0
$\begingroup$

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

My question is how we prove that $$\dfrac{\partial u}{\partial t}−\Delta u=0 ,\in D′(]0,+\infty [\times \Omega) ?$$

i.e. For all $\psi \in C^{\infty}_c(I \times \Omega),$ $$\langle \dfrac{\partial u}{\partial t}-\Delta u ,\psi \rangle_{D',D} = 0$$

Thank's for the help

$\endgroup$
2
$\begingroup$

Let $u_n=e^{-\lambda_n t}(g,e_n)e_n$, so that $u=\sum_n u_n$. Clearly $(\partial/\partial t-\Delta)u_n=0$. Next notice that $\sum_n u_n$ does converge in $\mathcal D'(]0,\infty[\times\Omega)=:\mathcal D'$, as it converges in $L^2(]0,\infty[\times\Omega)$. Let $T=\partial/\partial t-\Delta$. The beauty of distributions is that $T$ (as any linear differential operator with smooth coefficients) is a continuous map $\mathcal D'\to\mathcal D'$, and thus $T(\sum_n u_n)=\sum_n (T u_n)=0$.

PS: Continuity is seen as follows ($S$ denotes the adjoint of $T$, here $S=-\partial/\partial t-\Delta$): if $f_n\to f$ in $\mathcal D'$ then $$\langle Tf_n,\phi\rangle=\langle f_n,S\phi\rangle\to\langle f,S\phi\rangle=\langle Tf,\phi\rangle.$$

$\endgroup$
6
  • $\begingroup$ Thank you for the answer. I have two questions please: 1- How you justify that $(\dfrac{\partial}{\partial t} - \Delta)u_n=0$? and this equality is in $\mathcal{D}'$? And my second question is 2- How you justify the continuity? And Linearity to prouve that $T$ is an distribution? The distribution must be defined : $C^{\infty}_c--> \mathbb{C}$ so why you tell that $T:D'-->D'$$? And Thank's for all. $\endgroup$
    – varphi
    May 26 '14 at 21:46
  • $\begingroup$ @varphi for 1: $(\partial/\partial t-\Delta)u_n=(-\lambda_n+\lambda_n)u_n$ (supposing $\Delta e_n=-\lambda_n e_n$). For 2: $T$ is an operator $\mathcal D'\to\mathcal D'$, not a distribution; its continuity is proven in the answer $\endgroup$
    – user8268
    May 26 '14 at 21:51
  • $\begingroup$ But is an operator $D-->\mathbb{C}$ I don't understand why it's an operator $D'-->D'$. $\endgroup$
    – varphi
    May 26 '14 at 21:55
  • $\begingroup$ @varphi: $T=\partial/\partial t-\Delta$ $\endgroup$
    – user8268
    May 26 '14 at 21:58
  • $\begingroup$ Sorry, but i dont understand. First, we have $\sum_n u_n$ converge in $L^2(]0,+\infty[\times \Omega)$, okay. But how you choising $T$ and how we define $T:D' -->D'$? And why we need to define an operator $D'-->D'$? $\endgroup$
    – varphi
    May 26 '14 at 22:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.