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), \quad \langle \dfrac{\partial u}{\partial t}-\Delta u \rangle_{D',D} = 0$$ We write $u_t$ for $u(.,t)$.
I tried this:
$$\Delta u(t)=\Delta \sum_{n=1}^{+\infty} [e^{-\lambda_n t} (g,e_n)e_n)=\sum_{n=1}^{+\infty} e^{-\lambda_n t} (g,e_n) (-\lambda_n) e_n$$
Let $\psi \in C^{\infty}_c(I \times \Omega)$; We have: $$(\dfrac{d}{dt} u(t),\psi)=-(u(t),\dfrac{\partial \psi}{\partial t})$$ My problem is to conclude that $\dfrac{\partial u}{\partial t} - \Delta u = 0$ in $\mathcal{D}'(I \times \Omega)$.
Thanks for the help.