Past exam of the differential equations 2 course of the University of Oxford / 2015 (A6, A13758W1)

Source paper: released examination archive, member 2014/T/a13758w1.pdf.

Source limitation: the held cover has a grey loss across its right-hand portion. The 2015 year is a high-confidence inference from question-specific correspondence with the official examiner report, not a year read from an intact cover. Full cover wording and time remain unrecovered; the two question pages are readable in full.^[ref]

On an interval with $x>0$, use the power-law ansatz $y=x^r$. Substitution gives $r(r-1)+2r+1=r^2+r+1=0$, whose roots are $r=-1/2\pm i\sqrt{3}/2$. Taking real and imaginary parts of $x^r$ givesThese two real solutions are linearly independent: their Wronskian is $(\sqrt{3}/2)x^{-2}\ne 0$. They therefore span the solution space of the second-order linear ordinary differential equation on that interval.

For $x<0$, replace $x^{-1/2}$ by $|x{|}^{-1/2}$ and $\log x$ by $\log |x|$, with independent real constants on the negative interval. This follows equally by the change of variable $t=\log |x|$, which turns the Cauchy-Euler equation into $d^{2}y/d{t}^2+dy/dt+y=0$. The coefficient of $y^{\prime \prime }$ vanishes at zero, so the general formula is stated separately on either side. Every nonzero mode has an unbounded oscillatory envelope as $x\to 0$; a continuous extension through zero requires the solution to vanish identically on both sides.

Use the usual inner product $⟨u,v⟩={\int }_0^1\overline{u}vdx$. Moving four derivatives by integration by parts leaves $L^{\ast }=D^4+1$ and gives the fourth-order boundary identityThus the differential expression is formally self-adjoint. If both functions satisfy the clamped boundary conditions, every displayed endpoint product vanishes, proving symmetry on the specified domain.

To check full self-adjointness, also identify the adjoint domain. Allow $v$ to run through the clamped domain. At either endpoint, $v^{\prime \prime }$ and $v^{\prime \prime \prime }$ can be chosen independently, for instance using a polynomial near that endpoint multiplied by a smooth cutoff function. The remaining boundary form is $[\overline{u}{v}^{\prime \prime \prime }-\overline{u^{\prime }}{v}^{\prime \prime }{]}_0^1$. It vanishes for every such $v$ precisely when $u=u^{\prime }=0$ at both endpoints. These are exactly the original boundary conditions. On the regular finite interval, the adjoint domain consequently equals $\{u\in H^4(0,1):u(0)=u^{\prime }(0)=u(1)=u^{\prime }(1)=0\}$, rather than a larger domain. Here $H^4$ is the Sobolev space of functions whose weak derivatives through order four are square-integrable.

The operator together with these boundary conditions is fully self-adjoint. As a further consistency check, integration by parts gives $⟨y,Ly⟩={\int }_0^1(|y^{\prime \prime }{|}^2+|y{|}^2)dx>0$ for every nonzero admissible $y$. The reusable calculation is the Lagrange identity for a fourth-order differential operator.

Start with a fundamental solution of the second derivative operator:Here $H$ is the Heaviside step function and the last equality holds in the sense of distributions. The factor $x-\xi$ removes a jump in the function itself, while its first derivative has a unit jump. Adding any homogeneous solution preserves the Dirac delta source.

Choose $y_1(x)=x-2$ and $y_2(x)=x-3$. They are independent homogeneous solutions, with $B_1{y}_1=0$, $B_2{y}_2=0$, $B_1{y}_2=-1$, and $B_2{y}_1=1$. Since $1<\xi <2$, the fundamental solution has $B_1\tilde{g}=1$ and $B_2\tilde{g}=2-\xi$. Imposing the two boundary conditions on the proposed combination therefore gives $-A_2+1=0$ and $A_1+2-\xi =0$. ConsequentlyEquivalently,At $x=\xi$ both expressions equal ${\xi }^2-3\xi +1$ and the right slope minus the left slope is $\xi -(\xi -1)=1$. At the endpoints, $g(1,\xi )=-\xi$ and $g_x(2,\xi )=\xi$, while $g_x(1,\xi )=\xi -1$ and $g(2,\xi )=1-\xi$. These explicitly verify both coupled boundary conditions. The homogeneous equation has only affine solutions; its two boundary conditions give $2a+b=3a+b=0$, hence $a=b=0$, proving uniqueness of the Green function.

The reciprocity defect is $g(x,\xi )-g(\xi ,x)=2(x-\xi )$. It records how the coupled boundary conditions affect self-adjointness, even though the differential expression $D^2$ is formally self-adjoint. The construction is an instance of the boundary-matrix construction of an ordinary differential Green function.

Divide by $2x$ away from zero to obtain $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, with $P(x)=x/2-1/(2x)$ and $Q(x)=\lambda /2$. At zero, the quantities $xP(x)=(x^2-1)/2$ and $x^{2}Q(x)=\lambda x^2/2$ are analytic. The pole in $P$ makes zero singular, so $x=0$ is a regular singular point. Its indicial equation isAt $x=1$, both normalized coefficients are analytic functions, so $x=1$ is an ordinary point. These are the standard normalized-coefficient and indicial tests.^[ref]

For infinity, set $t=1/x$ and $Y(t)=y(1/t)$. The chain rule gives $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, so the transformed normalized equation isThe product of the first-derivative coefficient with $t$ still has a second-order pole at zero. Hence the criterion for a regular singular point fails and $x=\infty$ is an irregular singular point, including when $\lambda =0$. The primes on $Y$ here refer to differentiation with respect to $t$.

Seek a Frobenius series $y=x^r{\sum }_{n\ge 0}{a}_n{x}^n$, with $a_0\ne 0$, on the positive side of zero. The lowest coefficient gives the indicial equation from part (a); the next gives $(r+1)(2r-1)a_1=0$. For either permitted indicial exponent, $a_1=0$. Matching the subsequent powers givesThe two indicial exponents differ by a noninteger, so there are two independent Frobenius solutions without a logarithmic companion. Normalizing each leading coefficient to one, their forms areThus the general local solution is $C_0{f}_0+C_1{f}_{3/2}$. The endpoint condition at zero forces $C_0=0$. For a nontrivial solution, $C_1\ne 0$, so $y(x)\sim C_1{x}^{3/2}$ and $\alpha =3/2$, with the arbitrary nonzero amplitude understood.

The condition at one selects values of $\lambda$ for which this branch also vanishes there; it does not change the selected local exponent. The zero solution, available for every $\lambda$, has no nonzero leading coefficient and is excluded when asking for a leading exponent. This is a direct application of the Frobenius method and the noninteger-separated-root case.^[ref]

First classify nonzero polynomial solutions of the differential equation itself. If $y$ has degree $d$ and leading coefficient $a_d\ne 0$, the coefficient of $x^{d+1}$ is $(d+\lambda )a_d$. Hence $\lambda =-d$ is necessary.

An analytic solution at zero belongs to the exponent-zero Frobenius branch. Its odd-power coefficients vanish, because $a_1=0$ and the recurrence relation preserves parity. For the even-power coefficients,All denominators are nonzero for $j\ge 0$. With $\lambda =-2m$, the coefficients through $a_{2m}$ are nonzero when $a_0\ne 0$, the next coefficient vanishes, and every subsequent coefficient vanishes. This proves sufficiency as well as the restriction to even polynomial degree:For example, setting $a_0=1$ gives $1$ when $\lambda =0$, $1+x^2$ when $\lambda =-2$, and $1+2x^2+x^4/5$ when $\lambda =-4$. A polynomial cannot instead begin with a positive integer power: its lowest power must have an indicial exponent, and the only integer indicial exponent is zero. This also proves that no odd-degree family was missed. The finite-series mechanism is a terminating Frobenius series.

If the additional endpoint conditions from part (b) are imposed on this part too, there is no nonzero polynomial solution: every nonzero polynomial just classified has $y(0)=a_0\ne 0$. The displayed family and degree answer therefore concern the differential equation without those extra endpoint conditions.

By the definition of little o notation, the required quotient isAs $ϵ\to 0^{+}$, write this as $\log ϵ/(1/ϵ)$. L'Hôpital's rule applies to this infinite-over-infinite ratio and givesTherefore ${ϵ}^2\log ϵ=o(ϵ)$. The logarithm is negative on this approach, which is consistent with a quotient tending to zero from below; little o notation concerns the limiting ratio, not its sign.^[ref]

Take $ϵ\to 0^{+}$. A root on the smaller scale has $x=ϵu$, giving $u+2+ϵu^3=0$. Put $u=u_0+ϵu_1+O({ϵ}^2)$. Successive coefficients give $u_0=-2$ and $u_1+u_0^3=0$, hence $u_1=8$. The real root is thereforeThe other dominant balance is between the cubic and linear terms, requiring $x={ϵ}^{1/2}z$. Set $t={ϵ}^{1/2}$; the equation becomes $z^3+z+2t=0$. At $t=0$ the two nonzero roots are $z_0=\pm i$. With $z=z_0+t{z}_1+O(t^2)$, the first correction satisfies $(3z_0^2+1)z_1+2=0$, so $-2z_1+2=0$ and $z_1=1$. ThusThe limiting roots in both rescalings are simple, so the implicit function theorem justifies these regular expansions in their respective small parameters. The fractional powers in the original variable are a fractional-power root expansion. The real derivative $3x^2+ϵ$ is strictly positive; the cubic therefore has exactly one real root. Its other roots are the displayed complex conjugate pair, accounting for all three roots. Their leading real corrections sum to $-2ϵ+ϵ+ϵ=0$, as required by the absence of a quadratic coefficient.

For a positive eigenvalue write $\mu =\sqrt{\lambda }>0$. The general solution of the limiting constant-coefficient differential equation is $y=A\cos (\mu x)+B\sin (\mu x)$. The endpoint at zero gives $A=0$. A nonzero eigenfunction then requires $B\ne 0$, and the endpoint at $\pi$ gives $\sin (\mu \pi )=0$. ConsequentlyFor real eigenfunctions the constants may be any nonzero real amplitudes. Requiring unit L2 norm would set $|C_k|=\sqrt{2/\pi }$, but that normalization is not required by the question. The lowest eigenvalue is ${\lambda }_1=1$ and its eigenspace is spanned by $\sin x$. For completeness, zero yields only an affine solution annihilated by both endpoints, and negative eigenvalues yield hyperbolic functions whose two zero endpoint values also force the zero solution.

Choose the real leading eigenfunction $y_1=C\sin x$, keeping its amplitude $C\ne 0$ fixed. At first order in the regular perturbation expansion, the equation becomesThe operator $L=D^2+1$ with these homogeneous Dirichlet boundary conditions is self-adjoint, as allowed in the question, and its null space is spanned by $y_1$. The Fredholm solvability condition requires the forcing to be orthogonal to that null space:The same condition follows directly by integrating $y_{1}L{y}_{11}$ by parts: the endpoint term $[y_1{y}_{11}^{\prime }-y_1^{\prime }{y}_{11}{]}_0^{\pi }$ vanishes, and $L{y}_1=0$. Since a nonzero real $y_1$ has a positive squared integral,This is the requested expression and is an instance of the leading nonlinear eigenvalue shift in a Sturm-Liouville problem. In the amplitude convention above, ${\int }_0^{\pi }{\sin }^{2}xdx=\pi /2$ and ${\int }_0^{\pi }{\sin }^{3}xdx=4/3$, so ${\lambda }_{11}=-8C/(3\pi )$. For the particular choice $C=1$, it is $-8/(3\pi )$.

The amplitude dependence of a nonlinear eigenvalue correction follows here from the quadratic term: rescaling the eigenfunction changes the eigenvalue shift. There is no single numerical correction independent of the chosen leading amplitude. Once the solvability condition is met, $y_{11}$ is determined up to addition of a multiple of $y_1$; a normalization such as ${\int }_0^{\pi }{y}_1{y}_{11}dx=0$ can fix that remaining freedom.