Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2005 (AC2, XMAT 4302, XMST 4302)

Expanding gives $(I-E{)}^2-(I-E)=E^2-E$, so either operator is idempotent exactly when the other is. For an idempotent $E$,If $w\in \ker E\cap \mathrm{Im}E$, write $w=Eu$; then $0=Ew=E^{2}u=Eu=w$. Hence $V=\ker E\oplus \mathrm{Im}E$.

Next,If the right side vanishes, left multiplication by $E_1$ gives $E_1{E}_2+E_1{E}_2{E}_1=0$, while right multiplication gives $E_1{E}_2{E}_1+E_2{E}_1=0$. Subtraction shows $E_1{E}_2=E_2{E}_1$, and their original sum now gives $2E_1{E}_2=0$. Over the real or complex numbers this implies $E_1{E}_2=E_2{E}_1=0$.

For three idempotents summing to $I$, each pair sum equals the complementary idempotent $I-E_k$. Thus every product of distinct operators vanishes. Every vector equals $E_{1}v+E_{2}v+E_{3}v$. If $v_1+v_2+v_3=0$ with $v_i\in \mathrm{Im}{E}_i$, applying $E_j$ gives $v_j=0$. ThereforeTake a basis of each summand and unite the three bases. A vector in the $j$th basis has eigenvalue one for $E_j$ and zero for the other two, so this is the requested common eigenbasis. Zero summands simply contribute no basis vectors.

A real inner product is a symmetric bilinear map $⟨\cdot ,\cdot ⟩:V\times V\to \mathbb{R}$ with $⟨v,v⟩>0$ for $v\ne 0$. It defines $\Vert v\Vert =\sqrt{⟨v,v⟩}$. Set $W=\mathrm{span}\{e_1,\dots ,e_k\}$ and $Pv={\sum }_i⟨v,e_i⟩e_i$. Orthonormality gives $v-Pv\perp W$, hence the Pythagorean identity givesThis proves Bessel's inequality. Equality holds exactly when $v\in W$.

If $v=0$, equality is immediate. If $v\ne 0$, the nonnegative squared normproves the Cauchy-Schwarz inequality. Equality holds precisely when the displayed vector vanishes. Thus equality is equivalent to linear dependence of $u,v$, including the cases with a zero vector.

The sum in question is $⟨Pu,Pv⟩$. Applying Cauchy-Schwarz and then Bessel's inequality givesIf either original vector is zero, equality holds. If both are nonzero, equality forces both projection norms to equal their original norms and also forces dependence of the projected vectors. Thus apart from the zero-vector cases, equality holds exactly when $u,v\in W$ and $u,v$ are linearly dependent.

For the final unnumbered request, apply Cauchy-Schwarz in ${\mathbb{R}}^k$ to $p_i=a^{i/2}$ and $q_i=a^{-i/2}$. Their inner product is $k$, soEquality requires the positive vectors $p,q$ to be proportional, equivalently $a^i$ constant for $i=1,\dots ,k$. For $k\ge 2$ this is $a=1$. If $k=1$, equality holds for every $a>0$.

An integral domain is a nonzero commutative ring with identity and no zero divisors. A principal ideal domain is an integral domain in which every ideal is generated by one element.

The leading coefficient of a product of nonzero real polynomials is nonzero, so $\mathbb{R}[x]$ is an integral domain. The zero ideal is principal. In any nonzero ideal $J$, choose a nonzero polynomial $g$ of least degree. Divide an arbitrary $f\in J$ by $g$: $f=qg+r$ with $r=0$ or $\mathrm{deg}r<\mathrm{deg}g$. Since $r\in J$, minimality forces $r=0$. Thus $J=(g)$ and $\mathbb{R}[x]$ is a principal ideal domain.

In $\mathbb{Z}[x]$ consider $J=(2,x)$. If $J=(h)$, then $h$ divides the constant polynomial 2, so $h$ is constant, and divisibility of $x$ forces $h=\pm 1$. That would give $J=\mathbb{Z}[x]$. But the constant coefficient of every element $2p+xq$ is even, so $1\notin J$. This contradiction proves $\mathbb{Z}[x]$ is not a principal ideal domain.

Let $J={\bigcup }_n{I}_n$. It contains zero. If $a,b\in J$, both belong to $I_N$ for some common $N$, so $a-b\in I_N\subseteq J$. If $r\in R$, then $ra\in I_N\subseteq J$. These are the ideal axioms, proving $J$ is an ideal.

In a principal ideal domain write $J=(g)$. Since $g\in J$, there is $n_0$ with $g\in I_{n_0}$. For $n\ge n_0$,All inclusions are equalities. Hence the chain stabilizes at $I_{n_0}$. This includes $J=(0)$ and is the ascending chain condition for ideals.

If $f$ is holomorphic on a neighborhood of the closed disc, the differentiated Cauchy integral formula iswith positive orientation. On the circle the integrand has modulus at most $M/r^{k+1}$ and the contour has length $2\pi r$. Thus

Take a circle of radius $R\ge S$ about zero. The Cauchy estimate gives, for each integer $k>n$,Therefore all these derivatives vanish. The Taylor series of an entire function converges to it everywhere, and henceAs usual $n$ is a nonnegative integer; if a real exponent is allowed, the degree is at most its integer part. A negative growth exponent forces $f=0$.

For a bounded entire function, the preceding result with $n=0$ gives Liouville's theorem. If a nonconstant polynomial $p$ had no complex root, $1/p$ would be entire. The leading term gives $|p(z)|\to \infty$ at infinity, so $1/p$ is bounded outside a large disc and is also bounded on the remaining compact disc. Liouville's theorem would make $1/p$, and hence $p$, constant. This proves the fundamental theorem of algebra.

Finally, away from the zeros of $\cos z$, the holomorphic quotient $h=f/\cos z$ has modulus at most $C$. At each zero it therefore has a removable singularity. These isolated points have no finite accumulation, so removing all of them gives a bounded entire function. Liouville's theorem gives a constant $c$ with $|c|\le C$, andConversely every such multiple satisfies the assumed bound.

Integrate $e^{iz}/z$ around the upper half-disc with the upper half-disc of radius $\epsilon$ about zero removed. The real segments run from $-R$ to $-\epsilon$ and $\epsilon$ to $R$; the small semicircle is clockwise. There are no poles inside this indented contour.

The large arc contribution tends to zero: its modulus is at most ${\int }_0^{\pi }{e}^{-R\sin \theta }d\theta$, which tends to zero by dominated convergence. On the small arc $z=\epsilon e^{i\theta }$, $\theta$ decreases from $\pi$ to zero, so its integral tends to ${\int }_{\pi }^{0}id\theta =-i\pi$. The contour theorem consequently givesTaking imaginary parts is legitimate for each truncated integral. The imaginary integrand $\sin x/x$ is continuous at zero after assigning its limiting value one, so the inner indentation can be removed. Therefore

Let $\omega =e^{2\pi i/3}$ and integrate $1/(1+z^3)$ around the positively oriented sector from argument zero to $2\pi /3$, with outer radius $R>1$. The single pole inside is $z_0=e^{i\pi /3}$, with residue $1/(3z_0^2)=1/(3\omega )$. The outward real radial segment contributes $I_R={\int }_0^{R}dx/(1+x^3)$; the return radial segment contributes $-\omega I_R$ because $(\omega x{)}^3=x^3$.

On the circular arc, $|1+z^3|\ge R^3-1$, so its integral has modulus at most $(2\pi R/3)/(R^3-1)\to 0$. Thus the residue theorem yieldsSince $\omega (1-\omega )=\omega -{\omega }^2=i\sqrt{3}$,The ordinary improper integral converges at both endpoints, so no principal-value interpretation is needed here.

For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, a singular point at zero is regular when $xp(x)$ and $x^{2}q(x)$ extend analytically to zero. Hereso these products are analytic at zero. The equation remains singular there because $q$ has a pole.

Substitute $y={\sum }_{n\ge 0}{a}_n{x}^{n+\sigma }$. The lowest power gives the indicial equation $\sigma (\sigma -1+\lambda )=0$. Comparing the next powers givesFor $0<\lambda <1$ the denominators do not vanish. With $\sigma =0$, $a_1=-a_0/\lambda$ and $a_2=0$, soWith $\sigma =1-\lambda$, the recurrence becomesConsequently $y_2=a_0{x}^{1-\lambda }(1-x{)}^{\lambda }$ on $0<x<1$, with the binomial series for its last factor. The nonzero coefficient ratios tend to one, giving radius one for that power-series factor. The factor $x^{1-\lambda }$ specifies the Frobenius branch; the different leading powers prove independence.

When $\lambda =1$, both indicial roots become zero and the analytic Frobenius solution in both cases is a multiple of $1-x$. To obtain the missing solution use reduction of order. Away from zero and one, $p=1/x$, henceThus the real general solution on any interval avoiding the singular points isIts Wronskian with $1-x$ is $1/x\ne 0$, proving independence. For complex solutions choose a logarithm branch on a suitable simply connected region.

On $[-a,a]$ let $F=\Vert f{\Vert }_{\infty }$ and bound $|K|$ by $M$ on the compact integration region. Write $Th(x)={\int }_0^{x}K(x,s)h(s)ds$. The iteration gives $y_n={\sum }_{j=0}^n{T}^{j}f$. Induction, integrating absolute values also when $x<0$, givesThe exponential series is summable uniformly for $|x|\le a$. The Weierstrass M-test therefore proves uniform convergence to a continuous function $y$. Also $\Vert T(y_n-y){\Vert }_{\infty }\le Ma\Vert y_n-y{\Vert }_{\infty }\to 0$, so passing to the limit yields $y=f+Ty$.

For the indicated kernel, twice differentiating its integral givesConversely integrating the differential equation twice, with these initial conditions, gives the integral equation. This is valid also for negative $x$, either by oriented integration or by differentiating the double-integral identity.

For every finite $a$ the continuous kernel $(x-s)s$ meets the preceding hypotheses. Solutions on overlapping intervals agree: if $d=Td$ is their difference, iteration gives $\Vert d{\Vert }_{\infty }\le \Vert d{\Vert }_{\infty }(Ma{)}^j/j!$ for every $j$, forcing $d=0$. The solutions therefore patch to one solution defined on all of $\mathbb{R}$.

PutLeibniz differentiation gives $z^{\prime }=-{\int }_0^x\cos (x-s)f(s)ds$ and $z^{\prime \prime }=-f(x)+{\int }_0^x\sin (x-s)f(s)ds=-f-z$. Thus $z$ satisfies both zero initial conditions and the inhomogeneous equation. Every solution with $y(0)=0$ is $y=A\sin x+z(x)$.

On $[0,1]$, the second endpoint determines $A=-z(1)/\sin 1$ uniquely, since $\sin 1\ne 0$. Collecting the integrals, and using the sine addition formula for $s<x$, givesThis continuous kernel vanishes at the endpoints, satisfies $G_{xx}+G=0$ off $x=s$, and has derivative jump $-1$, as required for $-y^{\prime \prime }-y=f$. The preceding construction proves existence, and a homogeneous solution with both endpoints zero must vanish, proving uniqueness.

On $[0,\pi ]$, instead, the second boundary value becomesThus solvability is equivalent to ${\int }_0^{\pi }f(s)\sin sds=0$. When it holds, all solutions areThere is one free homogeneous mode because the endpoint length is resonant.

For a nonzero principal part the discriminant $\Delta =b^2-ac$ classifies the equation: it is hyperbolic for $\Delta >0$, parabolic for $\Delta =0$, and elliptic for $\Delta <0$. These are the three definitions requested by the inline roman labels.

In the hyperbolic case the characteristic equation $a(dy{)}^2-2bdxdy+c(dx{)}^2=0$ gives two distinct real tangent directions. Locally, with suitably smooth coefficients, solve the corresponding characteristic ordinary differential equations to obtain two independent families of curves. Label their level sets $\varphi =\text{constant}$ and $\psi =\text{constant}$. Their gradients satisfyand the coordinate Jacobian is nonzero. The chain rule therefore eliminates $u_{\varphi \varphi }$ and $u_{\psi \psi }$. The remaining mixed coefficient $2[a{\varphi }_x{\psi }_x+b({\varphi }_x{\psi }_y+{\varphi }_y{\psi }_x)+c{\varphi }_y{\psi }_y]$ is nonzero: otherwise the invertible coordinate change would send the nondegenerate principal quadratic form to zero. Divide by this coefficient and absorb all first-order terms into $g$, giving $u_{\varphi \psi }=g(\varphi ,\psi ,u,u_{\varphi },u_{\psi })$. This canonical reduction is local; a global chart is not automatic.

For the specific equation $a=y,b=0,c=-1$, so it is hyperbolic exactly on $y>0$, elliptic for $y<0$, and degenerate parabolic on $y=0$. For $\varphi =3x-2y^{3/2}$ and $\psi =3x+2y^{3/2}$, the Jacobian is $18\sqrt{y}\ne 0$. Direct differentiation givesSince $\psi -\varphi =4y^{3/2}$, the canonical form isThe characteristic families are $3x-2y^{3/2}=c$ and $3x+2y^{3/2}=c$, with slopes $dx/dy=\pm \sqrt{y}$. They are distinct for $y>0$ and meet the degeneracy line with coincident vertical tangent directions.