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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Yuval
    Mar 16, 2021 at 4:12
  • $\begingroup$ @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. $\endgroup$
    – ktoi
    Mar 16, 2021 at 19:33
  • $\begingroup$ Thanks, that helps a lot! $\endgroup$
    – Yuval
    Mar 17, 2021 at 5:34

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