Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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.

share|cite|improve this question
Can $g$ approach $0$ and/or change sign, or do you have positive bounds for it? – user53153 Dec 27 '12 at 4:26
@PavelM $g$ is bounded above and below by positive numbers. – soup Dec 27 '12 at 10:15
up vote 2 down vote accepted

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.

share|cite|improve this answer
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? – soup Jan 1 '13 at 21:47
@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 $H^{-1}$. – timur Jan 2 '13 at 3:30
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). – soup Jan 2 '13 at 13:27
@soup: Can you please elaborate on how you would prove it if $g\equiv1$ or if $g$ was time independent? – timur Jan 3 '13 at 16:33
@soup: I mean Gronwall's inequality, not Galerkin.'s_inequality – timur Jan 7 '13 at 18:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.