Solution via Green's Function

The most delightful aspect about this problem is that its solution can readily be expressed in terms of the Green's function for the given linear system.

The reasoning leading to this solution is an extension into two dimensions of the 1-dimensional problem considered in Sections 4.2 (p. ) and 4.7 (p. ). As in that case, the solution is easily given in terms of the associated Green's function $G(\vec x;\vec x_0)$ . It satisfies

$$(\nabla^2+k^2)G(\vec x;\vec x_0)=-\delta^2(\vec x-\vec x_0)\equiv... ...heta-\theta_0)& \parbox{.9in}{\textrm{in polar coordinates}} \end{array}\right.$$ (561)

and

$$\left[ a(\vec x)G(\vec x;\vec x_0)+\vec n\cdot \vec\nabla G (\vec x;\vec x_0) \right]_{\partial R}=0~.$$ (562)

Figure 5.19: Integration region $R$ with boundary $\partial R$ having outward pointing normal $\vec n$ perpendicular to each boundary element $ds$ .

The solution process to this 2-dimensional problem parallels the one for one dimension. First of all, use Lagrange's identity

$$\psi\nabla^2 G - G \nabla^2\psi= \nabla\cdot(\psi\vec\nabla G - G \vec\nabla\psi)$$

Its integral over the region $R$ yields the Green's identity

$$\int_R\int (\psi\nabla^2 G - G \nabla^2\psi)d^2x= \oint_{\partial R} (\psi\vec\nabla G - G \vec\nabla\psi)\cdot \vec n~ ds$$

Secondly, applying the inhomogeneous Helmholtz equation, Eq.(5.59), and Eq.(5.61) to the left hand side, one obtains

$$\int_R\int\left[ \psi(\vec x)(-)\delta^2(\vec x-\vec x_0)+ G(\vec... ...2x= \oint_{\partial R} (\psi\vec\nabla G - G \vec\nabla\psi)\cdot \vec n~ ds~.$$

Finally, substituting the two boundary conditions, Eqs.(5.60) and (5.62) into the right hand side, one finds that

$$\psi(\vec x_0)= \int_R\int f(\vec x)G(\vec x;\vec x_0)d^2x -\oint_{\partial R}\left. g(\vec x)G(\vec x;\vec x_0)\right\vert _{\partial R}ds~.$$

Thus, knowledge of the Green's function $G$ , automatically yields the response amplitude $\psi$ in terms of its values on the boundary $\partial R$ and in terms of the source distribution $f$ in the region $R$ .