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?


Your Answer

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

Browse other questions tagged or ask your own question.