Past exam of the differential equations 2 course of the University of Oxford / 2021 (A6, A13758W1)

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 $r^2-a^2=0$.

For the first Frobenius solution, take the larger indicial root $r=a$ and normalize the first coefficient to one:Substitution and comparison of the coefficient of $x^{n+a}$ giveEvery denominator is nonzero because $a\ge 0$.

For $\beta =1$, the recurrence relation is $c_n=-b{c}_{n-1}/[n(n+2a)]$, soFor $\beta =2$, $c_1=0$ and all odd coefficients vanish; the even coefficients satisfy $c_{2k}=-b{c}_{2k-2}/[2k(2k+2a)]$. HenceFor any positive integer $\beta$, induction in the recurrence relation shows that only indices divisible by $\beta$ survive:The empty product for $k=0$ equals one. Successive terms in the power series factor have ratio of magnitudesThus the ratio test gives convergence for every finite $x$; a nonintegral $a$ still requires a chosen side of zero or a complex branch for the factor $x^a$.

For a putative second Frobenius solution with the smaller indicial root $-a$, set $y_{-}=x^{-a}{\sum }_{n\ge 0}{d}_n{x}^n$, with $d_0=1$ and $d_j=0$ for $j<0$. Its recurrence relation isThe Frobenius case classification concerns the gap $2a$ between the indicial roots. In the course convention, Case IIIa is the obstructed recurrence requiring a Logarithmic Frobenius solution; Case IIIb permits two pure Frobenius series.

If $a>0$ and $2a\notin \mathbb{Z}$, this is Case I: no denominator vanishes. There are two linearly independent pure Frobenius solutions, withTheir different leading powers establish linear independence.

If $a=0$, this is Case II, with a repeated indicial root. There is only one pure Frobenius series with a nonzero leading coefficient. To see the second solution's form, reduction of order givesHere $y_{+}(t)=1+O(t^{\beta })$, so the integrand is $t^{-1}+O(t^{\beta -1})$. ConsequentlyThe logarithm has a nonzero coefficient and cannot be removed by adding a multiple of $y_{+}$.

If $N=2a$ is a positive integer, the only potentially obstructed coefficient is $n=N$. Below that index, $d_n=0$ unless $\beta$ divides $n$. At the exceptional index the equation readsIf $N=m\beta$ for some positive integer $m$, all coefficients $d_0,d_{\beta },\dots ,d_{(m-1)\beta }$ are nonzero because $b\ne 0$. The exceptional equation is inconsistent. This is Case IIIa, and the independent second solution has the formThe Frobenius resonance for an Euler equation with monomial potential explains why divisibility, rather than the integer gap alone, decides this outcome. For example, with $d_0=1$, applying the original differential operator to $y_{+}\log x$ gives $2x{y}_{+}^{\prime }$. Cancellation of the coefficient of $x^a$ requires $K=-b{d}_{N-\beta }/(2a)\ne 0$.

If $\beta$ does not divide $N$, then $d_{N-\beta }=0$; this includes $N<\beta$ through the negative-index convention. The exceptional equation is $0=0$, so $d_N$ is free. This is Case IIIb: set $d_N=0$ to select an independent pure Frobenius series starting with $x^{-a}$. Subsequent coefficients again follow the recurrence relation. Changing $d_N$ simply adds a multiple of $y_{+}$, the undetermined coefficient at Frobenius resonance.

The complete answer is thereforeIn particular, for $\beta =1$ every positive integer gap gives Case IIIa. For $\beta =2$, positive integer $a$ gives Case IIIa, while $a=1/2,3/2,5/2,\dots$ gives Case IIIb; the remaining positive $a$ give Case I. The condition $b\ne 0$ matters: at $b=0$ the two Cauchy-Euler equation powers already give independent solutions whenever $a>0$.

Normalize both leading coefficients to one. For the indicial root $a=1/2$, the recurrence relation from part (b) givesRecognizing the Taylor series of the sine gives the first Frobenius solutionFor the smaller indicial root $-1/2$, $n(n-1)d_n=-d_{n-2}$. The equation at $n=1$ is $0=0$; choose the free coefficient $d_1=0$ to avoid adding the first Frobenius solution. The remaining coefficients areThus the Taylor series of the cosine givesTheir leading terms are $x^{1/2}$ and $x^{-1/2}$, as required for Case IIIb. A direct check uses $y=x^{-1/2}u$: the differential equation becomes $x^{3/2}(u^{\prime \prime }+u)=0$, so these two functions solve it on $x>0$. Their Wronskian is $-1/x\ne 0$, proving linear independence. The general solution is $(C_1\sin x+C_2\cos x)/\sqrt{x}$.

The function $F$ is an odd function, passes through $(0,0)$, and has the horizontal asymptote $F=0$ at both infinities. Its derivative isIt decreases for $|x|>1$ and increases for $|x|<1$, with its lowest value $F(-1)=-1/2$ and highest value $F(1)=1/2$. These features determine the graph of a function shown below.

For $0<ϵ<1/2$, the equation is the quadratic equation $ϵx^2-x+ϵ=0$. The quadratic formula gives exactly two real equation roots:The binomial series gives $\sqrt{1-4{ϵ}^2}=1-2{ϵ}^2-2{ϵ}^4-4{ϵ}^6+O({ϵ}^8)$, hence the two requested asymptotic expansions areTheir product is exactly one. Both are positive, with $x_{-}<1<x_{+}$; there are no negative equation roots because $F(x)<0$ for $x<0$. At $ϵ=1/2$ the two real equation roots coincide at one, and for $ϵ>1/2$ there are none. In the stated limit $ϵ\to 0^{+}$, these two branches are therefore exhaustive.

The power-law ansatz $y=x^r$ givesSince $ϵ>0$, the equation roots $1$ and $-1/ϵ$ are distinct. The Cauchy-Euler equation therefore has general solutionFor $C_1=C_2=1$, the first derivative and second derivative areThe strictly convex function diverges to $+\infty$ as $x\to 0^{+}$, passes through $(1,2)$, and has a unique lowest point atAs $x\to \infty$, $y-x\to 0$, so the line $y=x$ is an asymptote. When $0<ϵ<1$, the lowest point is to the right of one; at $ϵ=1$ it is at one, and for $ϵ>1$ it is to the left of one.

The ratio of the two terms isThus $x^{-1/ϵ}$ is larger for $0<x<1$, the terms are equal at $x=1$, and $x$ is larger for $x>1$. In the stronger asymptotic sense as $ϵ\to 0^{+}$, $x^{-1/ϵ}$ dominates at each fixed $x<1$, and $x$ dominates at each fixed $x>1$. Their crossover has $|\log x|=O(ϵ)$, equivalently $x=1+O(ϵ)$, as described by the power-law crossover in a singular perturbation. In that region $x=1+ϵX$ with bounded $X$ gives $y\sim 1+e^{-X}$.

Substitute the proposed regular perturbation $y=y_0+ϵy_1+\cdots$ into the differential equation. The leading and subsequent equations areConsequently $y_0=Cx$, and in fact each successive coefficient has the form $y_j=C_{j}x$, because the preceding second derivative is zero. The leading approximation is thereforeIt has only one adjustable constant and generally cannot impose two independent boundary conditions. Those assumed in the question require a nonzero $x^{-1/ϵ}$ coefficient, which a power series in $ϵ$ cannot recover: at fixed $x>1$ it is exponentially small, and at fixed $x<1$ it grows without bound as $ϵ\to 0^{+}$. Thus the regular perturbation describes the specified solution only in the region to the right of the crossover, away from $x=1$; it is not a uniform approximation on $(0,\infty )$.

Setting $ϵ=0$ gives the reduced first-order linear differential equation $y^{\prime }=y$. Its nonzero solutions cannot satisfy the zero value at $x=2$, so one endpoint needs a boundary layer. To determine which endpoint, freeze the coefficient at $x=x_b>0$ in the rapidly varying part: $ϵy^{\prime \prime }-x_b{y}^{\prime }=0$ has fast mode $e^{x_b(x-x_b)/ϵ}$. It decays into the interval from the right endpoint but grows into the interval from the left endpoint. HenceThe dominant balance $ϵ/{\delta }^2\sim x_b/\delta$ gives $\delta \sim ϵ/x_b$, so a convenient local thickness is $ϵ/2$. The left boundary condition must select the outer solution. Equivalently, a putative left inner variable $X=(x-1)/ϵ$ would give $Y_{XX}-Y_X=0$, whose nonconstant solution grows like $e^X$ and cannot satisfy the asymptotic matching condition as $X\to \infty$.

The reduced equation and the left boundary condition give the leading outer solutionIntroduce the right inner variable $X=(2-x)/ϵ\ge 0$ and write $Y(X)=y(2-ϵX)$. The chain rule gives $y^{\prime }=-Y_X/ϵ$ and $y^{\prime \prime }=Y_{XX}/{ϵ}^2$. Multiplying the rescaled differential equation by $ϵ$ gives the exact equationAt leading order, $Y_{0,XX}+2Y_{0,X}=0$, with general solution $Y_0=A+B{e}^{-2X}$. The right boundary condition gives $Y_0(0)=0$. The asymptotic matching condition in the overlap region $ϵ\ll 2-x\ll 1$ givesThus $A=e$ and $B=-e$, yielding the leading inner expansionEquivalently, the inner variable $\xi =2(2-x)/ϵ$ gives $Y_0=e(1-e^{-\xi })$. Both normalizations describe the same boundary layer.

The common value in the overlap region is $e$. The additive composite expansion therefore givesThis leading approximation has $y_{\mathrm{comp},0}(2)=0$ and $y_{\mathrm{comp},0}(1)=1-e^{1-2/ϵ}$; the left endpoint error is exponentially small. It rises from near one following the outer solution $e^{x-1}$, approaches a value close to $e$, then drops rapidly to zero in the right boundary layer. Its first derivative isFor small $ϵ$ the largest value lies slightly inside the interval. For this additive composite expansion, the first derivative vanishes atThis locates the rounded top of the sketch; the rapid final drop has the $O(ϵ)$ scale found above. The additive composite expansion is a leading approximation, so its exponentially small endpoint error does not imply an exponentially small error throughout the interval: neglected algebraic corrections to the outer expansion and inner expansion remain.