Past exam of the differential equations 2 course of the University of Oxford

Solutions to all three questions of the University of Oxford Differential Equations II paper, sat on 22 June 2023. Source: released examination PDF, parent PDF pages 118–120; questions on pages 119–120. Question statements are omitted. In Question 2(b), the usual interpretation that the forcing extends continuously to the closed interval is distinguished from continuity only in the interior.

The characteristic equation of the linear ordinary differential equation is $r^2+2r+2=0$, with roots $-1\pm i$. A real pair of homogeneous solutions is thereforeTheir Wronskian, with the convention $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, isThus the two functions are linearly independent throughout the interval and span its homogeneous solutions.

Use the unweighted inner product $⟨w,y⟩={\int }_0^a\overline{w}ydx$. Two applications of integration by parts giveThe given boundary conditions set $y=0$ at both ends, but leave $y^{\prime }(0)$ and $y^{\prime }(a)$ independently variable. The remaining boundary term vanishes for every admissible $y$ precisely when $w(0)=w(a)=0$. Consequently the formal adjoint and the boundary conditions specifying the adjoint operator are

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.

At $a=\pi$, the nonzero homogeneous solution $e^{-x}\sin x$ satisfies both zero boundary conditions. Thus the boundary value problem has a nontrivial kernel and cannot have an inverse for arbitrary forcing.

More directly, the adjoint operator has the zero-endpoint homogeneous solution $w(x)=e^x\sin x$. If an ordinary Green's function existed, the integration by parts identity would implyBut $w(\xi )\ne 0$ for $0<\xi <\pi$. This contradiction rules out the Green's function for the unmodified operator. It is the Fredholm alternative obstruction.

The nonzero endpoint data contribute to the Fredholm alternative condition. With $w=e^x\sin x$, the same integration by parts identity givesFor the specified forcing the left side is $A{\int }_0^{\pi }{\sin }^{2}xdx=A\pi /2$. HenceTo verify sufficiency, set $y=e^{-x}u$. Direct differentiation gives $\mathcal{L}y=e^{-x}(u^{\prime \prime }+u)$. At this value of $A$, a particular solution of $u^{\prime \prime }+u=(2/\pi )\sin x$ is $-(x/\pi )\cos x$. The complete family obeying the two boundary conditions isThe arbitrary constant records the one-dimensional kernel; existence does not imply uniqueness here.

Put $N=n(n+1)$. After division by $1-x^2$, the Legendre differential equation has coefficientsBoth are analytic functions at zero, so $\boxed{x=0\text{ is an ordinary point}}$. At either $x_0=\pm 1$, $p$ has a genuine simple pole; hence these are singular points even when $n=0$ makes $q$ vanish. However,and both products extend analytically across $x_0$. By the regular singular point criterion for a second-order equation,None of the three specified points is an irregular singular point.

Use $t=x+1$. The Legendre differential equation becomesSubstituting a Frobenius series $y=t^r{\sum }_{k\ge 0}{a}_k{t}^k$, $a_0\ne 0$, gives the indicial equation $2r^2=0$: the indicial root zero is repeated. For $r=0$, the coefficient of $t^k$ yieldsorFor $0\le k<n$ the multiplying factor is nonzero. At $k=n$ it vanishes, so $a_{n+1}$ and all later coefficients are zero. Thus one solution is a polynomial of exact polynomial degree $n$, including a nonzero constant when $n=0$. Denote a chosen normalization by $P_n$.

The linear recurrence relation determines every analytic coefficient from $a_0$, so there cannot be a second independent analytic solution at $t=0$. To see the required logarithm explicitly, reduction of order gives the second solution locally asNear $t=0$, the integrand is $1/(2a_0^{2}t)$ plus an analytic function. After rescaling the second solution, its form is thereforeThe analytic remainder converges locally; its constant term may be set to zero by adding a multiple of $P_n$. The nonzero logarithmic coefficient makes $y_2$ linearly independent of $P_n$. This Frobenius expansion of the Legendre equation at an endpoint gives a Logarithmic Frobenius solution; the same endpoint calculation at $x=1$ explains why the independent Legendre function of the second kind is excluded when boundedness at both ends is required.

Write the operator in divergence form:The Legendre differential equation therefore givesThese are the eigenvalues of $\mathcal{L}$. With the common positive-sign convention for a Sturm-Liouville problem, use $-\mathcal{L}$ instead:This is a singular Sturm-Liouville problem, because $p$ vanishes at the endpoints. Bounded eigenfunctions select the nonlogarithmic branches and satisfy $p{P}_n^{\prime }\to 0$ there. The Lagrange identity for a Sturm-Liouville operator givesThe eigenvalues are distinct, proving the orthogonality conditionThe weight is one. No choice of normalization of the Legendre polynomials is needed for this conclusion.

First take the usual endpoint-regular interpretation $f\in C([-1,1])$. Integrating the divergence form shows that $F=(1-x^2)y^{\prime }$ has finite endpoint limits. A nonzero limit would make $y^{\prime }$ proportional to $1/(1\mp x)$ near that endpoint, forcing a logarithmically unbounded $y$. Thus boundedness requires $F(-1)=F(1)=0$, and consequentlyThis is orthogonality to the constant eigenfunction $P_0$, hence the Fredholm alternative condition. It is also sufficient under the stated endpoint regularity: putIf $M=\max |f|$, the zero-mean condition gives $|F(s)|\le M\min (1+s,1-s)$. Hence the last integrand is bounded, so $y$ extends continuously to both endpoints. Any two bounded solutions differ by a constant: their difference solves $((1-x^2)z^{\prime }{)}^{\prime }=0$, whose other solution is $\frac{1}{2}\log ((1+x)/(1-x))$.

For the requested eigenfunction expansion, setThe Legendre polynomials span all polynomials, so their completeness follows from the Weierstrass approximation theorem. Projecting $\mathcal{L}y=f$ onto each eigenfunction gives $-k(k+1)y_k=f_k$ for $k\ge 1$. The zero mode is free, while $f_0=0$. Thus the full family isThe formula is unchanged by rescaling any $P_k$. For the usual normalization $P_k(1)=1$, the facts $|P_k(x)|\le 1$ on $[-1,1]$ and $N_k=2/(2k+1)$, together with the Cauchy-Schwarz inequality, bound each summand by a constant times $k^{-3/2}$. The series therefore converges absolutely and uniformly. Its identification with the preceding integral solution avoids unjustified termwise second differentiation.

There is a source ambiguity: continuity only on $(-1,1)$ does not supply the endpoint control used above. Even for integrable forcing, zero mean alone is insufficient. For example, letThis $f$ is continuous in the interior, absolutely integrable, and has zero mean because $F(\pm 1)=0$. Yet every candidate with zero endpoint values of $(1-x^2)y^{\prime }$ haswhose integral diverges logarithmically in $H$ at each endpoint. Thus no bounded solution exists. For a merely interior-continuous $f$, the exact criterion is that some constant $B$ makesbounded as $x$ approaches both endpoints. The Bounded solvability for the Legendre operator criterion above covers the intended regular forcing and makes this endpoint limitation explicit.

When $n=0$, the forcing has nonzero integral ${\int }_{-1}^{1}1dx=2$, violating the Fredholm alternative condition. There is no bounded solution. Directly, integration gives $(1-x^2)y^{\prime }=x+B$; cancellation of its logarithmic divergence at $x=-1$ requires $B=1$, while cancellation at $x=1$ requires $B=-1$.

For $n\ge 1$, the constant $1/[n(n+1)]$ is a particular solution. Subtract it from any solution. The remainder satisfies the Legendre differential equation, whose general solution is a combination of $P_n$ and the logarithmically unbounded Legendre function of the second kind. Boundedness removes the latter coefficient, givingThe constant forcing is orthogonal to $P_n$ for $n\ge 1$, consistently with Orthogonality of Legendre polynomials and the Fredholm alternative.

Consider real roots. Every zero at $ϵ=0$ is $k\pi$, $k\in \mathbb{Z}$, and its derivative $\cos (k\pi )=(-1{)}^k$ is nonzero. For a regular perturbation, write $x=k\pi +\delta$. The Taylor seriesgives $\delta =(-1{)}^kϵ+O({ϵ}^3)$. Thus, through the requested order,There is no quadratic term. The exact parametrization $x_k=k\pi +(-1{)}^k\arcsin ϵ$ also verifies completeness and the remainder order for $|ϵ|<1$.

Take $ϵ\to 0^{+}$. There are two real branches, on different scales. For a root tending to zero, the Taylor series givesThe dominant balance is $x^3\sim -6ϵ$, so a fractional-power root expansion yieldsMore precisely $x_{-}=-(6ϵ{)}^{1/3}+O(ϵ)$. An ordinary power series in integer powers of $ϵ$ would miss this scale.

There is also a root escaping to positive infinity. Boundedness of the sine term makes the leading dominant balance $x-ϵx^2=0$. The nonzero branch isIndeed the equation rearranges to $ϵx_{+}^2-x_{+}=ϵ-\sin x_{+}$, which yields $x_{+}={ϵ}^{-1}+O(1)$ on this branch. The bounded oscillatory correction need not have a constant limiting value.

To check that these are all the real solutions, solve for the parameter:Positive values occur only on $(-1,0)$ and $(1,\infty )$; $x=\pm 1$ are not roots. A useful exact derivative isIt vanishes only at isolated points, so $E$ is strictly decreasing on each of these intervals. Its endpoint limits are $+\infty$ and zero on each interval. This monotone parametrization of sine perturbation roots gives exactly one root in each interval for every $ϵ>0$, proving that both asymptotic branches, and no others, have been included. If negative small parameters are allowed, the transformation $(x,ϵ)\mapsto (-x,-ϵ)$ reflects these two branches and gives all real solutions for that sign as well.

For the positive coefficient of $y^{\prime \prime }$, setting $ϵ=0$ gives the reduced equation $y_0^{\prime }+x{y}_0=0$, with general solution $y_0=C{e}^{-x^2/2}$. A fast decaying homogeneous solution of $ϵy^{\prime \prime }+y^{\prime }=0$ is $e^{-x/ϵ}$, placing the boundary layer at the left endpoint. Thus use the right boundary condition for the outer solution: $y_0(1)=0$ impliesThe dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives width $\delta =ϵ$. With the inner variable $X=x/ϵ$ and $Y(X)=y(ϵX)$, the exact rescaled equation isAt leading order, $Y_{0,XX}+Y_{0,X}=0$, so $Y_0=A+B{e}^{-X}$. The asymptotic matching condition $Y_0(\infty )=y_{\mathrm{out},0}(0)=0$ sets $A=0$, and $Y_0(0)=1$ sets $B=1$. ThereforeThe additive composite expansion has zero overlap to subtract. Its graph starts at one, decreases steeply over $x=O(ϵ)$, and then stays close to zero; its initial slope is $-1/ϵ$. The right endpoint discrepancy $e^{-1/ϵ}$ is exponentially small. The left panel of the figure below sketches this profile. A right-end inner expansion would instead contain a growing $e^X$ term and could not repair the left endpoint mismatch.

With the negative coefficient of $y^{\prime \prime }$, the reduced equation is unchanged, but the fast mode now decays away from the right endpoint. Apply the left boundary condition to the outer solution, givingThe boundary layer still has width $O(ϵ)$. Put $X=(1-x)/ϵ$ and $Y(X)=y(1-ϵX)$. The rescaled equation isIts leading inner expansion has $Y_0=A+B{e}^{-X}$. The asymptotic matching condition is $Y_0(\infty )=e^{-1/2}$, while $Y_0(0)=0$. ThusSubtracting the overlap $e^{-1/2}$ in the additive composite expansion givesIts graph starts exponentially close to one, follows the gently decreasing outer solution, and drops from near $e^{-1/2}$ to zero in a right boundary layer of width $O(ϵ)$. The right endpoint value is exactly zero; the left endpoint error is $-e^{-1/2-1/ϵ}$. Its right endpoint slope is $-e^{-1/2}(1+1/ϵ)$, showing the rapid final descent.

Solutions to all three questions of the University of Oxford Differential Equations II paper, sat on 25 June 2021. Source: released examination PDF, bundle pages 111–113; questions on pages 112–113.

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.

Work first on $x>0$. For $b=0$, the power-law ansatz $y=x^r$ gives the indicial equation $r(r-1)+r-a^2=r^2-a^2=0$. Therefore the Cauchy-Euler equation has general solutionFor $a=0$ the repeated indicial root supplies the logarithm rather than two copies of the constant solution. Alternatively, the change of variables $t=\log x$ turns the differential equation into $Y^{\prime \prime }-a^{2}Y=0$. On $x<0$ the corresponding real formulas use $|x|$ and $\log |x|$, with independent constants on the two sides of zero.

For general $b$, normalize the second-order linear differential equation asThe coefficient $P$ is singular at zero, but $xP(x)=1$ and $x^{2}Q(x)=-a^2+b{x}^{\beta }$ are analytic functions there because $\beta$ is a positive integer. The regular singular point criterion for a second-order equation therefore proves that zero is a regular singular point for every allowed $a,b,\beta$. The indicial equation remains $r^2-a^2=0$.