Linear inhomogeneous PDE - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:09:59Z http://mathoverflow.net/feeds/question/95674 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95674/linear-inhomogeneous-pde Linear inhomogeneous PDE unknown (google) 2012-05-01T15:44:47Z 2012-05-01T17:55:51Z <p>I am trying to understand the behavior of the following linear PDE:</p> <p>$$\partial_t u(x,t) = \partial_{xx} u(x,t) + f(x) u(x,t)$$</p> <p>where I set $f(x)=\lambda e^{-x^2} -1$ and with:</p> <p>IC : $u(x,0)= 1$ on $[-1,1]$ and $0$ elsewhere.</p> <p>BC : Neumann $\partial_x u(\pm 1,t) = 0$ </p> <p>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 $u(x,t)\to 0$.</p> <p>Can you explain me how to find this critical value ?</p> <p>THANK YOU !!!</p> http://mathoverflow.net/questions/95674/linear-inhomogeneous-pde/95681#95681 Answer by Robert Israel for Linear inhomogeneous PDE Robert Israel 2012-05-01T17:55:51Z 2012-05-01T17:55:51Z <p>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:</p> <blockquote> <p>ode:= diff(u(x),x\$2)+(lambda*exp(-x^2)-1)*u(x) = 0;</p> <p>S:= dsolve({ode,u(0)=1,D(u)(0)=0},numeric,parameters=[lambda],output=listprocedure);</p> <p>Du:= subs(S,diff(u(x),x));</p> <p>Sp:= proc(p) Du(parameters=[p]); Du(1) end proc;</p> <p>fsolve(Sp, 1.3 .. 1.4);</p> </blockquote> <p>1.329543556</p>