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?

  • 8
    $\begingroup$ 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). $\endgroup$ Jan 5 '16 at 21:25
  • 1
    $\begingroup$ But there is only one solution. $\endgroup$ Jan 6 '16 at 0:40
  • 6
    $\begingroup$ 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. $\endgroup$
    – 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/.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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