Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 1 / 11D (Methods)

Away from the source point, the Green function obeys the homogeneous equation and must satisfy the appropriate endpoint condition. Thus it has the form $A\sin (kx)$ to the left of $\xi$ and $B\sin (k(\pi -x))$ to the right. It must be continuous at $\xi$: otherwise its second distributional derivative would contain an unwanted derivative of a Dirac delta. Integrating across $\xi$ gives the jump condition $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$.

The left and right homogeneous solutions $u=\sin (kx)$, $v=\sin (k(\pi -x))$ have constant WronskianThis is nonzero for real noninteger $k$. Dividing the product of the left and right solutions by that Wronskian therefore gives the Dirichlet Helmholtz Green function on an intervalThe formula vanishes at both endpoints, is continuous at the source and has derivative jump one, verifying all distributional and boundary conditions. A difference of two such Green functions would be a homogeneous Dirichlet solution; it vanishes because $\sin (k\pi )\ne 0$, proving uniqueness. Although zero is excluded by the stated noninteger condition, the removable $k\to 0$ limit is $-\min (x,\xi )(\pi -\max (x,\xi ))/\pi$.

First use a homogeneous solution to meet the nonzero endpoint values:It satisfies $y_h^{\prime \prime }+k^2{y}_h=0$, $y_h(0)=a$ and $y_h(\pi )=b$. The Green-function representation then givesApplying the differential operator to the integral gives $f(x)$ by the defining Dirac delta property of $G$, and that integral vanishes at the endpoints. For continuous $f$ this is a classical solution; integrable $f$ also gives the corresponding weak solution. The same nonresonant homogeneous argument as in part (a) proves uniqueness.