# Parabolic equations without compatibility conditions

Consider $$\left\{\begin{array}{rrrr} u_t-Lu&=&f\\ u|_{\partial\Omega}&=&0\\ u(0)&=&g \end{array}\right.$$ where $$L$$ is an elliptic operator; for convenience assume $$L$$ has smooth coefficients and $$f, g$$ are all smooth. Now (for example following the PDE book of Evans) by Galerkin method one gets a weak solution $$u\in L^2(0, T; H_0^1(\Omega))$$ with $$u'\in L^2(0, T; H^{-1}(\Omega))$$. If $$f, g$$ satisfy compatibility conditions of all orders, one can prove $$u$$ is smooth on $$\overline{\Omega}\times[0, T]$$, as done in the book of Evans. Without the compatibility conditions, I can prove that $$u$$ is smooth on $$\overline{\Omega}\times (0, T]$$, namely the time $$0$$ is excluded.

MY QUESTION: is there an easy way (say along the line of the Evans book) to prove that, without compatibility conditions, $$u$$ is smooth on $$\Omega'\times [0, T]$$ with $$\Omega'\subset\subset\Omega$$? Note time $$0$$ is included.

Yes, and the idea is fairly simple: multiply $$u$$ by a cutoff function that vanishes near $$\partial\Omega$$ and apply the global regularity result, noting the compatibility conditions are satisfied for this modified equation. The rest is just keeping track of the details.

Given $$\Omega' \Subset \Omega,$$ choose an intermediate subdomain $$\Omega''$$ such that $$\Omega' \Subset \Omega'' \Subset \Omega$$ and let $$\eta,\tilde \eta \in C^{\infty}(\Omega)$$ such that $$1_{\Omega''} \leq \eta \leq 1_{\Omega}$$ and $$1_{\Omega'} \leq \tilde\eta \leq 1_{\Omega''}.$$ Then $$v(x,t) = \eta(x) u(x,t)$$ solves the modified equation $$\begin{cases} v_t - Lv = f - [L,\eta]u & \text{ in } \Omega \times (0,T), \\ v(x,t) = 0 & \text{ on } \partial\Omega \times (0,T), \\ v(x,0) = \eta(x)g(x) & \text{ on } \Omega \times \{0\}, \end{cases}$$ where $$[L,\eta]u = L(\eta u) - \eta Lu$$ is the commutator. Note that since $$\eta g$$ is compactly supported in $$\Omega,$$ this equation satisfies compatibility conditions up to all orders and so the global higher regularity theorems apply.

We know from the improved regularity result Chapter 7 Theorem 5(i) that if $$g \in H^1_0(\Omega)$$ and $$f \in L^2(0,T;L^2(\Omega)),$$ then $$u \in L^2(0,T;H^2(\Omega))$$ and $$\partial_tu \in L^2(0,T;L^2(\Omega)).$$ Then $$v = \eta u$$ lies in the same regularity class (noting $$\partial_tv = \eta \partial_tu$$) so we deduce that $$[L,\eta]u \in L^2(0,T;H^1_0(\Omega)).$$

Hence* $$\nabla v$$ satisfies the equation $$\begin{cases} (\partial_t-L)\nabla v = \nabla \tilde f + [L,\nabla]v & \text{ in } \Omega \times (0,T), \\ \nabla v(x,t) = 0 & \text{ on } \partial\Omega \times (0,T), \\ \nabla v(x,0) = \nabla(\eta(x)g(x)) & \text{ on } \Omega \times \{0\}, \end{cases}$$ and note the commutator term also lies in $$L^2(0,T;L^2(\Omega))$$ so we can apply the improved regularity result one again to deduce that $$v \in L^2(0,T;H^3(\Omega))$$ and $$\partial_tv \in L^2(0,T;H^1(\Omega)).$$ Since $$v \equiv u$$ on $$\Omega' \times (0,T),$$ the same holds for $$u$$ on $$\Omega'' \times (0,T).$$ Now if we let $$\tilde v = \tilde\eta \nabla u$$ then the associated commutator term satisfies $$\partial_t[L,\tilde\eta]\nabla u = [L,\tilde\eta]\partial_t \nabla u \in L^2(0,T;L^2(\Omega)$$ and similarly $$\partial_t[L,\nabla]u \in L^2(0,T;L^2(\Omega))$$ so assuming $$g,f$$ are sufficiently regular by applying (ii) of the same theorem to $$\tilde v$$ we get \begin{align*} u &\in L^2(0,T;H^3(\Omega')) \\ \partial_tu &\in L^2(0,T;H^2(\Omega'), \\ \partial_t^2u &\in L^2(0,T;L^2(\Omega'). \end{align*} Now the idea is to iterate this and argue by induction to show that $$\partial_t^ku \in L^2(0,T;H^{2m+2-2k}(\Omega'))$$ for all $$m \geq 0$$ and each $$\Omega'\Subset \Omega.$$ The argument is essentially the same as the proof of Theorem 6, notably the compatibility conditions are vacuously satisfied thanks to the cutoff argument above.

*Added later: To see this note we have the weak formulation $$\langle \partial_tv(t),\phi \rangle + B[v,\phi;t] = \langle f(t),\phi \rangle$$ for almost every $$t \in (0,T),$$ with $$\phi \in C^{\infty}_c(\Omega).$$ Then taking $$\phi = \nabla\varphi$$ we can integrate by parts to get $$\langle \partial_tv(t),\nabla\varphi\rangle + B[\nabla v,\phi;t] = \langle \nabla \tilde f(t),\phi \rangle + \left( B[\nabla v, \phi ;t] - B[v,\nabla\phi;t]\right).$$ Now we can extend by density to take $$\phi \in H^1_0(\Omega);$$ the only term that needs justification is the $$\partial_t\nabla v(t)$$ term; for this we know that $$\partial_tv \in L^2(0,T;L^2(\Omega))$$ and $$\partial_t\nabla v = \nabla \partial_t v$$ by definition of weak deriavtives and Fubini, noting that $$\int_{\Omega} \left(\int_0^T\eta'(t) \nabla v(t,x)\,\mathrm{d}t\right) \varphi(x) \,\mathrm{d} x = \int_0^T \eta(t) \left(\int_{\Omega} \partial_tv(t,x) \nabla \varphi(x) \,\mathrm{d}x\right) \,\mathrm{d}t.$$ This verifies the weak formulation, as required.

• Thanks! However I was stuck at "Hence $\nabla v$ satisfies the equation...": The issue is, I don't know the existence of the weak derivative $(\nabla v)_t$, because we only know $u_t\in L^2(0, T; L^2(\Omega))$. At least for the definition of weak solution in Evans book, one would require something like $(\nabla v)_t\in L^2(0, T; H^{-1}(\Omega))$ which I don't know how to get. Can you provide some more details on how to check $\nabla v$ is a solution to that equation? Mar 16, 2021 at 4:12
• @Yuval I've edited my answer to add more justification; one can show that $(\nabla v)_t \in L^2(0,T;H^{-1}(\Omega))$ from which the weak formulation follows.
– ktoi
Mar 16, 2021 at 19:33
• Thanks, that helps a lot! Mar 17, 2021 at 5:34