Past exam of the differential equations 2 course of the University of Oxford / 2023 / 1 / b

Set $a=\pi /2$. To construct the Green's function from a Dirac delta function, require ${\mathcal{L}}_{x}g(x,\xi )=\delta (x-\xi )$ and zero endpoint values. Away from $x=\xi$, the homogeneous solutions satisfying the left and right boundary conditions are respectivelyWrite $g=A(\xi )u(x)$ for $x<\xi$ and $g=B(\xi )v(x)$ for $x>\xi$. A discontinuity in $g$ would generate a derivative of the Dirac delta function, which is absent from the desired equation. Thus $g$ is continuous. Integrating that equation through $\xi$ then gives the jump conditionThe resulting two equations are $Au(\xi )=Bv(\xi )$ and $B{v}^{\prime }(\xi )-A{u}^{\prime }(\xi )=1$. Since the Wronskian $u{v}^{\prime }-u^{\prime }v=-e^{-2\xi }$, they giveBoth endpoint values vanish. At $x=\xi$, subtracting the two first derivatives gives ${\sin }^2\xi +{\cos }^2\xi =1$, verifying the sign and normalization of the Dirac delta function.

The homogeneous solution $h(x)=e^{-x}\cos x$ already has the prescribed endpoint values. Thus $z=y-h$ has zero boundary conditions and satisfies $\mathcal{L}z=f$. The Green's function givesEquivalently, splitting the integral at $\xi =x$ gives the explicit formulaAny difference between two solutions would be $e^{-x}(C\cos x+D\sin x)$ with zero endpoint values, forcing $C=D=0$. The displayed solution is therefore unique.