Past exam of the differential equations and applications course of the University of Oxford / 2010 / A / 1 / i

The characteristic polynomial is $r^2+2r+2=(r+1{)}^2+1$, with distinct complex roots $r=-1\pm i$. Taking the real and imaginary parts of the corresponding exponential functions gives the real fundamental system of solutions $e^{-x}\cos x$ and $e^{-x}\sin x$. Consequently the general real solution isTheir Wronskian is $e^{-2x}\ne 0$, so they span the two-dimensional real solution space.

For each $0<\xi <2\pi$, the Green's function is a distribution in $x$ satisfyingEquivalently, $g$ is a classical homogeneous solution on each side of $\xi$, satisfies the same endpoint conditions, and obeys the Green function jump conditionsTo see equivalence, let $[g]$ and $[g_x]$ denote its jumps. The singular part of $L_{x}g$ is $[g]{\delta }^{\prime }(x-\xi )+([g_x]+2[g])\delta (x-\xi )$. Matching a unit Dirac delta distribution with no derivative first forces continuity, then the unit derivative jump. Conversely these conditions produce exactly that distribution. The solution for a regular forcing is $y(x)={\int }_0^{2\pi }g(x,\xi )f(\xi )d\xi$. The homogeneous endpoint problem has only the zero solution, which ensures uniqueness of this kernel.

Choose a left-end solution $u(x)=e^{-x}\sin x$ and a right-end solution $v(x)=e^{-x}(\sin x+\cos x)$. They satisfy $u(0)=0$ and $v^{\prime }(2\pi )=0$. Their Wronskian, with the order used here, isThe Green's function is $u(x_{<})v(x_{>})/W(\xi )$, where $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. Explicitly,The two expressions agree at the source. Their derivative difference is $[u(\xi )v^{\prime }(\xi )-u^{\prime }(\xi )v(\xi )]/W(\xi )=1$, checking both normalization and sign. The first-order derivative term means the kernel need not be symmetric under exchange of $x$ and $\xi$.