Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2009 / 7 / Solution

If two solutions exist, their difference satisfies $d^{\prime \prime }+d=0$ with $d(0)=d(L)=0$, where $L=\pi /4$. The general solution is $d=a\sin x+b\cos x$. The first endpoint gives $b=0$ and the second gives $a\sin L=0$, hence $a=0$. Thus the solution is unique if it exists.

For the operator $-D^2-1$, choose the homogeneous solutions $\sin x$ and $\sin (L-x)$ satisfying the left and right endpoint conditions. Continuity at $x=s$ and derivative jump $G_x(s^{+},s)-G_x(s^{-},s)=-1$ yield the Dirichlet Green functionIndeed it vanishes at both endpoints and satisfies $-G_{xx}-G={\delta }_s$. Splitting the integral at $s=x$ and differentiating verifies that $y={\int }_0^{L}G(x,s)f(s)ds$ is twice continuously differentiable for continuous $f$ and solves the boundary-value problem, establishing existence as well.

At zero, $G_x(0,s)=\sin (L-s)/\sin L=\cos s-\sin s$, since $\cot L=1$. Thus