# Green's function using method of images

I'm working on the exercises 1.42 in the textbook "Green's functions and boundary value problems" by Stakgold and Holst, used in a graduate class "Applied Mathematics".

1. Show that the electrostatic potential for a line source of uniform unit density is $u = (1/2 \pi)\log(1/\rho),$ where $\rho$ is the cylindrical coordiate measured from the line source.
2. Consider the two-dimensional problem $$\tag{1.4.49} \frac{\partial^2 u}{\partial x_1^2 }+ \frac{\partial^2 u}{\partial x_2^2}=0, \qquad x_2 > 0, -\infty < x_1 < \infty;\qquad u(x_1,0)=h(x_1),$$ where $h(x_1)$ is a given function. First find Green's function $g(x,\xi)$ for $0$ boundary data by the method of images [using part (a)]. Then write the solution of $(1.4.49)$ by using $(1.1.5)$ as $$\tag{1.4.50} u(x_1,x_2) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{x_2}{x_2^2 + (x_1 - \xi_1)^2}h(\xi_1)d\xi_1.$$

I've solved part $(a)$ by reformulating the equation $$-\Delta u + q^2 u = f(x)$$using cylindrical polar coordinates, and simplifying it the modified Bessel function, which then leads to the claimed formula.

My problem lies with part (b), I'm not sure how to apply the method of images here to extend my result from the line source of uniform unit density to the two-dimensional problem stated above.

Consider the domain $D = \{(x_1,x_2): x_1 \in (-\infty,\infty), x_2 >0\}.$ We use the method of images to find the Green's function $g(x,\xi)$. Consider the point $(x_1,x_2)$ and it's image point $(x_1,-x_2)$. From (a), we add the function $$h = -\frac{1}{2\pi}\log \rho' = -\frac{1}{2\pi}\log \sqrt{(\xi - x_1)^2 + (\eta + x_2)^2}$$ to make $g = 0$ on the boundary. Since the image point is by definition not in the domain, $h$ is twice differentiable and satisfies Laplace's equation, $$\Delta h = \frac{\partial^2 h}{\partial \xi^2}+\frac{\partial^2 h}{\partial \eta^2}=0$$ for $(x_1,x_2)\in D$. Therefore, $g$ is defined by $$g = \frac{1}{2\pi}\log \rho + h = \frac{1}{2\pi}\log \rho - \frac{1}{2\pi}\log \rho' = \frac{1}{2\pi}\log{\frac{\rho}{\rho'}} = \frac{1}{4\pi}\log \frac{(\xi-x_1)^2 + (\eta - x_2)^2}{(\xi - x_1)^2 + (\eta+x_2)^2}.$$
On the boundary C, we have $\eta = 0$, therefore $g = 0$ and $$\nabla g = \frac{1}{\pi}\frac{x_2}{(x_1 - \xi)^2 + x_2^2}.$$ Inserting this solution of the two-dimensional problem into (1.1.5), we finally obtain $$u(x_1,x_2) = \frac1\pi \int_{-\infty}^{\infty}\frac{x_2 h(\xi)}{(x_1 - \xi)^2 + x_2^2}d\xi.$$
Note that (1.1.5) is given in the book as: $$u(x) = \int_\Omega g(x,\xi)f(\xi)d\xi - \int_\Gamma \frac{\partial g}{\partial n}h(\xi)dS_\xi.$$