Stability of eigenfunctions: Difference between revisions

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which can be viewed as infinitesimal variants of the estimates in the previous section.
which can be viewed as infinitesimal variants of the estimates in the previous section.
== A Sobolev inequality ==
:'''Lemma''' Suppose that <math>u: B(0,R) \to {\mathbf R}</math> obeys the inhomogeneous eigenfunction equation <math>-\Delta u = \lambda u + F</math>.  Suppose also that <math>\eta: {\mathbf R} \to {\mathbf R}</math> is a smooth function that equals 1 at 0 and vanishes on <math>[R,+\infty)</math>.  Then
: <math>|u(0)| \leq \frac{\sqrt{\pi}}{2\sqrt{2}} (\|F\|_{L^2(B(0,R)} (\int_0^\infty \eta(r)^2 Y_0(\sqrt{\lambda} r)^2\ r dr)^{1/2} + \lambda^{1/2} \|u\|_{L^2(B(0,R))} (\int_0^R \eta'(r)^2 Y'_0(\sqrt{\lambda}r)^2\ r dr)^{1/2}</math>
: <math> + \|\nabla u\|_{L^2(B(0,R))} (\int_0^R \eta'(r)^2 Y_0(\sqrt{\lambda}r)^2\ r dr)^{1/2} ).</math>
'''Proof'''  We introduce the averaged functions
:<math> \bar u(r) := \frac{1}{2\pi} \int_0^{2\pi} u(r,\theta)\ d\theta</math>
:<math> \bar F(r) := \frac{1}{2\pi} \int_0^{2\pi} F(r,\theta)\ d\theta</math>
and observe that <math>\bar u</math> obeys the inhomogeneous Bessel equation
:<math> -\bar u''(r) - \frac{1}{r} \bar u'(r) = \lambda \bar u(r) + \bar F(r)</math>
with initial condition <math>\bar u(0) = u(0)</math>.
We can solve this equation by separation of variables, writing
:<math> \bar{u}(r) = a(r) J_0(\sqrt{\lambda} r) + b(r) Y_0(\sqrt{\lambda} r)</math>
:<math> \bar{u}'(r) = \sqrt{\lambda} a(r) J_0'(\sqrt{\lambda} r) + \sqrt{\lambda} b(r) Y_0'(\sqrt{\lambda} r)</math>.
Using the standard Wronskian identity
:<math> J_0(\sqrt{\lambda} r) Y'_0(\sqrt{\lambda} r) - Y_0(\sqrt{\lambda} r) J'_0(\sqrt{\lambda} r) = \frac{2}{\sqrt{\lambda} \pi r}</math>
one soon arrives at the equations
:<math> a(r) = \frac{\pi r}{2} ( \sqrt{\lambda} \bar u(r) Y'_0(\sqrt{\lambda} r) - \bar u'(r) Y_0(\sqrt{\lambda} r) )</math>
and
:<math> a'(r) = -\frac{\pi r}{2} \bar F(r) Y_0(\sqrt{\lambda} r)</math>.
Meanwhile, by integration by parts
:<math> a(0) = -\int_0^R a'(r) \eta(r)\ dr - \int_0^R a(r) \eta'(r)\ dr</math>
and thus
:<math>|a(0)| \leq \frac{\pi}{2}
( \int_0^R |\bar F(r)| |Y_0(\sqrt{\lambda} r)|\ r dr + \int_0^R \sqrt{\lambda} |\bar u(r)| |Y'_0(\sqrt{\lambda} r)|\ r dr
+ \int_0^R |\bar u'(r)| |Y_0(\sqrt{\lambda} r)|\ r dr).</math>
From Cauchy-Schwarz one has
:<math> 2\pi \int_0^R |\bar u(r)|^2\ r dr \leq \| u \|_{L^2(B(0,R))}^2</math>
and
:<math> 2\pi \int_0^R |\bar u'(r)|^2\ r dr \leq \| \nabla u \|_{L^2(B(0,R))}^2</math>
and similarly
:<math> 2\pi \int_0^R |\bar F(r)|^2\ r dr \leq \| F \|_{L^2(B(0,R))}^2.</math>
The claim then follows from further applications of the Cauchy-Schwarz inequality. []
In principle, the above inequality, when combined with the previous L^2 and H^1 estimates, gives L^infty control on <math>\dot u</math> in the interior of the triangle.  One should also be able to get control near the interior of an edge by reflection.  One needs to modify the argument a bit though to handle what is going on at vertices.

Revision as of 16:40, 26 June 2012

In [CZ1994] some bounds for the Neumann heat kernels [math]\displaystyle{ P_t(x,y) }[/math] on a general domain are given; in particular, one can bound this kernel by a multiple of the Euclidean heat kernel for t small enough. Using this bound, [BP2008] showed uniform stability of the second eigenfunction (in the uniform topology) and eigenvalue with respect to uniform perturbations of the domain.

Formal theory

Suppose one has a one-parameter family [math]\displaystyle{ t \mapsto L(t) }[/math] of self-adjoint operators (on some Hilbert space, e.g. [math]\displaystyle{ L^2(\Omega) }[/math], though for this formal computation the domain will not be important), and one-parameter families [math]\displaystyle{ t \mapsto u(t) }[/math], [math]\displaystyle{ t \mapsto \lambda(t) }[/math] to the eigenfunction equation

[math]\displaystyle{ L(t) u(t) = \lambda(t) u(t) }[/math]. (1)

We normalise [math]\displaystyle{ u(t) }[/math] to have norm 1:

[math]\displaystyle{ \langle u(t), u(t) \rangle = 1 }[/math]. (2)

Formally, if we differentiate the norm equation (2) at time zero, we get

[math]\displaystyle{ \langle \dot u(0), u(0) \rangle = 0 }[/math] (3)

while if we differentiate (1) at time zero we obtain

[math]\displaystyle{ \dot L(0) u(0) + L(0) \dot u(0) = \dot \lambda(0) u(0) + \lambda(0) \dot u(0) }[/math]. (4)

Taking the inner product of (4) with u(0) and using (2), (3) we conclude the Hadamard first variation formula for eigenvalues:

[math]\displaystyle{ \dot \lambda(0) = \langle \dot L(0) u(0), u(0) \rangle }[/math]. (5)

If we instead take the orthogonal projection [math]\displaystyle{ \pi_{u(0)}^\perp }[/math] onto the orthogonal complement of [math]\displaystyle{ u(0) }[/math], we obtain the Hadamard first variation formula for eigenfunctions:

[math]\displaystyle{ (L(0)-\lambda) \dot u(0) = - \pi_{u(0)}^\perp( \dot L(0) u(0) ) }[/math]. (6)

Formally, if [math]\displaystyle{ \lambda }[/math] is a simple eigenvalue, then [math]\displaystyle{ L(0)-\lambda }[/math] is invertible on the orthogonal complement of [math]\displaystyle{ u(0) }[/math], and so (6) and (3) allow one to solve for [math]\displaystyle{ \dot u(0) }[/math].


L^2 and H^1 theory

Suppose that one has a domain [math]\displaystyle{ \Omega }[/math] with second Neumann eigenvalue [math]\displaystyle{ \lambda_2 }[/math], and third Neumann eigenvalue (not counting multiplicity) [math]\displaystyle{ \lambda_3 \gt \lambda_2 }[/math]. Thus, one has

[math]\displaystyle{ \int_\Omega |\nabla u|^2 \geq \lambda_2 \int_\Omega |u|^2 }[/math] (7)

for all mean zero u, with equality when u lies in the second eigenspace [math]\displaystyle{ V_2 }[/math], and one can improve this to

[math]\displaystyle{ \int_\Omega |\nabla u|^2 \geq \lambda_3 \int_\Omega |u|^2 }[/math] (8)

when u is orthogonal to [math]\displaystyle{ V_2 }[/math].

Now consider a perturbation [math]\displaystyle{ B\Omega }[/math] of [math]\displaystyle{ \Omega }[/math], where B is an invertible linear transformation. Then a second eigenfunction of [math]\displaystyle{ B\Omega }[/math], after change of variables, becomes a function u on [math]\displaystyle{ \Omega }[/math] that minimizes the modified Rayleigh quotient

[math]\displaystyle{ \int_\Omega M \nabla u \cdot \nabla u / \int_\Omega |u|^2 }[/math]

where [math]\displaystyle{ M := (B^{-1}) (B^{-1})^T }[/math]. We may normalize this eigenfunction as [math]\displaystyle{ u = u_2 + v }[/math], where u_2 is a unit eigenfunction in V_2 and v is orthogonal to V_2, so that [math]\displaystyle{ \|u\|_2^2 = 1 + \|v\|_2^2 }[/math]. Then the modified Rayleigh quotient of u is less than or equal to that of u_2, and hence

[math]\displaystyle{ \int_\Omega M \nabla u \cdot \nabla u \leq (\int_\Omega M \nabla u_2 \cdot \nabla u_2) (1 + \|v\|_2^2 ). }[/math]

Expanding out u as u_2+v and rearranging, we end up at

[math]\displaystyle{ \int_\Omega M \nabla v \cdot \nabla v \leq \|v\|_2^2 \int_\Omega M \nabla u_2 \cdot \nabla u_2 - 2 \int_\Omega M \nabla u_2 \cdot \nabla v. }[/math]

Note that [math]\displaystyle{ \nabla v }[/math] is orthogonal to [math]\displaystyle{ \nabla V_2 }[/math] by integration by parts, and so we may replace [math]\displaystyle{ M \nabla u_2 }[/math] on the RHS by the orthogonal projection [math]\displaystyle{ \pi_{\nabla V_2}^\perp(M \nabla u_2) }[/math]. Letting [math]\displaystyle{ \sigma_1(M) = \sigma_2(B)^{-2} }[/math] be the least singular value of M, we now apply Cauchy-Schwarz and conclude that

[math]\displaystyle{ \sigma_1(M) \|\nabla v \|_2^2 \leq \|v\|_2^2 \int_\Omega M \nabla u_2 \cdot \nabla u_2 + 2 \| \pi_{\nabla V_2}^\perp(M \nabla u_2) \|_2 \|\nabla v\|_2 }[/math].

By (8) we may bound [math]\displaystyle{ \|v\|_2^2 \leq \frac{1}{\lambda_3} \|\nabla v\|_2^2 }[/math], and so we conclude that

[math]\displaystyle{ (\sigma_1(M) - \frac{\int_\Omega M \nabla u_2 \cdot u_2}{\lambda_3}) \|\nabla v \|_2 \leq 2 \| \pi_{\nabla V_2}^\perp(M \nabla u_2) \|_2 }[/math]. (9)

This gives an H^1 bound on the error v between the perturbed eigenfunction u and the original eigenfunction u_2. To understand this bound, note that we may upper bound

[math]\displaystyle{ \int_\Omega M \nabla u_2 \cdot \nabla u_2 \leq \sigma_2(M) \int_\Omega |\nabla u_2|^2 = \lambda_2 \sigma_2(M) }[/math]

and also [math]\displaystyle{ \pi_{\nabla V_2}^\perp(M \nabla u_2) = \pi_{\nabla V_2^\perp}((M-\frac{\sigma_1(M)+\sigma_2(M)}{2}) \nabla u_2) }[/math] so that

[math]\displaystyle{ \|\pi_{\nabla V_2^\perp}((M-\frac{\sigma_1(M)+\sigma_2(M)}{2}) \nabla u_2)\|_2 \leq \frac{\sigma_2(M)-\sigma_1(M)}{2} \|\nabla u_2\|_2 = \frac{\sigma_2(M)-\sigma_1(M)}{2} \lambda_2^{1/2} }[/math]

and so one has

[math]\displaystyle{ \|\nabla v \|_2 \leq \frac{(\sigma_2(M)-\sigma_1(M)) \lambda_2^{1/2}}{\sigma_1(M) - \sigma_2(M) \frac{\lambda_2}{\lambda_3}} }[/math]

provided that the denominator is positive. In terms of the condition number [math]\displaystyle{ \kappa := \sigma_2(B)/\sigma_1(B) }[/math] of B, this becomes

[math]\displaystyle{ \|\nabla v \|_2 \leq \frac{(\kappa^2-1) \lambda_2^{1/2}}{1 - \kappa^2 \frac{\lambda_2}{\lambda_3}} }[/math]

which is a non-trivial bound when [math]\displaystyle{ \kappa \lt (\lambda_3/\lambda_2)^{1/2} }[/math], and is of the order of [math]\displaystyle{ O(\lambda_2^{1/2} (\kappa-1)) }[/math] when [math]\displaystyle{ \kappa }[/math] is close to 1. Using (8), we conclude in particular that

[math]\displaystyle{ \|v \|_2 \leq \frac{(\kappa^2-1) (\lambda_2/\lambda_3)^{1/2}}{1 - \kappa^2 \frac{\lambda_2}{\lambda_3}} }[/math]

For these calculations performed on a third reference triangle, http://www.math.sfu.ca/~nigam/polymath-figures/Perturbation.pdf

An alternate approach to the L^2 and H^1 theory

Let us keep the reference triangle [math]\displaystyle{ \Omega }[/math] fixed, and view the matrix M=M(t) as being smoothly time dependent, so that the eigenvalues [math]\displaystyle{ \lambda_k = \lambda_k(t) }[/math] and L^2-normalised eigenfunctions [math]\displaystyle{ u_k = u_k(t) }[/math] are also time dependent. Assume for the sake of argument that eigenvalues stay simple and all functions depend smoothly on t. We have the eigenfunction equation

[math]\displaystyle{ -\nabla \cdot M \nabla u_k = \lambda_k u_k }[/math]

and the Neumann boundary condition

[math]\displaystyle{ -n \cdot M \nabla u_k = 0 }[/math]

and the L^2 normalisation

[math]\displaystyle{ \langle u_k, u_k \rangle = 1 }[/math].

Differentiating, we conclude that

[math]\displaystyle{ (-\nabla \cdot M \nabla - \lambda_k) \dot u_k = \nabla \cdot \dot M \nabla u_k + \dot \lambda_k u_k }[/math] (10)

and

[math]\displaystyle{ -n \cdot M \nabla \dot u_k = n \cdot \dot M \nabla u_k }[/math] (11)

and

[math]\displaystyle{ \langle \dot u_k, u_k \rangle = 0 }[/math]. (12)

Taking the inner product of (10) with u_k and using (12) yields

[math]\displaystyle{ \langle -\nabla \cdot M \nabla \dot u_k, u_k \rangle = \langle \nabla \cdot \dot M \nabla u_k, u_k \rangle + \dot \lambda_k }[/math].

Integrating by parts using the Neumann condition and (11) yields

[math]\displaystyle{ \langle -\nabla \cdot M \nabla \dot u_k, u_k \rangle = \langle \dot u_k, -\nabla \cdot M \nabla u_k \rangle + \int_{\partial \Omega} (n \cdot \dot M \nabla u_k) u_k }[/math].

By the eigenfunction equation and (12), the first inner product on the RHS vanishes. By Stokes theorem one has

[math]\displaystyle{ \int_{\partial \Omega} (n \cdot \dot M \nabla u_k) u_k = \langle \nabla \cdot \dot M \nabla u_k, u_k \rangle + \langle \dot M \nabla u_k, \nabla u_k \rangle }[/math]

and thus we have the variation formula

[math]\displaystyle{ \dot \lambda_k = \langle \dot M \nabla u_k, \nabla u_k \rangle }[/math]. (14)

Next, if we take the inner product of (10) with u_l for some l distinct from k, one has

[math]\displaystyle{ \langle (-\nabla \cdot M \nabla - \lambda_k) \dot u_k, u_l \rangle = \langle \nabla \cdot \dot M \nabla u_k, u_l \rangle }[/math].

Integrating by parts as before, we have

[math]\displaystyle{ \langle -\nabla \cdot M \nabla \dot u_k, u_l \rangle = \langle \dot u_k, -\nabla \cdot M \nabla u_l \rangle + \int_{\partial \Omega} (n \cdot \dot M \nabla u_k) u_l }[/math].

By the eigenfunction equation, the first inner product on the RHS is [math]\displaystyle{ \lambda_l \langle \dot u_k, u_l \rangle }[/math]. By Stokes theorem we have

[math]\displaystyle{ \int_{\partial \Omega} (n \cdot \dot M \nabla u_k) u_l = \langle \nabla \cdot \dot M \nabla u_k, u_l \rangle + \langle \dot M \nabla u_k, \nabla u_l \rangle }[/math]

and thus

[math]\displaystyle{ (\lambda_l - \lambda_k) \langle \dot u_k, u_l \rangle = \langle \dot M \nabla u_k, \nabla u_l \rangle }[/math]

and thus by eigenfunction expansion (and (12))

[math]\displaystyle{ \dot u_k = \sum_{l \neq k} \frac{1}{\lambda_l-\lambda_k} \langle \dot M \nabla u_k, \nabla u_l \rangle u_l }[/math]

where the convergence is in an unconditional L^2 sense. (Note that [math]\displaystyle{ \nabla u_l / \lambda_l^{1/2} }[/math] is an orthonormal system and so from Bessel's inequality we know that [math]\displaystyle{ \sum_l \langle \dot M \nabla u_k, \nabla u_l \rangle^2 / \lambda_l \lt \infty }[/math], which is enough decay to justify the L^2 convergence of the RHS.) In fact we may differentiate and conclude that

[math]\displaystyle{ \dot \nabla u_k = \sum_{l \neq k} \frac{1}{\lambda_l-\lambda_k} \langle \dot M \nabla u_k, \nabla u_l \rangle \nabla u_l }[/math]

where the convergence is again in the unconditional L^2 sense (i.e. the previous convergence was in the unconditional H^1 sense). From the Bessel inequality we see in particular that

[math]\displaystyle{ \| \dot \nabla u_k \|_{L^2}^2 = \sum_{l \neq k} \frac{\lambda_l^2}{(\lambda_l-\lambda_k)^2} |\langle \dot M \nabla u_k, \nabla u_l \rangle|/\lambda_l }[/math]
[math]\displaystyle{ \leq \| \dot M \nabla u_k \|_{L^2}^2 \sup_{l \neq k} \frac{\lambda_l^2}{(\lambda_l - \lambda_k)^2} }[/math];

in particular, we have

[math]\displaystyle{ \| \dot \nabla u_2 \|_{L^2}^2 \leq \frac{\lambda_3}{\lambda_3 - \lambda_2} \| \dot M \nabla u_2 \|_{L^2} }[/math]

and thus

[math]\displaystyle{ \| \dot u_2 \|_{L^2}^2 \leq \frac{1}{\lambda_3 - \lambda_2} \| \dot M \nabla u_2 \|_{L^2} }[/math]

which can be viewed as infinitesimal variants of the estimates in the previous section.

A Sobolev inequality

Lemma Suppose that [math]\displaystyle{ u: B(0,R) \to {\mathbf R} }[/math] obeys the inhomogeneous eigenfunction equation [math]\displaystyle{ -\Delta u = \lambda u + F }[/math]. Suppose also that [math]\displaystyle{ \eta: {\mathbf R} \to {\mathbf R} }[/math] is a smooth function that equals 1 at 0 and vanishes on [math]\displaystyle{ [R,+\infty) }[/math]. Then
[math]\displaystyle{ |u(0)| \leq \frac{\sqrt{\pi}}{2\sqrt{2}} (\|F\|_{L^2(B(0,R)} (\int_0^\infty \eta(r)^2 Y_0(\sqrt{\lambda} r)^2\ r dr)^{1/2} + \lambda^{1/2} \|u\|_{L^2(B(0,R))} (\int_0^R \eta'(r)^2 Y'_0(\sqrt{\lambda}r)^2\ r dr)^{1/2} }[/math]
[math]\displaystyle{ + \|\nabla u\|_{L^2(B(0,R))} (\int_0^R \eta'(r)^2 Y_0(\sqrt{\lambda}r)^2\ r dr)^{1/2} ). }[/math]

Proof We introduce the averaged functions

[math]\displaystyle{ \bar u(r) := \frac{1}{2\pi} \int_0^{2\pi} u(r,\theta)\ d\theta }[/math]
[math]\displaystyle{ \bar F(r) := \frac{1}{2\pi} \int_0^{2\pi} F(r,\theta)\ d\theta }[/math]

and observe that [math]\displaystyle{ \bar u }[/math] obeys the inhomogeneous Bessel equation

[math]\displaystyle{ -\bar u''(r) - \frac{1}{r} \bar u'(r) = \lambda \bar u(r) + \bar F(r) }[/math]

with initial condition [math]\displaystyle{ \bar u(0) = u(0) }[/math].

We can solve this equation by separation of variables, writing

[math]\displaystyle{ \bar{u}(r) = a(r) J_0(\sqrt{\lambda} r) + b(r) Y_0(\sqrt{\lambda} r) }[/math]
[math]\displaystyle{ \bar{u}'(r) = \sqrt{\lambda} a(r) J_0'(\sqrt{\lambda} r) + \sqrt{\lambda} b(r) Y_0'(\sqrt{\lambda} r) }[/math].

Using the standard Wronskian identity

[math]\displaystyle{ J_0(\sqrt{\lambda} r) Y'_0(\sqrt{\lambda} r) - Y_0(\sqrt{\lambda} r) J'_0(\sqrt{\lambda} r) = \frac{2}{\sqrt{\lambda} \pi r} }[/math]

one soon arrives at the equations

[math]\displaystyle{ a(r) = \frac{\pi r}{2} ( \sqrt{\lambda} \bar u(r) Y'_0(\sqrt{\lambda} r) - \bar u'(r) Y_0(\sqrt{\lambda} r) ) }[/math]

and

[math]\displaystyle{ a'(r) = -\frac{\pi r}{2} \bar F(r) Y_0(\sqrt{\lambda} r) }[/math].

Meanwhile, by integration by parts

[math]\displaystyle{ a(0) = -\int_0^R a'(r) \eta(r)\ dr - \int_0^R a(r) \eta'(r)\ dr }[/math]

and thus

[math]\displaystyle{ |a(0)| \leq \frac{\pi}{2} ( \int_0^R |\bar F(r)| |Y_0(\sqrt{\lambda} r)|\ r dr + \int_0^R \sqrt{\lambda} |\bar u(r)| |Y'_0(\sqrt{\lambda} r)|\ r dr + \int_0^R |\bar u'(r)| |Y_0(\sqrt{\lambda} r)|\ r dr). }[/math]

From Cauchy-Schwarz one has

[math]\displaystyle{ 2\pi \int_0^R |\bar u(r)|^2\ r dr \leq \| u \|_{L^2(B(0,R))}^2 }[/math]

and

[math]\displaystyle{ 2\pi \int_0^R |\bar u'(r)|^2\ r dr \leq \| \nabla u \|_{L^2(B(0,R))}^2 }[/math]

and similarly

[math]\displaystyle{ 2\pi \int_0^R |\bar F(r)|^2\ r dr \leq \| F \|_{L^2(B(0,R))}^2. }[/math]

The claim then follows from further applications of the Cauchy-Schwarz inequality. []

In principle, the above inequality, when combined with the previous L^2 and H^1 estimates, gives L^infty control on [math]\displaystyle{ \dot u }[/math] in the interior of the triangle. One should also be able to get control near the interior of an edge by reflection. One needs to modify the argument a bit though to handle what is going on at vertices.