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        riddles >> hard >> PDE
        
(Message started by: trusure on Apr 10^th, 2010, 9:20pm)

Title: PDE
Post by trusure on Apr 10^th, 2010, 9:20pm

I'm trying to find a solution to the Laplacian eigenvalue problem
Lu + ku(x,t)=0 , x in R^n, t>0

(Lu= u_x1x1 + .... + u_xnxn)

using the method of Fourier Analysis in two cases; for k>0 and k<0.

I started by taking fourier transform:

-|s|^2 u^(s,t)+k u^(s,t)=0

so: (k-|s|^2) u^(s,t)=0

how we can continue to get a nonzero solution
u(x,t) ??

thanks

Title: Re: PDE
Post by Obob on Apr 10^th, 2010, 10:02pm

This argument only works if u is in L^2, since otherwise the Fourier transform isn't defined. For instance, taking k = 0, the solutions to Lu = 0 are just the harmonic functions on R^n. But the only square-integrable harmonic function is 0, since the value of the function at any point is the average of the function on a ball of any radius centered at that point. Likewise, for other values of k, your argument shows there are no nonzero solutions in L^2.

Title: Re: PDE
Post by trusure on Apr 11^th, 2010, 2:32pm

thank you, but I want to show that there is a solution of polynomial growth if k>0, and no solution of polynomial growth if k<0 ?\
Note: if k>0 this is called the "Homogeneous Helmholtz equation"

Title: Re: PDE
Post by Obob on Apr 11^th, 2010, 5:28pm

If you are looking for polynomial growth solutions, you have to be a bit more careful and use distributions. The key is that if f(x)u = 0 where f(x) is a function and u is a distribution, this does NOT mean that u = 0: this conclusion only follows if u is not entirely supported on the locus where f = 0. For instance, if u is the delta "function" d supported at the origin, then xd = 0: if p is any test function,

<xd,p> = <d,xp> = (xp)(0) = 0 * p(0) = 0.

The flaw in your argument can be remedied by addressing this issue.

Here is a more detailed analysis:

[hide]To do this using Fourier analysis you'll have to use distributions. The tempered distributions are the distributions for which Fourier transform makes sense. A function with at most polynomial growth gives rise to a tempered distribution.

First of all notice that the equation Lu + ku = 0 is time-independent, in the sense that a function u(x,t) satisfies the equation iff the function u(x,t_0) satisfies the equation for all fixed times t_0. So we might as well drop the time parameter.

Say u is a tempered distribution such that Lu + ku = 0 in the sense of distributions, so that <Lu + ku,p> = 0 for any test function p. We can take our space of test functions to be the Schwartz space of rapidly decreasing functions (this is essentially what it means for u to be a tempered distribution: it extends to a continuous functional on the Schwartz space). The Fourier transform gives a bijection from the space of tempered distributions to itself, so Lu + ku = 0 iff (Lu + ku)^ = 0 as a distribution. Now

<(Lu+ku)^,p> = <Lu + ku, p^> = <u, (L+k)p^> = <u, ((k-|x|^2)p)^> = <u^, (k-|x|^2)p>,

so u satisfies Lu + ku = 0 if and only if u^ vanishes on all functions (k-|x|^2)p, as p ranges over the Schwartz space.

If k<0, then for any test function p, p/(k-|x|^2) is again a test function, so (k-|x|^2)p ranges over the whole Schwartz space as p does. Thus when k<0, if u is a tempered distribution with Lu + ku = 0 then u^ = 0 and hence u = 0.

On the other hand, if k >= 0, then as p ranges over the Schwartz space the functions (k-|x|^2)p range over a proper subspace, namely, the image of the map S -> S given by multiplication by k-|x|^2, where S is the Schwartz space. We know u^ must vanish on this subspace.

To find actual polynomial growth solutions, we should try to guess what the Fourier transform u^ is. It is some bounded linear functional on the Schwartz space which vanishes on all functions divisible by k - |x|^2. Any Schwartz function divisible by k - |x|^2 must in particular vanish on the sphere of radius sqrt(k). So we can try to look for a way of "integrating" Schwartz functions on R^n which will give zero any time our function is identically zero on the sphere of radius k. A couple possibilities come to mind:

1) Pick any point x_0 on the sphere of radius sqrt(k), and take u^ to be a delta function supported at the point. Then u(x) = exp(i (x . x_0)), where . is dot product of vectors. Then the real and imaginary parts cos(x . x_0) and sin(x . x_0) are both bounded real solutions.

2) Another (more interesting) solution can be found as follows: given a function p on R^n, restrict it to the sphere of radius sqrt(k), and then integrate your function over this sphere with respect to the volume form on the sphere. This gives a continuous linear functional u^ on the Schwarz space, whose inverse Fourier transform u is a tempered distribution solving the equation Lu + ku = 0. I'm guessing that u is actually a function, not just a distribution, given by u(x) = integral exp(i (x . y)) dy, where the integral is over the sphere of radius sqrt(k). An explicit form for this function should be findable.

3) More generally, given a finite measure on the sphere, you can use it to find a u^ as in (2) by integrating with respect to that measure. I think this should still give a function, not just a distribution. This generalizes both (1) and (2), since delta functions correspond to measures with atoms.

It's possible that there are solutions not coming from the construction in (3), since a function p being divisible by k - |x|^2 isn't the same thing as saying that p vanishes on the sphere of radius sqrt(k). For instance, if we are working in R^2 and we have k=0, we can get polynomial growth solutions by looking at the real or imaginary part of a complex polynomial, identifying R^2 with C.[/hide]

Title: Re: PDE
Post by trusure on Apr 15^th, 2010, 10:01am

" [hide]If k<0, then for any test function p, p/(k-|x|^2) is again a test function, so (k-|x|^2)p ranges over the whole Schwartz space as p does [/hide]"

why it is true that: " [hide]p/(k-|x|^2) is again a test function[/hide]"

Title: Re: PDE
Post by Obob on Apr 15^th, 2010, 10:14am

Because if k<0 then k - |x|^2 is never zero.

Title: Re: PDE
Post by trusure on Apr 15^th, 2010, 10:22am

so: this means that the solution if k>=0 will be

u(x) = exp(i (x.x0) ) , and
u(x)=integral ( exp(i (x.y) ) dy ) , and both of these are of polynomial growth ?

!! what about the variable t ?
sorry I'm not sure that I got the idea for this case :-[

Title: Re: PDE
Post by Obob on Apr 15^th, 2010, 10:43am

I already addressed what happens with t in my post: since there aren't any t-derivatives, t is essentially irrelevant.

I wouldn't say "the solution" is those things. There are lots and lots of solutions; those are just particular ones. And of course for the first one it matters what x_0 is, and for the second the integral is over a sphere. I'm not totally positive the second one works, but I think it does. The first one definitely works.

The first solution u = exp(i(x.x0)) is clearly polynomial growth; it is bounded in absolute value by 1 everywhere. I'm not sure about the second one, but I suspect it is.

Title: Re: PDE
Post by trusure on Apr 15^th, 2010, 10:50am

I realy thank you for your solution and replay, but is this enough to prove that the PDE has a solution of polynomial growth for k>0 ?

Title: Re: PDE
Post by Obob on Apr 15^th, 2010, 10:56am

Yes, it does far far more than that. Case in point: take

u(x1,...,xn,t) = cos(sqrt(k)*x1)

(this is essentially the simplest of all the solutions)

The Laplacian of this u is -k cos(sqrt(k)*x1), so Lu + ku = 0. And u has polynomial growth: it is bounded by 1!

Title: Re: PDE
Post by trusure on Apr 15^th, 2010, 11:00am

I'm lost.... how did you come with this u ?

Do you know any reference where I can study this equation in details ?

Title: Re: PDE
Post by Obob on Apr 15^th, 2010, 11:54am

This u is the real part of exp(i(x.x0)), where x0 = (sqrt(k),0,...,0).

I don't have any references, aside from the basics of distribution theory.