Green's Function for Two-Point Boundary Value Problem

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SUMMARY

The discussion focuses on determining the Green's function for the boundary value problem defined by the differential equation $$-u'' + cu = f$$ in the domain D = (0,1) with boundary conditions u(0)=u(1)=0. The Green's function is expressed as $$G(x,y)$$ based on the Wronskian of the fundamental solutions $$u_0$$ and $$u_1$$, which are derived from the homogeneous equation $$-u'' + cu = 0$$. The conversation also addresses the complexity of finding the Green's function when the coefficient $$c$$ is not constant, specifically when $$c = c(x)$$, and highlights that there is no general method for arbitrary functions of $$c(x)$$.

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Brian T
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Homework Statement


Determine the Green's function for the following problem
$$-u'' + cu = f$$ in the domain D = (0,1) with BC's u(0)=u(1)=0

Homework Equations


The Green's function is given by
$$G(x,y) =\begin{cases}
\frac{1}{W} u_0(x)u_1(y), &\text{for } 0 \leq y \leq x \leq 1, \\
\frac{1}{W} u_0(x)u_1(y), &\text{for } 0 \leq x \leq y \leq 1.
\end{cases} $$
Where W is the Wronskian of the solutions to the homogeneous equation u_0, u_1, where u_0, u_1 satisfy the boundary conditions u_0(0)=1, u_0(1)=0, u_1(0)=0, u_1(1)=1

The Attempt at a Solution


I am able to find the Green's function assuming c = const.
To find the fundamental solutions, we solve the homogenous equation -u'' + cu = 0.
The fundamental solutions are e^{\sqrt{c}x}, e^{-\sqrt{c}x}. To satisfy the boundary conditions, choose the appropriate linear combination of these fundamental solutions. After a little algebra, I get:
$$u_0(x) = \frac{1}{1 - e^{2\sqrt{c}}}e^{\sqrt{c}x} + \frac{-e^{2\sqrt{c}}}{1-e^{2\sqrt{c}}))}e^{-\sqrt{c}x}$$
$$u_1(x) = \frac{1}{e^{\sqrt{c}} - e^{-\sqrt{c}}}e^{\sqrt{c}x} + \frac{1}{e^{-\sqrt{c}} - e^{\sqrt{c}}}e^{-\sqrt{c}x}$$

I know this is all correct (plugged in and verified my solutions).
The question I have is for c not a constant, i.e. c = c(x). Since the problem does not specify whether c is constant or a function, I would assume that it could be a function... The DE becomes
$$ -u'' + c(x)u = f $$
This would require getting the fundamental solutions by solving
$$ -u'' + c(x)u = 0 $$
which I am unsure of how to do for an arbitrary function c(x). There are certain methods for certain types of functions of c(x), but does anyone have a clue on how to solve this ODE for a general c(x)? If not do you think the problem could be referring to a constant c (in the actual chapter the author usually takes the coefficient to be an arbitrary function)? By the way, the book is numerical PDE's by Larrson
 
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There is no general way of solving that for any function c(x). By the way, finding Green's function for the case c constant can by simplified by using cosh(cx) and sinh(cx) rather than e^{cx} and e^{-cx} .
 
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HallsofIvy said:
There is no general way of solving that for any function c(x). By the way, finding Green's function for the case c constant can by simplified by using cosh(cx) and sinh(cx) rather than e^{cx} and e^{-cx} .

Ohh alright, thanks. I'll try that instead :).
 

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