# In which space does my (weak) solution exist?

I am reading through some numerical analysis papers, and was wondering what characteristics of a problem define which function space the weak solution of a problem belongs to. For example, for the standard Laplacian equation of the form

$$-\nabla\cdot\nabla u = f, \quad x \in \Omega \\ u = 0, \quad x \in \partial \Omega$$

where $\Omega$ is some domain we want to solve in, and $u = u(x)$ for $x \in \Omega$. $\partial\Omega$ is the boundary of $\Omega$. The weak formulation is given by

$$\displaystyle \int_{\Omega}\nabla u \nabla v \text{ d}x = \int_{\Omega}fv \text{ d}x, \quad \forall v \in H^{1}_{0}(\Omega) \tag{1}$$

if we are using a Galerkin approach. Here the solution $u$ is in the space $H_{0}^{1}(\Omega)$, which is the usual Hilbert space of square integrable functions with square integrable first derivatives and compact support in $\Omega$. More generally for the $p$-Laplacian, given by

$$-\nabla\cdot\left( |\nabla u|^{p-2}\nabla u \right) = f, \quad x \in \Omega\\ u = 0, \quad x \in \partial\Omega$$

we have that the weak formulation is

$\displaystyle \int_{\Omega}|\nabla u|^{p-2}\nabla u \nabla v \text{ d}x = \int_{\Omega}fv \text{ d}x, \quad \forall v \in W^{1,p}_{0} \tag{2}$

where the solution is now in the Sobolev space $W^{1,p}_{0}$, which is the usual $W^{1,p}$ space with compact support in $\Omega$. The first part of my question is how do I know that the solution should be in the spaces $W^{1,p}_{0}$ from the weak formulations (1) and (2)?

I understand that in this case the problems are equivalent to minimizing the functional

$$\mathscr{J}(u) \equiv \int_{\Omega}\frac{1}{p}|\nabla u|^{p} \text{ d}x + \int_{\Omega}fu \text{ d}x$$

subject to the constraint that $u$ vanishes on $\partial\Omega$. I can see that the functions in the minimizer need to be in $W^{1,p}_{0}$ such that if the problems are equivalent the solution of (2) also needs to be in $W_{0}^{1,p}$. However, I would not think to do this myself, and I am confused as to which space a weak solution should belong to if we don't have an associated minimizing functional. For example consider the equation

$$-\nabla \cdot(u^{m}\nabla u) = f, \quad x \in \Omega \\ u = 0, \quad x \in \partial\Omega$$

and the weak formulation

$$\int_{\Omega}u^{m}\nabla u \nabla v \text{ d}x = \int_{\Omega}fv \text{ d}x, \quad \forall v \in \mathcal{U} \tag{3}$$

where the solution is now in the space $\mathcal{U}$, and I am not sure what it is. My guess would be that $\mathcal{U} \equiv W^{1,m+1}_{0}$, but this is just a guess. Would someone be able to tell me what the important characteristic of the weak formulation is such that I know in which space the solution exists for problems (2) and (3)?

• There is no general method to know such a thing. Each problem has its particularities. For $(3)$ its reasonable to start by asking that the left hand side is well defined. – Tomás Nov 9 '13 at 18:44
• @Tomás Thanks for the information. Would you be able to give me a hint how to show the well definedness of the left hand side. I am assuming that what I want to show is $|\int u \nabla u \nabla v \text{ d}x| \leq M < \infty$ for all $u, v$ (i.e. that the left hand side is bounded), and to find a suitable norm for $u$ in which this holds true? – Keeran Brabazon Nov 9 '13 at 19:21
• What is $m$? Is it a real number bigger than $1$? – Tomás Nov 9 '13 at 19:30
• Also we have to know whether $\Omega$ is bounded or not. In which space are you working? in $\mathbf R^2$? in $\mathbf R^3$ or $\mathbf R^n$ for $n\geq 3$. Since you will need the Sobolev embedding. – daulomb Nov 9 '13 at 21:23
• @Tomás Yeah, sorry guys. We can assume that $\Omega$ is bounded, polygonal subset of $R^{2}$. If necessary it can be Lipschitz as well. $m$ can be any number that it wants to be. For practical applications it is usually an integer greater than or equal to one, but for my experiments I am giving it fractional values greater than zero to see the effect. If it makes things easier I will consider here the case $m=2$ and the domain is a subset of $R^{2}$ – Keeran Brabazon Nov 10 '13 at 15:44