Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 8

For the linear differential operator $L=D^2+4D+5$, the characteristic equation is $m^2+4m+5=0$, with roots $-2\pm i$. Take the fundamental system of solutionsThe Wronskian is never zero, so variation of parameters is valid on the whole interval. Write a particular solution as $a(x)u(x)+b(x)v(x)$, and impose the auxiliary condition $a^{\prime }u+b^{\prime }v=0$. Substitution then gives $a^{\prime }{u}^{\prime }+b^{\prime }{v}^{\prime }=f$. These two equations giveChoosing both integrals to start at zero makes the resulting particular solution and its first derivative vanish there:A general solution is $Au+Cv+y_p$. The boundary condition at zero sets $A=0$. The condition at $\ell =\pi /2$ givesbecause $\sin \ell =1$ and $\sin (\ell -t)=\cos t$. Combining the two integrals yields the Green's function representationIndeed, for $t<x$ the kernel is $-e^{2(t-x)}\sin t\cos x$, since $\sin (x-t)-\sin x\cos t=-\cos x\sin t$; for $x<t$ it is $-e^{2(t-x)}\sin x\cos t$.

For fixed interior $t$, the Green's function vanishes at both endpoints, is continuous at $x=t$, solves the homogeneous equation away from $t$, and satisfiesThis derivative jump is the normalization required by the leading coefficient one in $L$. Finally, a homogeneous solution satisfying the first boundary condition is $C{e}^{-2x}\sin x$, and the second boundary condition at $\pi /2$ forces $C=0$. Thus the boundary value problem has exactly one solution for every continuous forcing, justifying the unique inverse represented by this kernel.

For this forcing the integral in part (a) simplifies toConsequently the unique boundary value problem solution isIt vanishes at zero and at $\pi /2$. To check the differential equation directly, set $y=e^{-2x}w$. Then $Ly=e^{-2x}(w^{\prime \prime }+w)$. For $w=1-\sin x-\cos x$, one has $w^{\prime \prime }+w=1$, as required.

The same variation of parameters construction works on $[0,\pi ]$. All solutions satisfying the boundary condition at zero have the formAt $x=\pi$, the homogeneous term vanishes for every $C$; it can no longer be chosen to correct the final boundary value. Since $\sin (\pi -t)=\sin t$, that boundary value is zero exactly whenThis condition is necessary by the displayed general solution, and sufficient by the same formula. When it holds, the full answer isFor complex-valued forcing, allow $C\in \mathbb{C}$ instead. The freedom is addition of the nonzero homogeneous Dirichlet solution $e^{-2x}\sin x$. At this endpoint separation, the homogeneous boundary problem has a nonzero solution. The resulting Fredholm solvability condition restricts the forcing to the compatible functions described above. In the language of the Fredholm alternative, the condition is orthogonality to the adjoint null solution $e^{2x}\sin x$ of $D^2-4D+5$ with zero endpoint values.