# Why find weak solutions of PDE, and how to go from weak to strong?

In some math books, instead of finding a strongly differentiable solution to some PDEs, they find weak solutions (weakly differentiable). How can we say that the weak solution satisfies that PDE? This is an example: Consider the eigenvalue problem $$$$\tag{1} \begin{split} \Delta u+m(x)u=\lambda u; \Omega\\ \frac{\partial u}{\partial\eta}+\beta(x)u=0;\partial \Omega \end{split}$$$$

where $$\lambda$$ is a parameter, $$m(x)\in L^\infty(\Omega)$$, $$\beta(x)\in L^\infty(\partial\Omega)$$, $$\frac{\partial u}{\partial\eta}$$ is the outward normal derivative, and $$\Omega$$ is a bounded domain with $$\partial\Omega$$ smooth boundary.

This is how the principal eigenvalue $$\lambda_1$$ of the above problem is defined (and consequently the eigenfunction $$\psi$$),

$$$$\tag{2} \lambda_1=\max_{\psi\in W^{1,2}(\Omega), \psi\neq 0}\big[\frac{-\int_\Omega|\nabla\psi|^2dx+\int_\Omega m(x)\psi^2dx-\int_{\partial\Omega}\beta(x)\psi^2ds}{\int_\Omega\psi^2dx}\big]$$$$

My concern is that, even if we have a function $$\psi$$ that satisfies (2), it may not twice differentiable (strongly) so how can we guarantee that it is a solution to (1)

• Perhaps, in your math books, you should continue reading after the part they prove existence of weak solutions. My hunch is that they prove that under some certain conditions, the solutions are indeed strongly twice differentiable Aug 1, 2022 at 1:28
• For certain sorts of PDEs (elliptic in particular), weak solutions are automatically genuine solutions. For other sorts of PDEs, strong solutions may fail to exist while weak solutions accurately model physical phenomena (e.g., the development of shocks). Aug 1, 2022 at 1:42
• Weak solutions often exist even when strong ones don't, representing a sort of degeneracy. I struggled a lot with this, so allow me to plug my own related question about weak solutions: math.stackexchange.com/questions/3314557/… Aug 1, 2022 at 2:47