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I am trying to understand the behavior of the following linear PDE:

$$\partial_t u(x,t) = \partial_{xx} u(x,t) + f(x) u(x,t)$$

where I set $f(x)=\lambda e^{-x^2} -1$ and with:

IC : $u(x,0)= 1$ on $[-1,1]$ and $0$ elsewhere.

BC : Neumann $\partial_x u(\pm 1,t) = 0$

Numerically, it seems that there is a critical value $\lambda^* \approx 1.33$ such that: if $\lambda>\lambda*$ then $u(x,t)\to \infty$ and otherwise $u(x,t)\to 0$.

Can you explain me how to find this critical value ?


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You need to solve the associated eigenvalue problem and find the lowest eigenvalue. There is little hope that this can be done in closed form. – Michael Renardy May 1 '12 at 16:01
@MichaelRenardy thank you ! Now I can sleep in peace ... Do you think it is easier if $f(x)$ is piecewise constant ? – user16215 May 1 '12 at 16:30
It is easy if $f$ is constant. – timur May 1 '12 at 17:41

You want to find the value of $\lambda$ that makes $0$ an eigenvalue. That is, find $\lambda$ so that the solution of the ode $\dfrac{d^2 u}{dx^2} + (\lambda e^{-x^2} - 1) u = 0$ with $u(0) = 1$, $u'(0)=0$ has $u'(1) = 0$. As Michael Renardy commented, you can't hope for a closed form, but a numerical solution is possible. Here's some Maple 16 code and its result:

ode:= diff(u(x),x$2)+(lambda*exp(-x^2)-1)*u(x) = 0;

S:= dsolve({ode,u(0)=1,D(u)(0)=0},numeric,parameters=[lambda],output=listprocedure);

Du:= subs(S,diff(u(x),x));

Sp:= proc(p) Du(parameters=[p]); Du(1) end proc;

fsolve(Sp, 1.3 .. 1.4);


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