Past exam of the differential equations 2 course of the University of Oxford / 2021 / 1

Take $l>0$ and use the inner product $⟨u,v⟩={\int }_0^l\overline{u}vdx$ on the Hilbert space $L^2(0,l)$. The differential operator has domainwhere $H^2$ is a Sobolev space. Twice applying integration by parts gives the Lagrange identity for a Sturm-Liouville operatorThe Dirichlet boundary conditions make this expression zero for $u,v\in D(\mathcal{L})$, and the formal adjoint is ${\mathcal{L}}^{\ast }=d^2/d{x}^2+1$. To obtain the domain of the adjoint operator, initially impose the boundary conditions only on $u$. The remaining expression is $-[\overline{u^{\prime }}v{]}_0^l$. Since $u^{\prime }(0)$ and $u^{\prime }(l)$ vary independently, it vanishes for all such $u$ exactly when $v(0)=v(l)=0$. The adjoint domain is therefore the original domain: this is full self-adjointness, including the boundary conditions.

Consequently the eigenvalues are real, eigenfunctions for distinct eigenvalues are orthogonal, and regular Sturm-Liouville theory supplies a complete orthonormal eigenbasis. Here these objects can be exhibited explicitly:Indeed $\mathcal{L}{\varphi }_n={\lambda }_n{\varphi }_n$, and the Fourier sine basis is complete in $L^2(0,l)$. There are no additional eigenvalues: for a nonzero eigenfunction $u$, integration by parts givesso $u^{\prime \prime }+(1-\lambda )u=0$ and the endpoint values force $\sqrt{1-\lambda }l=n\pi$. All these eigenvalues are simple. The eigenvectors are precisely the nonzero multiples of these eigenfunctions, regarded as vectors in the Hilbert space.

The homogeneous solution is $A\cos x+B\sin x$. The left boundary condition removes $A$, and the right boundary condition gives $B\sin l=0$. Thus the kernel of a linear map associated with this boundary value problem isThe Fredholm alternative now gives the complete classification:For necessity at $l=m\pi$, multiply by the homogeneous solution $\sin x$ and use the Lagrange identity for a Sturm-Liouville operator to obtain ${\int }_0^{l}f\sin xdx=0$. For sufficiency, the particular solutionsatisfies $z(0)=0$ and $z^{\prime \prime }+z=f$. At the other endpoint,which is zero under the compatibility condition. All solutions are then $z+B\sin x$, with $B$ arbitrary. The assumed Lipschitz continuity of $f$ in particular makes $f$ continuous, so the displayed integral defines a classical solution. The length $l=0$ is excluded: it would not define a nonempty interval for this boundary value problem.

For $\sin l\ne 0$, use variation of parameters with the homogeneous solutions $\cos x,\sin x$, whose Wronskian is one. Write $y=A(x)\cos x+B(x)\sin x$ and impose $A^{\prime }\cos x+B^{\prime }\sin x=0$. The differential equation gives $-A^{\prime }\sin x+B^{\prime }\cos x=f$, henceAfter integrating from zero and using the left boundary condition,The right boundary condition determinesso the unique solution isCombining the two integrals gives $y(x)={\int }_0^{l}G(x,t)f(t)dt$, with Green's functionIt is symmetric in $x,t$, as expected from self-adjointness, and has zero endpoint values. At $x=t$ it is continuous, while its first derivative obeys the jump conditionThus $({\partial }_x^2+1)G=\delta (x-t)$, with the positive normalization of the Dirac delta function required for $\mathcal{L}y=f$.

Expand in the Fourier sine basis, using unnormalized eigenfunctions:Projection of $\mathcal{L}y=f$ onto each eigenfunction givesWhen $l\notin \pi {\mathbb{N}}_{>0}$, every eigenvalue is nonzero and the eigenfunction expansion is uniquely determined:When $l=m\pi$, the exceptional index is $n=m$, for which the eigenfunction is $\sin x$. If $f_m\ne 0$, its coefficient equation is $0=f_m$, an inconsistency: there is no solution. If $f_m=0$, the coefficient equation is $0=0$, and $c_m$ is arbitrary. The full family isThis is the Fredholm alternative directly in the coefficient equations: nonzero eigenvalues fix their coefficients; a zero eigenvalue requires zero forcing in that direction and leaves the corresponding coefficient free. An inconsistent equation at a zero eigenvalue cannot be repaired by assigning an infinite coefficient.

For convergence, Lipschitz continuity implies absolute continuity of $f$, with bounded derivative almost everywhere. One integration by parts gives $f_n=O(n^{-1})$; since the denominators grow quadratically, $c_n=O(n^{-3})$ apart from the single free coefficient. The solution series and its first derivative therefore converge uniformly. The second derivative converges in $L^2$, and the resulting Sobolev space solution satisfies $y^{\prime \prime }=f-y$. As $f-y$ is a continuous function, this is a classical $C^2$ solution, with both boundary conditions satisfied. The forcing's Fourier sine series is interpreted on the open interval or in $L^2$; its endpoint values need not equal the endpoint values of $f$.

At $l=\pi$, the Fredholm solvability condition isby orthogonality of the Fourier sine basis. Thus a non-unique solution exists exactly when $a=0$, with any real $b$. Since $(d^2/d{x}^2+1)\sin 2x=-3\sin 2x$, the complete family isIf $a\ne 0$ there is no solution, for any $b$. In particular, when $a=b=0$, the family reduces to the kernel $C\sin x$; it is still non-unique. There is no choice of $a,b$ producing a unique solution for this length.

Putting both data at zero makes this an initial value problem. The uniqueness theorem for ordinary differential equations applies, and one can also prove uniqueness directly: the difference of two solutions is a homogeneous solution $A\cos x+B\sin x$ whose value and first derivative at zero are both zero, forcing $A=B=0$. In fact, for every prescribed continuous function $f$,satisfies both initial data and $y^{\prime \prime }+y=f$. Consequently non-uniqueness is impossible for every $l>0$, including $l=m\pi$. In this setting the kernel is trivial. The boundary conditions at a single endpoint do not give the same self-adjoint operator domain as the two separated Dirichlet boundary conditions, so the earlier exceptional lengths do not apply.