# Regularity of Parabolic pde

In Evans' pde Book, In Theorem 5, p. 360 (old edition) which concern regularity of parabolic pdes. he consider the case where the coefficients $a_{ij},b_i,c$ of the uniformly parabolic operator (divergent form) $L$ coefficients are all smooth and don't depend on the time parameter $t$ \begin{cases} \mathbf{u}_t + L \mathbf{u}= \mathbf{f} &\text{ in } U\times [0,T] \\ \mathbf{u} = 0 &\text{ in } \partial U \times [0,T]\\ \mathbf{u}(0) = g &\text{ in } U \end{cases} where $\mathbf{f} \in L^{2}(0,T; L^2(U))$, $g \in H^1_0(U)$ and $U$ is a open, bounded subset of $\mathbb{R}^N$ with smooth boudary.

The second assertion (ii) of theorm is the following

if in addition $g \in H^1_0(U)\cap H^2(U)$ and $\mathbf{f} \in H^{1}(0,T; L^2(U))$ then we have

(ii) $u' \in L^\infty(0,T; L^2(U) \cap L^2(0,T; H_1^0(U))$ and $u'' \in L^\infty(0,T; H^{-1}(U)$.

For proving (ii) Evans use Galerkin approxitamtion.

My question is, is there any generalization for proving regularty of time deivative u' in general abstract Cauchy problem, or can we obtain, at least, the previous result by semigroups theory ? Thank you for any help.

Yes, there is a generalization of this result. Let us again consider the abstract Cauchy problem

$$\begin{cases} u_t + L u= f &\text{ in } U\times [0,T] \\ u(0) = g &\text{ in } U \end{cases}$$

equipped with some boundary conditions. We consider the Gelfand tripel $V \hookrightarrow H \hookrightarrow V'$. In your case we had $V=H_0^1(U)\cap H^2(U), H=L^2(U)$ with homogeneous Dirichlet boundary. This is the first generalization.

First of all, the general existence result gives us a unique solution $$u \in L^2(0,T;V) \cap H^1(0,T;V')$$ if we assumed $f\in L^2(0,T;V'), g \in H$. See e.g. Wloka's book "Partielle Differentialgleichungen".

Now, you are interested in regularity results. And yes, we can generalize your result. For the solution $u$ we get the regularity

$$u \in H^k(0,T;V) \cap H^{k+1}(0,T;V')$$

if we assumed $f \in H^k(0,T;V')$, $\frac{d^j u}{dt^j}(0) \in V$ for all $j=0,...,k-1$ and $\frac{d^k y}{dt^k}(0)\in H$.

You can also check that for $\mathbf{k=0}$ the existence result generalizes to the one above and for $\bf{k=1}$ we have

$$u \in H^1(0,T;V) \cap H^2(0,T;V')$$

if we assumed $f \in H^1(0,T;V'), g \in V, u_t(0) \in H$. This is what Evans has. Indeed, he additionally assumed $f\in H^1(0,T;H) \hookrightarrow C([0,T];H)$ which implies $u_t(0) \in H$. Further, the $L^\infty$ spaces can be achieved from several embedding results, e.g. $$u' \in L^2(0,T;V) \cap H^1(0,T;V') \hookrightarrow C([0,T];H).$$

• Thank you very much Mr. Marvin for your rigorous answer, sometimes some autors write $u' \in L^2(0,T;V) \cap H^1(0,T;H)$ instead of $u' \in L^2(0,T;V) \cap H^1(0,T;V')$ , i.e. $H$ instead of $V'$ in the right hand side. Is the assumption $f\in H^1(0,T;H)$ guarantee this addition result. Thanks agian. – S. Maths May 24 '18 at 13:44
• @S.Cho Yes, indeed the additional assumption guarantees this result. We have $u'=f-Lu$ and hence $u''=f'-Lu'$. Now $f' \in L^2(0,T;H)$ and in case of Evan's book $L: L^2(0,T;V) \to L^2(0,T;H)$. Thus $u'' \in L^2(0,T;H)$. – Fritz May 24 '18 at 13:54
• Thank you so much for all clarifications :) – S. Maths May 24 '18 at 14:05
• I'm trying to apply the general theorem in Wloka's book in the case of Evans theorem as you said, for $Lu=-\Delta u +u$, $V=H_0^1(U)\cap H^2(U), H=L^2(U)$, we have to work with the sesquilinear form $a(u,v)=\int_U (-\Delta u v +u v) dx$, it most verify the three conditions $(a), (b)$ and $(c)$. For $(c) : a(u,u)+ k_0 \|u\|_H^2 \geq \alpha \|u\|_V$ (for some $k_0,\alpha \geq 0$). I think, this estimate don't holds with the norm chosen in $V$. Is there any solution to this problem ? – S. Maths May 24 '18 at 22:53
• @S.Cho It really seems that for the standard Laplacian we have to take $V=H_0^1$ and $H=L^2$ to apply Wloka's result. I don't have the book at my side right now. You cannot take $V,H$ as proposed since $H_0^1 \cap H^2$ is not dense in $H^2$, see for a similar question. I think about your problem, but time is short right now. – Fritz May 26 '18 at 8:54