4
$\begingroup$

Let $\zeta, u_0\in L^2(\Omega)$, with $\zeta \geq 0$ and $\Omega\subset \Bbb R^d$ open and bounded.
\begin{equation}\label{Star-3.7} \begin{cases} \partial_t u -\Delta u + \zeta u=0 &\mbox{ in }\; \Omega\times (0, T),\\ u = 0 &\mbox{ in }\; \partial\Omega\times (0, T), \\ u(\cdot,0) = u_{0}, &\mbox{ in }\; \Omega, \end{cases} \end{equation} We say that $u: \Omega\times (0, T)\to \Bbb R$ is a weak solution if: $u \in L^2(0,T; H_0^1(\Omega) \cap L^2(\Omega, \zeta dx))$ and we have $$\int_\Omega\partial_t uv +\int_\Omega\nabla u\nabla v+ \int_\Omega\zeta uv \qquad\text{ for all $\quad v \in L^2(0,T; H_0^1(\Omega) \cap L^2(\Omega, \zeta dx))$}. $$

Question: Is it possible to show the existence and uniqueness of a weak solution?

If Yes which method could be suitable here or what are the right references for such equations?

$\endgroup$

2 Answers 2

7
$\begingroup$

Here is a functional analytic approach (Kato's book on perturbation theory is a good reference):

Let $$ a\colon D(a)\times D(a)\to \mathbb{R},\,(u,v)=\int \nabla u\cdot \nabla v+\int \zeta uv. $$ with $D(a)=H^1_0(\Omega)\cap L^2(\zeta\,dx)$. It is not hard to see that $a$ is closed, that is, $D(a)$ endowed with the inner product $\langle\cdot,\cdot\rangle_2+a$ is complete. Therefore there exists a positive self-adjoint operator $A$ on $L^2(\Omega)$ with $D(A)\subset D(a)$ such that $a(u,v)=\langle Au,v\rangle$ for $u\in D(A)$, $v\in D(a)$.

Let $u(t,\cdot)=e^{-tA}u_0$. By functional calculus, $u\in C^\infty([0,\infty);H^1_0(\Omega)\cap L^2(\zeta\,dx))$ and $$ \partial_t u=-Au. $$ In particular, $u$ is a weak solution of your PDE by the definition of $A$.

To prove uniqueness, you can apply the usual energy method, i.e., show that $a(u,u)$ is decreasing along solutions.

$\endgroup$
4
$\begingroup$

Since you use it as measure, I guess $\zeta$ is non-negative ? Also your fortmulation needs to be integrated in time or you should add a regularity assumption for the time derivative.

I would first solve the equation in $L^2(0,T;H^1_0(\Omega))$ replacing $\zeta$ by $\zeta_R:=\mathbf{1}_{|\zeta|\leq R} \zeta$. In that case this just follows from standard parabolic theory (see Evans for instance).

Your then get a family of solution $(u_R)_R$ which solves a weak formulation in $L^2(0,T;H^1_0(\Omega))$. In particular, choosing $u_R$ as a test function in its own formulation you get after integration by parts

\begin{align*} \|u_R\|_{L^\infty(0,T;L^2(\Omega)}^2 + \|\nabla u_R\|_{L^2(0,T;L^2(\Omega)}^2 + \|u_R\|_{L^2(0,T;L^2(\Omega,\zeta_R \mathrm{d}x))}^2 \leq \|u_0\|_{L^2(\Omega)}^2. \end{align*}

Up to a subsequence $(u_R)_R$ is $(\star$-)weakly converging towards some $u$ in $L^\infty(0,T;L^2(\Omega))\cap L^2(0,T;H^1_0(\Omega))$. Since $(\xi_R)_R$ converges in $L^2(0,T;L^2(\Omega))$, you can pass to the limit the equation, at least in the distribution sense.

To recover the weighted estimate, you can first get strong convergence for $(u_R)_R$ in $L^2(0,T;L^2(\Omega))$ by the Aubin-Lions lemma, because since $(u_R\xi_R)_R$ is bounded in $L^1(0,T;L^1(\Omega))$. Then, using \begin{align*} \int_0^T \int_\Omega \zeta_R |u_R|^2 \leq \int_\Omega |u_0|^2 \end{align*} you recover the weighted integrability by Fatou's lemma. The extended formulation follows by density.

$\endgroup$
2
  • $\begingroup$ What is the aim behind the truncation argument? Do you intend to get $\zeta_R$ bounded? because the one you use is not bounded. $\endgroup$
    – Guy Fsone
    Dec 19, 2020 at 15:30
  • $\begingroup$ Sorry, I corrected the typo. $\endgroup$ Dec 19, 2020 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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