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?


2 Answers 2


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.


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.

  • $\begingroup$ 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
    Commented Dec 19, 2020 at 15:30
  • $\begingroup$ Sorry, I corrected the typo. $\endgroup$ Commented Dec 19, 2020 at 16:28

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