Past exam of the differential equations 2 course of the University of Oxford / 2017 (A13758W1, A6)

Source paper: released examination archive, member 2016/T/a13758w1.pdf. The original cover dates this examination to 2017-06-23.

The complete four-page replacement is in the Mathematics Part A FOI release, member FOI-20210708-1 - Data/a13758w1 (2).pdf. Its cover confirms the same A6 paper, formal code A13758W1, sitting on 23 June 2017. This copy supplies the previously missing printed pages 3–4; the original source above is retained for provenance.

Extend the given functions continuously to zero by $w_1(0)=w_2(0)=0$. This is forced by their stated differentiability. Their derivatives at zero are both zero, since the corresponding difference quotients tend to zero.

For $x<0$, $w_2=w_2^{\prime }=0$, and for $x>0$, $w_1=w_1^{\prime }=0$. At zero all four quantities vanish. Thus the Wronskian isNevertheless, if the constant linear combination $A{w}_1+B{w}_2$ vanishes identically, evaluation at $x=-1$ gives $-A=0$, and evaluation at $x=1$ gives $B=0$. Therefore the two functions are linearly independent.

This illustrates why a vanishing Wronskian does not imply linear dependence for arbitrary differentiable functions. Here both functions are continuously differentiable, but neither has a second derivative at zero. The common regular differential equation hypothesis in the next part is absent.

Differentiate the Wronskian using the product rule and substitute the common homogeneous linear differential equation:Multiplying this first-order equation by the integrating factor $\exp ({\int }_{x_0}^{x}p(t)dt)$ and integrating gives Abel's identity:The exponential factor is finite and nonzero for every $x$ in the given interval, because $p$ is continuous on the compact interval between $x_0$ and $x$. Hence $W(x_0)\ne 0$ implies $W(x)\ne 0$ everywhere in that interval. Using an integrating factor avoids dividing by $W$, which would assume the conclusion.

Interpret the Dirac delta coefficient as a point interaction in a second-order differential equation. The solution is continuous at $x=1$ and is twice differentiable on either side, where $y^{\prime \prime }=f$. Its transmission conditions for a delta coefficient areIndeed, the distributional second derivative across a jump contributes $[y^{\prime }{]}_1\delta (x-1)$, which must cancel the delta coefficient because $f$ has no point source.

Put $s=y^{\prime }(0)$ and $F(x)={\int }_0^x(x-t)f(t)dt$. Integrating the ordinary equation and inserting the required derivative jump giveswhere $(x-1{)}_{+}=\max (x-1,0)$. This formula is continuous, has the stated jump, satisfies the equation, and parametrizes every solution with the left endpoint value. Imposing the right endpoint value gives the scalar equationConsequently a unique $s$, and hence a unique solution for every $f,c_1,c_2$, exists precisely whenAt the exceptional value, the nonzero homogeneous solution with zero endpoint values isIt has derivative jump $-\pi /(\pi -1)=-{\alpha }_{\ast }\varphi (1)$. Thus uniqueness fails. More precisely, there are infinitely many solutions ifand none otherwise. This follows either from the scalar equation or by integration by parts against $\varphi$. It is the compatibility obstruction at Dirichlet resonance for a point interaction.

Use a Green function satisfying $L_{x}G(x,\xi )=\delta (x-\xi )$ with homogeneous Dirichlet boundary conditions $G(0,\xi )=G(\pi ,\xi )=0$, for $0<\xi <\pi$. Since $G_{xx}=0$ away from $x=1$ and $x=\xi$, it is continuous and piecewise affine, with possible derivative jumps at those two locations.

For $\xi \ne 1$, the transmission conditions for a delta coefficient and the unit source condition areIf $\xi =1$, there is a single junction, and the correct combined condition isFor example, a suitable piecewise form when $0<\xi <1$ iswhere the coefficients depend on $\xi$. The two continuity conditions, $B-A=1$ and $-D-B=-G(1,\xi )$ determine them. When $1<\xi <\pi$, use the same affine form on the intervals cut at $1,\xi$, with $B-A=-G(1,\xi )$ and $-D-B=1$ instead. This gives all requested boundary and jump conditions without needing the coefficients.

An explicit compact kernel also makes the construction transparent. The left and right homogeneous solutions at $\alpha =1$ areThey obey the interaction jump and have Wronskian $u{v}^{\prime }-u^{\prime }v=-1$ on either side. ThereforeDifferentiating its two branches at the source gives a unit derivative jump whenever $\xi \ne 1$. At $\xi =1$, $G(1,1)=-(\pi -1)$ and $[G_x{]}_1=\pi$, so the combined condition is also satisfied. This is the Dirichlet Green function for a point interaction specialized to the present parameters.

The Green kernel itself has zero endpoint values. The prescribed nonzero endpoint values enter through the homogeneous lift:Indeed, $v(0)=u(\pi )=1$ and $u(0)=v(\pi )=0$. The kernel accounts for the forcing; the lift accounts for the boundary data. Uniqueness follows from part (i), since $1\ne \pi /(\pi -1)$.

Use integration by parts with a complex conjugate, so that the argument also covers complex-valued eigenfunctions. The coefficient functions are understood to be real. For a nonzero eigenfunction $u$ with Dirichlet boundary conditions,The denominator is strictly positive and the numerator is real. Consequently the eigenvalue satisfiesThis is the Rayleigh quotient for the sign convention used here. Positivity of $a$ is unnecessary for this particular reality argument; real coefficients and disappearance of the endpoint term are what matter.

The Lagrange identity for a Sturm-Liouville operator givesIn the intended class of functions whose values and first derivatives stay finite at the endpoints, the endpoint factor $a$ makes this entire boundary term vanish. Thus no separately prescribed endpoint values are needed to eliminate the boundary form. The same integration by parts argument then proves that the eigenvalues are real.

The admissible endpoint behaviour is part of this conclusion: it is the vanishing endpoint coefficient and the Sturm-Liouville boundary form, together with endpoint regularity, that replaces explicit boundary conditions. If singular functions are allowed, a vanishing coefficient alone need not suffice: $a(x)=x(1-x)$ and $v(x)=\log x$ have $a(x)v^{\prime }(x)\to 1$ as $x\downarrow 0$. A self-adjoint differential operator therefore still needs a specified domain of admissible functions.

The useful change of independent variable in a second-order ODE is $t=\log x$, followed by $y(x)=xv(t)$. The chain rule givesThus the eigenvalue equation becomes $v_{tt}=\lambda v$ on $0<t<1$, and the endpoint values become $v(0)=v(1)=0$. At the specified eigenvalues,The first endpoint forces $A=0$; the second holds for every positive integer $k$. Choosing a convenient normalization yieldsEach nonzero constant multiple is the same eigenfunction up to normalization. This logarithmic conjugacy of a repeated-root Euler operator turns the Cauchy-Euler equation into the familiar Fourier sine series problem on a unit interval.

Match the coefficient of $y^{\prime }$ with the derivative of the coefficient of $y^{\prime \prime }$. For Sturm-Liouville form this requiresA positive integrating factor is therefore $r(x)=x^{-3}$, with any positive constant multiple equally valid. Direct calculation givesThis uses the plus-divergence sign convention of the question. The usual minus-divergence Sturm-Liouville operator is its negative; no extra sign is inserted into the requested eigenvalue equation.

Because $r$ never vanishes on the interval, multiplying the original eigenvalue equation by $r$ leaves the requested weighted eigenvalue problem unchanged:The weight in the inner product is essential. The substitution $t=\log x$ givesThus orthogonality of Sturm-Liouville eigenfunctions holds for the weighted inner product with weight $x^{-3}$, and the squared norm is $N_k=1/2$.

Use the normalization from part (ii), and writeThe Sturm-Liouville eigenfunction expansion coefficient is $c_k=N_k^{-1}{\int }_1^{e}ry{\hat{y}}_{k}dx$. Although ${\hat{y}}_k$ has zero endpoint values, $y$ generally does not. Retain the endpoint term in the Lagrange identity for a Sturm-Liouville operator:The positive sign comes from subtracting the contribution at $x=1$. The differential equation also gives $\hat{L}y=r{e}^x-(\beta -1)ry$. Substitution into the preceding identity yieldsHence, away from resonance in a differential equation,This is the effect of boundary data in a Sturm-Liouville eigenfunction expansion: omitting the endpoint contribution would discard the dependence on $\alpha$.

If $\beta =1+n^2{\pi }^2$ for a positive integer $n$, the Fredholm solvability condition isWhen it fails, there is no solution. When it holds, $c_n$ is arbitrary and the other coefficients retain the displayed formula, giving infinitely many solutions differing by a multiple of ${\hat{y}}_n$. Sufficiency follows by subtracting the smooth lift $h(x)=\alpha x(1-\log x)$, which has the required endpoint values and satisfies $Lh=0$, and applying the Fredholm alternative to the resulting homogeneous Dirichlet boundary conditions.

Finally, for $\alpha \ne 0$, the displayed Sturm-Liouville eigenfunction expansion represents the solution on $1<x<e$ and in the weighted inner product sense. Every term is zero at $x=1$, so that series cannot give the nonzero prescribed value by substitution at the endpoint. The boundary trace must be imposed separately. Equivalently, expand $y-h$ with homogeneous Dirichlet boundary conditions and add $h$ back afterwards.

Move the expansion point to the origin with $z=x+1$. The transformed ordinary differential equation isAfter division by $z$, the functions $z(1-1/z)=z-1$ and $z^2(c/z)=cz$ are analytic functions at zero, so this is a regular singular point. For a Frobenius solution with leading term $a_0{z}^{\rho }$, $a_0\ne 0$, the lowest power is $z^{\rho -1}$ and its coefficient isTherefore the indicial equation and its indicial exponents areThe exponent $\rho$ denotes the quantity called $\alpha$ in the paper, and avoids confusion with other uses of that letter.

Substitute a Frobenius series $y={\sum }_{n\ge 0}{a}_n{z}^{n+\rho }$ into the transformed equation. For $n\ge 1$, equating the coefficient of $z^{n+\rho -1}$ gives the recurrence relationAt the larger indicial root $\rho =2$, the coefficient of $a_n$ is $n(n+2)\ne 0$, so a Frobenius solution always exists. At the smaller indicial root $\rho =0$, the first two equations areSince $a_0\ne 0$, the exceptional equation is consistent exactly whenThese are both admissible values, including the zero value; the singular wording of the question does not remove either. This is the undetermined coefficient at Frobenius resonance: at either value $a_2$ is free, whereas otherwise the second independent solution requires a Logarithmic Frobenius solution.

Explicit pairs demonstrate sufficiency and linear independence. In the variable $z=x+1$ one may takeBoth displayed $u_2$ have leading term $z^2/2$, while $u_0(0)=1$. They therefore give independent Frobenius solutions with the two indicial exponents, and direct differentiation verifies the equation. The point is an apparent regular singular point in these two cases.

For the requested closed forms, particularly short choices areIn addition, $(x+2)e^{-x}$ solves the zero-parameter equation and $e^{-x}$ solves the equation with $c=-1$.

The leading regular perturbation equation has $c=0$. Its homogeneous solutions are $1$ and $(x+2)e^{-x}$, soThe boundary data must also be expanded: $2e^{ϵ}=2+2ϵ+\cdots$, while the right endpoint value is independent of $ϵ$. Thus $y_0(0)=2$ and $y_0(1)=3/e$, givingSince $2-3/e\ne 0$, the unique values are $A=0$, $B=1$. ThereforeIts derivatives are $y_0^{\prime }=-(x+1)e^{-x}$ and $y_0^{\prime \prime }=x{e}^{-x}$, so the leading differential equation and both leading endpoint values can be checked directly.

Let $L_{0}u=(x+1)u^{\prime \prime }+x{u}^{\prime }$. Matching powers in the regular perturbation expansion and in the Taylor series of the left endpoint value gives the complete recursive boundary value problemThe higher-order left endpoint values are not zero. The parameter-dependent boundary data in a regular perturbation hierarchy rule applies here.

First prove uniqueness for any one of these forced problems. The difference $v$ of two solutions obeys $L_{0}v=0$ and homogeneous Dirichlet boundary conditions. The positive integrating factor $r_0=e^x/(x+1{)}^2$ givesMultiplication by $\overline{v}$ and integration by parts yieldHence $v^{\prime }=0$, and the endpoint values force $v=0$. The kernel is trivial: zero is not an eigenvalue of this homogeneous Dirichlet boundary value problem.

Existence is also explicit. For arbitrary smooth forcing $g$ and endpoint values $u(0)=a$, $u(1)=b$, two integrations of $(p{u}^{\prime }{)}^{\prime }=r_{0}g$ givewhereSince $H(1)=2-3/e>0$, the unique choicealways meets the second endpoint value. Starting from the smooth $y_0$, induction with $g=-y_{n-1}$, $a=2/n!$, $b=0$ proves existence and uniqueness at every order.

Equivalently, the Fredholm alternative has no compatibility obstruction because the homogeneous kernel vanishes; the corresponding adjoint operator has no zero eigenvalue either. The direct construction above proves this without computing any particular $y_n$ beyond the leading term.