Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2004 (AC1, XMAT 4301, XMST 4301)

${\dim }_{\mathbb{R}}V=4$. For arbitrary matrices,This is bilinear and symmetric. Also $⟨A\mid A⟩={\sum }_{i,j}{a}_{ij}^2\ge 0$, with equality exactly when every entry vanishes. Thus it is an inner product, the Frobenius inner product.

If $S^t=S$ and $K^t=-K$, invariance of trace under transposition and cyclicity giveFurthermore $⟨I\mid A⟩=\mathrm{tr}A$, so every traceless matrix is orthogonal to the identity. One convenient orthonormal basis adapted to these observations isTheir entrywise inner products are zero in distinct pairs and one on the diagonal; four orthonormal vectors in this four-dimensional space form a basis.

Self-adjointness means $⟨Tv,w⟩=⟨v,Tw⟩$ for all $v,w\in V$, or $T=T^{\ast }$. Take the inner product linear in its first variable. If $Tv=\lambda v$ with $v\ne 0$, thenTherefore every eigenvalue is real.

If $U$ is invariant and $w\in U^{\perp }$, then for every $u\in U$, $Tu\in U$ and $⟨Tw,u⟩=⟨w,Tu⟩=0$. Hence $Tw\in U^{\perp }$, proving invariance of the orthogonal complement.

Induct on $d=\dim V$. The zero-dimensional case is empty. For $d>0$, choose an eigenvector and normalize it to a unit vector $e$. Its span is invariant, so its orthogonal complement is invariant as just proved. The restriction of $T$ to that complement is again self-adjoint. Its dimension is $d-1$, so induction supplies an orthonormal basis of eigenvectors there. Adjoining $e$ gives an orthonormal eigenbasis of $V$. This proves the finite-dimensional spectral theorem without assuming it in the argument.

A ring is a set with an abelian-group addition, associative multiplication and both distributive laws; here we use rings with multiplicative identity. A field is a nonzero commutative ring in which every nonzero element has a multiplicative inverse.

The positive integers form neither a ring nor a field. They do not contain an additive identity zero, and they also lack additive inverses.

$\mathbb{Z}/4\mathbb{Z}$ is a ring but not a field. Quotient operations inherit associativity and distributivity from the integers, and residues have additive inverses. However the nonzero residue $\overline{2}$ satisfies $\overline{2}\overline{2}=\overline{0}$, so it cannot be invertible. A field has no nonzero zero divisors.

$\mathbb{Z}/3\mathbb{Z}$ is a field, and consequently also a ring. Its nonzero elements are $\overline{1}$ and $\overline{2}$, with inverses $\overline{1}$ and $\overline{2}$ respectively. The quotient ring axioms are inherited as in the preceding part.

This is the Gaussian integer ring $\mathbb{Z}[i]$. It contains zero and one and is closed under addition, additive inverses and multiplication, sinceIts operations inherit the ring axioms from $\mathbb{C}$. It is a ring but not a field: the nonzero element 2 has complex inverse $1/2$, which is not a Gaussian integer.

Writing $z=x+iy$ and $f=u+iv$, the Cauchy-Riemann equations areWith continuous second derivatives, mixed derivatives commute. ThusBoth real and imaginary parts are harmonic.

For $g(z)=\mathrm{Re}z$, the real and imaginary parts are $x$ and zero, so $u_x=1\ne 0=v_y$. For $h(z)=\bar{z}$, they are $x$ and $-y$, so $u_x=1\ne -1=v_y$. The necessary Cauchy-Riemann equations fail at every point. Hence both functions are nowhere holomorphic, indeed nowhere complex differentiable.

A removable singularity at $a$ means that there is a holomorphic function on the whole disc which agrees with $f$ away from $a$. Equivalently, the Laurent series has no negative powers. The value at $a$ is filled in by its finite limit.

A pole of order $m\ge 1$ means that $(z-a{)}^{m}f(z)$ extends holomorphically to $a$ and has a nonzero value there. Equivalently, the lowest nonzero power in the Laurent expansion is $(z-a{)}^{-m}$.

An essential singularity is an isolated singularity which is neither removable nor a pole; equivalently its Laurent expansion has infinitely many nonzero negative-power terms.

For the final unnumbered classification, the only possible finite singularities are the zeros of the denominator, namely $z=\pm i$ and $z=k\pi$ for integers $k$. At zero,so zero is removable, with extension value one. At each nonzero $k\pi$, $\sin z$ has a simple zero, the other denominator factor is nonzero and the numerator is nonzero. These points are poles of order two, with leading coefficient $(k\pi {)}^2/((k\pi {)}^2+1)$.

At $z=\pm i$, the polynomial denominator has a simple zero. Since $\sin (\pm i)=\pm i\sinh 1\ne 0$ and the numerator is $-1$, these are simple poles. There are no essential singularities on $\mathbb{C}$ and no other finite singularities.

After division by $z(z+2)$, the coefficients areTheir only finite singularities are $z=0$ and $z=-2$. At either point $z_0$, the products $(z-z_0)p$ and $(z-z_0{)}^{2}q$ are analytic, so both points are regular singular points. All other finite points are ordinary.

The lowest power after substituting the Frobenius series gives $\sigma (2\sigma -1)=0$. The coefficient of $z^{n+\sigma }$ then givesFor both exponents the denominator is nonzero for every $n\ge 0$, so each arbitrary nonzero $a_0$ determines a series. Unless it terminates, its coefficient ratio tends to $-1/2$, proving convergence for $|z|<2$. For the half-integer exponent choose a square-root branch. Their leading powers $1$ and $z^{1/2}$ show that the two local solutions are linearly independent.

If $\alpha =m$ is a positive integer, choose $\sigma =0$. The recurrence numerator is nonzero for $0\le n<m$ and vanishes at $n=m$. Thus $a_m\ne 0$, $a_{m+1}=0$, and all later coefficients vanish: a polynomial solution of exact degree $m$ exists. It is proportional to the Chebyshev polynomial $T_m(z+1)$, since the substitution $t=z+1$ transforms the equation into $(t^2-1)y^{\prime \prime }+t{y}^{\prime }-m^{2}y=0$.

The first iterate isOn the prescribed box, $F(x,y)=1+x^2+y^2$ is continuous, $|F|\le 102$, and $|F(x,y)-F(x,v)|\le 20|y-v|$. The initial value is distance nine from the nearer horizontal edge. Thus on $|x|\le h=9/102=3/34$, the Picard map keeps functions within the box: its values differ from one by at most $102h=9$.

Successive differences satisfy, by induction,The Weierstrass M-test proves uniform convergence on this local interval. The uniform limit is continuous, and the Lipschitz estimate allows passage to the limit inside the integral, giving the displayed integral equation. Differentiation gives $y^{\prime }=1+x^2+y^2$, $y(0)=1$. The same local Lipschitz bound proves local uniqueness.

Direct differentiation gives $u^{\prime }={\sec }^2(x+\pi /4)=1+u^2$, $u(0)=1$, and $u\to +\infty$ as $x\uparrow \pi /4$. On the common positive existence interval put $w=y-u$. It satisfies $w^{\prime }=(y+u)w+x^2$, $w(0)=0$, so the integrating-factor formula givesThus $y\ge u$. Let $x_{\ast }>0$ be the maximal forward existence endpoint. The solution is strictly increasing because $y^{\prime }\ge 1$. If $x_{\ast }$ were finite and $y$ remained bounded, its finite limit and the smooth right side would allow local continuation, contradicting maximality. Comparison with $u$ prevents existence through $\pi /4$. ConsequentlyThere is no contradiction with Picard's theorem, which gives local existence and uniqueness. Its box estimate gives only a short guaranteed interval, and the Lipschitz constant for $y^2$ is not globally bounded in $y$.

First solve for $v=y^{\prime }$. Multiplication by $e^x$ and use of $v(1)=0$ giveIntegrating from zero to $x$, and using $y(0)=0$, yields $y(x)=-{\int }_0^x{e}^{-s}{\int }_s^1{e}^{t}f(t)dtds$. Reversing the order of integration givesThus $G(x,t)$ equals $G_1$ for $t<x$ and $G_2$ for $x<t$. It is continuous at $x=t$, vanishes at $x=0$, and has $G_x(1,t)=0$. Its derivative jumps from $-1$ to zero across $x=t$, so the jump is $+1$, matching the coefficient of $y^{\prime \prime }$. Both pieces satisfy $G_{xx}+G_x=0$ away from the jump. These facts also verify the Green representation directly. The homogeneous solutions are $A+B{e}^{-x}$, and the two boundary conditions force $A=B=0$, proving uniqueness.

For $f(t)=e^{-t}$, the formula givesHence $y(x)=-x{e}^{-x}$. Indeed $y^{\prime }=(x-1)e^{-x}$ and $y^{\prime \prime }=(2-x)e^{-x}$, which verify the equation and both boundary conditions.

This is the heat equation, with $y$ as time. The stated zero solution is unique. Define $E(y)=\frac{1}{2}{\int }_0^{1}u(x,y{)}^{2}dx$. For $0<y<1$, integration by parts and $u_y=u_{xx}$ giveThe boundary term is zero because $u(0,y)=0$ and $u_x(1,y)=0$. Also $E(0)=0$. Nonnegativity forces $E(y)=0$, and continuity gives $u=0$ throughout the square. The unknown final-time boundary values are determined by evolution and need not be prescribed. Under the stated regularity the integrations can first be made on an interior interval and then passed to the endpoints using continuity of $u,u_x,u_y$.

This is Laplace's equation. Zero is not unique: with $k=\pi /2$,is nonzero and harmonic. It vanishes on $y=0$ and $x=0$, while $u_x(1,y)=k\cos (k)\sinh (ky)=0$.

A sufficient additional condition is $u(x,1)=0$ on the top edge. Green's first identity then givesOn the bottom, left and top edges $u=0$; on the right edge the normal derivative is zero. Thus the gradient vanishes and $u$ is constant; the bottom value makes that constant zero. For $C^2(D)\cap C^1(\overline{D})$, apply the identity on inset rectangles and pass to the boundary using continuity of the function and its gradient.

This is the wave equation, again taking $y$ as time. Zero is not unique: for $k=\pi /2$,satisfies $u_{xx}=u_{yy}$ and all the stated boundary conditions. The initial displacement is zero but the initial velocity is not prescribed.

A sufficient extra condition is $u_y(x,0)=0$. For the resulting homogeneous data use wave energyThe equation and integration by parts give $E^{\prime }(y)=[u_x{u}_y{]}_0^1=0$. At $x=1$ the factor $u_x$ vanishes, and at $x=0$ the factor $u_y$ vanishes because $u(0,y)=0$. The initial displacement implies $u_x(x,0)=0$, so the extra initial velocity condition gives $E(0)=0$. Consequently $u_x=u_y=0$ and $u=0$. With the stated boundary regularity, integrate the local energy conservation law over inset rectangles and let their edges approach the boundary; the energy and flux terms converge by continuity of the first derivatives.