# Double layer potential in 1d?

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

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

Expressing the solution as a double layer potential, I get

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

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

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

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 $$a(1-x) + bx$$.

I can place a source with strength $$q$$ outside of the domain, at say $$y=2$$ to get the representation
$$$$u(x) = \mu(1)\left(\frac{1}{2} - H(x-1)\right) - \mu(0)\left(\frac{1}{2} - H(x)\right) + \frac{q}{2}|x - 2|$$$$
Imposing an additional constraint $$q = \mu(1) - \mu(0)$$ results in a solvable system for $$\mu(0)$$, $$\mu(1)$$ and $$q$$.