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How does one directly construct the Green's function to the following problem:

$y''-k^2y(x)=f(x)$ subject to $y(0)=y(L)=0$ ?

I beleive the correct method is $G(x,\xi) = \frac{1}{c}$$ \left\{\begin{aligned} &u_1(x)u_2(\xi) &&: x \lt \xi\\ &u_1(\xi)u_2(x) &&: \xi \lt x \end{aligned} \right.$

I've found that the answer is $G(x,\xi)=\frac{-sinh(kx)sinh(k(L-\xi))}{ksinh(kL)}$

but do not see how it's arrived at. Furthermore, I am asked to show that the solution to $G''-k^2G=\delta(x-\xi)$ using a sine series is equivalent to the above Green's function. I not sure how to handle that.

CINA
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1 Answers1

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So you consider $G''-k^2G=\delta(x-\xi)$, noting that $\delta(x-\xi)=0$ for all $x\ne\xi$ so we have the following problems: $$u_1''-k^2u_1=0\text{ for }x\lt\xi\text{ with BC }u_1(0)=0,$$ and $$u_2''-k^2u_2=0\text{ for }x\gt\xi\text{ with BC }u_2(L)=0,$$

so $u_i=A_ie^{kx}+B_ie^{-kx},\,i=1,2$ applying the BC's: $u_1 = A(e^{kx}-e^{-kx})$, $u_2=B(e^{kx}-e^{2kL+kx})$, for for some constants $A,B$.

Now we want to determine the constants $A,B$, remember that $G''-K^2G=\delta(x-\xi)$, integrating from $0$ to $L$, and noting the properties of the delta function under integration we obtain:

Jump condition: $G'(\xi^+)-G'(\xi^-)=1$, i.e. $u_2'(\xi)-u_1'(\xi) = 1$

and continuity condition: $G(\xi^+)-G(\xi^-)=0$, i.e. $u_2(\xi)-u_1(\xi) = 0$

This gives us $u_1$ and $u_2$ (I will leave it to you to calculate the constants), and note that we can write $\sinh(x)=\frac{1}{2}(e^x-e^{-x})$.

Ellya
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