= Solution
{c}
If two solutions exist, their difference satisfies $d''+d=0$ 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 function>
$$
\boxed{G(x,s)=\frac{\sin(\min\{x,s\})\sin(L-\max\{x,s\})}{\sin L}.}
$$
Indeed it vanishes at both endpoints and satisfies $-G_{xx}-G=\delta_s$. Splitting the integral at $s=x$ and differentiating verifies that $y=\int_0^LG(x,s)f(s)ds$ 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
$$
\boxed{y'(0)=\int_0^{\pi/4}f(s)(\cos s-\sin s)ds.}
$$
Back to article page
