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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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    Are you sure $y_2(x)$ is a solution of the homogeneous equation? For the uniqueness: which function(s) solve the homogeneous equation with $y(0)= 0 = y(\pi/4)$.2012-01-10
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    @Fabian: OK, so $y_2$ doesn't satisy the homogeneous case; but if we need linearly independent solutions $y_1,y_2$ then I can't think of any that do?2012-01-11
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    $\sin x$ and $\cos x$ solve the homogeneous differential equation (not caring about the boundary conditions). Taking linear combinations of those should enable you to implement the required conditions...2012-01-11
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    In the statement of the problem, do you mean that $f$ is a continuous function (instead of "where $y$ is a continuous function...")?2012-01-11
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    @Fabian: OK, so taking $y_1(x):=sin(x)$ and $y_2(x)=cos(x)$, then using the Variation of Parameters method to find a solution to the non-homogeneous case in the form $y(x)=c_1(x)y_1(x)+c_2(x)y_2(x)$, we find $c_1(x)=\int^{\frac{\pi}{4}}_{x} f(s)cos(s) ds$ and $c_2(x)=\int^{\frac{\pi}{4}}_{x} f(s)sin(s) ds$. Now, subbing these into $y:=c_1(x)y_1(x)+c_2(x)y_2(x)$, we can determine $G$, which works out to be: $G(x,s)=sin(s)cos(x)$ for $0 \leq s \leq x \leq \frac{\pi}{4}$ and $G(x,s)=cos(s)sin(x)$ for $0 \leq x \leq s \leq \frac{\pi}{4}$. Is this right?2012-01-11
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    @WillieWong: Yes, it should be $f$, I've edited. Thanks.2012-01-11
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    @MiamiMaths: You can get proper sine and cosine functions without italics using `\sin` and `\cos`. (I'd fixed them in the question but you've produced some more in the comments.)2012-01-11
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    One method to derive the uniqueness of the BVP is to use [Wirtinger's inequality](http://en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions), which in your case states that $16 \int_0^{\pi/4} y^2 \leq \int_0^{\pi/4} (y')^2$. So supposing $y$ is the difference between two solutions (and so solve the homogeneous equation $y'' + y = 0$), multiply the equation by $y$ and integrate by parts you get $\int (y')^2 - y^2 = 0$. Combined with Wirtinger's inequality this implies that $\int y^2 = 0$.2012-01-11
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    @WillieWong: Thanks! I've never seen that before; offers a very neat way of showing uniqueness. Are you able to advise on whether my construction of Green's function (see comments) is correct? Regards.2012-01-11
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    The general idea looks alright, but you may have made a typo: from the boundary conditions $c_2(0) = 0$ since $y_2(0) = 1$. But your computation asks $c_2(0) = \int_0^{\pi/4} f(x) \sin(x)$ which may or may not be 0.2012-01-11
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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

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