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Consider the heat equation:

$$u_t -\Delta u=0 \quad \text{in} \quad Q_T=\Omega \times(0,T) $$ $$u(\sigma,t)=0 $$ $$u(x,0)=g(x) \quad \text{in} \quad \Omega $$

a weak formulation is: find $u \in H^1(0,T;H^1_0(\Omega),H^-1(\Omega))$ s.t.

$$\lt \dot{u(t)},v\gt_* +(\nabla u(t),\nabla v)_{L^2}=0$$ $$ ||u(t)-g ||_{L_2} \rightarrow 0 \quad \text{as} \quad t \rightarrow0^+$$

From general theory, we have an apriori estimate on $u,\nabla u:$

$$ ||u(t)||_{L_2}^2 + \int_0^t||\nabla u(t)||_{L_2}^2dt \leq C ||g|||_{L_2}^2e^t $$

However, through the Faedo-Galerkin approximations, choosing as orthonormal basis in $L^2$ the "Dirichelet" eivengectors of $-\Delta$ we have a more specific esimate:

$$||\nabla u(t) ||_{L_2}^2 \leq \frac{1}{2et}||g||_{L_2}^2$$

and if $g \in H^1_0(\Omega)$

$$||\nabla u(t) ||_{L_2}^2 \leq e^{-2 \lambda_1 t}||\nabla g||_{L_2}^2 $$

This is very fine for the moment, but my book asserts that these results let us conclude that, if $$f=f(x) \in L^2(\Omega) $$ the solution $u$ of:

$$u_t -\Delta u=f(x) \quad \text{in} \quad Q_T=\Omega \times(0,T) $$ $$u(\sigma,t)=0 $$ $$u(x,0)=g(x) \quad \text{in} \quad \Omega $$

converges wrt $H^1_0$ norm to $u_{\infty}$ solution of:

$$ - \Delta u_{\infty} = f \quad \text{in} \quad \Omega$$

$$ u_{\infty}=0 \quad \text{on} \quad \partial \Omega$$

This further point is not very clear to me, because in order to find those estimates, I solved a linear sistem of ODEs on the Fourier's coefficients of $u$, which was homogeneous because $f=0$! Moreover, the exact formulation of these coefficient was the key to get those estimates, so how can I prove the latter result?

Finally, is that convergence always true?

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