Advanced Topics (Elite Level)

Q1 (10)

(a) Let $y(x)+\varepsilon\eta(x)$, $\eta(a)=\eta(b)=0$. Define $\Phi(\varepsilon)=\int_a^b F(x,y+\varepsilon\eta, y'+\varepsilon\eta')dx$. (1) Necessary condition for extremum: $\Phi'(0)=0$. Differentiate under integral: $\Phi'(0)=\int_a^b\left(\frac{\partial F}{\partial y}\eta + \frac{\partial F}{\partial y'}\eta'\right)dx=0.$ (2) Integrate the second term by parts: $\int_a^b \frac{\partial F}{\partial y'}\eta'\,dx = \Big[\frac{\partial F}{\partial y'}\eta\Big]_a^b - \int_a^b \frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)\eta\,dx.$ Boundary term vanishes since $\eta(a)=\eta(b)=0$. (2) Hence $\int_a^b\left(\frac{\partial F}{\partial y} - \frac{d}{dx}\frac{\partial F}{\partial y'}\right)\eta\,dx=0$ for all admissible $\eta$. By the fundamental lemma (integrand continuous, arbitrary $\eta$), the bracket is identically zero: (1) $\boxed{\frac{\partial F}{\partial y} - \frac{d}{dx}\frac{\partial F}{\partial y'} = 0.}$ (1)

(b) Compute $\frac{d}{dx}\!\left(F - y'\frac{\partial F}{\partial y'}\right) = F_x + F_y y' + F_{y'}y'' - y''F_{y'} - y'\frac{d}{dx}F_{y'}$ $= F_x + y'\big(F_y - \frac{d}{dx}F_{y'}\big)$. (2) The bracket is zero by E-L, so if $F_x=0$ then the derivative is $0 \Rightarrow F - y'F_{y'}=$const. (1)

Q2 (12)

(a) Use upper-half-plane semicircle: segment $[-R,R]$ plus arc of radius $R$. On the arc $|z^4+1|\ge R^4-1$ so $|\int_{arc}|\le \pi R/(R^4-1)\to 0$ as $R\to\infty$. (4)

(b) Poles: $z^4=-1 \Rightarrow z=e^{i\pi/4}, e^{i3\pi/4}, e^{i5\pi/4}, e^{i7\pi/4}$. In UHP: $z_1=e^{i\pi/4}$, $z_2=e^{i3\pi/4}$. (2) Residue at simple pole $z_k$: $\dfrac{1}{4z_k^3} = \dfrac{z_k}{4z_k^4} = \dfrac{z_k}{-4}$ (since $z_k^4=-1$). (2) $\text{Res}_{z_1}+\text{Res}_{z_2} = -\tfrac14(e^{i\pi/4}+e^{i3\pi/4}) = -\tfrac14(\tfrac{\sqrt2}{2}(1+i) + \tfrac{\sqrt2}{2}(-1+i))$ $= -\tfrac14\cdot i\sqrt2 = -\tfrac{i\sqrt2}{4}$. (2)

(c) $I = 2\pi i \sum \text{Res} = 2\pi i\cdot(-\tfrac{i\sqrt2}{4}) = \dfrac{2\pi\sqrt2}{4} = \boxed{\dfrac{\pi}{\sqrt2} = \dfrac{\pi\sqrt2}{2}}.$ (2)

Q3 (10)

(a) CR: $u_x=v_y$, $u_y=-v_x$. (2) Complex differentiability requires the limit $\frac{f(z+h)-f(z)}{h}$ be independent of direction of $h$. Equating limits along real and imaginary directions yields CR; with continuity of partials these become sufficient for differentiability. (2)

(b) $f=|z|^2 = x^2+y^2$, so $u=x^2+y^2$, $v=0$. Then $u_x=2x$, $v_y=0$: CR needs $2x=0$; $u_y=2y$, $-v_x=0$: needs $2y=0$. Both hold only at $(0,0)$. So $f$ is complex-differentiable only at $z=0$, nowhere analytic (analyticity needs a neighborhood). (3)

(c) $u=x^2-y^2$. $v_y=u_x=2x \Rightarrow v=2xy+g(x)$; $v_x=2y+g'(x)$ must equal $-u_y=2y \Rightarrow g'=0$. So $v=2xy$ (+const). $f=x^2-y^2+2ixy=(x+iy)^2=z^2$. (3)

Q4 (10)

(a) Solve $\pi P=\pi$: with $\pi=(a,b,c)$:

• $a = 0.5a+0.25b$
• $b = 0.5a+0.5b+0.5c$
• $c = 0.25b+0.5c$ From first: $0.5a=0.25b\Rightarrow b=2a$. From third: $0.5c=0.25b\Rightarrow b=2c\Rightarrow a=c$. (3) Normalize $a+b+c=1$: $a+2a+a=4a=1\Rightarrow a=1/4$, $b=1/2$, $c=1/4$. $\boxed{\pi=(\tfrac14,\tfrac12,\tfrac14).}$ (3)

(b)

import numpy as np
P = np.array([[.5,.5,0],[.25,.5,.25],[0,.5,.5]])
pi = np.array([1/3,1/3,1/3])
for _ in range(1000):
    pi = pi @ P
print(pi)  # -> [0.25 0.5 0.25]

Converges because the chain is irreducible and aperiodic (regular), so $P^n$ tends to a rank-1 matrix with rows $\pi$; power iteration converges at rate governed by second-largest $|\lambda|$. (4)

Q5 (10)

(a) $g_{ij}=\begin{pmatrix}1&0\\0&r^2\end{pmatrix}$, $g^{ij}=\begin{pmatrix}1&0\\0&1/r^2\end{pmatrix}$. (3)

(b) $A^r=g^{rr}A_r=A_r$, $A^\theta=g^{\theta\theta}A_\theta=A_\theta/r^2$. (3)

(c) $\Gamma^r_{\theta\theta}=\tfrac12 g^{rl}(\partial_\theta g_{l\theta}+\partial_\theta g_{l\theta}-\partial_l g_{\theta\theta})$. Only $l=r$ contributes ($g^{rr}=1$): $=\tfrac12(2\partial_\theta g_{r\theta}-\partial_r g_{\theta\theta}) = \tfrac12(0 - \partial_r r^2)=\tfrac12(-2r)=\boxed{-r}.$ (4)

Q6 (8)

(a) Pointwise: $\forall x\,\forall\varepsilon>0\,\exists N(x,\varepsilon)$ s.t. $n\ge N\Rightarrow|f_n(x)-f(x)|<\varepsilon$. (1) Uniform: $\forall\varepsilon>0\,\exists N(\varepsilon)$ (independent of $x$) s.t. $n\ge N\Rightarrow \sup_x|f_n(x)-f(x)|<\varepsilon$. (1) Difference: $N$ works uniformly across all $x$. (1) $f_n(x)=x^n$ on $[0,1]$: pointwise limit $f(x)=0$ for $x<1$, $f(1)=1$; $\sup_x|f_n-f|=1$ for all $n$ (near $x=1^-$), so not uniform (and limit is discontinuous). (1)

(b) KKT for min $f$ s.t. $g_i\le0$, $h_j=0$: (4, 1 each)

1. Stationarity: $\nabla f + \sum\mu_i\nabla g_i + \sum\lambda_j\nabla h_j=0$.
2. Primal feasibility: $g_i(x)\le0$, $h_j(x)=0$.
3. Dual feasibility: $\mu_i\ge0$.
4. Complementary slackness: $\mu_i g_i(x)=0$.

[
 {"claim":"Integral of 1/(x^4+1) over R equals pi/sqrt2","code":"x=symbols('x'); I=integrate(1/(x**4+1),(x,-oo,oo)); result = simplify(I - pi/sqrt(2))==0"},
 {"claim":"Sum of UHP residues of 1/(x^4+1) is -i*sqrt2/4","code":"z=symbols('z'); z1=exp(I*pi/4); z2=exp(I*3*pi/4); r=z1/(-4)+z2/(-4); result = simplify(r + I*sqrt(2)/4)==0"},
 {"claim":"Stationary distribution is (1/4,1/2,1/4)","code":"P=Matrix([[Rational(1,2),Rational(1,2),0],[Rational(1,4),Rational(1,2),Rational(1,4)],[0,Rational(1,2),Rational(1,2)]]); pi=Matrix([[Rational(1,4),Rational(1,2),Rational(1,4)]]); result = (pi*P)==pi"},
 {"claim":"Christoffel Gamma^r_theta_theta = -r for polar metric","code":"r=symbols('r'); g_theta_theta=r**2; Gamma = -Rational(1,2)*diff(g_theta_theta,r); result = simplify(Gamma + r)==0"}
]