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


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 $$\langle Tf_n,\phi\rangle=\langle f_n,S\phi\rangle\to\langle f,S\phi\rangle=\langle Tf,\phi\rangle.$$

  • $\begingroup$ 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

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