)} K ( t ; x , y ) ≤ L ∣ x − y ∣ − n − α . (44) A local variational subsolution with bounded source is locally bounded above. Proof. Use the preceding iteration in a unit cylinder compactly contained in the equation domain. Split its exterior integral into B 1 ∖ B r j B_1 \setminus B_{r_j} B 1 ∖ B r j B 1 c B_1^c B 1 c x x x
∫ B 1 ∖ B r j u + ( t , y ) K ( t ; x , y ) d y ≤ C 2 ( n + α ) j ∫ B 1 u + ( t , y ) d y . \int_{B_1 \setminus B_{r_j}} u_+(t,y)K(t;x,y)\,\mathrm{d}y \le C2^{(n+\alpha)j}\int_{B_1}u_+(t,y)\,\mathrm{d}y. ∫ B 1 ∖ B r j u + ( t , y ) K ( t ; x , y ) d y ≤ C 2 ( n + α ) j ∫ B 1 u + ( t , y ) d y . Write u = w + c ( t ) u=w+c(t) u = w + c ( t ) Definition 4.2 . On the compact time interval under consideration, the quantity
B 2 = ess sup t ∥ w ( t ) ∥ 2 + sup t ∣ c ( t ) ∣ B_2=\operatorname*{ess\,sup}_t\|w(t)\|_2+\sup_t|c(t)| B 2 = t ess sup ∥ w ( t ) ∥ 2 + t sup ∣ c ( t ) ∣ is finite and controls the last local integral. For x ∈ B 3 / 4 x\in B_{3/4} x ∈ B 3/4 y ∉ B 1 y\notin B_1 y ∈ / B 1 (44) give
∫ B 1 c ∣ w ( t , y ) ∣ K ( t ; x , y ) d y ≤ L ∥ w ( t ) ∥ 2 ( ∫ ∣ z ∣ > 1 / 4 ∣ z ∣ − 2 n − 2 α d z ) 1 / 2 ≤ C ∥ w ( t ) ∥ 2 . \int_{B_1^c}|w(t,y)|K(t;x,y)\,\mathrm{d}y\le L\|w(t)\|_2\left(\int_{|z|>1/4}|z|^{-2n-2\alpha}\,\mathrm{d}z\right)^{1/2}\le C\|w(t)\|_2. ∫ B 1 c ∣ w ( t , y ) ∣ K ( t ; x , y ) d y ≤ L ∥ w ( t ) ∥ 2 ( ∫ ∣ z ∣ > 1/4 ∣ z ∣ − 2 n − 2 α d z ) 1/2 ≤ C ∥ w ( t ) ∥ 2 . Assumption (1.5) controls the contribution of c ( t ) c(t) c ( t ) j j j C 2 ( n + α ) j B 2 C2^{(n+\alpha)j}B_2 C 2 ( n + α ) j B 2 (38) and using (41) gives
S j ≤ C 2 ( n + α + 1 ) j ( 1 + ∥ b ∥ ∞ + B 2 / k ) A j − 1 . S_j\le C2^{(n+\alpha+1)j}\left(1+\|b\|_\infty+B_2/k\right)A_{j-1}. S j ≤ C 2 ( n + α + 1 ) j ( 1 + ∥ b ∥ ∞ + B 2 / k ) A j − 1 . The recurrence (43) follows with a larger finite exponent. Taking k k k A 1 / 2 − 1 + B 2 A_{1/2}^{-1}+B_2 A 1/2 − 1 + B 2
The weak Harnack inequality
We work first in C 1 = ( − 4 , 1 ) × B 4 C_1=(-4,1)\times B_4 C 1 = ( − 4 , 1 ) × B 4 ( 0 , 0 ) (0,0) ( 0 , 0 )
h ( t ) = ε + ∫ − 4 t ( ∥ f − ∥ L ∞ ( C 1 ) + Θ ( u − ; τ , 3 , 4 ) ) d τ , v = u + h ( t ) , ε > 0. (45) h(t)=\varepsilon+\int_{-4}^{t}\left(\|f_-\|_{L^\infty(C_1)}+\Theta(u_-;\tau,3,4)\right)\,\mathrm{d}\tau,\qquad v=u+h(t),\qquad\varepsilon>0. \tag*{(45)} h ( t ) = ε + ∫ − 4 t ( ∥ f − ∥ L ∞ ( C 1 ) + Θ ( u − ; τ , 3 , 4 ) ) d τ , v = u + h ( t ) , ε > 0. (45) The shift satisfies h ∈ W 1 , 1 ( − 4 , 1 ) h\in W^{1,1}(-4,1) h ∈ W 1 , 1 ( − 4 , 1 ) v ≥ ε v\ge\varepsilon v ≥ ε C 1 C_1 C 1 u u u B 4 B_4 B 4 u u u h h h
Power inequalities and moment iteration
Lemma 5.1 (A weighted power inequality). Fix p ∗ ∈ ( 0 , 1 ) p_*\in(0,1) p ∗ ∈ ( 0 , 1 ) c ∗ , C ∗ > 0 c_*,C_*>0 c ∗ , C ∗ > 0 p ∗ p_* p ∗ p ∈ ( − ∞ , 0 ) ∪ ( 0 , p ∗ ] p\in(-\infty,0)\cup(0,p_*] p ∈ ( − ∞ , 0 ) ∪ ( 0 , p ∗ ] a , b > 0 a,b>0 a , b > 0 A , B ≥ 0 A,B\ge0 A , B ≥ 0
( a − b ) ( A 2 a p − 1 − B 2 b p − 1 ) ≤ − c ∗ ∣ p ∣ ( A a p / 2 − B b p / 2 ) 2 + C ∗ 1 + ∣ p ∣ ∣ p ∣ ( A − B ) 2 ( a p + b p ) . (46) (a-b)\left(A^2a^{p-1}-B^2b^{p-1}\right)\le-\frac{c_*}{|p|}\left(Aa^{p/2}-Bb^{p/2}\right)^2+C_*\frac{1+|p|}{|p|}(A-B)^2(a^p+b^p). \tag*{(46)} ( a − b ) ( A 2 a p − 1 − B 2 b p − 1 ) ≤ − ∣ p ∣ c ∗ ( A a p /2 − B b p /2 ) 2 + C ∗ ∣ p ∣ 1 + ∣ p ∣ ( A − B ) 2 ( a p + b p ) . (46) Proof. By symmetry and homogeneity, assume b = 1 b=1 b = 1 a = t ≥ 1 a=t\ge1 a = t ≥ 1 q = 1 − p > 0 q=1-p>0 q = 1 − p > 0 d = A − B d=A-B d = A − B
D = ( t − 1 ) ( 1 − t − q ) , E = ( t − 1 ) t − q . D=(t-1)(1-t^{-q}),\qquad E=(t-1)t^{-q}. D = ( t − 1 ) ( 1 − t − q ) , E = ( t − 1 ) t − q . For t > 1 t>1 t > 1
( t − 1 ) ( A 2 t − q − B 2 ) = − B 2 D + 2 B d E + d 2 E ≤ − 1 2 B 2 D + d 2 ( E + 2 E 2 / D ) . (47) (t-1)(A^2t^{-q}-B^2)=-B^2D+2BdE+d^2E\le-\frac{1}{2}B^2D+d^2(E+2E^2/D). \tag*{(47)} ( t − 1 ) ( A 2 t − q − B 2 ) = − B 2 D + 2 B d E + d 2 E ≤ − 2 1 B 2 D + d 2 ( E + 2 E 2 / D ) . (47) Since 1 − t − q = q ∫ 1 t r − q − 1 d r ≥ q ( t − 1 ) t − q − 1 1-t^{-q}=q\int_1^t r^{-q-1}\,\mathrm{d}r\ge q(t-1)t^{-q-1} 1 − t − q = q ∫ 1 t r − q − 1 d r ≥ q ( t − 1 ) t − q − 1
E ≤ t p , E 2 / D ≤ q − 1 t p . E\le t^p,\qquad E^2/D\le q^{-1}t^p. E ≤ t p , E 2 / D ≤ q − 1 t p . Cauchy–Schwarz applied to t p / 2 − 1 = ( p / 2 ) ∫ 1 t r p / 2 − 1 d r t^{p/2}-1=(p/2)\int_1^t r^{p/2-1}\,\mathrm{d}r t p /2 − 1 = ( p /2 ) ∫ 1 t r p /2 − 1 d r
D ≥ 4 q p 2 ( t p / 2 − 1 ) 2 . D\ge\frac{4q}{p^2}(t^{p/2}-1)^2. D ≥ p 2 4 q ( t p /2 − 1 ) 2 . Throughout the stated range of p p p q / ∣ p ∣ ≥ c ∗ : = 1 − p ∗ q/|p| \ge c_* := 1-p_* q /∣ p ∣ ≥ c ∗ := 1 − p ∗ q ≥ 1 − p ∗ q \ge1-p_* q ≥ 1 − p ∗ (47) is therefore at most
− 2 c ∗ ∣ p ∣ B 2 ( t p / 2 − 1 ) 2 + ( 1 + 2 1 − p ∗ ) d 2 t p -\frac{2c_*}{|p|}B^2(t^{p/2}-1)^2+\left(1+\frac{2}{1-p_*}\right)d^2t^p − ∣ p ∣ 2 c ∗ B 2 ( t p /2 − 1 ) 2 + ( 1 + 1 − p ∗ 2 ) d 2 t p The bound ( A t p / 2 − B ) 2 ≤ 2 B 2 ( t p / 2 − 1 ) 2 + 2 d 2 t p (At^{p/2}-B)^2 \le2B^2(t^{p/2}-1)^2+2d^2t^p ( A t p /2 − B ) 2 ≤ 2 B 2 ( t p /2 − 1 ) 2 + 2 d 2 t p C ∗ C_* C ∗ t = 1 t=1 t = 1
Consider cylinders J ′ × B r ⊂ J × B ρ ⊂ ( − 4 , 1 ) × B 3 J' \times B_r \subset J \times B_\rho\subset(-4,1) \times B_3 J ′ × B r ⊂ J × B ρ ⊂ ( − 4 , 1 ) × B 3 d = ρ − r d=\rho-r d = ρ − r δ > 0 \delta>0 δ > 0 J ′ J' J ′ J J J p < 0 p<0 p < 0 p > 0 p>0 p > 0
Proposition 5.2 (Power energy estimates). Let η = 1 \eta=1 η = 1 B r B_r B r B ( r + ρ ) / 2 B_{(r+\rho)/2} B ( r + ρ ) /2 ∣ ∇ η ∣ ≤ C / d |\nabla\eta|\le C/d ∣∇ η ∣ ≤ C / d
ess sup t ∈ J ′ ∫ B r v p d x + ∫ J ′ E t , B ρ ( η v p / 2 , η v p / 2 ) d t \operatorname*{ess\,sup}_{t\in J'}\int_{B_r}v^p\,dx+\int_{J'}\mathcal{E}_{t,B_\rho}(\eta v^{p/2},\eta v^{p/2})\,dt t ∈ J ′ ess sup ∫ B r v p d x + ∫ J ′ E t , B ρ ( η v p /2 , η v p /2 ) d t ≤ C ( 1 + ∣ p ∣ ) ( d − α + ∥ b ∥ ∞ d − 1 + δ − 1 ) ∫ J ∫ B ρ v p d x d t (48) \le C(1+|p|)\left(d^{-\alpha}+\|b\|_\infty d^{-1}+\delta^{-1}\right)\int_J\int_{B_\rho}v^p\,dx\,dt \tag*{(48)} ≤ C ( 1 + ∣ p ∣ ) ( d − α + ∥ b ∥ ∞ d − 1 + δ − 1 ) ∫ J ∫ B ρ v p d x d t (48) The constant is uniform as p → 0 p\to0 p → 0 ( − ∞ , 0 ) ∪ ( 0 , p ∗ ] (-\infty,0)\cup(0,p_*] ( − ∞ , 0 ) ∪ ( 0 , p ∗ ]
Proof. Test by η 2 v p − 1 \eta^2v^{p-1} η 2 v p − 1 B ρ × B ρ B_\rho\times B_\rho B ρ × B ρ x ∈ supp η x\in\operatorname{supp}\eta x ∈ supp η y ∉ B ρ y\notin B_\rho y ∈ / B ρ
( u ( x ) − u ( y ) ) v ( x ) p − 1 ≤ v ( x ) p + u − ( y ) v ( x ) p − 1 . (u(x)-u(y))v(x)^{p-1}\le v(x)^p+u_-(y)v(x)^{p-1}. ( u ( x ) − u ( y )) v ( x ) p − 1 ≤ v ( x ) p + u − ( y ) v ( x ) p − 1 . By (1.5), the first term contributes at most C d − α ∫ B ρ v p Cd^{-\alpha}\int_{B_\rho}v^p C d − α ∫ B ρ v p y ∈ B 4 y\in B_4 y ∈ B 4 u ≥ 0 u\ge0 u ≥ 0 y ∉ B 4 y\notin B_4 y ∈ / B 4 Θ ( u − ; t , 3 , 4 ) ∫ η 2 v p − 1 \Theta(u_-;t,3,4)\int\eta^2v^{p-1} Θ ( u − ; t , 3 , 4 ) ∫ η 2 v p − 1 h ′ h' h ′
B b ( u , η 2 v p − 1 ) = − 1 p ∫ v p b ⋅ ∇ ( η 2 ) d x (49) \mathcal{B}_b(u,\eta^2v^{p-1})=-\frac{1}{p}\int v^p b\cdot\nabla(\eta^2)\,dx \tag*{(49)} B b ( u , η 2 v p − 1 ) = − p 1 ∫ v p b ⋅ ∇ ( η 2 ) d x (49) Its absolute value is at most C ∣ p ∣ − 1 ∥ b ∥ ∞ d − 1 ∫ B ρ v p C|p|^{-1}\|b\|_\infty d^{-1}\int_{B_\rho}v^p C ∣ p ∣ − 1 ∥ b ∥ ∞ d − 1 ∫ B ρ v p
Write Y ( t ) = ∫ η 2 v p Y(t)=\int\eta^2v^p Y ( t ) = ∫ η 2 v p ∣ p ∣ |p| ∣ p ∣
Y ′ ( t ) + c E t , B ρ ( η v p / 2 , η v p / 2 ) ≤ C ( 1 + ∣ p ∣ ) ( d − α + ∥ b ∥ ∞ d − 1 ) ∫ B ρ v p . Y'(t)+c\mathcal{E}_{t,B_\rho}(\eta v^{p/2},\eta v^{p/2})\le C(1+|p|)(d^{-\alpha}+\|b\|_\infty d^{-1})\int_{B_\rho}v^p. Y ′ ( t ) + c E t , B ρ ( η v p /2 , η v p /2 ) ≤ C ( 1 + ∣ p ∣ ) ( d − α + ∥ b ∥ ∞ d − 1 ) ∫ B ρ v p . when p < 0 p<0 p < 0 p > 0 p>0 p > 0 − Y ′ -Y' − Y ′ Y ′ Y' Y ′ p < 0 p<0 p < 0 J J J p > 0 p>0 p > 0 C δ − 1 ∫ J ∫ B ρ v p C\delta^{-1}\int_J\int_{B_\rho}v^p C δ − 1 ∫ J ∫ B ρ v p
To justify the test, extend r p − 1 r^{p-1} r p − 1 [ ε , ∞ ) [\varepsilon,\infty) [ ε , ∞ ) r p / p r^p/p r p / p v p − 1 ≤ ε p − 1 v^{p-1}\le\varepsilon^{p-1} v p − 1 ≤ ε p − 1 h ′ ∈ L 1 h'\in L^1 h ′ ∈ L 1
Define, for 0 ≤ a ≤ 1 0\le a\le1 0 ≤ a ≤ 1
E a = ( − 3 − a / 4 , − 1 + a / 4 ) × B 2 + a / 4 , G a = ( − a / 4 , 1 ) × B 1 + a / 4 . (50) E_a=(-3-a/4,-1+a/4)\times B_{2+a/4},\qquad G_a=(-a/4,1)\times B_{1+a/4}. \tag*{(50)} E a = ( − 3 − a /4 , − 1 + a /4 ) × B 2 + a /4 , G a = ( − a /4 , 1 ) × B 1 + a /4 . (50) Thus E 0 = P 1 E_0=P_1 E 0 = P 1 G 0 = F 1 G_0=F_1 G 0 = F 1
Lemma 5.3 (Moment estimates). Put p 0 = 1 / 4 p_0 = 1/4 p 0 = 1/4 For 0 ≤ a < c ≤ 1 0 \leq a < c \leq1 0 ≤ a < c ≤ 1 and 0 < p ≤ p 0 / 2 0 < p \leq p_0/2 0 < p ≤ p 0 /2
∥ v ∥ L p 0 ( E a ) ≤ ( C ( c − a ) N ) 1 / p ∥ v ∥ L p ( E c ) , (51) \lVert v\rVert_{L^{p_0}(E_a)} \leq\left(\frac{C}{(c-a)^N}\right)^{1/p}\lVert v\rVert_{L^p(E_c)}, \tag*{(51)} ∥ v ∥ L p 0 ( E a ) ≤ ( ( c − a ) N C ) 1/ p ∥ v ∥ L p ( E c ) , (51) ∥ v − 1 ∥ L p 0 ( G a ) ≤ ( C ( c − a ) N ) 1 / p ∥ v − 1 ∥ L p ( G c ) . (52) \lVert v^{-1}\rVert_{L^{p_0}(G_a)} \leq\left(\frac{C}{(c-a)^N}\right)^{1/p}\lVert v^{-1}\rVert_{L^p(G_c)}. \tag*{(52)} ∥ v − 1 ∥ L p 0 ( G a ) ≤ ( ( c − a ) N C ) 1/ p ∥ v − 1 ∥ L p ( G c ) . (52) Moreover,
∥ v ∥ L 1 ( E 0 ) ≤ C ∥ v ∥ L p 0 ( E 1 / 2 ) , (53) \lVert v\rVert_{L^1(E_0)} \leq C\lVert v\rVert_{L^{p_0}(E_{1/2})}, \tag*{(53)} ∥ v ∥ L 1 ( E 0 ) ≤ C ∥ v ∥ L p 0 ( E 1/2 ) , (53) ∥ v − 1 ∥ L ∞ ( G 0 ) ≤ C ∥ v − 1 ∥ L p 0 ( G 1 / 2 ) . (54) \lVert v^{-1}\rVert_{L^\infty(G_0)} \leq C\lVert v^{-1}\rVert_{L^{p_0}(G_{1/2})}. \tag*{(54)} ∥ v − 1 ∥ L ∞ ( G 0 ) ≤ C ∥ v − 1 ∥ L p 0 ( G 1/2 ) . (54) Proof. Choose p ∗ = ( 1 + 1 / κ ) / 2 < 1 p_* = (1 + 1/\kappa)/2 < 1 p ∗ = ( 1 + 1/ κ ) /2 < 1 Lemma 5.1 . Combining (48) with (37) gives the single-step bound
∥ v ∥ L p κ ( E a ) ≤ ( C ( 1 + p ) ( c − a ) N ) 1 / p ∥ v ∥ L p ( E c ) , 0 < p ≤ p ∗ . (55) \lVert v\rVert_{L^{p\kappa}(E_a)} \leq\left(\frac{C(1+p)}{(c-a)^N}\right)^{1/p}\lVert v\rVert_{L^p(E_c)}, \qquad0 < p \leq p_*. \tag*{(55)} ∥ v ∥ L p κ ( E a ) ≤ ( ( c − a ) N C ( 1 + p ) ) 1/ p ∥ v ∥ L p ( E c ) , 0 < p ≤ p ∗ . (55) On G a ⊂ G c G_a \subset G_c G a ⊂ G c p = − β p=-\beta p = − β (48) gives the corresponding estimate for v − 1 v^{-1} v − 1 β \beta β β κ \beta\kappa β κ β > 0 \beta> 0 β > 0 (50) is at the initial endpoint for G a ⊂ G c G_a \subset G_c G a ⊂ G c E a ⊂ E c E_a \subset E_c E a ⊂ E c
For the inverse iteration use β j = κ j p \beta_j = \kappa^j p β j = κ j p ( c − a ) 2 − j − 1 (c-a)2^{-j-1} ( c − a ) 2 − j − 1
∑ j = 0 ∞ log C + log ( 1 + κ j p ) + N j log 2 + N log ( 1 / ( c − a ) ) κ j p ≤ C + N ′ log ( 1 / ( c − a ) ) p \sum_{j=0}^{\infty}\frac{\log C+\log(1+\kappa^j p)+Nj\log2+N\log(1/(c-a))}{\kappa^j p}
\leq\frac{C+N'\log(1/(c-a))}{p} j = 0 ∑ ∞ κ j p log C + log ( 1 + κ j p ) + N j log 2 + N log ( 1/ ( c − a )) ≤ p C + N ′ log ( 1/ ( c − a )) for p ≤ p 0 p \leq p_0 p ≤ p 0 L ∞ L^\infty L ∞ v − 1 v^{-1} v − 1 L p L^p L p (52) and (54) .
For (51) , stop when the exponent first reaches or exceeds p 0 p_0 p 0 (53) , start at p 0 p_0 p 0 1 / κ 1/\kappa 1/ κ 1 / κ 1/\kappa 1/ κ (55) reaches 1. All input exponents are at most max ( p 0 , 1 / κ ) < p ∗ \max(p_0,1/\kappa) < p_* max ( p 0 , 1/ κ ) < p ∗
A weighted logarithmic inequality
Choose a radial Lipschitz cutoff η \eta η B 5 / 2 B_{5/2} B 5/2 B 3 B_3 B 3 θ = η 2 \theta=\eta^2 θ = η 2 m θ = ∫ θ m_\theta=\int\theta m θ = ∫ θ
D θ ( w ) = 1 2 ∬ B 3 × B 3 min ( θ ( x ) , θ ( y ) ) ∣ w ( x ) − w ( y ) ∣ 2 K ( t ; x , y ) d x d y , w θ = m θ − 1 ∫ θ w . D_\theta(w)=\frac{1}{2}\iint_{B_3\times B_3}\min(\theta(x),\theta(y))|w(x)-w(y)|^2K(t;x,y)\,dx\,dy,\qquad w_\theta=m_\theta^{-1}\int\theta w. D θ ( w ) = 2 1 ∬ B 3 × B 3 min ( θ ( x ) , θ ( y )) ∣ w ( x ) − w ( y ) ∣ 2 K ( t ; x , y ) d x d y , w θ = m θ − 1 ∫ θ w . Dyda and Kassmann [9 ] use radial level sets to derive weighted Poincaré inequalities from unweighted estimates. Here the level sets of θ \theta θ (3) applies directly.
Lemma 5.4 (Weighted Poincaré inequality). For this cutoff,
∫ θ ∣ w − w θ ∣ 2 d x ≤ C D θ ( w ) . (56) \int\theta|w-w_\theta|^2\,dx \leq CD_\theta(w). \tag*{(56)} ∫ θ ∣ w − w θ ∣ 2 d x ≤ C D θ ( w ) . (56) At scale R R R the right side is C R α D θ ( w ) CR^\alpha D_\theta(w) C R α D θ ( w )
Proof. Use
min ( θ ( x ) , θ ( y ) ) = ∫ 0 1 1 { θ > q } ( x ) 1 { θ > q } ( y ) d q . \min(\theta(x),\theta(y))=\int_0^1\mathbf{1}_{\{\theta>q\}}(x)\mathbf{1}_{\{\theta>q\}}(y)\,dq. min ( θ ( x ) , θ ( y )) = ∫ 0 1 1 { θ > q } ( x ) 1 { θ > q } ( y ) d q . The superlevel sets are balls. Applying (3) on each such ball and integrating in q q q
D θ ( w ) ≥ λ ∬ B 3 × B 3 min ( θ ( x ) , θ ( y ) ) ∣ w ( x ) − w ( y ) ∣ 2 ∣ x − y ∣ n + α d x d y . D_\theta(w)\geq\lambda\iint_{B_3\times B_3}\min(\theta(x),\theta(y))\frac{|w(x)-w(y)|^2}{|x-y|^{n+\alpha}}\,dx\,dy. D θ ( w ) ≥ λ ∬ B 3 × B 3 min ( θ ( x ) , θ ( y )) ∣ x − y ∣ n + α ∣ w ( x ) − w ( y ) ∣ 2 d x d y . Restrict y y y B 5 / 2 B_{5/2} B 5/2 x x x B 3 B_3 B 3 θ ( x ) \theta(x) θ ( x ) ∣ x − y ∣ ≤ 6 |x-y|\le6 ∣ x − y ∣ ≤ 6 c ∫ θ ∣ w − w B 5 / 2 ∣ 2 c\int\theta|w-w_{B_{5/2}}|^2 c ∫ θ ∣ w − w B 5/2 ∣ 2 w w w w w w
Lemma 5.5 (Weighted logarithmic mean). For v v v
W ( t ) = m θ − 1 ∫ θ ( x ) log v ( t , x ) d x . W(t)=m_\theta^{-1}\int\theta(x)\log v(t,x)\,dx. W ( t ) = m θ − 1 ∫ θ ( x ) log v ( t , x ) d x . Then W W W
W ′ ( t ) ≥ c D θ ( log v ) − C ( 1 + ∥ b ∥ ∞ 2 ) ≥ c ′ ∫ θ ∣ log v − W ( t ) ∣ 2 d x − C ( 1 + ∥ b ∥ ∞ 2 ) . (57) W'(t)\ge cD_\theta(\log v)-C(1+\|b\|_\infty^2)\ge c'\int\theta|\log v-W(t)|^2\,dx-C(1+\|b\|_\infty^2). \tag*{(57)} W ′ ( t ) ≥ c D θ ( log v ) − C ( 1 + ∥ b ∥ ∞ 2 ) ≥ c ′ ∫ θ ∣ log v − W ( t ) ∣ 2 d x − C ( 1 + ∥ b ∥ ∞ 2 ) . (57) Proof. For a , b > 0 a,b>0 a , b > 0 A , B ≥ 0 A,B\ge0 A , B ≥ 0
( a − b ) ( A 2 / a − B 2 / b ) ≤ − 1 2 min ( A , B ) 2 ∣ log ( a / b ) ∣ 2 + 3 ( A − B ) 2 . (58) (a-b)(A^2/a-B^2/b)\le-\frac{1}{2}\min(A,B)^2|\log(a/b)|^2+3(A-B)^2. \tag*{(58)} ( a − b ) ( A 2 / a − B 2 / b ) ≤ − 2 1 min ( A , B ) 2 ∣ log ( a / b ) ∣ 2 + 3 ( A − B ) 2 . (58) By symmetry we may assume A ≥ B A\ge B A ≥ B d = A − B d=A-B d = A − B t = a / b t=a/b t = a / b
− B 2 ( t + t − 1 − 2 ) + ( 2 B d + d 2 ) ( 1 − t − 1 ) . -B^2(t+t^{-1}-2)+(2Bd+d^2)(1-t^{-1}). − B 2 ( t + t − 1 − 2 ) + ( 2 B d + d 2 ) ( 1 − t − 1 ) . Write Z = t − 1 / t Z=\sqrt{t}-1/\sqrt{t} Z = t − 1/ t Z 2 = t + t − 1 − 2 ≥ ∣ log t ∣ 2 Z^2=t+t^{-1}-2\ge|\log t|^2 Z 2 = t + t − 1 − 2 ≥ ∣ log t ∣ 2 t ≤ 1 t\le1 t ≤ 1 t ≥ 1 t\ge1 t ≥ 1 1 − t − 1 ≤ min ( Z , 1 ) 1-t^{-1}\le\min(Z,1) 1 − t − 1 ≤ min ( Z , 1 ) 2 B d Z ≤ B 2 Z 2 / 2 + 2 d 2 2BdZ\le B^2Z^2/2+2d^2 2 B d Z ≤ B 2 Z 2 /2 + 2 d 2
Test the supersolution by η 2 / v \eta^2/v η 2 / v B 4 × B 4 B_4\times B_4 B 4 × B 4 − c D θ ( log v ) + C -cD_\theta(\log v)+C − c D θ ( log v ) + C
C + Θ ( u − ; t , 3 , 4 ) ∫ η 2 v − 1 d x . C+\Theta(u_-;t,3,4)\int\eta^2v^{-1}\,dx. C + Θ ( u − ; t , 3 , 4 ) ∫ η 2 v − 1 d x . The shift h ′ h' h ′ f − f_- f −
− ∫ log v b ⋅ ∇ θ d x = − 2 ∫ ( log v − W ) η b ⋅ ∇ η d x . -\int\log v\,b\cdot\nabla\theta\,dx=-2\int(\log v-W)\eta b\cdot\nabla\eta\,dx. − ∫ log v b ⋅ ∇ θ d x = − 2 ∫ ( log v − W ) η b ⋅ ∇ η d x . The equality uses ∫ b ⋅ ∇ θ = 0 \int b\cdot\nabla\theta=0 ∫ b ⋅ ∇ θ = 0
C ∥ b ∥ ∞ D θ ( log v ) 1 / 2 ≤ c 2 D θ ( log v ) + C ∥ b ∥ ∞ 2 . C\|b\|_\infty D_\theta(\log v)^{1/2}\le\frac{c}{2}D_\theta(\log v)+C\|b\|_\infty^2. C ∥ b ∥ ∞ D θ ( log v ) 1/2 ≤ 2 c D θ ( log v ) + C ∥ b ∥ ∞ 2 . Lemma 2.6 with a Lipschitz extension of r ↦ r − 1 r\mapsto r^{-1} r ↦ r − 1 [ ε , ∞ ) [\varepsilon,\infty) [ ε , ∞ ) W W W m θ W ′ m_\theta W' m θ W ′ h ′ h' h ′ m θ m_\theta m θ
Lemma 5.6 (One-sided logarithmic distribution estimates). There is a ∈ R a\in\mathbb{R} a ∈ R ℓ > 0 \ell>0 ℓ > 0
∣ { E 1 : log v > a + ℓ } ∣ + ∣ { G 1 : log v < a − ℓ } ∣ ≤ C / ℓ . (59) |\{E_1:\log v>a+\ell\}|+|\{G_1:\log v<a-\ell\}|\le C/\ell. \tag*{(59)} ∣ { E 1 : log v > a + ℓ } ∣ + ∣ { G 1 : log v < a − ℓ } ∣ ≤ C / ℓ . (59) Proof. Let c 0 = C ( 1 + ∥ b ∥ ∞ 2 ) c_0=C(1+\|b\|_\infty^2) c 0 = C ( 1 + ∥ b ∥ ∞ 2 )
z ( t , x ) = log v ( t , x ) + c 0 t , Z ( t ) = W ( t ) + c 0 t . z(t,x)=\log v(t,x)+c_0t,\qquad Z(t)=W(t)+c_0t. z ( t , x ) = log v ( t , x ) + c 0 t , Z ( t ) = W ( t ) + c 0 t . Then Z ′ ≥ c ∫ θ ∣ z − Z ∣ 2 ≥ 0 Z'\ge c\int\theta|z-Z|^2\ge0 Z ′ ≥ c ∫ θ ∣ z − Z ∣ 2 ≥ 0 τ ∗ = − 1 / 2 \tau_*=-1/2 τ ∗ = − 1/2 a = Z ( τ ∗ ) a=Z(\tau_*) a = Z ( τ ∗ ) Z Z Z t < τ ∗ t<\tau_* t < τ ∗ H ( t ) = a − Z ( t ) ≥ 0 H(t)=a-Z(t)\ge0 H ( t ) = a − Z ( t ) ≥ 0 { x ∈ B 5 / 2 : z ( t , x ) > a + ℓ } \{x\in B_{5/2}:z(t,x)>a+\ell\} { x ∈ B 5/2 : z ( t , x ) > a + ℓ } z − Z > ℓ + H z-Z>\ell+H z − Z > ℓ + H
∣ { x ∈ B 5 / 2 : z ( t , x ) > a + ℓ } ∣ ≤ − H ′ ( t ) c ( ℓ + H ( t ) ) 2 . |\{x\in B_{5/2}:z(t,x)>a+\ell\}|\le\frac{-H'(t)}{c(\ell+H(t))^2}. ∣ { x ∈ B 5/2 : z ( t , x ) > a + ℓ } ∣ ≤ c ( ℓ + H ( t ) ) 2 − H ′ ( t ) . Since − H ′ / ( ℓ + H ) 2 -H'/(\ell+H)^2 − H ′ / ( ℓ + H ) 2 ( ℓ + H ) − 1 (\ell+H)^{-1} ( ℓ + H ) − 1 1 / ( c ℓ ) 1/(c\ell) 1/ ( c ℓ ) H = Z − a H=Z-a H = Z − a { x ∈ B 5 / 2 : z ( t , x ) < a − ℓ } \{x\in B_{5/2}:z(t,x)<a-\ell\} { x ∈ B 5/2 : z ( t , x ) < a − ℓ } H ′ / ( c ( ℓ + H ) 2 ) H'/(c(\ell+H)^2) H ′ / ( c ( ℓ + H ) 2 ) E 1 E_1 E 1 G 1 G_1 G 1 B 5 / 2 B_{5/2} B 5/2
sides of τ ∗ \tau_* τ ∗ z z z log v \log v log v ∣ c 0 t ∣ ≤ 4 c 0 |c_0t| \le4c_0 ∣ c 0 t ∣ ≤ 4 c 0 C 1 C_1 C 1 ℓ > 8 c 0 \ell> 8c_0 ℓ > 8 c 0 ℓ / 2 \ell/2 ℓ /2 ℓ \ell ℓ (59) after increasing C C C
Proof of weak Harnack
Proof of Theorem 1.2. At unit scale, let a 0 a_0 a 0 e − a 0 v e^{-a_0v} e − a 0 v ( E r ) (E_r) ( E r ) e a 0 v − 1 e^{a_0v^{-1}} e a 0 v − 1 ( G r ) (G_r) ( G r ) (51) and (52) , respectively. The two level-set bounds in (59) give the logarithmic hypotheses. Hence
∥ v ∥ L p 0 ( E 1 / 2 ) ≤ C e a 0 , ∥ v − 1 ∥ L p 0 ( G 1 / 2 ) ≤ C e − a 0 . \|v\|_{L^{p_0}(E_{1/2})} \le Ce^{a_0}, \qquad\|v^{-1}\|_{L^{p_0}(G_{1/2})} \le Ce^{-a_0}. ∥ v ∥ L p 0 ( E 1/2 ) ≤ C e a 0 , ∥ v − 1 ∥ L p 0 ( G 1/2 ) ≤ C e − a 0 . Equations (53) and (54) imply
∫ P 1 v ≤ C e a 0 ≤ C ess inf F 1 v . \int_{P_1} v \le Ce^{a_0} \le C\operatorname*{ess\,inf}_{F_1}v. ∫ P 1 v ≤ C e a 0 ≤ C F 1 ess inf v . Since u ≤ v u \le v u ≤ v C 1 C_1 C 1 h ( t ) ≤ ε + 5 ∥ f − ∥ ∞ + T R − ( u ) h(t) \le\varepsilon+ 5\|f_-\|_\infty+ T_R^-(u) h ( t ) ≤ ε + 5∥ f − ∥ ∞ + T R − ( u ) ε ↓ 0 \varepsilon\downarrow0 ε ↓ 0
u ~ ( τ , ξ ) = u ( t 0 + R α τ , x 0 + R ξ ) , K ~ = R n + α K , b ~ = R α − 1 b , f ~ = R α f \widetilde{u}(\tau,\xi)=u(t_0+R^\alpha\tau,x_0+R\xi),\qquad\widetilde{K}=R^{n+\alpha}K,\qquad\widetilde{b}=R^{\alpha-1}b,\qquad\widetilde{f}=R^\alpha f u ( τ , ξ ) = u ( t 0 + R α τ , x 0 + R ξ ) , K = R n + α K , b = R α − 1 b , f = R α f preserves (1.3)–(1.5), with the coefficients on the right evaluated at the transformed arguments. Since ∫ Θ ~ d τ = T R − ( u ) \int\widetilde{\Theta}\,d\tau=T_R^-(u) ∫ Θ d τ = T R − ( u ) ∥ b ~ ∥ L ∞ ( C 1 ) = M R \|\widetilde{b}\|_{L^\infty(C_1)}=MR ∥ b ∥ L ∞ ( C 1 ) = M R
Corollary 5.7 (Growth from a positive-measure set). In the setting of Theorem 1.2, if
∣ { u ≥ k } ∩ P R ∣ ≥ ν ∣ P R ∣ , k > 0 , 0 < ν ≤ 1 , |\{u\ge k\}\cap P_R|\ge\nu|P_R|,\qquad k>0,\qquad0<\nu\le1, ∣ { u ≥ k } ∩ P R ∣ ≥ ν ∣ P R ∣ , k > 0 , 0 < ν ≤ 1 , then
ess inf F R u ≥ c ν k − C ( R α ∥ f − ∥ ∞ + T R − ( u ) ) . (60) \operatorname*{ess\,inf}_{F_R}u\ge c\nu k-C\left(R^\alpha\|f_-\|_\infty+T_R^-(u)\right). \tag*{(60)} F R ess inf u ≥ c ν k − C ( R α ∥ f − ∥ ∞ + T R − ( u ) ) . (60) Proof. The average on the left of (1.14) is at least ν k \nu k ν k
Oscillation decay and Hölder continuity
Proof of Theorem 1.3. Translate and scale to Q 1 = ( − 1 , 0 ) × B 1 Q_1=(-1,0)\times B_1 Q 1 = ( − 1 , 0 ) × B 1
Θ ( t ) = Θ ( u ; t , 1 / 2 , 1 ) , F = ∥ f ∥ L ∞ ( Q 1 ) , A = 2 U + ∥ Θ ∥ L q ( − 1 , 0 ) + F . \Theta(t)=\Theta(u;t,1/2,1),\qquad F=\|f\|_{L^\infty(Q_1)},\qquad A=2U+\|\Theta\|_{L^q(-1,0)}+F. Θ ( t ) = Θ ( u ; t , 1/2 , 1 ) , F = ∥ f ∥ L ∞ ( Q 1 ) , A = 2 U + ∥Θ ∥ L q ( − 1 , 0 ) + F . When A = 0 A=0 A = 0 c c c c ≤ 1 c\le1 c ≤ 1
a 0 = c / 16 , θ = 1 − a 0 . a_0=c/16,\qquad\theta=1-a_0. a 0 = c /16 , θ = 1 − a 0 . We choose a radius ratio ρ ∈ ( 0 , 1 / 8 ) \rho\in(0,1/8) ρ ∈ ( 0 , 1/8 ) r j = ρ j r_j=\rho^j r j = ρ j ω j = A θ j \omega_j=A\theta^j ω j = A θ j [ m j , M j ] ⊂ [ − U , U ] [m_j,M_j]\subset[-U,U] [ m j , M j ] ⊂ [ − U , U ]
m j ≤ u ≤ M j on Q r j , M j − m j ≤ ω j , (61) m_j\le u\le M_j\ \text{ on }Q_{r_j},\qquad M_j-m_j\le\omega_j, \tag*{(61)} m j ≤ u ≤ M j on Q r j , M j − m j ≤ ω j , (61) and such that each endpoint changes by at most a 0 ω j a_0\omega_j a 0 ω j j j j j + 1 j+1 j + 1 m 0 = − U m_0=-U m 0 = − U M 0 = U M_0=U M 0 = U
Suppose (61) holds through scale j j j M j − m j ≤ θ ω j M_j-m_j\le\theta\omega_j M j − m j ≤ θ ω j d j = M j − m j > θ ω j d_j=M_j-m_j>\theta\omega_j d j = M j − m j > θ ω j r = r j + 1 r=r_{j+1} r = r j + 1 − r α -r^\alpha − r α
( − 5 r α , 0 ) × B 4 r ⊂ Q r j (-5r^\alpha,0)\times B_{4r}\subset Q_{r_j} ( − 5 r α , 0 ) × B 4 r ⊂ Q r j provided 5 ρ α ≤ 1 5\rho^\alpha\le1 5 ρ α ≤ 1 Q r Q_r Q r
{ u − m j ≥ d j / 2 } , { M j − u ≥ d j / 2 } \{u-m_j\ge d_j/2\},\qquad\{M_j-u\ge d_j/2\} { u − m j ≥ d j /2 } , { M j − u ≥ d j /2 } has relative measure at least 1 / 2 1/2 1/2 (60) contains the term c d j / 4 ≥ 2 a 0 ω j cd_j/4\ge2a_0\omega_j c d j /4 ≥ 2 a 0 ω j Q r j Q_{r_j} Q r j B r j c B^c_{r_j} B r j c B 1 c B^c_1 B 1 c B r i ∖ B r i + 1 B_{r_i}\setminus B_{r_{i+1}} B r i ∖ B r i + 1 0 ≤ i < j 0 \le i < j 0 ≤ i < j i i i ( − 5 r α , 0 ) (-5r^\alpha,0) ( − 5 r α , 0 )
max ( m j − m i , M i − M j ) ≤ a 0 ∑ k = i j − 1 ω k ≤ ω i . \max(m_j-m_i,M_i-M_j) \le a_0\sum_{k=i}^{j-1}\omega_k \le\omega_i. max ( m j − m i , M i − M j ) ≤ a 0 k = i ∑ j − 1 ω k ≤ ω i . For x ∈ B 3 r x \in B_{3r} x ∈ B 3 r y y y ∣ x − y ∣ ≥ r i + 1 − 3 r ≥ r i + 1 / 2 |x-y| \ge r_{i+1}-3r \ge r_{i+1}/2 ∣ x − y ∣ ≥ r i + 1 − 3 r ≥ r i + 1 /2
T j a n n ≤ C ∑ i = 0 j − 1 ( r / r i + 1 ) α ω i ≤ C ω j ∑ ℓ = 1 j ( ρ α / θ ) ℓ ≤ C ω j ρ α θ − ρ α . (62) \begin{aligned}
T_j^{\mathrm{ann}} \le C\sum_{i=0}^{j-1}(r/r_{i+1})^\alpha\omega_i \\
&\le C\omega_j\sum_{\ell=1}^{j}(\rho^\alpha/\theta)^\ell\le C\omega_j\frac{\rho^\alpha}{\theta-\rho^\alpha}. \tag*{(62)}
\end{aligned} T j ann ≤ C i = 0 ∑ j − 1 ( r / r i + 1 ) α ω i ≤ C ω j ℓ = 1 ∑ j ( ρ α / θ ) ℓ ≤ C ω j θ − ρ α ρ α . (62) All time intervals used here are contained in the earlier Q r i Q_{r_i} Q r i 5 ρ α ≤ 1 5\rho^\alpha\le1 5 ρ α ≤ 1
For y ∉ B 1 y\notin B_1 y ∈ / B 1 U + ∣ u ( t , y ) ∣ U+|u(t,y)| U + ∣ u ( t , y ) ∣ [ − U , U ] [-U,U] [ − U , U ]
T j f a r ≤ C U r α + C ∥ Θ ∥ L q ( − 1 , 0 ) r α ( 1 − 1 / q ) . (63) T_j^{\mathrm{far}}\le CU r^\alpha+C\|\Theta\|_{L^q(-1,0)}r^{\alpha(1-1/q)}. \tag*{(63)} T j far ≤ C U r α + C ∥Θ ∥ L q ( − 1 , 0 ) r α ( 1 − 1/ q ) . (63) The source contributes at most C r α F Cr^\alpha F C r α F β = α ( 1 − 1 / q ) > 0 \beta=\alpha(1-1/q)>0 β = α ( 1 − 1/ q ) > 0 ρ \rho ρ
ρ α < θ , ρ β < θ , C ( ρ α θ − ρ α + ρ α + ρ β ) ≤ a 0 . \rho^\alpha<\theta,\qquad\rho^\beta<\theta,\qquad C\left(\frac{\rho^\alpha}{\theta-\rho^\alpha}+\rho^\alpha+\rho^\beta\right)\le a_0. ρ α < θ , ρ β < θ , C ( θ − ρ α ρ α + ρ α + ρ β ) ≤ a 0 . Since U , ∥ Θ ∥ q , F ≤ A U,\|\Theta\|_q,F\le A U , ∥Θ ∥ q , F ≤ A a 0 ω j a_0\omega_j a 0 ω j Q r Q_r Q r a 0 ω j a_0\omega_j a 0 ω j d j − θ ω j ≤ a 0 ω j d_j-\theta\omega_j\le a_0\omega_j d j − θ ω j ≤ a 0 ω j j + 1 j+1 j + 1
Let
γ = − log θ log ( 1 / ρ ) . \gamma=\frac{-\log\theta}{\log(1/\rho)}. γ = log ( 1/ ρ ) − log θ . Decreasing ρ \rho ρ 0 < γ < min ( 1 , β ) 0<\gamma<\min(1,\beta) 0 < γ < min ( 1 , β ) r j + 1 < r ≤ r j r_{j+1}<r\le r_j r j + 1 < r ≤ r j
ess osc Q r u ≤ ω j ≤ C A r γ . (64) \underset{Q_r}{\operatorname{ess\,osc}}\,u\le\omega_j\le CAr^\gamma. \tag*{(64)} Q r ess osc u ≤ ω j ≤ C A r γ . (64) The constants in (5.16) are uniform in j j j r j α − 1 ∥ b ∥ ∞ ≤ ∥ b ∥ ∞ r_j^{\alpha-1}\|b\|_\infty\le\|b\|_\infty r j α − 1 ∥ b ∥ ∞ ≤ ∥ b ∥ ∞ α ≥ 1 \alpha\ge1 α ≥ 1
Repeat the argument on a backward cylinder of radius 1 / 8 1/8 1/8 Q 1 / 4 Q_{1/4} Q 1/4 B 1 B_1 B 1 C U CU C U B 1 B_1 B 1 Θ ( t ) \Theta(t) Θ ( t ) B 1 / 2 B_{1/2} B 1/2 L q L^q L q C ( U + ∥ Θ ∥ q ) C(U+\|\Theta\|_q) C ( U + ∥Θ ∥ q )
The oscillation bound makes averages on shrinking backward cylinders converge at each center. Their limits agree with u u u 0 < d < 1 / 64 0<d<1/64 0 < d < 1/64 z z z Q 2 d ( z ) Q_{2d}(z) Q 2 d ( z ) C A d γ CAd^\gamma C A d γ 2 U 2U 2 U Q 1 / 4 Q_{1/4} Q 1/4 □ \square □
Corollary 6.1 (Globally bounded solutions and positive times). If u u u Q R Q_R Q R C ∥ u ∥ L ∞ ( I × R n ) C\|u\|_{L^\infty(I\times\mathbb{R}^n)} C ∥ u ∥ L ∞ ( I × R n ) u 0 ∈ L 2 u_0\in L^2 u 0 ∈ L 2 ( 0 , T ) × R n (0,T)\times\mathbb{R}^n ( 0 , T ) × R n
Proof. The first assertion is (1.11). For every τ > 0 \tau>0 τ > 0 [ τ , T ] × R n [\tau,T]\times\mathbb{R}^n [ τ , T ] × R n τ \tau τ □ \square □
Positive tails and the parabolic Harnack inequality
The kernel upper bound and an earlier-tail estimate
Lemma 7.1. Symmetry, (5) , and (1.16) imply (44) , with L L L n n n α \alpha α Λ \Lambda Λ C J C_J C J
Proof. For d = ∣ x − y ∣ d=\lvert x-y\rvert d = ∣ x − y ∣ r = d / 4 r=d/4 r = d /4
K ( t ; x , y ) ≤ C J ∣ B d / 4 ∣ − 1 ∫ B d / 4 ( x ) K ( t ; y , z ) d z ≤ C d − n ∫ ∣ y − z ∣ ≥ 3 d / 4 K ( t ; y , z ) d z ≤ C d − n − α . K(t;x,y) \le C_J\lvert B_{d/4}\rvert^{-1}\int_{B_{d/4}(x)}K(t;y,z)\,\mathrm{d}z \le Cd^{-n}\int_{\lvert y-z\rvert\ge3d/4}K(t;y,z)\,\mathrm{d}z \le Cd^{-n-\alpha}. K ( t ; x , y ) ≤ C J ∣ B d /4 ∣ − 1 ∫ B d /4 ( x ) K ( t ; y , z ) d z ≤ C d − n ∫ ∣ y − z ∣ ≥ 3 d /4 K ( t ; y , z ) d z ≤ C d − n − α . Work at unit scale and let u u u (1) in C 1 C_1 C 1 u ≥ 0 u\ge0 u ≥ 0 ( − 4 , 1 ) × R n (-4,1)\times\mathbb{R}^n ( − 4 , 1 ) × R n Lemma 7.1 gives the pointwise upper bound needed in Lemma 4.6, so u u u C 1 C_1 C 1 F = ∥ f ∥ L ∞ ( C 1 ) F=\lVert f\rVert_{L^\infty(C_1)} F = ∥ f ∥ L ∞ ( C 1 ) ε > 0 \varepsilon>0 ε > 0
v = u + F ( t + 4 ) + ε , H = ess inf F 1 u + 5 F + ε . v=u+F(t+4)+\varepsilon,\qquad H=\underset{F_1}{\operatorname{ess\,inf}}u+5F+\varepsilon. v = u + F ( t + 4 ) + ε , H = F 1 ess inf u + 5 F + ε . Since f + F ≥ 0 f+F\ge0 f + F ≥ 0 v v v Theorem 1.2 and using ess inf F 1 v ≤ H \underset{F_1}{\operatorname{ess\,inf}}v\le H F 1 ess inf v ≤ H
∫ P 1 v ≤ C H . (65) \int_{P_1}v\le CH. \tag*{(65)} ∫ P 1 v ≤ C H . (65) For 0 ≤ a ≤ 1 0\le a\le1 0 ≤ a ≤ 1
U a = J a × B R a , J a = ( − 5 / 2 − a / 2 , − 2 + a / 2 ) , R a = 1 + a / 2 , M ( a ) = ess sup U a v . U_a=J_a\times B_{R_a},\qquad J_a=(-5/2-a/2,-2+a/2),\qquad R_a=1+a/2,\qquad M(a)=\underset{U_a}{\operatorname{ess\,sup}}v. U a = J a × B R a , J a = ( − 5/2 − a /2 , − 2 + a /2 ) , R a = 1 + a /2 , M ( a ) = U a ess sup v . These cylinders lie in P 1 P_1 P 1 U 0 = H 1 U_0=H_1 U 0 = H 1 M M M [ 0 , 1 ] [0,1] [ 0 , 1 ] 0 ≤ a < ξ ≤ 1 0\le a<\xi\le1 0 ≤ a < ξ ≤ 1 R m = ( R a + R ξ ) / 2 R_m=(R_a+R_\xi)/2 R m = ( R a + R ξ ) /2
S a , ξ = ∫ J ξ Θ ( u ; t , R m , R ξ ) d t . S_{a,\xi}=\int_{J_\xi}\Theta(u;t,R_m,R_\xi)\,\mathrm{d}t. S a , ξ = ∫ J ξ Θ ( u ; t , R m , R ξ ) d t . The spatial gap and (44) make this quantity finite by Cauchy–Schwarz and the global L t ∞ L x 2 L_t^\infty L_x^2 L t ∞ L x 2 u u u
Lemma 7.2 (Control of an earlier positive tail). For 0 ≤ a < ξ < c ≤ 1 0\le a<\xi<c\le1 0 ≤ a < ξ < c ≤ 1 q ∈ ( 0 , 1 ) q\in(0,1) q ∈ ( 0 , 1 )
S a , ξ ≤ C ( ( ξ − a ) ( c − ξ ) ) − N M ( c ) q H 1 − q . (66) S_{a,\xi}\le C((\xi-a)(c-\xi))^{-N}M(c)^qH^{1-q}. \tag*{(66)} S a , ξ ≤ C (( ξ − a ) ( c − ξ ) ) − N M ( c ) q H 1 − q . (66) Consequently, for every δ > 0 \delta>0 δ > 0
S a , ξ ≤ δ M ( c ) + C δ ( ( ξ − a ) ( c − ξ ) ) − N δ H . (67) S_{a,\xi}\le\delta M(c)+C_\delta((\xi-a)(c-\xi))^{-N_\delta}H. \tag*{(67)} S a , ξ ≤ δ M ( c ) + C δ (( ξ − a ) ( c − ξ ) ) − N δ H . (67) Proof. Put d = ( R ξ − R m ) / 4 d=(R_\xi-R_m)/4 d = ( R ξ − R m ) /4 0 ≤ η ≤ 1 0\le\eta\le1 0 ≤ η ≤ 1 C c 0 , 1 ( B R ξ − d ) C_c^{0,1}(B_{R_\xi-d}) C c 0 , 1 ( B R ξ − d ) B R m + d B_{R_m+d} B R m + d ∣ ∇ η ∣ ≤ C / d \lvert\nabla\eta\rvert\le C/d ∣ ∇ η ∣ ≤ C / d η 2 v − q \eta^2v^{-q} η 2 v − q
I ( t ) = ∫ B R ξ η ( x ) 2 v ( t , x ) − q ∫ B R ξ c u ( t , y ) K ( t ; x , y ) d y d x . I(t)=\int_{B_{R_\xi}}\eta(x)^2v(t,x)^{-q}\int_{B_{R_\xi}^{c}}u(t,y)K(t;x,y)\,\mathrm{d}y\,\mathrm{d}x. I ( t ) = ∫ B R ξ η ( x ) 2 v ( t , x ) − q ∫ B R ξ c u ( t , y ) K ( t ; x , y ) d y d x . The weighted power inequality with exponent 1 − q 1-q 1 − q (49) give, after discarding a nonnegative energy term,
− 1 1 − q d d t ∫ η 2 v 1 − q d x + I ( t ) ≤ C q ( d − α + ∥ b ∥ ∞ d − 1 ) ∫ B R ξ v 1 − q d x . (68) -\frac{1}{1-q}\frac{\mathrm{d}}{\mathrm{d}t}\int\eta^2v^{1-q}\,\mathrm{d}x+I(t)\le C_q(d^{-\alpha}+\lVert b\rVert_\infty d^{-1})\int_{B_{R_\xi}}v^{1-q}\,\mathrm{d}x. \tag*{(68)} − 1 − q 1 d t d ∫ η 2 v 1 − q d x + I ( t ) ≤ C q ( d − α + ∥ b ∥ ∞ d − 1 ) ∫ B R ξ v 1 − q d x . (68) In the cross interaction, (5) bounds the term containing v ( x ) v(x) v ( x ) C d − α ∫ B R ξ v 1 − q Cd^{-\alpha}\int_{B_{R_\xi}}v^{1-q} C d − α ∫ B R ξ v 1 − q v ( y ) v(y) v ( y ) I ( t ) I(t) I ( t ) v ( y ) ≥ u ( y ) v(y)\ge u(y) v ( y ) ≥ u ( y ) f + F f+F f + F t − = inf J ξ t_{-}=\inf J_{\xi} t − = inf J ξ t + = sup J c t_{+}=\sup J_{c} t + = sup J c ζ \zeta ζ J ξ J_{\xi} J ξ t + t_{+} t + ∣ ζ ′ ∣ ≤ C / ( c − ξ ) |\zeta'|\le C/(c-\xi) ∣ ζ ′ ∣ ≤ C / ( c − ξ ) Y ( t ) = ∫ η 2 v 1 − q Y(t)=\int\eta^{2}v^{1-q} Y ( t ) = ∫ η 2 v 1 − q
∫ t − t + − Y ′ ( t ) ζ ( t ) d t = Y ( t − ) + ∫ t − t + Y ( t ) ζ ′ ( t ) d t . \int_{t_{-}}^{t_{+}}-Y'(t)\zeta(t)\,\mathrm{d}t=Y(t_{-})+\int_{t_{-}}^{t_{+}}Y(t)\zeta'(t)\,\mathrm{d}t. ∫ t − t + − Y ′ ( t ) ζ ( t ) d t = Y ( t − ) + ∫ t − t + Y ( t ) ζ ′ ( t ) d t . The initial term Y ( t − ) Y(t_{-}) Y ( t − ) ζ ′ \zeta' ζ ′ − C ( c − ξ ) − 1 ∫ t − t + Y ( t ) d t -C(c-\xi)^{-1}\int_{t_{-}}^{t_{+}}Y(t)\,\mathrm{d}t − C ( c − ξ ) − 1 ∫ t − t + Y ( t ) d t ζ \zeta ζ I ≥ 0 I\ge0 I ≥ 0
∫ J ξ I ( t ) d t ≤ C q ( ξ − a ) ( c − ξ ) − N ∫ U c v 1 − q d x d t . (69) \int_{J_{\xi}}I(t)\,\mathrm{d}t\le C_{q}(\xi-a)(c-\xi)^{-N}\int_{U_{c}}v^{1-q}\,\mathrm{d}x\,\mathrm{d}t. \tag*{(69)} ∫ J ξ I ( t ) d t ≤ C q ( ξ − a ) ( c − ξ ) − N ∫ U c v 1 − q d x d t . (69) For z ∈ B R m z\in B_{R_{m}} z ∈ B R m y ∉ B R ξ y\notin B_{R_{\xi}} y ∈ / B R ξ d d d d ≤ ∣ z − y ∣ / 4 d\le|z-y|/4 d ≤ ∣ z − y ∣/4 η = 1 \eta=1 η = 1 B d ( z ) B_{d}(z) B d ( z )
I ( t ) ≥ M ( c ) − q ∫ B d ( z ) ∫ B R ξ c u ( t , y ) K ( t ; x , y ) d y d x ≥ C − 1 d n M ( c ) − q ∫ B R ξ c u ( t , y ) K ( t ; z , y ) d y . \begin{aligned}
I(t)\ge M(c)^{-q}\int_{B_{d}(z)}\int_{B_{R_{\xi}}^{c}}u(t,y)K(t;x,y)\,\mathrm{d}y\,\mathrm{d}x \\
&\ge C^{-1}d^{n}M(c)^{-q}\int_{B_{R_{\xi}}^{c}}u(t,y)K(t;z,y)\,\mathrm{d}y.
\end{aligned} I ( t ) ≥ M ( c ) − q ∫ B d ( z ) ∫ B R ξ c u ( t , y ) K ( t ; x , y ) d y d x ≥ C − 1 d n M ( c ) − q ∫ B R ξ c u ( t , y ) K ( t ; z , y ) d y . Taking the essential supremum over z z z J ξ J_{\xi} J ξ (69) give S a , ξ ≤ C ( ( ξ − a ) ( c − ξ ) ) − N M ( c ) − q ∫ U c v 1 − q S_{a,\xi}\le C((\xi-a)(c-\xi))^{-N}M(c)^{-q}\int_{U_{c}}v^{1-q} S a , ξ ≤ C (( ξ − a ) ( c − ξ ) ) − N M ( c ) − q ∫ U c v 1 − q U c ⊂ P 1 U_{c}\subset P_{1} U c ⊂ P 1 (65) bound the integral by ∣ U c ∣ q ( ∫ P 1 v ) 1 − q ≤ C H 1 − q |U_{c}|^{q}(\int_{P_{1}}v)^{1-q}\le CH^{1-q} ∣ U c ∣ q ( ∫ P 1 v ) 1 − q ≤ C H 1 − q (66) , and Young’s inequality yields (67) . □
Interpolation for the local supremum
Lemma 7.3 (Local interpolation with an integrated tail). For 0 ≤ a < ξ ≤ 1 0\le a<\xi\le1 0 ≤ a < ξ ≤ 1 and every δ ∈ ( 0 , 1 ) \delta\in(0,1) δ ∈ ( 0 , 1 )
ess sup U a u ≤ δ M ( ξ ) + C δ ( ξ − a ) − N δ ∥ u ∥ L 2 ( U ξ ) + C S a , ξ + C F . (70) \operatorname*{ess\,sup}_{U_{a}}u\le\delta M(\xi)+C_{\delta}(\xi-a)^{-N_{\delta}}\lVert u\rVert_{L^{2}(U_{\xi})}+CS_{a,\xi}+CF. \tag*{(70)} U a ess sup u ≤ δ M ( ξ ) + C δ ( ξ − a ) − N δ ∥ u ∥ L 2 ( U ξ ) + C S a , ξ + C F . (70) Proof. On J ξ J_{\xi} J ξ
ℓ ( t ) = ∫ inf J ξ t ( F + Θ ( u ; τ , R m , R ξ ) ) d τ , z = u − ℓ ( t ) . \ell(t)=\int_{\inf J_{\xi}}^{t}\bigl(F+\Theta(u;\tau,R_{m},R_{\xi})\bigr)\,\mathrm{d}\tau,\qquad z=u-\ell(t). ℓ ( t ) = ∫ i n f J ξ t ( F + Θ ( u ; τ , R m , R ξ ) ) d τ , z = u − ℓ ( t ) . By Lemma 2.6, we may test u u u ℓ ( t ) + k j \ell(t)+k_{j} ℓ ( t ) + k j z z z ℓ ′ \ell' ℓ ′ B R ξ c B_{R_{\xi}}^{c} B R ξ c R m R_{m} R m R a R_{a} R a J ξ J_{\xi} J ξ J a J_{a} J a B R ξ B_{R_{\xi}} B R ξ u ≤ M ( ξ ) u\le M(\xi) u ≤ M ( ξ ) C 2 α j ( ξ − a ) − α M ( ξ ) C2^{\alpha j}(\xi-a)^{-\alpha}M(\xi) C 2 α j ( ξ − a ) − α M ( ξ ) z + ≤ u z_{+}\le u z + ≤ u A j A_{j} A j
A j + 1 ≤ C ( ξ − a ) − N 2 N j k j − 2 ( κ − 1 ) ( 1 + M ( ξ ) / k ) κ A j κ . A_{j+1}\le C(\xi-a)^{-N}2^{Nj}k_{j}^{-2(\kappa-1)}(1+M(\xi)/k)^{\kappa}A_{j}^{\kappa}. A j + 1 ≤ C ( ξ − a ) − N 2 N j k j − 2 ( κ − 1 ) ( 1 + M ( ξ ) / k ) κ A j κ . The initial integral is at most ∥ u ∥ L 2 ( U ξ ) 2 \lVert u\rVert_{L^{2}(U_{\xi})}^{2} ∥ u ∥ L 2 ( U ξ ) 2
k = δ M ( ξ ) + C δ ( ξ − a ) − N δ ∥ u ∥ L 2 ( U ξ ) . k=\delta M(\xi)+C_{\delta}(\xi-a)^{-N_{\delta}}\lVert u\rVert_{L^{2}(U_{\xi})}. k = δ M ( ξ ) + C δ ( ξ − a ) − N δ ∥ u ∥ L 2 ( U ξ ) . Then M ( ξ ) / k < δ − 1 M(\xi)/k<\delta^{-1} M ( ξ ) / k < δ − 1 C δ C_{\delta} C δ N δ N_{\delta} N δ A − 1 / k 2 A_{-1}/k^{2} A − 1 / k 2 z ≤ k z\le k z ≤ k U a U_{a} U a 0 ≤ ℓ ( t ) ≤ S a , ξ + C F 0\le\ell(t)\le S_{a,\xi}+CF 0 ≤ ℓ ( t ) ≤ S a , ξ + C F J a J_{a} J a ℓ \ell ℓ k k k
Absorption on nested cylinders
Proof of Theorem 1.5. At unit scale fix 0 ≤ a < c ≤ 1 0 \le a < c \le1 0 ≤ a < c ≤ 1 ξ = ( a + c ) / 2 \xi= (a+c)/2 ξ = ( a + c ) /2 u ≤ v ≤ M ( c ) u \le v \le M(c) u ≤ v ≤ M ( c ) U ξ U_{\xi} U ξ (65) gives
∥ u ∥ L 2 ( U ξ ) ≤ ( M ( c ) ∫ U ξ u ) 1 / 2 ≤ C M ( c ) 1 / 2 H 1 1 / 2 . \lVert u\rVert_{L^2(U_{\xi})} \le\left(M(c)\int_{U_{\xi}} u\right)^{1/2} \le C M(c)^{1/2} H_1^{1/2}. ∥ u ∥ L 2 ( U ξ ) ≤ ( M ( c ) ∫ U ξ u ) 1/2 ≤ C M ( c ) 1/2 H 1 1/2 . Apply Lemma 7.3 and estimate S a , ξ S_{a,\xi} S a , ξ q = 1 / 2 q=1/2 q = 1/2 L 2 L^2 L 2 ( c − a ) − 1 (c-a)^{-1} ( c − a ) − 1 C δ M ( c ) 1 / 2 H 1 1 / 2 C\delta M(c)^{1/2}H_1^{1/2} C δ M ( c ) 1/2 H 1 1/2 δ = 1 / 4 \delta=1/4 δ = 1/4 δ M ( ξ ) ≤ M ( c ) / 4 \delta M(\xi) \le M(c)/4 δ M ( ξ ) ≤ M ( c ) /4 M ( c ) / 8 M(c)/8 M ( c ) /8 v − u ≤ 5 F + ε ≤ H v-u \le5F+\varepsilon\le H v − u ≤ 5 F + ε ≤ H
M ( a ) ≤ 1 2 M ( c ) + C ( c − a ) − N H ( 0 ≤ a < c ≤ 1 ) . (71) M(a) \le\tfrac{1}{2}M(c)+C(c-a)^{-N}H \qquad(0 \le a < c \le1). \tag*{(71)} M ( a ) ≤ 2 1 M ( c ) + C ( c − a ) − N H ( 0 ≤ a < c ≤ 1 ) . (71) By Lemma A.2, applied to the bounded function M M M [ 0 , 1 ] [0,1] [ 0 , 1 ] (71) implies M ( 0 ) ≤ C H M(0) \le CH M ( 0 ) ≤ C H ε ↓ 0 \varepsilon\downarrow0 ε ↓ 0
e s s s u p H 1 u ≤ C ( e s s i n f F 1 u + ∥ f ∥ ∞ ) . \mathop{\mathrm{ess\,sup}}_{H_1} u \le C\left(\mathop{\mathrm{ess\,inf}}_{F_1} u+\lVert f\rVert_{\infty}\right). ess sup H 1 u ≤ C ( ess inf F 1 u + ∥ f ∥ ∞ ) . Rescaling as in the proof of Theorem 1.2 gives (1.17). The dependence on C J C_J C J I ( t ) I(t) I ( t )
Kernel examples and exterior time dependence
A kernel supported on alternating annuli
Example 8.1. Let
A = ⋃ j ∈ Z [ 2 j , 3 2 2 j ] , K A ( x , y ) = 1 A ( ∣ x − y ∣ ) ∣ x − y ∣ − n − α . \mathcal{A}=\bigcup_{j\in\mathbb{Z}}[2^j,\tfrac{3}{2}2^j],\qquad K_{\mathcal{A}}(x,y)=\mathbf{1}_{\mathcal{A}}(|x-y|)|x-y|^{-n-\alpha}. A = j ∈ Z ⋃ [ 2 j , 2 3 2 j ] , K A ( x , y ) = 1 A ( ∣ x − y ∣ ) ∣ x − y ∣ − n − α . This kernel satisfies (3) –(5) and (1.16). For every j ∈ Z j\in\mathbb{Z} j ∈ Z 3 2 2 j < ∣ x − y ∣ < 2 j + 1 \tfrac{3}{2}2^j<|x-y|<2^{j+1} 2 3 2 j < ∣ x − y ∣ < 2 j + 1
Proof. Symmetry, the upper form bound, and the row bound follow from 0 ≤ K A ( x , y ) ≤ ∣ x − y ∣ − n − α 0\le K_{\mathcal{A}}(x,y)\le|x-y|^{-n-\alpha} 0 ≤ K A ( x , y ) ≤ ∣ x − y ∣ − n − α
∫ ∣ z ∣ ≥ r ∣ z ∣ − n − α d z = ∣ S n − 1 ∣ α r − α . \int_{|z|\ge r}|z|^{-n-\alpha}\,dz=\frac{|\mathbb{S}^{n-1}|}{\alpha}r^{-\alpha}. ∫ ∣ z ∣ ≥ r ∣ z ∣ − n − α d z = α ∣ S n − 1 ∣ r − α . For the lower form bound on a ball D D D
Q A ( v ; D ) = ∬ D × D 1 A ( ∣ x − y ∣ ) ∣ v ( x ) − v ( y ) ∣ 2 ∣ x − y ∣ n + α d x d y . Q_{\mathcal{A}}(v;D)=\iint_{D\times D}\mathbf{1}_{\mathcal{A}}(|x-y|)\frac{|v(x)-v(y)|^2}{|x-y|^{n+\alpha}}\,dx\,dy. Q A ( v ; D ) = ∬ D × D 1 A ( ∣ x − y ∣ ) ∣ x − y ∣ n + α ∣ v ( x ) − v ( y ) ∣ 2 d x d y . Suppose x , y ∈ D x,y\in D x , y ∈ D r = ∣ y − x ∣ ∈ ( 3 a / 2 , 2 a ) r=|y-x|\in(3a/2,2a) r = ∣ y − x ∣ ∈ ( 3 a /2 , 2 a ) a = 2 j a=2^j a = 2 j z = y − x z=y-x z = y − x w = x + 2 z / 3 w=x+2z/3 w = x + 2 z /3 D D D w ∈ D w\in D w ∈ D
∣ w − x ∣ = 2 r / 3 ∈ ( a , 4 a / 3 ) ⊂ A , ∣ y − w ∣ = r / 3 ∈ ( a / 2 , 2 a / 3 ) ⊂ A . |w-x|=2r/3\in(a,4a/3)\subset\mathcal{A},\qquad|y-w|=r/3\in(a/2,2a/3)\subset\mathcal{A}. ∣ w − x ∣ = 2 r /3 ∈ ( a , 4 a /3 ) ⊂ A , ∣ y − w ∣ = r /3 ∈ ( a /2 , 2 a /3 ) ⊂ A . By ∣ v ( x ) − v ( y ) ∣ 2 ≤ 2 ∣ v ( x ) − v ( w ) ∣ 2 + 2 ∣ v ( w ) − v ( y ) ∣ 2 |v(x)-v(y)|^2\le2|v(x)-v(w)|^2+2|v(w)-v(y)|^2 ∣ v ( x ) − v ( y ) ∣ 2 ≤ 2∣ v ( x ) − v ( w ) ∣ 2 + 2∣ v ( w ) − v ( y ) ∣ 2 ( x , z ) (x,z) ( x , z ) ( x , z 1 ) (x,z_1) ( x , z 1 ) z 1 = 2 z / 3 z_1=2z/3 z 1 = 2 z /3 ( w , z 2 ) (w,z_2) ( w , z 2 ) z 2 = z / 3 z_2=z/3 z 2 = z /3 ( 2 / 3 ) α (2/3)^\alpha ( 2/3 ) α ( 1 / 3 ) α (1/3)^\alpha ( 1/3 ) α D × D D\times D D × D A \mathcal{A} A
[ v ] s , D 2 ≤ ( 1 + 2 ( 2 / 3 ) α + 2 ( 1 / 3 ) α ) Q A ( v ; D ) . (72) [v]^2_{s,D}\le\left(1+2(2/3)^\alpha+2(1/3)^\alpha\right)Q_{\mathcal{A}}(v;D). \tag*{(72)} [ v ] s , D 2 ≤ ( 1 + 2 ( 2/3 ) α + 2 ( 1/3 ) α ) Q A ( v ; D ) . (72) Since E K A , D ( v , v ) = Q A ( v ; D ) / 2 \mathcal{E}_{K_{\mathcal{A}},D}(v,v)=Q_{\mathcal{A}}(v;D)/2 E K A , D ( v , v ) = Q A ( v ; D ) /2
λ = [ 2 ( 1 + 2 ( 2 / 3 ) α + 2 ( 1 / 3 ) α ) ] − 1 . \lambda=\left[2\left(1+2(2/3)^\alpha+2(1/3)^\alpha\right)\right]^{-1}. λ = [ 2 ( 1 + 2 ( 2/3 ) α + 2 ( 1/3 ) α ) ] − 1 . To verify (1.16), fix h h h d = ∣ h ∣ ∈ [ a , 3 a / 2 ] d=|h|\in[a,3a/2] d = ∣ h ∣ ∈ [ a , 3 a /2 ] 0 < r ≤ d / 4 0<r\le d/4 0 < r ≤ d /4 e = h / d e=h/d e = h / d h ∗ = h + ( r / 4 ) e h_* = h+(r/4)e h ∗ = h + ( r /4 ) e d ≤ 5 a / 4 d\le5a/4 d ≤ 5 a /4 h ∗ = h − ( r / 4 ) e h_* = h-(r/4)e h ∗ = h − ( r /4 ) e
B r / 8 ( h ∗ ) ⊂ B r ( h ) ∩ { z : a < ∣ z ∣ < 3 a / 2 } . (73) B_{r/8}(h_*) \subset B_r(h)\cap\{z:a<|z|<3a/2\}. \tag*{(73)} B r /8 ( h ∗ ) ⊂ B r ( h ) ∩ { z : a < ∣ z ∣ < 3 a /2 } . (73) The shift moves h h h r / 4 r/4 r /4 h ∗ h_* h ∗ a / 4 − r / 4 ≥ 5 a / 32 a/4-r/4\ge5a/32 a /4 − r /4 ≥ 5 a /32 r ≤ 3 a / 8 r\le3a/8 r ≤ 3 a /8 r / 8 r/8 r /8 ∣ h ∗ − h ∣ + r / 8 = 3 r / 8 < r |h_*-h|+r/8=3r/8<r ∣ h ∗ − h ∣ + r /8 = 3 r /8 < r (73) holds. On B r / 8 ( h ∗ ) B_{r/8}(h_*) B r /8 ( h ∗ ) ∣ z ∣ ≤ d + r ≤ 5 d / 4 |z|\le d+r\le5d/4 ∣ z ∣ ≤ d + r ≤ 5 d /4
∫ B r ( h ) 1 A ( ∣ z ∣ ) ∣ z ∣ − n − α d z ≥ 8 − n ( 5 / 4 ) − n − α d − n − α . \int_{B_r(h)} \mathbf{1}_{\mathcal{A}}(|z|)|z|^{-n-\alpha}\,dz \ge8^{-n}(5/4)^{-n-\alpha}d^{-n-\alpha}. ∫ B r ( h ) 1 A ( ∣ z ∣ ) ∣ z ∣ − n − α d z ≥ 8 − n ( 5/4 ) − n − α d − n − α . For d ∉ A d\notin\mathcal{A} d ∈ / A C J = 8 n ( 5 / 4 ) n + α C_J=8^n(5/4)^{n+\alpha} C J = 8 n ( 5/4 ) n + α □ \square □
Corollary 8.2 (Measurable coefficients on the annular kernel). Let a ( t ; x , y ) = a ( t ; y , x ) a(t;x,y)=a(t;y,x) a ( t ; x , y ) = a ( t ; y , x ) 0 < a 0 ≤ a ( t ; x , y ) ≤ a 1 < ∞ 0<a_0\le a(t;x,y)\le a_1<\infty 0 < a 0 ≤ a ( t ; x , y ) ≤ a 1 < ∞
K ( t ; x , y ) = a ( t ; x , y ) K A ( x , y ) K(t;x,y)=a(t;x,y)K_{\mathcal{A}}(x,y) K ( t ; x , y ) = a ( t ; x , y ) K A ( x , y ) satisfies the hypotheses of Theorem 1.5 . The constants in the kernel bounds depend only on n n n s s s a 0 a_0 a 0 a 1 a_1 a 1
Proof. Multiplication by a a a a 0 λ a_0\lambda a 0 λ a 1 Λ a_1\Lambda a 1 Λ
K ( t ; x , y ) ≤ a 1 C J ∫ B r ( x ) K A ( z , y ) d z ≤ ( a 1 / a 0 ) C J ∫ B r ( x ) K ( t ; z , y ) d z . K(t;x,y)\le a_1C_J\int_{B_r(x)}K_{\mathcal{A}}(z,y)\,dz\le(a_1/a_0)C_J\int_{B_r(x)}K(t;z,y)\,dz. K ( t ; x , y ) ≤ a 1 C J ∫ B r ( x ) K A ( z , y ) d z ≤ ( a 1 / a 0 ) C J ∫ B r ( x ) K ( t ; z , y ) d z . Thus ( a 1 / a 0 ) C J (a_1/a_0)C_J ( a 1 / a 0 ) C J □ \square □
Concentrated jumps
The next kernel satisfies the form and row bounds but fails upper jumping smoothness. Its added jumps have bounded total intensity at each point and connect pairs of balls of shrinking radius.
Example 8.3. There is a time-independent symmetric kernel satisfying (3) –(5) for which (1.16) fails for every finite constant C J C_J C J
Proof. Let e e e z j = 20 j e z_j=20^je z j = 2 0 j e 0 < ε j < 1 / 32 0<\varepsilon_j<1/32 0 < ε j < 1/32 ε j ↓ 0 \varepsilon_j\downarrow0 ε j ↓ 0
A j = B ε j ( z j ) , D j = B ε j ( z j + e ) , m j = ∣ A j ∣ = ∣ D j ∣ , A_j=B_{\varepsilon_j}(z_j),\qquad D_j=B_{\varepsilon_j}(z_j+e),\qquad m_j=|A_j|=|D_j|, A j = B ε j ( z j ) , D j = B ε j ( z j + e ) , m j = ∣ A j ∣ = ∣ D j ∣ , and set
K ( x , y ) = ∣ x − y ∣ − n − α + ∑ j = 1 ∞ m j − 1 ( 1 A j ( x ) 1 D j ( y ) + 1 D j ( x ) 1 A j ( y ) ) . (74) K(x,y)=|x-y|^{-n-\alpha}+\sum_{j=1}^{\infty}m_j^{-1}\bigl(\mathbf{1}_{A_j}(x)\mathbf{1}_{D_j}(y)+\mathbf{1}_{D_j}(x)\mathbf{1}_{A_j}(y)\bigr). \tag*{(74)} K ( x , y ) = ∣ x − y ∣ − n − α + j = 1 ∑ ∞ m j − 1 ( 1 A j ( x ) 1 D j ( y ) + 1 D j ( x ) 1 A j ( y ) ) . (74) The added kernel is nonnegative and symmetric. The first term gives (3) . Each row of the added kernel has total mass at most one, and all its jumps have length in [ 15 / 16 , 17 / 16 ] [15/16,17/16] [ 15/16 , 17/16 ] (5) is bounded by C r − α Cr^{-\alpha} C r − α
For the upper form bound, let V j = B 3 ( z j ) V_j=B_3(z_j) V j = B 3 ( z j ) c_j=\fint_{V_j}v . The added energy is
∑ j m j − 1 ∫ A j ∫ D j ∣ v ( x ) − v ( y ) ∣ 2 d y d x ≤ 2 ∑ j ∫ A j ∪ D j ∣ v − c j ∣ 2 d x ≤ C ∑ j [ v ] s , V j 2 ≤ C [ v ] s 2 . \sum_jm_j^{-1}\int_{A_j}\int_{D_j}|v(x)-v(y)|^2\,dy\,dx\le2\sum_j\int_{A_j\cup D_j}|v-c_j|^2\,dx
\le C\sum_j[v]_{s,V_j}^2\le C[v]_s^2. j ∑ m j − 1 ∫ A j ∫ D j ∣ v ( x ) − v ( y ) ∣ 2 d y d x ≤ 2 j ∑ ∫ A j ∪ D j ∣ v − c j ∣ 2 d x ≤ C j ∑ [ v ] s , V j 2 ≤ C [ v ] s 2 . Here the fractional Poincaré inequality on the fixed-radius balls follows directly from
∫ V j ∣ v − c j ∣ 2 d x = 1 2 ∣ V j ∣ ∫ V j × V j ∣ v ( x ) − v ( y ) ∣ 2 d x d y ≤ C [ v ] s , V j 2 . \int_{V_j}|v-c_j|^2\,dx=\frac{1}{2|V_j|}\int_{V_j\times V_j}|v(x)-v(y)|^2\,dx\,dy\le C[v]_{s,V_j}^2. ∫ V j ∣ v − c j ∣ 2 d x = 2∣ V j ∣ 1 ∫ V j × V j ∣ v ( x ) − v ( y ) ∣ 2 d x d y ≤ C [ v ] s , V j 2 . Disjointness of the V j V_j V j ∑ j [ v ] s , V j 2 ≤ [ v ] s 2 \sum_j[v]_{s,V_j}^2\le[v]_s^2 ∑ j [ v ] s , V j 2 ≤ [ v ] s 2 (4) .
Fix r 0 = 1 / 8 r_0 = 1/8 r 0 = 1/8 x ∈ A j x \in A_j x ∈ A j y ∈ D j y \in D_j y ∈ D j r 0 < ∣ x − y ∣ / 4 r_0 < |x-y|/4 r 0 < ∣ x − y ∣/4 K ( x , y ) ≥ m j − 1 K(x,y) \ge m_j^{-1} K ( x , y ) ≥ m j − 1 B r 0 ( x ) B_{r_0}(x) B r 0 ( x )
\fint_{B_{r_0}(x)} K(z,y)\,\mathrm{d}z \le C + |B_{r_0}|^{-1}.
The stable part is uniformly bounded on B r 0 ( x ) B_{r_0}(x) B r 0 ( x ) y y y m j − 1 ∣ A j ∣ = 1 m_j^{-1}|A_j| = 1 m j − 1 ∣ A j ∣ = 1 m j → 0 m_j \to0 m j → 0 K ( x , y ) K(x,y) K ( x , y )
The kernel in (74) defines a regular symmetric Dirichlet form on H s ( R n ) H^s(\mathbb{R}^n) H s ( R n ) C c ∞ ( R n ) C^\infty_c(\mathbb{R}^n) C c ∞ ( R n ) L 2 L^2 L 2 r α r^\alpha r α [5 Proposition 3.3] . The kernel in Example 8.3 therefore rules out a uniform full Harnack inequality under (1.3)–(1.5) alone, even for b = 0 b=0 b = 0
Time separation for exterior contributions
In Lemma 7.2 , the tail is integrated before the cylinder F 1 F_1 F 1
Example 8.4. Even for b = 0 b=0 b = 0 K ( x , y ) = ∣ x − y ∣ − n − α K(x,y)=|x-y|^{-n-\alpha} K ( x , y ) = ∣ x − y ∣ − n − α
Proof. Let D = B 8 D=B_8 D = B 8 g ( t , y ) g(t,y) g ( t , y ) ( τ , τ + 1 ) × ( B 11 ∖ B ‾ 10 ) (\tau,\tau+1)\times(B_{11}\setminus\overline{B}_{10}) ( τ , τ + 1 ) × ( B 11 ∖ B 10 ) L D L_D L D
V D = { w ∈ H s ( R n ) : w = 0 a.e. on D c } . V_D=\{w\in H^s(\mathbb{R}^n):w=0\ \text{a.e. on }D^c\}. V D = { w ∈ H s ( R n ) : w = 0 a.e. on D c } . For x ∈ D x\in D x ∈ D
F g ( t , x ) = ∫ D c g ( t , y ) ∣ x − y ∣ − n − α d y , w ( t ) = ∫ τ t e − ( t − r ) L D F g ( r ) d r F_g(t,x)=\int_{D^c}g(t,y)|x-y|^{-n-\alpha}\,\mathrm{d}y,\qquad
w(t)=\int_{\tau}^{t}e^{-(t-r)L_D}F_g(r)\,\mathrm{d}r F g ( t , x ) = ∫ D c g ( t , y ) ∣ x − y ∣ − n − α d y , w ( t ) = ∫ τ t e − ( t − r ) L D F g ( r ) d r for t > τ t>\tau t > τ w ( t ) = 0 w(t)=0 w ( t ) = 0 t ≤ τ t\le\tau t ≤ τ w w w D D D w ≥ 0 w\ge0 w ≥ 0 F g F_g F g w ∈ L t ∞ L x 2 ∩ L t 2 H x s w\in L^\infty_tL^2_x\cap L^2_tH^s_x w ∈ L t ∞ L x 2 ∩ L t 2 H x s g g g u = w + g u=w+g u = w + g
For a test function ϕ \phi ϕ D D D E t ( g , ϕ ) = − ∫ D F g ϕ d x \mathcal{E}_t(g,\phi)=-\int_DF_g\phi\,\mathrm{d}x E t ( g , ϕ ) = − ∫ D F g ϕ d x w w w u u u ∂ t u + L u = 0 \partial_tu+Lu=0 ∂ t u + Lu = 0 D D D τ \tau τ
∫ τ τ + 1 ess sup x ∈ B 2 ∫ R n ∖ B 4 u ( t , y ) ∣ x − y ∣ − n − α d y d t > 0 , \int_{\tau}^{\tau+1}\operatorname*{ess\,sup}_{x\in B_2}
\int_{\mathbb{R}^n\setminus B_4}u(t,y)|x-y|^{-n-\alpha}\,\mathrm{d}y\,\mathrm{d}t>0, ∫ τ τ + 1 x ∈ B 2 ess sup ∫ R n ∖ B 4 u ( t , y ) ∣ x − y ∣ − n − α d y d t > 0 , because u = g u=g u = g
Appendix A. Iteration lemmas
Superlinear and interpolation iterations
Lemma A.1 (Geometric convergence). Suppose that z j ≥ 0 z_j\ge0 z j ≥ 0
z j + 1 ≤ A B j z j 1 + δ , j ≥ 0 , A > 0 , B > 1 , δ > 0. z_{j+1}\le AB^jz_j^{1+\delta},\qquad j\ge0,\qquad A>0,\quad B>1,\quad\delta>0. z j + 1 ≤ A B j z j 1 + δ , j ≥ 0 , A > 0 , B > 1 , δ > 0. If z 0 ≤ A − 1 / δ B − 1 / δ 2 z_0\le A^{-1/\delta}B^{-1/\delta^2} z 0 ≤ A − 1/ δ B − 1/ δ 2
z j ≤ A − 1 / δ B − j / δ − 1 / δ 2 ( j ≥ 0 ) . z_j\le A^{-1/\delta}B^{-j/\delta-1/\delta^2}\qquad(j\ge0). z j ≤ A − 1/ δ B − j / δ − 1/ δ 2 ( j ≥ 0 ) . In particular z j → 0 z_j\to0 z j → 0 Y j + 1 ≤ A B j Y j − 1 1 + δ Y_{j+1}\le AB^jY_{j-1}^{1+\delta} Y j + 1 ≤ A B j Y j − 1 1 + δ j ≥ 0 j\ge0 j ≥ 0 Y − 1 Y_{-1} Y − 1 Y 0 Y_0 Y 0
Proof. Substituting the proposed bound for z j z_j z j
A B j ( A − 1 / δ B − j / δ − 1 / δ 2 ) 1 + δ = A − 1 / δ B − ( j + 1 ) / δ − 1 / δ 2 . AB^j\left(A^{-1/\delta}B^{-j/\delta-1/\delta^2}\right)^{1+\delta}=A^{-1/\delta}B^{-(j+1)/\delta-1/\delta^2}. A B j ( A − 1/ δ B − j / δ − 1/ δ 2 ) 1 + δ = A − 1/ δ B − ( j + 1 ) / δ − 1/ δ 2 . Induction proves the first assertion. For the two-step recurrence set Z m = max { Y 2 m − 1 , Y 2 m } Z_m=\max\{Y_{2m-1},Y_{2m}\} Z m = max { Y 2 m − 1 , Y 2 m } j = 2 m j=2m j = 2 m j = 2 m + 1 j=2m+1 j = 2 m + 1
Z m + 1 ≤ A B 2 m + 1 Z m 1 + δ . Z_{m+1}\le AB^{2m+1}Z_m^{1+\delta}. Z m + 1 ≤ A B 2 m + 1 Z m 1 + δ . Apply the first assertion with constants A B AB A B B 2 B^2 B 2
Lemma A.2 (Absorption on nested sets). Let F F F [ a , b ] [a,b] [ a , b ]
F ( r ) ≤ q F ( R ) + A ( R − r ) − N , a ≤ r < R ≤ b , F(r)\le qF(R)+A(R-r)^{-N},\qquad a\le r<R\le b, F ( r ) ≤ q F ( R ) + A ( R − r ) − N , a ≤ r < R ≤ b , where 0 < q < 1 0<q<1 0 < q < 1 A ≥ 0 A\ge0 A ≥ 0 N > 0 N>0 N > 0
F ( a ) ≤ C ( q , N ) A ( b − a ) − N . F(a)\le C(q,N)A(b-a)^{-N}. F ( a ) ≤ C ( q , N ) A ( b − a ) − N . Proof. Choose ϑ ∈ ( q 1 / N , 1 ) \vartheta\in(q^{1/N},1) ϑ ∈ ( q 1/ N , 1 ) r j = a + ( b − a ) ( 1 − ϑ j ) r_j=a+(b-a)(1-\vartheta^j) r j = a + ( b − a ) ( 1 − ϑ j )
F ( a ) ≤ q m F ( r m ) + A ( b − a ) − N ( 1 − ϑ ) − N ∑ j = 0 m − 1 ( q ϑ − N ) j . F(a)\le q^mF(r_m)+A(b-a)^{-N}(1-\vartheta)^{-N}\sum_{j=0}^{m-1}(q\vartheta^{-N})^j. F ( a ) ≤ q m F ( r m ) + A ( b − a ) − N ( 1 − ϑ ) − N j = 0 ∑ m − 1 ( q ϑ − N ) j . The first term tends to zero because F F F
The logarithmic iteration
A.2. The logarithmic iteration. The following form of the Bombieri–Giusti argument [3 ] combines a moment inequality with a bound on the upper level sets of the logarithm. Section 5 applies it separately to a positive function and its reciprocal.
Lemma A.3 (Moment control from logarithmic level sets). Let ( U r ) 1 / 2 ≤ r ≤ 1 (U_r)_{1/2\le r\le1} ( U r ) 1/2 ≤ r ≤ 1 μ \mu μ μ ( U 1 ) ≤ 1 \mu(U_1)\le1 μ ( U 1 ) ≤ 1 inf r μ ( U r ) > 0 \inf_r\mu(U_r)>0 inf r μ ( U r ) > 0 g > 0 g>0 g > 0 g ∈ L p 0 ( U 1 ) g\in L^{p_0}(U_1) g ∈ L p 0 ( U 1 ) p 0 > 0 p_0>0 p 0 > 0 A ≥ 1 A\ge1 A ≥ 1 ν > 0 \nu>0 ν > 0
∥ g ∥ L p 0 ( U r ) ≤ ( A ( R − r ) ν ) 1 / p ∥ g ∥ L p ( U R ) , 1 2 ≤ r < R ≤ 1 , 0 < p ≤ p 0 / 2. (75) \lVert g\rVert_{L^{p_0}(U_r)}\le\left(\frac{A}{(R-r)^\nu}\right)^{1/p}\lVert g\rVert_{L^p(U_R)},\qquad\frac12\le r<R\le1,\quad0<p\le p_0/2. \tag*{(75)} ∥ g ∥ L p 0 ( U r ) ≤ ( ( R − r ) ν A ) 1/ p ∥ g ∥ L p ( U R ) , 2 1 ≤ r < R ≤ 1 , 0 < p ≤ p 0 /2. (75) μ { x ∈ U 1 : log g ( x ) > ℓ } ≤ A ℓ , ℓ > 0. (76) \mu\{x\in U_1:\log g(x)>\ell\}\le\frac{A}{\ell},\qquad\ell>0. \tag*{(76)} μ { x ∈ U 1 : log g ( x ) > ℓ } ≤ ℓ A , ℓ > 0. (76) Then ∥ g ∥ L p 0 ( U 1 / 2 ) ≤ C ( A , ν , p 0 ) \lVert g\rVert_{L^{p_0}(U_{1/2})}\le C(A,\nu,p_0) ∥ g ∥ L p 0 ( U 1/2 ) ≤ C ( A , ν , p 0 )
Proof. Put
F ( r ) = max { 1 , log ∥ g ∥ L p 0 ( U r ) } . F(r)=\max\left\{1,\log\lVert g\rVert_{L^{p_0}(U_r)}\right\}. F ( r ) = max { 1 , log ∥ g ∥ L p 0 ( U r ) } . This is a bounded nondecreasing function. Fix r < R r<R r < R J = log ∥ g ∥ L p 0 ( U R ) J=\log\lVert g\rVert_{L^{p_0}(U_R)} J = log ∥ g ∥ L p 0 ( U R ) J J J A , p 0 A,p_0 A , p 0
p = log J 2 J ≤ p 0 2 . p=\frac{\log J}{2J}\le\frac{p_0}{2}. p = 2 J log J ≤ 2 p 0 . Split the integral of g p g^p g p log g = J / 2 \log g=J/2 log g = J /2
∫ U R g p d μ ≤ e p J / 2 + ( ∫ U R g p 0 d μ ) p / p 0 ( 2 A J ) 1 − p / p 0 \int_{U_R}g^p\,\mathrm{d}\mu\le e^{pJ/2}+\left(\int_{U_R}g^{p_0}\,\mathrm{d}\mu\right)^{p/p_0}\left(\frac{2A}{J}\right)^{1-p/p_0} ∫ U R g p d μ ≤ e p J /2 + ( ∫ U R g p 0 d μ ) p / p 0 ( J 2 A ) 1 − p / p 0 = J 1 / 4 + J 1 / 2 ( 2 A J ) 1 − p / p 0 ≤ 2 J 1 / 4 . =J^{1/4}+J^{1/2}\left(\frac{2A}{J}\right)^{1-p/p_0}\le2J^{1/4}. = J 1/4 + J 1/2 ( J 2 A ) 1 − p / p 0 ≤ 2 J 1/4 . Indeed, the second term equals 2 A J − 1 / 2 exp ( ( p / p 0 ) log ( J / ( 2 A ) ) ) 2AJ^{-1/2}\exp((p/p_0)\log(J/(2A))) 2 A J − 1/2 exp (( p / p 0 ) log ( J / ( 2 A ))) J J J ( log J ) 2 / J → 0 (\log J)^2/J\to0 ( log J ) 2 / J → 0
Let D = log A + ν log ( 1 / ( R − r ) ) D=\log A+\nu\log(1/(R-r)) D = log A + ν log ( 1/ ( R − r ))
log ∥ g ∥ L p 0 ( U r ) ≤ D + log 2 p + log J 4 p = J 2 + 2 J ( D + log 2 ) log J . \log\lVert g\rVert_{L^{p_0}(U_r)}\le\frac{D+\log2}{p}+\frac{\log J}{4p}=\frac{J}{2}+\frac{2J(D+\log2)}{\log J}. log ∥ g ∥ L p 0 ( U r ) ≤ p D + log 2 + 4 p log J = 2 J + log J 2 J ( D + log 2 ) . If log J ≥ 8 ( D + log 2 ) \log J\ge8(D+\log2) log J ≥ 8 ( D + log 2 ) 3 J / 4 3J/4 3 J /4 J J J
F ( r ) ≤ F ( R ) ≤ C exp ( 8 D ) ≤ C ( R − r ) − 8 ν . F(r)\le F(R)\le C\exp(8D)\le C(R-r)^{-8\nu}. F ( r ) ≤ F ( R ) ≤ C exp ( 8 D ) ≤ C ( R − r ) − 8 ν . Combining the two cases, and increasing C C C F ( r ) = 1 F(r)=1 F ( r ) = 1
F ( r ) ≤ 3 4 F ( R ) + C ( R − r ) − 8 ν . F(r)\le\frac{3}{4}F(R)+C(R-r)^{-8\nu}. F ( r ) ≤ 4 3 F ( R ) + C ( R − r ) − 8 ν . Since g ∈ L p 0 ( U 1 ) g\in L^{p_0}(U_1) g ∈ L p 0 ( U 1 ) F F F F ( 1 / 2 ) F(1/2) F ( 1/2 ) □ \square □
SCHOOL OF MATHEMATICS AND STATISTICS, NORTHEASTERN UNIVERSITY AT QINHUANGDAO, QINHUANGDAO 066004, CHINA
Email address: sunzhenyu@neuq.edu.cn
