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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



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 $y(0)= 0 = y(\pi/4)$. – Fabian Jan 10 '12 at 21:24
@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? – Mathmo Jan 11 '12 at 10:23
$\sin x$ 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... – Fabian Jan 11 '12 at 10:35
In the statement of the problem, do you mean that $f$ is a continuous function (instead of "where $y$ is a continuous function...")? – Willie Wong Jan 11 '12 at 10:47
One method to derive the uniqueness of the BVP is to use Wirtinger's inequality‌​, 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 $\int y^2 = 0$. – Willie Wong Jan 11 '12 at 11:41

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