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My question is related, but not a duplicate of Asymptotic convergence of the solution to a parabolic pde to the solution of an elliptic pde

Suppose I solve the parabolic PDE:

$u_t = \Delta u + f(x,t)$ on $\Omega \times(0,T]$

$\nabla u(x,t) \cdot n = 0 $ on $\partial\Omega$

$u(x,0) = g(x)$ on $\Omega$

and obtain the (unique) steady state solution $u^*(x)$. Assume that the source term in the parabolic PDE satisfies the compatibility condition for a pure neumann problem $\int_{\Omega}f(x,t)= \int_{\Omega}u_t$

  1. From the compatibility above, can I claim that it will also hold as $t \rightarrow \infty$ ? I suppose that the functions $f(x,t)$ and $u_t$ will need to be nice enough for me to take the limit inside the integral ?

Now consider the elliptic PDE: $0 = \Delta u + \tilde{f}(x)$ on $\Omega$

$\nabla u(x) \cdot n = 0 $ on $\partial\Omega$.

This is now a pure Neumann problem. Assume that the steady state value of the source is used, i.e., $\tilde{f}(x) = \lim_{t \rightarrow \infty} f(x,t)$ is used and that it satisfies the compatibility condition for this pure Neumann problem.

Now, I know that if $\hat{u}(x)$ be a solution, then $\hat{u}(x) +c$ for all real numbers $c$, also solves the PDE. I am aware that we fix the constant by demanding zero mean solution or prescribing the value at a point.

So my questions are:

  1. Is it possible to prescribe a condition for the elliptic problem a priori so that the solution to the elliptic PDE matches the steady state solution $u^*(x)$ of the parabolic PDE ?

  2. Can we say anything else about the relation between these two solutions ?

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    $\begingroup$ Is $f(x) = \lim_{t\rightarrow\infty} f(x,t)$ or something else? It can't be the "same" source function if one has $t$ dependence and the other doesn't. $\endgroup$ – Bill Barth Sep 18 '14 at 20:59
  • $\begingroup$ Thanks for pointing it out, that's what I meant to say but made a mistake in writing it. Have edited the question to reflect it. $\endgroup$ – me10240 Sep 19 '14 at 1:15
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I think that one key point to understand the answers is that, with the parabolic PDE that you wrote, we have some control on the "quantity of $u$", i.e. $\int_{\Omega} u $: If you integrate the PDE, you have

$$\int_{\Omega} u_t = \int_{\Omega} \Delta u + \int_{\Omega} f(x,t) = \int_{\Omega} f(x,t)$$

because $\int_{\Omega} \Delta u = \int_{\partial \Omega} \nabla u \cdot n = 0$ thanks to the boundary conditions. Now integrating in time and supposing that everything is nice enough to exchange the integrals, we have

$$ \int_{t_1}^{t_2} \int_{\Omega} f(x,t) = \int_{t_1}^{t_2} \int_{\Omega} u_t = \int_{\Omega} \int_{t_1}^{t_2} u_t = \int_{\Omega} u(t_2) - \int_{\Omega} u(t_1)$$

  1. Yes. If you want the solution $u^{*}$ to exist, you need $u_t \rightarrow 0$ (the solution converges in time). So you will need also $\int_{\Omega} u_t \rightarrow 0$ and for that, you need $\int_{\Omega} f(x,t) \rightarrow 0$. So you have $\int_{\Omega} \tilde{f} = 0$ for the stationnary solution to exist.

  2. Yes. Use the relation I wrote before with $t_1 =0$, you will get:

$$ \int_{\Omega} u(t_2) = \int_{\Omega} u_0 + \int_{0}^{t_2} \int_{\Omega} f(x,t) $$

If you let $t_2 \rightarrow \infty$ (and everything is sufficiently smooth), you obtain if the limits exist

$$ \int_{\Omega} u^{*} = \int_{\Omega} u_0 + \int_{0}^{\infty} \int_{\Omega} f(x,t) $$

This looks like the condition that you are looking for. And it looks rather intuitive: you want your stationary solution to contain as much "u" as would have been created through the whole evolution of the problem (remark that the condition from the first point is necessary for the integral to converge).

  1. I don't see what I could add.
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