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Problem

Consider the BVP y''+y+f=0, $ \quad$ $y(0)=0=y(\frac{\pi}{4})$ where $f$ is a continuous function defined on $[0,\frac{\pi}{4}].$

Show that the BVP has at most one solution, and construct Green's function $G$ such that the solution is given by

$y(x)=\int_{0}^{\frac{\pi}{4}} G(x,s)f(s)ds.$

Deduce that

y'(0)=\int^{\frac{\pi}{4}}_{0}f(s)(\cos(s)-\sin(s))ds

Progress

Thinking about uniqueness for the BVP: if the solution of the BVP is not unique then the difference between any two gives a solution of the homogeneous case satisfying both boundary conditions, which means vanishing at both ends. Not really sure if this is along the right lines, or how this can be formalised though.

To construct the Green's function, we find $y_1,y_2$ that satisfy the homogeneous case and so that $y_1(0)=0$ and $y_2(\frac{\pi}{4})=0$. The functions $y_1(x):=\sin(x)$ and $y_2(x):=\cos(2x)$ satisfy this condition.

We note then that the Wronskian is $-2\sin(x)\sin(2x)-\cos(x)\cos(2x)$, however, this will yield a Green's function that quite clearly won't satisfy the final part of the question.

Any help would be appreciated. Regards as always.

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    @MiamiMath: Did you take a look at [this](http://en.wikipedia.org/wiki/Green%27s_function#Example) example?2012-04-28

0 Answers 0