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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 ?

THANK YOU !!!

1 Answers 1

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The heuristic is probably that

$ \lambda^* = \frac{\int_{-1}^1 1 \mathrm{d}x}{\int_{-1}^1 e^{-x^2}\mathrm{d}x} = \frac{2}{\sqrt{\pi} \mathrm{erf}(1)} \approx 1.339 $

Note that if $\lambda^*$ is defined as the above you have that

$ \int_{-1}^1 \lambda^* e^{-x^2} - 1 \mathrm{d}x = 0 $

For larger $\lambda$ the integral is positive, while for smaller ones the integral is negative.

So integrating your equation from $-1$ to $1$ you have that

$ \partial_t \int_{-1}^1 u\mathrm{d}x = \int_{-1}^1 \partial_x^2u \mathrm{d}x + \int_{-1}^1 f(x) u(x) \mathrm{d}x $

The first term on the right hand side is 0, using Neumann's BC and the fundamental theorem of calculus.

Writing $\bar{u}$ for the mean value of $u$ on the interval $[-1,1]$, the equation becomes

$ \frac{d}{dt} \bar{u} = 2\bar{f}\bar{u} + \int_{-1}^1 (\bar{f}-f)(\bar{u}-u) \mathrm{d}x $

The second term is roughly speaking small: initially $\bar{u} - u$ is small, and the heat equation forces an exponential decay in all higher modes: the only mode that is not expected to decay is the 0 mode, which is captured by the mean $\bar{u}$. If we just heavyhandedly set the second term to be negligible, we have that the equation is roughly governed by the ODE

$ \frac{d}{dt}\bar{u} = \bar{f} \bar{u} $

from which we get that if $\bar{f}$ is positive ($\lambda > \lambda^*$) you get exponential increase, and if $\bar{f}$ is negative ($\lambda < \lambda^*$) you get exponential decrease.

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    Thnak you! I give you the answer.2012-05-02