Weak solution $u(x,t)$ of heat equation converges as $t \in \infty$

Where can I find a proof that the weak solution $u \in L^2(0,T;H^1) \cap H^1(0,T;H^{-1})$ of the heat equation $$u_t -\Delta u = f$$ converges as $t \to \infty$ to the solution of the elliptic PDE $$-\Delta u = f$$ ??

Any references greatly appreciated.

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I don't have it handy, but I would be surprised if it's not in Gilbarg-Trudinger or Evans. –  Neal Dec 5 '13 at 15:49
It's definitely not in Evans –  weasd Dec 5 '13 at 15:52

You can get an $L^2$ convergence result from basic energy methods. It seems that you're assuming that $f = f(x)$, i.e. $f$ is not time-dependent. Let's go ahead and take the slightly more general $f \in H^{-1}$. I also assume that you mean $H_0^1(\Omega)$ and $H^{-1} = (H_0^1)^*$.

Produce a weak solution $v \in H_0^1$ to the elliptic problem $$\begin{cases} -\Delta v = f & \text{in }\Omega \\ v = 0 &\text{on } \partial \Omega. \end{cases}$$

Next produce your weak solution $u \in L^2 H_0^1 \cap H^1 H^{-1}$ to $$\begin{cases} \partial_t u -\Delta u = f & \text{in }\Omega \\ u = 0 &\text{on } \partial \Omega \\ u = u_0 &\text{for }t=0. \end{cases}$$

Consider $w = u-v$. Since $v$ is time-independent, it's easy to see that $w \in L^2 H_0^1 \cap H^1 H^{-1}$ is a weak solution to $$\begin{cases} \partial_t w -\Delta w = 0 & \text{in }\Omega \\ w = 0 &\text{on } \partial \Omega \\ w = u_0 - v &\text{for }t=0. \end{cases}$$

Then use $w$ as a test function in the weak formulation: $$<\partial_t w,w>_* + (w,w)_1 = 0.$$ Here $<,>_*$ denotes the dual pairing between $H_0^1$ and $H^{-1}$ and $(,)_1$ is the $H^1_0$ inner product. From this we can deduce that $$\frac{d}{dt} \int_\Omega \frac{|w(x,t)|^2}{2}dx + \int_\Omega |\nabla w(x,t)|^2 dx =0.$$ Poincare's inequality in $H_0^1$ says there exists $C >0$ such that $$C \int_\Omega | w(x,t)|^2 \le \int_\Omega |\nabla w(x,t)|^2.$$ Plugging this in above shows $$\frac{d}{dt} \int_\Omega |w(x,t)|^2 dx + C \int_\Omega | w(x,t)|^2 \le 0$$ (here we trivially bound $C \ge C/2$ and then multiply everything by $2$).

This is a differential inequality of the form $\dot{z}(t) + Cz(t) \le 0$. Multiply by $e^{Ct}$: $$\frac{d}{dt}(z(t) e^{Ct}) \le 0.$$ Upon integrating, we find that $z(t) \le z(0)e^{-Ct}$ for all $t \ge 0$.

Using this above, we find that $$\int_\Omega |u(x,t) - v(x) |^2 dx = \int_\Omega |w(x,t)|^2 dx \le \exp(-Ct) \int_\Omega |u_0(x) -v(x)|^2 dx,$$ from which we deduce that $$\lim_{t \to \infty} || u(\cdot,t) - v||_{L^2(\Omega)} = 0.$$

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If somehow the constant $C$ was to be negative, do you know how to proceed? –  soup Dec 9 '13 at 18:10
The constant in the Poincare inequality is always positive. Otherwise its statement would be trivial. –  Glitch Dec 16 '13 at 16:35
I meant if the PDE was more complicated than stated and we somehow ended up with a negative constant. –  soup Dec 16 '13 at 17:46