# Green's function for 1D modified Helmoltz' equation

My equation is $$-k^2 \phi + \frac{\partial^2{\phi}}{\partial{z^2}} = -2 \delta(z)$$ with $$\phi =0$$ on $$z=\pm a$$

How do I show the Green's function is

$$\phi = \frac{\sinh(k(z+a))}{k\cosh(ka)}$$ if $$z<0$$

and $$\;\phi = \frac{\sinh(k(a-z))}{k\cosh(ka)}$$ if $$z>0$$

$$-k^2 \phi + \frac{\partial^2{\phi}}{\partial{z^2}} = 0$$

Solving this I get $$\phi = A\sinh(kz) + B\cosh(kz)$$

applying the BCs i get:

for $$z<0$$, $$0= A\sinh(-ka) + B\cosh(-ka)$$

and $$z>0$$, $$0= A\sinh(ka) + B\cosh(ka)$$

but am unsure how to proceed

For $$z > 0$$, you have $$B = - \sinh(ka) / \cosh (ka)$$, so

$$\phi = \frac{A}{\cosh (ka)} \left( \cosh (ka) \sinh (kz) - \sinh(ka) \cosh(kz) \right) = \frac{A}{\cosh(ka)} \sinh(k(z-a)).$$

[By the way, if you had written the general solution in the form $$\phi = C \sinh(k(z - a)) + D \cosh (k(z - a))$$, then it would have been obvious that $$D = 0$$ from the boundary condition at $$z = a$$, which would have led you immediately to this expression for $$\phi$$. Of course, $$C = A / \cosh(ka)$$.]

Similarly, for $$z < 0$$, we have $$\phi = \frac{A'}{\cosh (ka)}\sinh(k(z + a)).$$

All that remains is to find $$A$$ and $$A'$$. You do this by matching the boundary conditions at $$z = 0$$.

The original equation is $$-k^2 \phi + \frac{d^2\phi}{dz^2} = -2\delta(z).$$

If we integrate both sides over an infinitesimally thin interval around $$z = 0$$, we have

$$\lim_{z \to 0^+} \frac{d\phi }{dz} - \lim_{z \to 0^-} \frac{d\phi }{dz} = -2$$

So you need to choose $$A$$ and $$A'$$ such that $$d\phi / dz$$ has a discontinuity of $$-2$$ at $$z = 0$$ (but such that $$\phi$$ itself is continuous). In other words, you need

$$\frac{A}{\cosh (ka)} \times k\cosh(k(0-a)) - \frac{A'}{\cosh(ka)} \times k\cosh(k(0+a)) = -2$$ $$\frac{A}{\cosh (ka)} \times \sinh(k(0-a)) + \frac{A'}{\cosh(ka)} \times \sinh(k(0+a)) = 0$$ and this is solved by $$A = -1, \ \ \ A' = +1.$$

• Thank you mate appreciate it – pablo_mathscobar Apr 7 at 11:54