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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

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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 $$ A = -1, \ \ \ A' = +1.$$

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  • $\begingroup$ Thank you mate appreciate it $\endgroup$ – pablo_mathscobar Apr 7 at 11:54

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