An exact unstable mode of self-similar Yang–Mills solutionsin 5+1 dimensions

Self-similar solutions

The spherically symmetric Yang–Mills equation in five space dimensions is

In terms of the similarity coordinates

where $T$ is the blowup time, equation (1) becomes

By definition, self-similar solutions are independent of $\tau$: $w(\tau,y)=u(y)$. The profile $u$ therefore satisfies

It was proved in [1] that there exist infinitely many smooth profiles $u_n$, $n=0,1,2,\ldots$, where $u_n$ has exactly $n$ zeros on $(0,1)$. Their endpoint expansions are

corresponding to the regular center and the similarity horizon, respectively. The fundamental profile, found explicitly in [1], is

Numerical computations in [2] indicated that the fundamental profile $u_0$ acts as a universal attractor for generic blowup, while the first excited profile $u_1$ governs the threshold between dispersion and blowup. The nonlinear asymptotic stability of $u_0$ under small equivariant perturbations was subsequently established in [3, 4].

Further numerical computations revealed a remarkable feature: the least unstable mode has the same eigenvalue $\lambda= 5$ for every excited profile [1], raising the question of its structural origin. In this note we resolve this little spectral puzzle by constructing the corresponding eigenfunction explicitly.

Linearized eigenvalue problem

Substituting $w(\tau,y)=u_n(y)+\varepsilon e^{\lambda\tau}v(y)$ into (2) and retaining terms linear in $\varepsilon$ yields the eigenvalue equation

We seek eigenfunctions $v\in C^\infty([0,1])$.

One such eigenfunction arises from differentiating the family of self-similar solutions with respect to $T$:

which gives the gauge mode

It changes only the blowup time and is therefore not counted as a physical instability.

Self-adjoint formulation

For the spectral analysis and the exact-mode construction, it is convenient to introduce the reciprocal coordinate

In these variables, equation (3) becomes

At the similarity horizon $x=1$ and the center $x=\infty$, the profiles have the endpoint expansions

The fundamental profile (6) becomes

Equation (7) is transformed into

Substituting

into (13) we obtain the self-adjoint Sturm–Liouville equation for $\phi\in L^2((1,\infty),p^{-1}\,dx)$:

Hereafter we refer to $E$ as the energy and to $\lambda$ as the similarity eigenvalue. Although the transformation (14) relates the two differential equations for every $\lambda$, it identifies smooth similarity modes with square-integrable Sturm–Liouville eigenfunctions only when $\operatorname{Re}(\lambda)>1$. To see this, consider a smooth mode with $V(1)\ne0$. Near $x=1$, its transform satisfies

which is square integrable with weight $p^{-1}$ precisely for $\operatorname{Re}(\lambda)>1$. In that half-plane the second branch is not square integrable. At infinity, the regular branch $V=O(x^{-2})$ corresponds to $\phi=O(x^{-3})$, whereas the singular branch $V\sim Cx^3$ corresponds to $\phi\sim Cx^2$. Only the regular branch is square integrable with weight $p^{-1}$ at infinity.

Thus, for $\operatorname{Re}(\lambda)>1$, the similarity eigenmodes of (13) are in one-to-one correspondence with the eigenfunctions of the problem (15), allowing us to apply Sturm–Liouville theory. In particular, self-adjointness implies that $E$ is real and hence that $\lambda$ is real. The analogous correspondence for wave maps is discussed in [5]. This equivalence does not apply to similarity modes with $\operatorname{Re}(\lambda)<1$.

The gauge mode (8) becomes $V_g^{(n)}(x)=xU_n'(x)$, which under (14) gives the zero-energy resonance (a bounded but not square-integrable solution)

From the proof of existence in [1], it follows that the profile $U_n$ and its derivative each have exactly $n$ interior zeros, which interlace. Therefore, the Sturm oscillation theorem implies that the spectral problem (15) has exactly $n$ negative energy levels. We order them as

where $E_k^{(n)}$, $k=1,\ldots,n$, denote the negative energy levels for the profile $U_n$. The corresponding eigenfunctions are denoted by $\phi_k^{(n)}(x)$. The associated unstable similarity eigenvalues are ordered as

with the corresponding eigenfunctions denoted by $V_k^{(n)}(x)=v_k^{(n)}(y)$.

Numerical computations and asymptotic analysis in [1] indicated that the first eigenvalue in this sequence is independent of the profile:

Equivalently, every excited profile shares the same highest negative energy level $E_1^{(n)}=-16$, despite the dependence of the Sturm–Liouville operator on $U_n$. We now explain this spectral feature by constructing the corresponding eigenfunction explicitly and proving that it occupies this position in the spectrum.

Construction of the exact mode

The same energy level for every excited profile suggests seeking a corresponding eigenfunction given by a universal first-order expression in the profile (here we suppress the index $n$ on $U_n$):

where $A$ and $B$ are independent of the particular profile $U$. The gauge mode $\phi_g=U^{\prime}$ is the simplest example of such an expression. We now seek a nonzero energy for which an ansatz of this form solves the linearized equation for every profile.

Substituting the ansatz (18) into equation (15) and using the profile equation (9) and its derivative to eliminate $U^{\prime\prime}$ and $U^{\prime\prime\prime}$, we obtain

A sufficient condition for this expression to vanish for every profile is that the three coefficients vanish separately. Setting the coefficient of the cubic term to zero gives $B=A^{\prime}$, so the ansatz (18) takes the total-derivative form $\phi=(AU)^{\prime}$, while the remaining coefficient equations reduce to

Taking $A$ to be constant recovers the gauge mode with $E=0$. To find a nonconstant solution, we seek $A$ bounded at infinity and allow a simple pole at $x=1$, where every profile vanishes. The simplest rational ansatz in $p=x^{2}-1$ with these properties is

Substituting this into (20) and clearing denominators gives

Since $b\ne0$, the second equation yields $E=-16$, and then the first one gives $a=5b/8$. Normalizing $a=1$, we obtain

Consequently, for every excited profile $U_n$, $n\ge1$, the first-order ansatz yields the mode, provisionally denoted by $\phi_n$,

which satisfies the eigenvalue equation (15) with $E=-16$, corresponding to $\lambda=5$.

It remains to verify the behavior of $\phi_n$ at the endpoints. The endpoint expansions (10) and (11) give

Hence $\phi_n\in L^2((1,\infty),p^{-1}\,dx)$, completing the proof that $\phi_n$ is an eigenfunction of the spectral problem (15) with energy $E=-16$ for every $n\ge 1.

To locate this energy level in the spectrum, let us count the nodes of $\phi_n$. Since $U_0 > 0$ on $(1,\infty)$, the quotient $U_n/U_0$ has exactly $n$ zeros. Therefore, Rolle’s theorem gives at least $n-1$ distinct interior zeros of $\phi_n$. On the other hand, the spectral problem (15) has exactly $n$ negative energy levels, so the Sturm oscillation theorem allows at most $n-1$ interior zeros for an eigenfunction of negative energy. Thus $\phi_n$ has exactly $n-1$ nodes, and $E=-16$ is the highest negative energy level. In the notation (16), this reads

The corresponding similarity eigenfunction with $\lambda_1^{(n)}=5$, expressed in the original coordinate $y$, is

An underlying differential identity

It is instructive to recast the preceding calculation without imposing the profile equation. Writing

we obtain, for arbitrary sufficiently smooth functions $A$ and $U$,

With $A=U_0^{-1}$ and $E=-16$, the last two terms vanish by (20), yielding the identity

which is valid for any sufficiently smooth function $U(x)$. In particular, for every excited profile $U_n$, the identity shows that $(U_n/U_0)'$ satisfies the eigenvalue equation with $E=-16$, recovering the exact mode constructed above.

Comparison with reflected gauge modes in $d=7$ and $d=9$.

In dimension $d$, the transformed spectral equation depends on $\lambda$ through

and is therefore invariant under $\lambda\mapsto d-3-\lambda$. Consequently, the gauge mode $v_g=-yu'$ with $\lambda=1$ has the formal reflected partner

For smooth profiles in $d=7$ and $9$, the behavior $u(y)\sim b(1-y)^{\frac{d-3}{2}}$ makes the apparent singularity removable. The reflection therefore yields admissible exact modes with $\lambda=3$ and $\lambda=5$, respectively. Clearly, these modes can also be readily recovered from the first-order ansatz (18), which in these cases reduces to the derivative term alone ($B=0$). However, for $d=5$, the case considered here, the reflection $\lambda\mapsto2-\lambda$ fixes the gauge eigenvalue $\lambda=1$ and therefore produces no new mode.