I would like to illustrate the double layer potential idea with a simple 1d example, but seem to run into a situation where the resulting integral equation is singular.

The problem is $u''(x) = 0$ on $[0,1]$, subject to $u(0) = a$, $u(1) = b$. A free-space Green's function for this problem is given by $G_0(x,y) = \frac{1}{2}|x-y|$. This satisfies four desirable properties of the free-space Green's function :

  • $G_0(x,y)$ is continuous on $[0,1]\setminus y$.
  • $\partial^2 G_0(x,y)/\partial x^2 = 0$ on $[0,1]\setminus y$
  • $\left[\partial G_0(x,y)/\partial x\right]_{x=y} = 1$
  • $G_0(x,y) = G_0(y,x)$.

As associated dipole can be expressed as

\begin{equation} \lim_{\varepsilon \to 0}\frac{1}{2}\frac{|x-\frac{\varepsilon}{2}| - |x+\frac{\varepsilon}{2}|}{\varepsilon} = \frac{1}{2} - H(x) \equiv -\frac{\partial G_0(x,0)}{\partial y} \end{equation}

Expressing the solution as a double layer potential, I get

\begin{equation} u(x) = \mu(y)\left(-\frac{\partial G_0(x,y)}{\partial y}\right) \bigg\rvert_{y=0}^{y=1} = \mu(1)\left(\frac{1}{2} - H(x-1)\right) - \mu(0)\left(\frac{1}{2} - H(x)\right) \end{equation}

where $H(x)$ is the Heaviside function. To get an integral equation, I evaluate the above at the endpoints $x = 0^+$ and $x = 1^+$, where "+" indicates taking a limit as $x$ approaches boundary point from within the interval $[0,1]$. The resulting integral equation is given by

\begin{eqnarray} u(0^+) & = & \frac{\mu(0)}{2} + \frac{\mu(1)}{2} = a \\ u(1^+) & = & \frac{\mu(1)}{2} + \frac{\mu(0)}{2} = b \end{eqnarray}

which is clearly singular, and can only be solved if $a = b$. In fact, no linear combination of the dipoles will give the correct solution $a(1-x) + bx$ for general $a$ and $b$.

My question is, where did I go wrong? Or, if the above is correct, is there an explanation for why the 1d double layer potential doesn't exist for $a \ne b$?

I have considered the following ideas :

  • This is really a 2d problem in an infinite strip, and as such, maybe the "boundary" isn't really closed, and so therefore, the solution cannot be expressed as a double layer potential. This sounds dubious, however, since harmonic functions certainly exist in infinite and semi-infinite domains.
  • Design a different dipole expression by solving $w''(x) = -\delta'(x)$ and choosing constants of integration to satisfy jump conditions in the potential at $x=0$ and $x=1$. For example, $w(x) = -H(x) + \frac{1}{2}(x) + \frac{1}{2}$ works. This leads to the potential

\begin{equation} u(x) = -\frac{\mu(0)}{2} + \mu(0) H(x) + \frac{\mu(1) - \mu(0)}{2} x - \mu(1)H(x-1) \end{equation}

with $\mu(0) = 2a$ and $\mu(1) = 2b$. This satisfies necessary double-layer jump conditions, but the dipole representation is not obviously the derivative of a free-space Green's function.

  • The solvability issue goes away if $H(0)$ is defined to be $1/2$. In this case, the dipole densities become $\mu(0) = 2b$, and $\mu(1) = 2a$. But the solution is still not the harmonic function $a(1-x) + bx$.

1 Answer 1


I have since learned that it is not uncommon for integral equations to lead to singular systems. For example, the dipole representation of a Dirichlet problem in a multiply connected domain only supports constant solutions. (See this article).

(This is an updated solution which is much better than my previous answer).

I can place a source with strength $q$ outside of the domain, at say $y=2$ to get the representation

\begin{equation} u(x) = \mu(1)\left(\frac{1}{2} - H(x-1)\right) - \mu(0)\left(\frac{1}{2} - H(x)\right) + \frac{q}{2}|x - 2| \end{equation}

Imposing an additional constraint $q = \mu(1) - \mu(0)$ results in a solvable system for $\mu(0)$, $\mu(1)$ and $q$.

What is still not entirely clear to me, however, is why this trick is required for the 1d interior domain.


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