Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2006 (AC2, XMAT 4302, XMST 4302)

Put $p(t)={\prod }_{j=1}^m(t-{\mu }_j)$ and define the Lagrange interpolation polynomialsTheir denominators are nonzero and ${\ell }_j({\mu }_i)={\delta }_{ij}$. The polynomial ${\sum }_j{\ell }_j-1$ has degree at most $m-1$ and vanishes at all $m$ distinct scalars, so it is zero. Suppose condition (a) holds, and put $E_j={\ell }_j(T)$. Since $(t-{\mu }_j){\ell }_j(t)$ is a constant multiple of $p(t)$,Thus each $E_{j}v$ lies in $V_j$, and every vector is a sum of such vectors. If ${\sum }_j{v}_j=0$ with $v_j\in V_j$, applying $E_i$ gives $v_i=0$, because ${\ell }_i(T)v_j={\ell }_i({\mu }_j)v_j$. Hence the sum is direct, proving (b). Conversely, on each $V_j$ the factor $T-{\mu }_{j}I$ vanishes; the factors commute, so $p(T)$ vanishes on every summand and therefore on $V$. This proves the equivalence without invoking primary decomposition.

If $ST=TS$ and $v\in V_j$, then $(T-{\mu }_{j}I)Sv=S(T-{\mu }_{j}I)v=0$, so $S[V_j]\subseteq V_j$. Conversely, if each summand is preserved, then on $V_j$ both $ST$ and $TS$ act as ${\mu }_{j}S$. They agree on all summands and hence on $V$. Therefore

The dual space $V^{\ast }$ is the vector space of linear maps $V\to \mathbb{F}$, with pointwise operations. For the given basis define coordinate forms $e^i$ by $e^i({\sum }_j{a}_j{e}_j)=a_i$. These are linear and satisfy $e^i(e_j)={\delta }_{ij}$. If ${\sum }_i{c}_i{e}^i=0$, evaluation at $e_j$ gives $c_j=0$. For every linear form $\varphi$,so the coordinate forms span the dual and form its dual basis.

In the polynomial example write $p(X)=a+bX+c{X}^2$. ThenThese recover all three coefficients, so they span $V^{\ast }$ and, since the dual has dimension three, form a basis. Explicitly the basis dual to $1,X,X^2$ isEvaluation on the three monomials verifies the dual-basis identities directly.

An element $a$ of an integral domain is an irreducible element when it is nonzero and not a unit, and any factorization $a=bc$ has at least one unit factor. The exclusions of zero and units are part of the definition.

An element $a$ is a prime element when it is nonzero and not a unit, and $a\mid bc$ implies $a\mid b$ or $a\mid c$. If $a=bc$ and $a$ is a prime element, assume for example $b=ad$. Cancellation in $a=adc$ gives $dc=1$, so $c$ is a unit. This proves every prime element is an irreducible element.

In a Euclidean domain, each nonzero ideal has a generator of least Euclidean value: division of another member by that generator leaves a remainder in the ideal of smaller value, so the remainder must vanish. Let $a$ be an irreducible element and suppose $a\mid bc$ but $a\nmid b$. Write $(a,b)=(d)$. Since $d\mid a$, irreducibility says $d$ is a unit or is associate to $a$. The second case would give $a\mid b$, so $d$ is a unit. Hence $1=ra+sb$ for some $r,s$. Multiplying by $c$ shows $a\mid c$. Thus irreducibles in a Euclidean domain are prime.

For $R=\mathbb{Z}[i\sqrt{7}]$, use the multiplicative integer field norm $N(m+ni\sqrt{7})=m^2+7n^2$. Units have field norm one and are exactly $\pm 1$. If $2=uv$ were a product of nonunits, their norms would be integers greater than one multiplying to four, hence both would have field norm two. But $m^2+7n^2=2$ has no integer solution: $n\ne 0$ makes the left side at least seven, and $n=0$ would give $m^2=2$. Therefore 2 is an irreducible element.

On the other hand,is divisible by 2, while neither factor is, because divisibility by 2 requires both integer coefficients to be even. Thus 2 is not a prime element in this ring. It is also not a prime element in $\mathbb{Z}[i]$: $2\mid (1+i{)}^2=2i$, but $2\nmid 1+i$. In fact $2=(1+i)(1-i)$ is already a nontrivial factorization there.

A function is holomorphic on an open set if it is complex differentiable at every point of that set. At $z_0=x_0+i{y}_0$, take difference quotients first along real increments. Their limit is $f^{\prime }(z_0)$, so $u_x=\mathrm{Re}{f}^{\prime }(z_0)$ and $v_x=\mathrm{Im}{f}^{\prime }(z_0)$. Along imaginary increments,so $u_y=-\mathrm{Im}{f}^{\prime }(z_0)$ and $v_y=\mathrm{Re}{f}^{\prime }(z_0)$. These limits prove existence of all four partial derivatives and the Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$.

For the first example, the identity function and the reciprocal function are holomorphic away from zero, so $f$ is holomorphic on the given domain, with

Write $r=\sqrt{x^2+y^2}>0$. For $u=x/(1+r)$, $v=y/(1+r)$,The two Cauchy-Riemann equations would require $x^2=y^2$ and $xy=0$ simultaneously, impossible when $r>0$. Thus $f$ is not complex differentiable at any nonzero point.

At zero, however,So $f$ is complex differentiable at zero, with derivative one, but is not holomorphic on $\mathbb{C}$ or on any nonempty open subset. Complex differentiability at a single point is weaker than holomorphy in a neighborhood.

Choose a small positively oriented circle $C_{\epsilon }$ about $z_0$ inside the given region. Cauchy's theorem on the region between the two contours givesContinuity makes the last integrand converge uniformly to $f(z_0)$ as $\epsilon \downarrow 0$. Its limit is $2\pi if(z_0)$, proving the Cauchy integral formula.

The same formula holds with an interior point $w$ near $z_0$. Subtract its value at $z_0$ and divide by $w-z_0$:The contour is a positive distance from $z_0$, so the integrands converge uniformly as $w\to z_0$. Passing to the limit gives the differentiated formulaOn the unit circle, $\bar{z}=1/z$ and $\mathrm{Im}z=(z-\bar{z})/(2i)$, proving the requested identity. Hence the requested integral isThe first integral is $2\pi i/3$. For the second useThe differentiated formula applied to the constant function gives $\int z^{-2}dz=0$; the remaining two terms integrate to $-6\pi i+6\pi i=0$. Therefore

At a zero $a$, Taylor's theorem gives $f(z)={\sum }_{n\ge 0}{c}_n(z-a{)}^n$ near $a$. If some coefficient is nonzero, let $m$ be the least such index. Then $f(z)=(z-a{)}^{m}q(z)$ with $q(a)\ne 0$, and continuity makes $q$ nonzero on a smaller disc. Thus $a$ is an isolated zero. A limit point of zeros is also a zero by continuity and cannot be isolated, so all Taylor coefficients there vanish and $f$ vanishes on a nonempty open disc.

Let $A$ be the set of points having a neighborhood on which $f$ vanishes. It is nonempty and open. A point in its relative closure either already belongs to $A$ or is a limit point of distinct zeros; the preceding argument puts it in $A$ in either case. Thus $A$ is also relatively closed. Connectedness of $D$ gives $A=D$, proving $f$ is identically zero.

For conjugate reflection, the disc is invariant under conjugation, andHence $h$ is holomorphic, with derivative $\overline{g^{\prime }(\bar{z})}$. If $g$ is real on the real diameter, then for real $x\in (-1,1)$, $h(x)=\overline{g(x)}=g(x)$. The holomorphic difference $g-h$ therefore has zeros accumulating at zero, and the identity theorem just proved gives $g=h$ on the entire disc.

Assume the vector field and multiplier are continuously differentiable, as required for Green's theorem and uniqueness of trajectories. A nonconstant periodic trajectory of a planar autonomous system is a simple closed orbit: uniqueness prevents crossings, and one traversal can be taken over its least period. Let $\Omega$ be its bounded interior. Along the orbit $dx=fdt$, $dy=gdt$, soGreen's theorem identifies this with ${\iint }_{\Omega }[{\partial }_x(\varphi f)+{\partial }_y(\varphi g)]dxdy$, which is strictly positive by hypothesis, a contradiction. Thus there is no nonconstant periodic orbit. This standard periodic-orbit criterion does not by itself exclude stationary solutions.

For the first concrete system choose $\varphi =-1$. ThenThere are also no equilibria: $f=0$ requires $y=1$, but then $g=-e^{-x}-x^2<0$. Consequently it has no periodic solution, including constant ones.

For the second system the origin is outside the domain of the printed vector field. For $r>0$,Thus for an admissible initial point with $\rho =\sqrt{a^2+b^2}>0$,where $a=\rho \cos \alpha$, $b=\rho \sin \alpha$. These formulas are valid while $r>0$, in particular for every $t\ge 0$. Trajectories rotate counterclockwise, spiral outward when $0<r<1$, and inward when $r>1$, approaching the unit circle. On $r=1$ they give $x(t)=\cos (\alpha +t)$, $y(t)=\sin (\alpha +t)$, a periodic orbit with least period $2\pi$. It is an attracting limit cycle. There is no contradiction with the preceding criterion: this vector field is singular at the origin and has the exhibited closed orbit.

With parameter $t$ along a characteristic curve, the equations areTheir solutions are $y=t$, $x=s+t^2/2$ and $u=s{e}^t$. Thus the projected curve satisfies $y^2=2(x-s)$, as required. Eliminating the parameters givesIndeed $u_x=e^y$ and $u_y=e^y(x-y^2/2-y)$, so $y{u}_x+u_y=u$, and the initial values hold.

The parameter map $(s,t)\mapsto (s+t^2/2,t)$ has Jacobian one and is globally invertible. Every point whose characteristic curve meets the initial segment therefore lies in the stripThe uniquely determined open region is its interior, with the boundary values determined on the two endpoint characteristic curves. Both signs of $y$ are included. Outside the strip, a general solution is $u=e^{y}F(x-y^2/2)$ with $F(s)=s$ only required for $1\le s\le 2$. Smooth perturbations of $F$ supported outside that interval produce different solutions with the same prescribed data, proving the stated region is exact.

For a function for which the integral converges, the Laplace transform isFor a continuous function of exponential order this exists in a sufficiently far right half-plane. For the first input,This follows by evaluating the exponential antiderivative and using its decay at infinity.

Integration by parts, whose boundary term vanishes for $\mathrm{Re}p>0$, gives

Combining the exponential factors gives

The same integration by parts with $p$ replaced by $p+c$ givesFor the remaining unnumbered requests, the derivative and convolution formulas, under the usual convergence hypotheses, areApplying them to the general equation yields $p\bar{x}-a+\bar{k}\bar{x}=\bar{y}$ and thereforeFor the particular data, $\bar{y}=1/p$, $\bar{k}=1/(p+2)$ and $a=0$, soThe four transforms above giveAs a direct check, $x(0)=0$, $x^{\prime }=(1+t)e^{-t}$, andAdding the last two expressions gives one, verifying the original equation as well as the required limit.