Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2007 (AC2, XMAT 4302, XMST 4302)

Source: Oxford mathematics examination release, private bundle Part A 2007 11-1.pdf, pages 5–9. The printed cover identifies both mathematics and mathematics and statistics. The complete editable transcription is retained in the private examination archive.

Write $V^{\ast }=V^{\prime }$ for the dual space. Define $\theta (v)(f)=f(v)$. For fixed $v$, this is a linear functional on $V^{\ast }$; evaluation is also linear in $v$, so $\theta :V\to V^{\ast \ast }$ is a linear map. If $v\ne 0$, extend $v$ to a basis and take its first coordinate linear functional $f$; then $f(v)=1$. Thus $\theta$ is injective. A dual basis has the same number of elements as the original basis, so $\dim V=\dim V^{\ast }=\dim V^{\ast \ast }$. The rank-nullity theorem now makes $\theta$ an linear isomorphism.

The annihilator of a vector subspace of a linear subspace $U\le V$ isExtend a basis $v_1,\dots ,v_k$ of $U$ to a basis $v_1,\dots ,v_n$ of $V$, with dual basis $f_1,\dots ,f_n$. A linear functional belongs to $U^{\circ }$ exactly when its first $k$ coordinates in this dual basis vanish. Hence $U^{\circ }$ has basis $f_{k+1},\dots ,f_n$, provingThe dual map is ${\alpha }^{\ast }(h)=h\circ \alpha$; here this is the algebraic dual map, denoted ${\alpha }^{\prime }$ in the source. If $v\in \ker \alpha$, then $({\alpha }^{\prime }h)(v)=h(\alpha v)=0$. Consequently $\mathrm{im}{\alpha }^{\prime }\subseteq (\ker \alpha {)}^{\circ }$. Moreover,because coordinate linear functionals separate vectors. This proves $(\mathrm{im}{\alpha }^{\prime }{)}^{\circ }=\theta (\ker \alpha )$, including the requested second inclusion. Taking dimensions givesThe first inclusion has equal dimensions on both sides, soFor a matrix $A=(a_{ij})$ representing $\alpha$, the matrix of ${\alpha }^{\prime }$ in the corresponding dual bases is $A^{\mathsf{T}}$: its $ji$ entry is $h_i(\alpha v_j)=a_{ij}$. Thus the column rank of $A^{\mathsf{T}}$ is the row rank of $A$, while the column rank of $A$ is $\dim \mathrm{im}\alpha$. Applying the established equality to these two linear maps proves row rank equals column rank over every field.

The adjoint on an inner product space ${\alpha }^{\ast }:V\to V$ is characterized byHere is a finite-dimensional construction that supplies every required basis fact. If $v_1,\dots ,v_n$ is any basis, its coordinate linear functionals form a basis of $V^{\ast }$: a linear functional is uniquely specified by its values on these $n$ vectors. Hence $\dim V^{\ast }=n$. The linear mapis injective, because $J(w)=0$ implies $⟨w,w⟩=0$, and positive definiteness implies $w=0$. Equal dimensions make $J$ an linear isomorphism. If ${\alpha }^{\prime }$ is the dual map, thenThis establishes existence and linearity. If two possible vectors $u,w$ represent the same linear functional through the inner product, then $J(u-w)=0$, so $u=w$. Applied to every argument, this proves uniqueness of the adjoint on an inner product space.

For linear maps $S,T$, two applications of the defining inner product identity giveUniqueness therefore gives $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$. Likewise $(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$, $(cT{)}^{\ast }=c{T}^{\ast }$ for real $c$, and $I^{\ast }=I$. An induction gives $({\alpha }^n{)}^{\ast }=({\alpha }^{\ast }{)}^n$, and applying additivity to a polynomial with real coefficients givesSymmetry of the real inner product also gives $({\alpha }^{\ast }{)}^{\ast }=\alpha$. If $m$ is the minimal polynomial of $\alpha$, the displayed identity gives $m({\alpha }^{\ast })=0$, so the minimal polynomial of ${\alpha }^{\ast }$ divides $m$. Applying the same argument to ${\alpha }^{\ast }$ gives reverse divisibility. Both polynomials are monic, hence their minimal polynomials are equal.

The distinct eigenvalues are exactly the roots of the minimal polynomial in the chosen field. An eigenvector $v$ with eigenvalue $\lambda$ satisfies $m(\alpha )v=m(\lambda )v$, so $m(\lambda )=0$. Conversely, if $m(X)=(X-\lambda )q(X)$ and $\alpha -\lambda I$ were a linear isomorphism, then $m(\alpha )=0$ would imply $q(\alpha )=0$, contradicting minimal degree. Thus $\alpha -\lambda I$ is not injective, giving an eigenvector. It follows that $\alpha$ and ${\alpha }^{\ast }$ have the same distinct eigenvalues. This includes the complex eigenvalues after complexification of a real vector space; over the real field it states equality of their real eigenvalues.

A maximal ideal is a proper ideal $I$ of $R$ such that an ideal $J$ with $I\subseteq J\subseteq R$ is either $I$ or $R$. Suppose first that $I$ is a maximal ideal. If $a\notin I$, then $I+(a)=R$, so $1=i+ra$ for some $i\in I$ and $r\in R$. In the quotient ring, $(r+I)(a+I)=1+I$. Every nonzero element of $R/I$ therefore has a multiplicative inverse, so $R/I$ is a field. Conversely, if $R/I$ is a field and $J$ properly contains $I$, choose $a\in J\setminus I$. The element $a+I$ has a multiplicative inverse $r+I$, whence $1-ra\in I\subset J$, giving $1\in J$ and $J=R$. ThusThe squares in ${\mathbb{F}}_7$ are $0,1,2,4$, which do not include $-1=6$. Hence $X^2+1$ has no root in ${\mathbb{F}}_7$. A polynomial of degree two or three over a field is an irreducible polynomial exactly when it has no root: any proper polynomial factorization of such a degree has a factor of degree one. The polynomial ring over a field is a Euclidean domain, so the ideal generated by an irreducible polynomial is a maximal ideal. It follows thatis a field. Every coset has a unique representative $a+bX$, by polynomial division, so $|F|=7^2=49$.

Let $j$ be the coset of $X$ in this field. Since $j^2=-1$, the mapis a ring homomorphism. It is surjective by the representative description. Since $1,j$ are linearly independent over ${\mathbb{F}}_7$, its kernel of a ring homomorphism isThere is also a field with $343$ elements. The cubes in ${\mathbb{F}}_7$ are $0,1,-1$, so $X^3-2$ has no root and is an irreducible polynomial. Thereforeis a field whose unique representatives have degree below three. There are $7^3=343$ such representatives.

Use the following boundary version of Cauchy's integral theorem: if a function is holomorphic on a neighborhood of a compact region bounded by finitely many piecewise smooth simple closed curves, its integral around the positively oriented boundary is zero. On an annulus this means that its integrals around the two circles, both taken counterclockwise, are equal.

For $0<\rho <r$, the function $f(z)/(z-w)$ is holomorphic on a neighborhood of $\rho \le |z-w|\le r$. The theorem therefore givesAs $\rho \downarrow 0$, continuity of $f$ makes the integrand converge uniformly to $f(w)$. The right side tends to $f(w)$, proving Cauchy integral formula without assuming that formula in an intermediate step.

Let $I$ denote the requested boundary integral. Parametrize the unit circle by $z=e^{it}$. Since $dz=i{e}^{it}dt$,Taking complex conjugates and using $\overline{e^{it}}=e^{-it}$ givesThis is the conjugated Cauchy integral on a circle identity. The function $h(z)=f(z)/(1-\overline{a}z)$ is holomorphic on a neighborhood of the closed unit disc, because $|a|<1$. Applying Cauchy integral formula at zero gives $\overline{I}=h(0)=f(0)$. ThusThe complex conjugates in these formulas are essential; the boundary integrand contains $\overline{f(z)}$.

For a holomorphic function on $|z-a|<R$, the complex Taylor theorem stateswith uniform convergence on each smaller closed disc. Here is a proof. Choose $r<\rho <R$. For $|z-a|\le r$ and $|\zeta -a|=\rho$, the geometric seriesconverges uniformly. Substitute it into Cauchy integral formula and integrate term by term to getIf $M_{\rho }={\max }_{|\zeta -a|=\rho }|f(\zeta )|$, then $|c_n|\le M_{\rho }{\rho }^{-n}$. These bounds justify differentiating the power series any fixed number of times on a smaller closed disc, by comparison with convergent differentiated geometric series. Evaluation at $a$ gives $f^{(n)}(a)=n!c_n$. Since every $|z-a|<R$ belongs to some such smaller disc, the formula holds throughout the stated domain.

The identity theorem says that two holomorphic functions on a connected open subset of $\mathbb{C}$ are equal everywhere if their equality set has an accumulation point in that subset. Equivalently, a holomorphic function with an interior accumulation point of zeros is identically zero there.

The proposed Taylor coefficients at $i$, including the specified value for $n=0$, giveThe function $1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ and has exactly the given derivatives, so it supplies existence. Any other admissible holomorphic function has the same Taylor series on $|z-i|<1$. The punctured plane is connected, so the identity theorem gives uniqueness:

On the unit circle, put $z=e^{i\theta }$. Then $d\theta =dz/(iz)$ and $\cos \theta =(z+z^{-1})/2$, soWrite $r=2-\sqrt{3}$ and $s=2+\sqrt{3}$. Only $r$ lies inside the unit circle; the rational function $h(z)=z/[(z-r{)}^2(z-s{)}^2]$ has a double pole there. Its residue isBy the residue theorem,

Choose the branch $\mathrm{Log}z=\log |z|+i\arg z$ with $-\pi /2<\arg z<3\pi /2$. It is holomorphic off the downward imaginary ray and takes values $\log x$ on the positive real axis and $\log x+i\pi$ at $-x$ for $x>0$.

Take $R>2$ and $0<\epsilon <2$. The positively oriented contour consists of the real segments $[-R,-\epsilon ]$ and $[\epsilon ,R]$, the small upper semicircle from $-\epsilon$ to $\epsilon$ traversed clockwise, and the large upper semicircle from $R$ to $-R$ traversed counterclockwise. For $F(z)=\mathrm{Log}z/(4+z^2)$, the only enclosed pole is $2i$, withThe integral over the large arc is $O(\log R/R)$, since its length is $\pi R$ and $|4+z^2|\ge R^2-4$. The small arc contributes $O(\epsilon |\log \epsilon |)$, since its length is $\pi \epsilon$ and $|4+z^2|\ge 4-{\epsilon }^2$. Both tend to zero.

Let $J={\int }_0^{\infty }\log x/(4+x^2)dx$; it converges because $|\log x|$ is integrable near zero and $\log x/x^2$ is integrable at infinity. The two real segments, after substituting $z=-x$ on the negative one, tend toThe residue theorem therefore givesTaking real parts yields

Use Picard iteration on $[a-h,a+h]$:By induction, every iterate stays in the prescribed rectangle: if $u_n$ does, then $|u_{n+1}(x)-b|\le M|x-a|\le Mh\le k$. Continuity makes each next integral well defined. The Lipschitz condition gives, for $n\ge 0$,For $n=0$ this is the previous bound; integration of the bound multiplied by $A$ proves the next step, on either side of $a$.

The sum of these bounds converges uniformly on the entire interval. Thus $u_n$ converges uniformly to a continuous $u$, whose graph remains in the closed rectangle. The Lipschitz condition also makes $f(t,u_n(t))$ converge uniformly to $f(t,u(t))$, so passage to the limit givesThe fundamental theorem of calculus makes $u$ a continuously differentiable solution with $u(a)=b$.

For uniqueness, let $u,v$ be two solutions and put $D_0=\sup |u-v|$ on the interval. Repeated use of the Lipschitz condition in their integral equations givesfor every $n$. Letting $n\to \infty$ gives $u=v$. The global bound $M$ prevents any competing solution whose graph lies in $D$ from exiting the rectangle through a horizontal boundary before the end of the interval. Therefore a unique solution exists on the full stated interval. No extra restriction $Ah<1$ is needed; the factorial estimates replace that restriction.

At zero the right side is continuous but fails every local Lipschitz condition in its second variable, since ${\delta }^{1/4}/\delta ={\delta }^{-3/4}\to \infty$. For every $c\ge 0$, the functionis continuously differentiable and satisfies the differential equation, including at the joining point where both one-sided derivatives are zero. Each has $u_c(0)=0$. The identically zero function is another possibility. Thus solutions exist but are not unique, even locally: choosing $c=0$ and taking the zero function already gives two different solutions on every neighborhood of zero. These are instances of waiting-time solutions of a sublinear Bernoulli equation.

Near $y=1$, the right side is continuously differentiable in $y$, with derivative $1/(4y^{3/4})$ bounded on, for example, $1/2\le y\le 3/2$. It therefore satisfies a local Lipschitz condition; it is also bounded on an appropriate bounded neighborhood. The existence and uniqueness result applies there. Separation of variables givesThe unique extension to the whole real axis is $u(x)=[\max \{1+3x/4,0\}{]}^{4/3}$. It joins zero with zero derivative at $x=-4/3$. To see uniqueness of this extension, $u^{\prime }\ge 0$ makes every solution nondecreasing. Before it reaches zero it cannot be positive, while a negative interval would have $u^{\prime }=0$ and could not continuously reach zero. Thus it must be zero for $x\le -4/3$; the positive part is fixed by the separated formula.

Translation in $x$ leaves this autonomous differential equation unchanged. For every $c\ge 1$, the same formulasatisfies $u_c(1)=0$, and the zero function does too. Thus solutions exist and are not unique, exactly as for the initial point $(0,0)$. The relevant failure is the Lipschitz condition at $y=0$; moving the initial value of $x$ does not remove it.

First consider the usual local reduction with sufficiently smooth coefficients. The hyperbolic partial differential equation has two distinct characteristic directions because $S^2-RT>0$. Choose independent local characteristic coordinates $\xi ,\eta$ whose gradients satisfyLet $J={\xi }_x{\eta }_y-{\xi }_y{\eta }_x\ne 0$ andThe determinant of the transformed principal coefficient matrix gives $B^2=J^2(S^2-RT)>0$. By the chain rule, the left side of the original partial differential equation becomeswith the analogous expression for $A_{\eta }$. The original first derivatives are $z_x=u_{\xi }{\xi }_x+u_{\eta }{\eta }_x$ and $z_y=u_{\xi }{\xi }_y+u_{\eta }{\eta }_y$. Substitution and division by $2B$ yield the requested mixed-derivative canonical form. The construction is local, in a neighborhood where the two characteristic foliations give a diffeomorphism.

There is a regularity qualification in the printed hypotheses: merely continuous coefficients need not give twice continuously differentiable characteristic coordinates. Such coordinates are needed for the second-derivative chain rule above. For example, take $R=0$, $S=1/2$ and $T=|x|$ near the origin. The characteristic equation factors as ${\xi }_y({\xi }_x+|x|{\xi }_y)=0$. One family is $x=\text{constant}$ and the other has first integrals $\eta =h(y-x|x|/2)$. For an independent coordinate $h^{\prime }\ne 0$, but ${\eta }_{xx}$ has opposite one-sided limits at $x=0$. Hence no twice continuously differentiable characteristic coordinate system exists there. This illustrates that continuous hyperbolic coefficients need not admit smooth characteristic coordinates. The formal general argument requires additional smoothness, for example twice continuously differentiable coefficients. The concrete coefficients below are smooth, so this issue does not affect their solution.

For the concrete partial differential equation, putThe Jacobian determinant is $1/y^2\ne 0$. Direct use of the chain rule givesThereforeTo impose the data efficiently, let $v=z_x$. Then $2x{v}_x+y{v}_y=0$, whose characteristic curves preserve $x/y^2$, so $v=H(x/y^2)$. On the initial line $x=1$, the value $v(1,y)=y^2$ gives $H(\xi )=1/\xi$ for every $\xi >0$. Thus $z_x=y^2/x$, and integration in $x$ gives $z=y^2\log x+C(y)$. The remaining initial condition gives $C(y)=0$:Every point in this region is reached from the data by a characteristic with positive $x/y^2$. The line $x=0$ separates these characteristics from those with $x<0$, and the logarithmic divergence prevents a smooth extension of this solution through $x=0$. The original condition restricts us to $y>0$.

A sufficient condition for existence of the Laplace transform is that $y$ be piecewise continuous on each bounded interval and have exponential order: $|y(t)|\le C{e}^{at}$ for sufficiently large $t$. The defining integral then converges absolutely for real $p>a$, and more generally defines a holomorphic function for complex $p$ with $\mathrm{Re}p>a$. The boundary $p=a$ may or may not be included, depending on $y$.

Write $Y(p)=\mathcal{L}[y](p)$. To solve the differential equation by Laplace transform, initially seek a sufficiently regular solution with $y(0)=1$ and finite $y^{\prime }(0)=b$. The Laplace transform of a derivative givesUsing $\mathcal{L}[ty]=-Y^{\prime }$ from part (i), the transformed differential equation becomesorAfter division by $p(p+1)$, an integrating factor is $p^2(p+1)$, since $(3p+2)/(p(p+1))=2/p+1/(p+1)$. ConsequentlyFor a bounded $y$, $|Y(p)|\le \Vert y{\Vert }_{\infty }/p$ for $p>0$. Letting $p\downarrow 0$ forces $C=0$, an instance of how boundedness selects a Laplace transform at a singular endpoint. By the exponential formula in part (iii), this yieldsDirect substitution checks the differential equation, boundedness and the required limit.

To complete the uniqueness argument, no unproved endpoint regularity is needed. The two functions $e^{-t}$ and $t-1$ solve the homogeneous linear differential equation on $(0,\infty )$, and their Wronskian is $t{e}^{-t}\ne 0$. They form a fundamental system of solutions on that interval, so every classical solution is $A{e}^{-t}+B(t-1)$. Boundedness forces $B=0$ and the limit at zero forces $A=1$. Thus the bounded solution with the specified limit is unique. The limit alone would not suffice: $e^{-t}+C(t-1+e^{-t})$ has the same limit for every $C$, but is unbounded when $C\ne 0$.

For $\mathrm{Re}p>a$, differentiation under the integral is justified by the dominated convergence theorem: on a smaller closed half-plane $\mathrm{Re}p\ge a+\delta$, the additional factor $t$ is controlled by $t{e}^{-\delta t}$. HenceThis is the multiplication rule for the Laplace transform.

For $\mathrm{Re}p>0$, the starting integral is ${\int }_0^{\infty }{e}^{-pt}dt=1/p$. Repeated integration by parts, with vanishing boundary terms, givesThis also follows from the gamma function. The stated open half-plane ensures absolute convergence.

Integrating the exponential directly givesThe inequality is the absolute-convergence domain of this Laplace transform.