# Strong vs. weak solutions of PDEs

The strong form of a PDE requires that the unknown solution belongs in $H^2$. But the weak form requires only that the unknown solution belongs in $H^1$.

How do you reconcile this?

• The class of weak solutions is larger than the class of strong solutions (every strong solution is also a weak solution, but not every weak solution is also a strong solution). – Christian Clason Jan 5 '16 at 21:25
• But there is only one solution. – Mohamed Cheddadi Jan 6 '16 at 0:40
• There's one solution for every (appropriate) right hand side function or set of (appropriate) boundary conditions. The spaces of appropriate RHSes or BCs are bigger for weak solutions than strong ones. – Bill Barth Jan 6 '16 at 2:01

Let's look at the simplest case of Poisson's equation $$-\Delta u = f \tag{1}$$ on a domain $$\Omega\subset \mathbb{R}^n$$ together with homogeneous Dirichlet conditions $$u|_{\partial\Omega} = 0 \tag{2}$$ on the boundary $$\partial\Omega$$ of $$\Omega$$. We assume for now that $$\partial\Omega$$ is as smooth as we want (e.g., can be parametrized by a $$C^\infty$$ function) -- this will be important later.

The question now is how to interpret the (purely formal) PDE $$(1)$$. Usually, this is answered in terms of how to interpret the derivative $$\Delta$$, but for our purpose it is better to focus on how to interpret the equation.

1. The PDE $$(1)$$ is assumed to hold pointwise for every $$x\in\Omega$$. For this to make sense, the right-hand side $$f$$ must be continuous, otherwise we can't speak about pointwise values $$f(x)$$. This means that the second (classical) derivatives of the solution $$u$$ must be continuous, i.e., we have to look for $$u\in C^2(\Omega)$$.

A function $$u\in C^2(\Omega)$$ that satisfies $$(1)$$ together with the boundary condition $$(2)$$ pointwise is called a classical solution (sometimes, unfortunately, also strong solution).

2. The requirement that $$f$$ is continuous is much too restrictive for practical applications. If we only assume $$(1)$$ to hold pointwise for almost every $$x\in \Omega$$ (i.e., everywhere except for sets of Lebesgue measure zero), then we can get away with $$f\in L^2(\Omega)$$. This means that the second derivatives are functions in $$L^2$$, which makes sense if we take weak derivatives and hence look for $$u\in H^2(\Omega)\cap H^1_0(\Omega)$$. (Remember that for functions $$u$$ that are not continuous, we cannot take the boundary condition $$(2)$$ pointwise. Since $$\partial \Omega$$ has zero Lebesgue measure as a subset of $$\bar\Omega$$, pointwise almost everywhere doesn't make sense either.)

A function $$u\in H^2(\Omega)\cap H^1_0(\Omega)$$ that satisfies $$(1)$$ pointwise almost everywhere is called a strong solution. Note that it is in general necessary and non-trivial to show that such a solution exists and is unique (which is the case for the example here).

3. If we are already dealing with weak derivatives, we can also further relax the assumptions on $$f$$. If we take $$(1)$$ to hold as an abstract operator equation in $$H^{-1}(\Omega)$$, the dual space of $$H^1_0(\Omega)$$, then this makes sense for all $$f\in H^{-1}(\Omega)$$ (which is a larger space than $$L^2(\Omega)$$). Pretty much by definition of the dual space and the weak derivative, $$(1)$$ in this sense is equivalent to the variational equation $$\int_\Omega \nabla u(x)\cdot \nabla v(x)\,dx = \int_\Omega f(x)v(x)\,dx \qquad\text{for all }v\in H^1_0(\Omega)\tag{3}.$$
A function $$u\in H^1_0(\Omega)$$ that satisfies $$(3)$$ is called a weak solution. Again, it is in general necessary and non-trivial to show that such a solution exists and is unique (which is the case for the example here).

Now, since classical derivatives are also weak derivatives, every classical solution is also a strong solution. Similarly, by the embedding $$H^2(\Omega)\subset H^1(\Omega)$$, every strong solution is also a weak solution. The other directions are more subtle.

• If $$(3)$$ has a unique solution, which moreover satisfies $$u\in H^2(\Omega)$$ for $$f\in L^2(\Omega)$$ (rather than just $$H^{-1}(\Omega)$$), then the weak solution is also a strong solution (and for $$n=2$$ also a classical solution since in this case $$H^2(\Omega)$$ embeds into $$C(\bar\Omega)$$). This property is sometimes called maximal (elliptic) regularity, and holds for the Poisson equation assuming the boundary $$\partial\Omega$$ (and the boundary data) is smooth enough. (This is where the above assumption comes in.)

• Otherwise, it can happen even for $$f\in L^2(\Omega)$$ that the PDE has a weak solution but not a strong solution.

• If maximal regularity does not hold, it can also happen that the PDE has a unique strong solution (which is hence also a weak solution), but not a unique weak solution. This means that there exist many weak solution in, e.g., $$H^1_0(\Omega)$$, but only one of which is also in $$H^2(\Omega)$$ and hence a strong solution. (The actual examples require more complicated spaces; see, e.g., Meyer, Christian; Panizzi, Lucia; Schiela, Anton, Uniqueness criteria for the adjoint equation in state-constrained elliptic optimal control, Numer. Funct. Anal. Optim. 32, No. 9, 983-1007 (2011). ZBL1230.35041, or more complicated, nonlinear, equations; see, e.g., http://www.numdam.org/item/JEDP_2015____A10_0/.)

• I found this answer really useful. Can you provide a reference to your last part of your answer? I would like to see an example where a PDE has a unique strong solution but allows multiple weak solutions. Thanks! – induction601 Dec 7 '19 at 5:07