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The parabolic PDE $$\langle u', v \rangle + a(u,v) = \langle f, v \rangle \tag{*}$$ has a unique solution $u \in L^2(0,T; H^1)$ with $u' \in L^2(0,T;H^{-1})$ if $a$ is a bounded and coercive bilinear form (assuming $f$ is nice).

I want to know if the PDE $$\langle gu', v \rangle + a(u,v) = \langle f, v \rangle$$ has a unique solution for smooth functions $g$? I do not want to "divide by $g$" because that messes up my bilinear form and I can't show coercivity. Are there existence results for such equations? Alternatively, are there existence results for the equivalent PDE $$\langle u', gv \rangle + a(u,v) = \langle f, v \rangle?$$

Again please remember that I can't simply incorporate the $g$ into my bilinear form. Edit: at least I don't think so. The form $a$ is symmetric. I tried using $a(v,v) \geq \lVert v \rVert^2$ but this leads me nowhere unless I missed a trick.

Thanks.

Edit: (See Lions' Optimal control of systems governed by PDEs for the full details, page. 104)

For the case $g \equiv 1$, one uses a Galerkin method to prove existence. So define an approximate solution $u_m(t) = \sum_{i=1}^m g_{im}(t)w_i$ where the $w_i \in H^1$ are linearly independent basis, etc, and the $g_{im}$ satisfy $$(u_m'(t), w_j) + a(u_m(t), w_j) = (f(t), w_j) \tag{1}.$$ Letting $u = u_m$ and $v = u_m$ in the PDE (*), we can obtain an a-priori bound on the $u_m$ in the $L^2(0,T, H^1)$ norm which gives us a weakly convergent subsequence. Multiplying (1) by $\phi(t) \in C_c^1[0,T]$ and integrating over time, and passing to the limit gives us the result.

Comments for general $g$ I can't get the a-priori bound for a general $g$ because the term $(u', gu)$ doesn't just turn into $\frac{1}{2}\frac{d}{dt}\lVert gu \rVert^2$ but we get an extra negative term.

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  • $\begingroup$ Can $g$ approach $0$ and/or change sign, or do you have positive bounds for it? $\endgroup$
    – user53153
    Dec 27, 2012 at 4:26
  • $\begingroup$ @PavelM $g$ is bounded above and below by positive numbers. $\endgroup$
    – soup
    Dec 27, 2012 at 10:15

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You can use a weighted inner product instead of the standard $L^2$-inner product. In any case, the Galerkin method ought to work. There is also a direct method, that sometimes goes under the name of Lions' extension of the Lax-Milgram lemma.

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  • $\begingroup$ I tried the weighted inner product for the spatial spaces (so $H^1_w(\Omega)$) but I need to get the pairing $\langle u',gv \rangle_{H^{-1}(S), H^1(S)}$ to the form $\langle u', v \rangle_{H^{-1}_w(S), H^{1}_w(S)}$. Instead I only got $\langle f, v \rangle_{H^{-1}_w(S), H^{1}_w(S)}$ where $f$ is some functional (related to the Riesz map). We need $f$ to be distributional derivative of $u$ wrt. the new weighted spaces but I can't show this. Any ideas how to proceed? $\endgroup$
    – soup
    Jan 1, 2013 at 21:47
  • $\begingroup$ @soup: If I understood correctly, you don't need to weight the $H^1$ space. You only need to introduce a weight in the $L^2$ inner product. This would correspond to choosing an embedding of $L^2$ into $H^{-1}$. $\endgroup$
    – timur
    Jan 2, 2013 at 3:30
  • $\begingroup$ Thanks timur but I don't see it.. please can you expand your suggestions? (I should clarify just in case that $g$ depends on time and space). $\endgroup$
    – soup
    Jan 2, 2013 at 13:27
  • $\begingroup$ @soup: Can you please elaborate on how you would prove it if $g\equiv1$ or if $g$ was time independent? $\endgroup$
    – timur
    Jan 3, 2013 at 16:33
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    $\begingroup$ @soup: I mean Gronwall's inequality, not Galerkin. en.wikipedia.org/wiki/Gronwall's_inequality $\endgroup$
    – timur
    Jan 7, 2013 at 18:56

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