Past exam of the differential equations and applications course of the University of Oxford / 2008 / A / 2

Let $L=D^2+1$ on the mixed endpoint domain $y(0)=0$, $y^{\prime }(1)=0$ in $L^2(0,1)$. Integration by parts gives the boundary formAt zero, $y^{\prime }(0)$ is free, forcing $z(0)=0$; at one, $y(1)$ is free, forcing $z^{\prime }(1)=0$. Therefore the formal adjoint expression and full adjoint endpoint conditions coincide with the original ones:With the usual second-order operator domain this is a self-adjoint problem. The consequences for its eigenvalues and eigenfunctions follow below.

For $Ly=-\lambda y$, self-adjointness gives $-\lambda \Vert y{\Vert }^2=⟨y,Ly⟩=\overline{⟨y,Ly⟩}$. Since an eigenfunction is nonzero, $\boxed{\lambda \in \mathbb{R}}$. One may also see reality directly from $\lambda \int |y{|}^2=\int |y^{\prime }{|}^2-\int |y{|}^2$, whose boundary term vanishes under the mixed conditions.

If $L{y}_j=-{\lambda }_j{y}_j$, then $⟨y_1,L{y}_2⟩=⟨L{y}_1,y_2⟩$ gives $({\lambda }_1-{\lambda }_2)⟨y_1,y_2⟩=0$. Thus eigenfunctions belonging to distinct eigenvalues are orthogonal. Normalization can make them orthonormal. Equality of eigenvalues alone would not force arbitrary eigenvectors to be orthogonal, although in this particular scalar separated problem each eigenspace is one-dimensional.

The equation is $y^{\prime \prime }+(1+\lambda )y=0$. If $1+\lambda \le 0$, the linear or hyperbolic branches cannot meet both homogeneous endpoint conditions nontrivially. Otherwise $y=C\sin (kx)$, $k=\sqrt{1+\lambda }$, and $y^{\prime }(1)=Ck\cos k=0$ gives $k_n=(n+1/2)\pi$. ThereforeThe factor $\sqrt{2}$ makes the eigenfunctions unit normalized. All the eigenvalues are positive, in particular none is zero.

The difference of two solutions would satisfy $Ly=0$ with $y(0)=0$, $y^{\prime }(1)=0$. This would be a zero eigenvalue of the homogeneous problem, but part (b) shows there is none. Hence at most one solution exists. For continuous forcing the standard second-order existence construction also works: the free initial slope is fixed by an equation with coefficient $\cos 1\ne 0$. Thus for the stated regular boundary-value problem the solution is unique even with nonzero endpoint data.

Expand a forcing in the complete orthonormal basis ${\varphi }_n$. Since $L{\varphi }_n=-{\lambda }_n{\varphi }_n$, its inverse multiplies the $n$th coefficient by $-1/{\lambda }_n$. ConsequentlyThe negative sign follows from the eigenvalue convention $Ly=-\lambda y$. This kernel satisfies the mixed endpoint conditions in its first argument and $L_{x}g=\delta (x-t)$.

The integral equation says precisely $h=L^{-1}z$ with the homogeneous mixed endpoint conditions. Since the supplied $h$ satisfies those conditions, apply $L$ directly:No integration is required. The absence of a zero eigenvalue established in (b)–(c) also proves uniqueness of this recovered forcing.