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I am interested in any heuristic or formal necessary and sufficient conditions that a good definition of a weak solution of a PDE problem must satisfy.

This question is motivated by trying to show the existence of the wave equation $ \Box_{g} \phi = f $ in a compact subset, $ \Omega $, of a $ 4 $-dimensional manifold.

I can prove that I have an estimate of the form $$ (\| \omega \|_{K_{1}})^{2} \lesssim (\| \square_{g} \omega \|_{K_{1}})^{2}, $$ where $$ \| \phi \|_{K_{1}} = \left[ \int^{t}_{0} \left( \| \phi \|^{1}_{\Sigma_{\tau}} \right)^{2} ~ \mathrm{d}{\tau} \right]^{\frac{1}{2}} \quad \text{and} \quad \| \phi \|^{1}_{\Sigma_{\tau}} = \left[ \int_{\Sigma_{\tau}} \left( \frac{\partial \phi}{\partial \tau} \right)^{2} + \sum^{3}_{i = 1} \left( \frac{\partial \phi}{\partial x^{i}} \right)^{2} + \phi^{2} \mu_{h} \right]^{\frac{1}{2}}. $$ (I have denoted the volume form on each $ 3 $-surface by $ \mu_{h} $.)

Now, if I define a linear functional $ \displaystyle k(\Box \omega) = \int_{\Omega} f \omega $ with domain $V=\{\Box \omega \text{ s.t. } \omega\in C_{o}^{\infty} \}$.

Then I can prove that this is a bounded linear functional in $ K_{1} = {L^{2}}([0,t],{H^{1}}({\Sigma}_{\tau},\mu_{h})) $ consisting of the functions of $ \tau $ with values in the space $ {H^{1}}(\Sigma_{\tau},\mu_{h}) $.

Extending this linear functional using the Hahn-Banach Theorem to the closure of $V$ with respect the $K_{1}$ norm and the Riesz Representation Theorem, I can guarantee there is a $ \Psi \in K_{1} $ that satisfies: $$ (f,\omega)_{L^{2}(\Omega)} = \int_{\Omega} f \omega = \langle \Psi,\Box \omega \rangle_{K_{1}}. $$

for all $\Box\omega\in \overline{V}$.

Is this $ \Psi $ a good weak solution?

I can also prove that there is a $\xi$

$$ \langle f,\omega\rangle_{K_{1}} = \langle \xi,\Box \omega \rangle_{K_{1}}. $$

for all $\Box\omega\in \overline{V}$.

Is this $ \xi $ a good weak solution?

EDIT I will accept the answer from BaronVT as he answered the question before I made this edit. However, his answer make me think what if we have an estimate of the form

I can prove that I have an estimate of the form $$ (\| \omega \|_{K_{1}})^{2} \lesssim (\| \square_{g}^{*} \omega \|_{K_{1}})^{2}, $$

would then the weak solution

$$ \langle f,\omega\rangle_{K_{1}} = \langle \xi,\Box^{*} \omega \rangle_{K_{1}}. $$

for all $\Box\omega\in \overline{V}$.

Notice that by definition of the adjoint in $K_{1}$ any strong solution is satisfying: $$ \langle f,\omega\rangle_{K_{1}} = \langle \xi,\Box^{*} \omega \rangle_{K_{1}} =\langle \Box\xi,\omega \rangle_{K_{1}}. $$

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  • $\begingroup$ I'm not sure how good of an answer can be given to your original question. My professor this semester was fairly clear that about the best you can say is the following. A weak formulation of a strong equation is the pair of a function space and a weak equation, such that any strong solution is in the function space and satisfies the weak equation, and such that any weak solution which has all of the regularity in the strong equation is a strong solution. $\endgroup$ – Ian Apr 1 '15 at 18:52
  • $\begingroup$ If I recall correctly, the point about weak solutions is that they are easier to find, while one might end up with an awful lot of them and then has to choose the hopefully unique "good" one by other considerations then the weak formulation only. So, even though I do not remember the precise notion of weak solution, if I were in your place, I would try to find out, whether your $\Psi$ and $\xi$ agree in an appropriate sense, because if you got that issue straight, your understanding of the situation will have improved. $\endgroup$ – thomashennecke Apr 3 '15 at 18:00
  • $\begingroup$ @thomashennecke In many important cases weak solutions are unique. The main exceptions I can think of are in nonlinear problems (e.g. the eikonal equation). $\endgroup$ – Ian Apr 4 '15 at 16:36
  • $\begingroup$ @Ian Indeed, my main exposition to this was while trying to understand Hamilton-Jacobi type problems. $\endgroup$ – thomashennecke Apr 7 '15 at 17:06
  • $\begingroup$ @yess, I have addressed your edit $\endgroup$ – BaronVT Apr 7 '15 at 21:25
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Generally speaking, to define weak solutions, you define a bilinear form corresponding to your PDE that is equivalent to the strong formulation for classical solutions, but also admits weaker solutions (i.e. $H^1$ rather than $C^2$), and is such that any sufficiently regular weak solution is also a strong solution.

For instance for the PDE $-\Delta u = f; x \in \Omega$, the right bilinear form is $a(u,v) = \int_\Omega \nabla u \cdot \nabla v \,dx$. You can check that, for $u,v \in C_0^\infty$, you have $$ a(u,v) = (-\Delta u, v)_{L^2} $$ and so if $u$ is a strong solution $a(u,v) = (f,v)_{L^2}$, and this is how you define weak solutions: $u\in H^1_0$ that satisfy $a(u,v) = (f,v)_{L^2}$ for every $v \in H^1_0$. You can prove that $a(u,v)$ is symmetric and positive definite on $H^1_0$, thus you have a Hilbert space (with $a(u,v)$ as the inner product) and so you get existence of solutions (relative to that inner product) by fiat (i.e. Hahn-Banach/ Riesz representation theorem).

Now, the way I have usually seen existence of weak solutions for the wave equation $\square = \partial^2_{tt} - \Delta$ is to define a bilinear form

$$ b(u(t),v) = (u''(t),v)_{L^2} + a(u(t),v) $$

and look for $u$ so that $b(u(t),v) = (f(t),v)_{L^2}$ for all $t$ and all $v \in H^1$.

The way I have usually seen existence proved is by Galerkin approximation (see also Evans' PDE book), rather than what you're doing (Riesz representation / Lax-Milgram); I'm under the impression that you can't just pull the weak solution from the air like you want, and while I can't give you an entirely satisfactory explanation at the moment, I think essentially the problem is that the bilinear form isn't elliptic/bounded below, i.e. $(H^1_0,b(u,v))$ isn't a Hilbert space.

A few further notes about the weak solutions you've defined so far:

You can calculate

$$ \langle u, v \rangle_{K_1} = (\square u, v)_{L^2} - 2(u'',v)_{L^2} + (u,v)_{L^2} $$

for $u,v \in C^\infty_0$, so a strong solution would satisfy

$$ \langle u, v \rangle_{K_1} + 2(u'',v)_{L^2} - (u,v)_{L^2} = (f, v)_{L^2} $$

(the left hand side is how we would define the bilinear form; you can also see it isn't positive definite!) and so in particular, neither

$$ \langle \xi, \square w \rangle_{K_1} = \langle f, \square w \rangle_{K_1} $$

nor

$$ \langle \Psi, \square w \rangle_{K_1} = ( f, \square w )_{L^2} $$

are good weak formulations, because strong solutions won't satisfy the weak equation.

Regarding your edit, you would need the bilinear form $c(u,v) := \langle u, \square^* v \rangle_{K_1}$ to be an inner product (i.e. symmetric and positive definite). I think symmetry follows ($\square$ is self adjoint), but it is not positive definite (there are nontrivial $u$ solving the homogeneous wave equation).

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  • $\begingroup$ Hi, $c(u,v)$ is the inner product in $L^{2}([0,T],H^{1})$. Or am I missing something? $\endgroup$ – yess Apr 7 '15 at 23:01
  • $\begingroup$ The inner product in that space is just $\langle u,v \rangle_{K_1}$ (i.e. without the $\square$) $\endgroup$ – BaronVT Apr 7 '15 at 23:26
  • $\begingroup$ Basically, the method you are proposing would be good for finding solutions of $\partial^2_{tt} + \Delta$, rather than $\square$. $\endgroup$ – BaronVT Apr 7 '15 at 23:28
  • $\begingroup$ In particular, note that there are nontrivial $u \in L^2([0,T],H^1_0)$ so that $\square u =0$, thus $c(u,u) = 0$. $\endgroup$ – BaronVT Apr 7 '15 at 23:31

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