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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 !!!

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1 Answer 1

up vote 2 down vote accepted

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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Note that if you Fourier expand both $u$ and $f$, you get that $f$ has exponentially decaying coefficients, and that the initial value of $u$ only has the 0 mode. So the exponential mechanism from $\bar{f}$, at the first approximation, does not affect the higher modes (exponential evolution with 0 initial value stays 0 for all time). This gives further justification why the higher-mode terms can be neglected for the purpose of the heuristics. –  Willie Wong May 1 '12 at 16:12
    
Thank you very much. I think your value of $\lambda^*$ slightly overestimates the real one. This is maybe because $\bar{u}-u$ is not completely negligible as you mentioned. How could we improve your estimate ? –  mellow May 1 '12 at 16:13
    
You'd have to compute the Fourier coefficients for $f$, depending on $\lambda$. When $\bar{f}$ is sufficiently small (close to zero), the predominant contribution to the long-time behaviour of the equation will depend on the second/third order effects coming from $\int_{-1}^1 (\bar{f}-f)(\bar{u}-u)\mathrm{d}x$. The analysis would be slightly trickier, but you can try approximating in the regime $\lambda \approx 1.33$ the equation instead of by a single ODE but by a coupled system of 2, or 3, or 5 ODEs. –  Willie Wong May 2 '12 at 7:46
    
Basically you can try implement a "high frequency cut-off" for $u$, that is, just look at the first $k$ Fourier modes of $u$ and assume everything else is negligible. This reduces the problem to a linear system of ODEs with constant coefficients of the form $\frac{d}{dt}U = A_\lambda U$. The problem then becomes finding the Eigenspaces of $A_\lambda$ and see whether for $1.33 \leq \lambda \lesssim 1.339$ there is a positive eigenspace, and whether the projection of the initial data to the positive eigenspace is non-zero. –  Willie Wong May 2 '12 at 7:50
    
Thnak you! I give you the answer. –  mellow May 2 '12 at 7:59

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