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In Evans' PDE Book, in the first Theorem of regularity of parabolic PDE's there 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 $t$. Let $\{w_n\}$ be a basis of orthonormal eigenvectors of the operator $\Delta$. Finally, let $\{u_m\}$ be the sequence of Galerkin approximations for the equation $$ \begin{cases} \mathbf{u}_t + L \mathbf{u}= \mathbf{f} &\text{ in } U\times [0,T] \\ \mathbf{u} = 0 &\text{ in } \partial U \times [0,T]\\ \mathbf{u}(0) = g &\text{ in } U \end{cases} $$ where $\mathbf{f} \in H^{1}(0,T; L^2(U)) $, $g \in H^1_0(U)\cap H^2(U) $ and $U$ is a open, bounded subset of $\mathbb{R}^N $ with smooth boudary.

In p. 363 (of the first edition), Evans points out $$ (I) \qquad \|\mathbf{u}^\prime_{m}(0) \| \le C (\|\mathbf{f} \|_{H^{1}(0,T; L^2(U))} + \|\mathbf{u}(0) \|_{H^2(U)}) $$ (he actually claims something the term in the LHR plus something else is smaller than the RHS, so the term $\|\mathbf{f} \|_{H^{1}(0,T; L^2(U))}$ might not be required) which he justifies referring to the equation $$ (II)\qquad (\mathbf{u}^\prime_{m},w_k) + B[\mathbf{u}_m,w_k ]= (\mathbf{f},w_k). $$

Well, my attempt is to multiply $(II)$ by the right coefficients in both sides and evaluate at $0$ (which at this point I don't know how to justify that $(II)$ holds, as the equation only holds almost surely) to get $$ \|\mathbf{u}^\prime_{m}(0)\|^2_{L^2(U)} + B[\mathbf{u}_m,\mathbf{u}^\prime_m ]= (\mathbf{f},\mathbf{u}^\prime_m). $$ So I imagine the first step is to use the bounds of the coefficients of $L$ to get $$ (III) \qquad |B[\mathbf{u}_m(0),\mathbf{u}^\prime_m (0)]|\le C(\|\mathbf{u}_m(0)\|^2_{H^1_0(U)}+ \|\mathbf{u}^\prime_m(0) \|^2_{H^1_0(U)}) $$ and $$ (IV) \qquad |(\mathbf{f}(0),\mathbf{u}^\prime_m(0))|\le C(\|\mathbf{f}(0)\|^2_{L^2(U)}+ \|\mathbf{u}^\prime_m(0) \|^2_{L^2(U)}) $$

However, the fact that $(III)$ the RHS depends on $ \|\mathbf{u}^\prime_m(0) \|^2_{H^1_0(U)}$ and not on $ \|\mathbf{u}^\prime_m(0) \|^2_{L^2(U)}$ is rather annoying, as if it was the case, I could just use $\varepsilon$-Cauchy to get the desired result. Unfortunately, the Poincaré Inequality would point to the 'wrong' side of what I want. Moreover, as I said, I don't know how to justify that $(II)$ holds at $t=0$. How could I solve those problems?

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    $\begingroup$ Instead of (III) and (IV), use the bound $B[u,v]\leq C\|u\|_{H^1_0} \|v\|_{H^1_0}$ and use Cauchy-Schwarz in (IV). Then you can divide both sides by $\|u_m'(0)\|$ to get the estimate. $\endgroup$ – Jeff Feb 17 '18 at 2:33
  • $\begingroup$ Could you develop a bit further? The problem for me is that the inequality you suggested $B[u(0),u^\prime (0)]\le C \|u(0)\|_{H^1_0} \|u_m^\prime(0)\|_{H^1_0} $. On the other hand Cauchy-Schwarz gets $|(\mathbf{f}(0),u_m^\prime(0))|\le C(\|f(0)\|^2_{L^2} \|u_m^\prime(0) \|^2_{L^2}). $ Then you would have $\|u^\prime_m(0)\|^2_{L^2} \le \|f(0)\|_{L^2} \|u^\prime_m(0)\|_{L^2} + C \|u_m(0)\|_{H^1_0} \|u_m^\prime(0)\|_{H^1_0}$. So if divide by either $ \|u_m^\prime(0)\|_{H^1_0} $ or $ \|u_m^\prime(0)\|_{L^2}$ you don't really get the expected result. $\endgroup$ – Kernel Feb 20 '18 at 12:34
  • $\begingroup$ No, Cauchy Schwarz gives $|(u,v)|\leq \|u\| \|v\|$. $\endgroup$ – Jeff Feb 20 '18 at 14:03
  • $\begingroup$ Sorry, that I was me being lazy and copying from the question without correcting it. In any case, in the last inequality of my previous comment, I had applied CS properly, so how should I proceed? $\endgroup$ – Kernel Feb 20 '18 at 16:21
  • $\begingroup$ It should work if you use $B[u_m,u_m']=(Lu_m,u_m')\leq C\|u_m\|_{H^2}\|u_m'\|_{L^2}$. $\endgroup$ – Jeff Feb 20 '18 at 23:02

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