mersenneforum.org Asymptotic Behavior of a Differential Equation
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 2005-12-22, 11:16 #1 jinydu     Dec 2003 Hopefully Near M48 2·3·293 Posts Asymptotic Behavior of a Differential Equation Here's a question that I've been trying to answer for a while, with no real progress: Suppose that $\psi(r,\theta,\phi)$ is a complex-valued function that satisfies: $\Large\nabla^2\psi=f(r,\theta,\phi)\psi$ where $\nabla^2$ is the Laplacian operator (in spherical coordinates) and $f(r,\theta,\phi)$ is a real-valued function that is always positive. Prove that $\lim_{r\rightarrow\infty}\psi\neq0$, or give a counterexample (exclude the trivial case where $\psi=0$ at all points in space). Last fiddled with by jinydu on 2005-12-22 at 11:19
 2005-12-22, 14:07 #2 jinydu     Dec 2003 Hopefully Near M48 2·3·293 Posts If it helps, I can split the differential equation up into components to derive relations. In Cartesian coordinates: $\Large{Re(\nabla^2\psi)Re(\psi)+Im(\nabla^2\psi)Im(\psi)>0}$ $\Large{Re(\psi)Im(\nabla^2\psi)=Re(\nabla^2\psi)Im(\psi)}$ In Polar coordinates: $\Large{|\nabla^2\psi|=f|\psi|}$ (which in fact tells us nothing, since $f$ is an arbitrary positive function; and the absolute values are non-negative) $\Large{arg(\nabla^2\psi)=arg(\psi)}$ If we use the polar coordinate relation (presumably because it is the simplest), the problem becomes to show that: $\Large{\lim_{r\rightarrow\infty}[|\psi|e^{i\cdot{}arg(\nabla^2\psi)}]\neq0}$ Last fiddled with by akruppa on 2005-12-22 at 16:29 Reason: fixed tex
 2005-12-22, 14:23 #3 jinydu     Dec 2003 Hopefully Near M48 2·3·293 Posts Argh! I ran out of time editing my last message. The last line should read: $\Large{\lim_{r\rightarrow\infty}[|\psi|e^{i\cdot{}arg(\nabla^2\psi)}]\neq0}$ Last fiddled with by jinydu on 2005-12-22 at 14:23
 2005-12-22, 20:16 #4 ewmayer ∂2ω=0     Sep 2002 República de California 2D4216 Posts Solutions of such elliptic PDEs are well-known to be entirely dependent on their boundary conditions, so specifying the equations sans BCs is not meaningful. In the case of an infinite domain ("free boundary conditions"), I believe it is possible to show that so so long as the forcing function f({coordinates}) is L2-integrable (which implies that it must decay at infinity), the solutions of the corresponding elliptic must also be L2-integrable, i.e. must also decay at infinity - in fact I believe in the complex case the solutions must be entire if f is. I'll see if I can find a reference...
 2005-12-23, 00:36 #5 jinydu     Dec 2003 Hopefully Near M48 6DE16 Posts This problem comes from my trying to prove a theorem in quantum mechanics: That there are no normalizable solutions to the time-independent Schrodinger equation where the total energy is less than the potential energy at every point in space. I've managed to show, after some algebra, that it is sufficient to prove the claim in this thread, since a wavefunction that doesn't go to zero at infinity can't possibly be normalized. Last fiddled with by jinydu on 2005-12-23 at 00:36
 2006-01-04, 04:08 #6 jinydu     Dec 2003 Hopefully Near M48 2·3·293 Posts Bump. Please?
 2006-01-05, 20:17 #7 ewmayer ∂2ω=0     Sep 2002 República de California 1158610 Posts Ah - I thought you were indicating in your 22 Dec post that you'd solved the problem to your own satisfaction. I suggest assuming a separable forcing function, i.e. replace f(r,theta,phi) by F(r)*G(theta)*H(phi), and a separable solution psi(r,theta,phi) = a(r)*b(theta)*c(phi). Plugging into the equation should yield a fairly simple ODE for a(r) in terms of F(r) and the radial terms of the spherical Laplacian. If you assume F(r) can be expanded as a power series in r, you should be able to derive general expressions for the coefficients of the series that then represents the solution function a(r), and say something pretty general about decay at infinity. Turning that into a general bound on generic (i.e. not necessarily separable f and psi) is trickier, but perhaps your radial-only analysis will allow you to bound a general solution, e.g. show that if a general forcing function f(r,theta,phi) can be bounded above by a separable function F(r)*G(theta)*H(phi), then the corresponding (usually unknown, since it would require an exact solution of the nonseparable equation) solution psi(r,theta,phi) can similarly be bounded above by the separable-case solution function a(r)*b(theta)*c(phi). Sorry I don't have more time to help with the details, but I really think in terms of asymptotic-decay-at-infinity behavior, it should suffice to study a radial-only ODE analog of the full equation - angular variations must always remain bounded (at any constant r) by the mere fact of their periodicity in the angular coordinates (i.e. psi(r,theta+2*k*pi,phi) = psi(r, theta, phi), similar for f, similarly for the second angle phi.)
 2006-01-21, 20:58 #8 jinydu     Dec 2003 Hopefully Near M48 2·3·293 Posts Unfortunately, even the 1D case has me stumped: $\Large{arg(\frac{d^2\psi(x)}{dx^2})=arg(\psi(x))}$ Now prove that $\Large{\int_{-\infty}^{\infty}|\psi(x)|^2dx}$ is not finite and nonzero. Last fiddled with by jinydu on 2006-01-21 at 21:00

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