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Let $\Omega$ be a bounded subset of $\mathbb R^d$ and let $g\in L^2\left(\Omega \right)$. Let $\left(\lambda_n\right)_{n\in \mathbb{N}}$ be the eigenvalues of the Laplacian operator $\Delta$, and $\left(e_n\right)_n$ the eigenvectors associated to the eigenvalues. There exist a unique $u=(u_t)\in C^0([0,+\infty [,L^2(\Omega ))$ that solves the problem $$\dfrac{\partial u}{\partial t}−\Delta u=0 ,\in D′\left(\right]0,+\infty \left[\times \Omega\right),\quad u_0=g,\quad u_t\in H^1_0\left(\Omega \right)$$ with $u_t=\sum^{+\infty }_{n=1}e^{−λ_nt}\left(g,e_n\right) e_n$

My question is how we prove that $$\dfrac{\partial u}{\partial t}−\Delta u=0 ,\in D′\left(\right]0,+\infty \left[\times \Omega\right) ?$$ i.e. For all $\psi \in C^{\infty}_c\left(I \times \Omega\right), \quad \left\langle \dfrac{\partial u}{\partial t}-\Delta u \right\rangle_{D',D} = 0$. We write $u_t$ for $u\left(.,t\right)$.

I tried this:

$$\Delta u\left(t\right)=\Delta \sum_{n=1}^{+\infty} e^{-\lambda_n t} \left(g,e_n\right)e_n=\sum_{n=1}^{+\infty} e^{-\lambda_n t} \left(g,e_n\right) \left(-\lambda_n\right) e_n$$

Let $\psi \in C^{\infty}_c\left(I \times \Omega\right)$; We have: $$\left(\dfrac{d}{dt} u\left(t\right),\psi\right)=-\left(u\left(t\right),\dfrac{\partial \psi}{\partial t}\right)$$ My problem is to conclude that $\dfrac{\partial u}{\partial t} - \Delta u = 0$ in $\mathcal{D}'\left(I \times \Omega\right)$.

Thanks for the help.

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I won't deal with the issue of distributional uniqueness. But I can offer some insight into this problem along the lines of what else I've written for you. You should be able to tackle uniqueness for classical settings afterwards. I'd have to look into distribution theory more to deal with uniqueness there, and I probably won't have time to do that.

Background: Assume $\Omega$ is a bounded, open region with a regular (smooth) boundary (or something similar.) Then the Laplacian has as a unique selfadjoint extension from $C_{c}(\Omega)$ to $\Delta : \mathcal{D}(\Delta)\subset L^{2}(\Omega)\rightarrow L^{2}(\Omega)$. The operator $\Delta$ satisfies $((-\Delta) f,f)_{L^{2}(\Omega)}\ge \lambda_{0}\|f\|^{2}_{L^{2}(\Omega)}$ for some $\lambda_{0} > 0$ and for all $f \in \mathcal{D}(\Delta)$. $\Delta$ has a compact resolvent and, therefore, also has a complete orthonormal basis $\{ e_{n}(x)\}_{n=1}^{\infty}$ of eigenfunctions with $(-\Delta)e_{n}=\lambda_{n}e_{n}$ and real eigenvalues $\lambda_{j}$, $j \ge 1$, satisfying $$ 0 < \lambda_{0} \le \lambda_{1}\le \lambda_{2} \le \lambda_{3} \le \cdots . $$ By the spectral theorem, $f \in L^{2}(\Omega)$ satisfies $\sum_{n}\lambda_{n}^{2k}|(f,e_{n})|^{2} < \infty$ iff $f \in \mathcal{D}((-\Delta)^{k})$. In that case, $\|(-\Delta)^{k}f\|^{2}=\sum_{n}\lambda_{n}^{2k}|(f,e_{n})|^{2}$. It can be shown that $\bigcap_{k=1}^{\infty}\mathcal{D}((-\Delta)^{k})\subseteq C_{0}^{\infty}(\Omega)$. Therefore, if $f \in L^{2}(\Omega)$ satisfies $\sum_{n}\lambda_{n}^{2k}|(f,e_{n})|^{2} < \infty$ for each $k=1,2,3,\cdots$, then $f \in C_{0}^{\infty}(\Omega)$.

Using these ideas, you can show that $u(t,x)=\sum_{n}e^{-\lambda_{n}t}(f,e_{n})e_{n}$ is infinitely differentiable in $x$ for all $t > 0$; indeed, for a fixed $t > 0$ and fixed positive integer $k$, the function $r(\lambda)=\lambda^{k}e^{-\lambda t}$ is bounded by some constant $M_{k,t}$ for all $\lambda \in [0,\infty)$, which is enough to prove that your function $u=\sum_{n}e^{-\lambda_{n}t}(g,e_{n})e_{n}$ satisfies $$ \sum_{n}\lambda_{n}^{2k}|(u,e_{n})|^{2} = \sum_{n}\lambda_{n}^{2k}e^{-2\lambda_{n}t}|(g,e_{n})|^{2} \le M_{2t,2k}\sum_{n}|(g,e_{n})|^{2} < \infty \mbox{ for all } k \ge 1,\; t > 0. $$ So, for each $t > 0$, the function $u(t,x)$ is in $C_{0}^{\infty}(\Omega)$. In fact, you can take as many derivatives with respect to $t$ as you want for $t > 0$, and $\frac{\partial^{k}u}{\partial t^{k}}$ is infinitely differentiable in $x$. Actually, $u \in C^{\infty}((0,\infty)\times\Omega)$. That definitely makes it easier to show that $$ \frac{\partial u}{\partial t}-\Delta u = 0 \in \mathcal{D'}((0,\infty)\times \Omega). $$ I'll leave the demonstration to you, knowing that $u \in C^{\infty}((0,\infty)\times\Omega)$.

The Lebesgue dominated convergence theorem applied to sums can be used to show that $u : [0,\infty)\rightarrow L^{2}(\Omega)$ is continuous with $u_{0}=g$; for example, because $1-e^{-\lambda_{n}t} \le 1$ for all $n \ge 1$ and for $t \ge 0$, then dominated convergence gives $$ \lim_{t\downarrow 0}\|u_{t}-u_{0}\|^{2}_{L^{2}(\Omega)}=\sum_{n}\lim_{t\downarrow 0}(1-e^{-\lambda_{n}}t)^{2}|(g,e_{n})|^{2}=0. $$ So $u \in C^{0}([0,\infty),L^{2}(\Omega))$. It's not hard to show that $u$ is infinitely differentiable as a map from $(0,\infty)$ to $L^{2}(\Omega)$ (or even into $C^{k}$ types of spaces), but $u$ is not differentiable at $t=0$ unless the initial data $g$ is in $\mathcal{D}(-\Delta)$. If $g \in \mathcal{D}((-\Delta)^{k})$, then $u : [0,\infty)\rightarrow L^{2}(\Omega)$ is k times differentiable at $t=0$. However, $u$ is always nicely differentiable for $t > 0$ because the $e^{-t\lambda_{n}}$ terms make $u(t)$ infinitely differentiable in $x$ for $t > 0$.

Continuity: Start by considering the function $u(t) = \sum_{n}e^{-\lambda_{n}t}(g,e_{n})e_{n}$ is a function from $[0,\infty)$ to $L^{2}(\Omega)$, where $g \in L^{2}(\Omega)$ is given. This function is a continuous vector function because $$ \|u(t)-u(t')\|_{L^{2}(\Omega)}^{2}=\sum_{n}|e^{-t\lambda_{n}}-e^{-t'\lambda_{n}}|^{2}|(g,e_{n})|^{2},\;\;\; t,t' \ge 0. $$ The Lebesgue bounded convergence theorem gives you continuity because $$ |e^{-t\lambda_{n}}-e^{-t'\lambda_{n}}|^{2}|(g,e_{n})|^{2} \le |(g,e_{n})|^{2}, \;\; t,t' \ge 0, $$ and $\sum_{n}|(g,e_{n})|^{2}=\|g\|_{L^{2}(\Omega)} < \infty$.

A similar argument can be used to show that, for $k \ge 0$, the vector function $u : (0,\infty)\rightarrow H^{k}(\Omega)$ is continuous. This is because $\mathcal{D}((-\Delta)^{k})\subseteq H^{k}(\Omega)$ and because $$ \lambda^{2k}e^{-2t\lambda} \le e^{-2k}(k/t)^{2k},\;\;\; t > 0, k=1,2,3,\cdots,\;\; \lambda \ge 0. $$ (This is a basic Calculus problem to find the minimum of the above for $\lambda \ge 0$.) To see how this follows, let $k \ge 1$ and $t > 0$, and consider that $$ \lambda_{n}^{2k}|(u(t),e_{n})|^{2}=\lambda_{n}^{2k}e^{-2\lambda_{n}t}|(g,e_{n})|^{2} \le e^{-2k}(k/t)^{2k}|(g,e_{n})|^{2}. $$ It follows that $u(t) \in \mathcal{D}((-\Delta)^{k})\subseteq H^{2k}(\Omega)$ for all $k$, and $$ (-\Delta)^{k}u(t) = \sum_{n}\lambda_{n}^{k}e^{-\lambda_{n}t}(g,e_{n})e_{n},\;\; t > 0. $$ Using the Lebesgue dominated convergence theorem, $(-\Delta)^{k}u(t)$ is continuous as a map from $(0,\infty)$ to $L^{2}(\Omega)$, and it remains continuous at $t=0$ iff $g \in \mathcal{D}((\Delta)^{k})$. This is enough to show that $u : (0,\infty)\rightarrow H^{2k}(\Omega)\cap H^{1}_{0}(\Omega)$ for $k\ge 1$, and $u$ is continuous at $t=0$ if $g \in \mathcal{D}((-\Delta)^{k})$.

Distributional Solution Using the same type of argument as before, assume $\psi \in \mathcal{C}_{c}((0,\infty)\times\Omega)$. That way $0$ is excluded from consideration, and $\psi$ vanishes for $t$ outside some interval $I=[a,b]\subseteq (0,\infty)$. Let $u(t)$ be as above. Then $u(t)=\sum_{n}e^{-t\lambda_{n}}(g,e_{n})e_{n}(x)$ converges in $L^{2}(I\times\Omega)$ by the same arguments I gave you before. So, $$ \begin{align} \left((\frac{\partial}{\partial t}-\Delta)T_{u}\right)(\psi) & = T_{u}\left(-\frac{\partial}{\partial t}\psi -\Delta\psi\right) \\ & = \int_{I\times\Omega}u\left(-\frac{\partial}{\partial t}\psi-\Delta\psi\right)\,dxdt \\ & = (u,-\frac{\partial}{\partial t}\psi-\Delta\psi)_{L^{2}(I\times\Omega)} \\ & = (\sum_{n}e^{-t\lambda_{n}}(g,e_{n})_{L^{2}(\Omega)}e_{n},-\frac{\partial}{\partial t}\psi-\Delta\psi)_{L^{2}(I\times\Omega)} \\ & = \sum_{n}(g,e_{n})_{L^{2}(\Omega)}(e^{-t\lambda_{n}}e_{n}(x),-\frac{\partial}{\partial t}\psi-\Delta\psi)_{L^{2}(I\times\Omega)} \\ & = \sum_{n}(g,e_{n})_{L^{2}(\Omega)}(\frac{\partial}{\partial t}(e^{-\lambda_{n}t}e_{n})-\Delta(e^{-\lambda_{n}t}e_{n}),\psi)_{L^{2}(I\times\Omega)} \\ & = \sum_{n}(g,e_{n})_{L^{2}(\Omega)}(0,\psi)_{L^{2}(I\times\Omega)}=0. \end{align} $$

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  • $\begingroup$ Please, how you prouve that $\dfrac{\partial u}{\partial t} - \Delta u =0$? $\endgroup$
    – varphi
    May 27 '14 at 15:37
  • $\begingroup$ My demonstration isn't complete. How you prouve that $$\dfrac{\partial u}{\partial t} - \Delta u = 0 \quad \mbox{in} \mathcal{D}'(I \times \Omega)$$ please $\endgroup$
    – varphi
    May 27 '14 at 16:10
  • $\begingroup$ Have you tried to adapt what I showed you before for the wave operator $\Box$? $\endgroup$ May 27 '14 at 16:21
  • $\begingroup$ Yes of corse. We have $$(\Delta)u(t,.)=(\Delta)\sum_{n=1}^{+\infty} [e^{-\lambda_n t} (g,e_n)e_n] = \dfrac{\partial}{\partial t} \sum_n e^{-\lambda_n t} (g,e_n)e_n$$ but my problem is to compute $(\dfrac{d}{dt} u,\psi)$ and to conclude. $\endgroup$
    – varphi
    May 27 '14 at 16:25
  • $\begingroup$ + i think that there is an mistake: we say that so, for each $t>0$, the function $u(t,x)$ is ic $C^{\infty}_c(\Omega)$.... and $\dfrac{\partial^k u}{\partial t^k}$ is infinitly differentiable in $t$ not in $x$. No? $\endgroup$
    – varphi
    May 27 '14 at 17:20

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