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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?

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    $\begingroup$ 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). $\endgroup$ – Christian Clason Jan 5 '16 at 21:25
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    $\begingroup$ But there is only one solution. $\endgroup$ – Mohamed Cheddadi Jan 6 '16 at 0:40
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    $\begingroup$ 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
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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.)

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