I'm currently reading about solutions to boundary-value problems for Laplace's equation, and I'm a bit confused with regards to the discontinuity properties of double-layer potentials. So the text says that for a bounded domain $\Omega \subset \mathbb{R}^n$ we can look for a solution to the Dirichlet problem \begin{align*} \Delta u &= 0 ~~~~~x \in \Omega\\ u &=f ~~~~~x \in \partial \Omega \end{align*} in the form of a double-layer potential $$u(x)=\int_{\partial \Omega} h(y) \frac{\partial \Phi}{\partial \nu_y}(x-y) \, dS_y$$ where $h \in \mathcal{C}(\partial \Omega)$.

The double-layer potential is discontinuous so that as $x$ approaches the boundary $\partial \Omega$ \begin{align*} u_-&=\frac{h}{2} + u \\ u_+&=-\frac{h}{2} + u \\ u_- - u_+ &=h \end{align*} where $u_-$ is the limit as we go to the boundary from inside, and $u_+$ is the limit as we go to the boundary from the outside.

I'm really confused. Shouldn't a harmonic function be continuous on $\overline{\Omega}\,$? If we use this method to solve electrostatics problems then the electrostatic potential will be discontinuous across the surface. However, this should not be the case.


For each fixed $y\in \partial \Omega$ the function $v_{y,\nu}(x): = \frac{\partial\Phi}{\partial\nu}\Phi(x-y)$ is harmonic in $\mathbb R^n\setminus \{y\}$. This is the potential function of a unit dipole placed at $y$ and oriented along $\nu$. As such, it has a fairly complicated singularity at $y$, approaching $+\infty$, $-\infty$, and everything in between as $x\to y$. You may want to write down an explicit formula for $v_{y,\nu}(x)$ to check this.

Next, we form the integral $u(x)=\int_{\partial\Omega} h(y)\,v_y(x)\,dy$ which puts together the contributions of dipoles distributed over the surface $\partial \Omega$. The function $u$ is harmonic in $\mathbb R^n\setminus \partial\Omega$, which is an open set with at least two connected components, one of which is $\Omega$. Integration against the continuous function $h$ tones down the singularities at each $y\in\partial \Omega$, but they do not go away entirely: their residual effect is discontinuity at every $y$ such that $h(y)\ne 0$.

There is no contradiction here, either mathematical or physical. Harmonic functions need not be continuous on the closure of their domain: this is exactly what allows us to solve boundary value problems with discontinuous boundary data. As for physical interpretation, note that we consider a surface of dipoles, not a uniformly charged surface (the latter would be described by a single layer potential).

  • $\begingroup$ Ah ok I think I get it. Since we usually want to solve Laplace's equation so that $u \in \mathcal{C}(\overline{\Omega}) \cap \mathcal{C}^2(\Omega)$, I assume we just define $u=\int_{\partial \Omega} h(y) \frac{\partial \Phi}{\partial \nu_y}(x-y) \, dS_y$ if $x \in \Omega$ and $u=f$ for $x \in \partial \Omega$, right? $\endgroup$ – B0112358 Dec 18 '12 at 7:52
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    $\begingroup$ @BartekEwertowski Yes. Another way to put it: $u$ was already defined for us on $\partial \Omega$, we only need to fill in the interior with a harmonic function so that the result is continuous. Of course, the main question is how to find $h$ that works for our $f$, but I suppose your book covers that. $\endgroup$ – user53153 Dec 18 '12 at 15:40
  • $\begingroup$ Ok thanks! Yes, the textbook covers the standard integral equation methods, but it's a bit vague about the properties of the solution. For the Neumann problem I assume we just use the same trick and define a $C^1$ extension of the single layer potential? $\endgroup$ – B0112358 Dec 19 '12 at 19:53
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    $\begingroup$ @BartekEwertowski I would not say that we "define a $C^1$ extension"; we define a continuous extension and then prove that it's $C^1$. Since $\Phi(x-y)$ has a relatively mild singularity as $x\to y$, the integral converges even for $x\in \partial \Omega$. As a consequence, the single layer potential is obtained as a continuous function on $\mathbb R^n$. The issue is in proving that it has one-sided normal derivatives at the boundary (inner and outer), the inner derivative is what we want to satisfy the Neumann condition. $\endgroup$ – user53153 Dec 19 '12 at 20:16

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