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### Messages - Jessica Chen

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1
##### Final Exam / Re: FE Problem 1
« on: April 14, 2015, 11:40:50 PM »

2
##### Final Exam / Re: FE Problem 2
« on: April 14, 2015, 11:39:04 PM »
\begin{align}
u(x,t)&=\frac{1}{\sqrt{4k\pi t}}[\int_0 ^\infty e^{-\frac{(x-y)^2}{4kt}}f(y) dy
+\int_0 ^\infty e^{-\frac{(x+y)^2}{4kt}}f(y)]\\
&=\frac{1}{\sqrt{4k\pi t}}[\int_0 ^1 e^{-\frac{(x-y)^2}{4kt}}+ e^{-\frac{(x+y)^2}{4kt}}dy]\\
&=\frac{1}{\sqrt{12\pi t}}[\int_0 ^1 e^{-\frac{(x-y)^2}{12t}}+ e^{-\frac{(x+y)^2}{12t}}dy]\\

\end{align}
Let $I_1(x) = \int_0 ^1 e^{-\frac{(x-y)^2}{2}}$ then $I_2(x)= I_!(-x)$\\
Let $z_1 = (x-y)$, we get
\begin{align}
I_1(x) &= \int_0 ^1 e^{-\frac{(x-y)^2}{2}}dy\\
&=\int_{x-1} ^{x} e^{-\frac{z^2}{2}}dz\\
&=\int_{0} ^{x-1} e^{-\frac{z^2}{2}}dz+\int_{0} ^{x} e^{-\frac{z^2}{2}}dz\\
&=\sqrt{\frac{\pi}{2}}(erf(x-1)+erf(x)))\\
I_2(x)&=\sqrt{\frac{\pi}{2}}(erf(-x-1)+erf(-x)))

\end{align}
Then, we get
\begin{align}
u(x, t) &= \frac{1}{\sqrt{12\pi t}}e^{\frac{1}{6t}}[\sqrt{\frac{\pi}{2}}(erf(x-1)+erf(x)))+\sqrt{\frac{\pi}{2}}(erf(-x-1)+erf(-x)))]\\
&= \frac{1}{\sqrt{24t}}e^{\frac{1}{6t}}(erf(x-1)+erf(x)+erf(-x-1)+erf(-x)]\\
\end{align}

3
##### HA10 / Re: HA10 Problem 1
« on: April 02, 2015, 08:43:07 PM »
\begin{align*}
&\Phi(u):=\frac{1}{2}\int_0^2 u_{xx}^2\,dx,\\
&\Psi(u)= \frac{1}{2}\int_0^2 u_{x}^2\,dx =1,\\
&u(0)=u'(0)=u(2)=u'(2)=0.
\end{align*}

a. Write down Eulerâ€“Lagrange equation $\delta (\Phi-\lambda \Psi)=0$.
Let $L (x, u, u', u'') = u_{xx}^2$
then
\begin{align}
\delta \Phi&=
\iiint_\Omega\Bigl(\frac{\partial L}{\partial u} - \sum_{1\le j\le n} \frac{\partial\ }{\partial x_j} \frac{\partial L}{\partial u_{x_j}} \Bigr)\delta u \,dx \\
&=\frac{\partial^2}{\partial x^2}2u_{xx}\\
&=2u_{xxxx}
\end{align}
similarly,
\begin{align}
L_2(x,u,u') = u_x^2\\
\delta\Psi &=\frac{\partial L}{\partial u}-\frac{\partial}{\partial x}\frac{\partial L}{\partial u_x}\\
&=-\frac{\partial}{\partial x}2u_x\\
&=-2u_{xx}
\end{align}
Thus

\delta(\Phi-\lambda\Psi) = u_{xxxx}+\lambda u_{xx}=0

b. Under boundary conditions (\ref{eq-10.3}) solve it and find out eigenvalues $\lambda$ for each solution exists.
Characteristic Equation:

k^4+\lambda k^2 = 0\\
k = 0,\sqrt{\lambda}i, -\sqrt{\lambda}i

Let $\lambda = \omega ^2$
Then $u = A + Bx+C\cos(\omega x)+D\sin(\omega x)$
By hint, we have $u(-1)=u'(-1)=u(1)=u'(-1) = 0$
Even eigenfunction is:
\begin{align}
X &= A+C\cos(\omega x)\\
X(1) &= A+C\cos(\omega)=0\\
X'(1) &= -C\sin(\omega) = 0\\
\sin(\omega) &=0 \implies \omega = n\pi\\
\lambda_n &= \omega ^2=(n\pi)^2\\
A+C\cos(\omega) = 0\\
A = 1\\
C = (-1)^{n+1}\\
\end{align}
$X_n = 1+ (-1)^{n+1}cos(nx)$ is the even eigenfunction

Odd function
\begin{align}
X &= Bx + D \sin (\omega x)\\
X(1) &=B+D\cos(\omega ) = 0\\
X'(1) &=B+D\cos(\omega)\omega = 0\\

\cos(\omega)\omega &= \sin(\omega)\\
\omega&=\tan(\omega)\\
\end{align}
Then the graph $\omega$ and $\tan(\omega)$ intersections are the eigenvalues

4
##### HA9 / Re: HA9 Problem 1
« on: March 26, 2015, 11:26:42 PM »
Q1
From lecture section 28.1, we have

\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho +
\frac{1}{\rho^2}\Lambda

with

\Lambda:=
\bigl(\partial_{\phi}^2 + \cot(\phi)\partial_\phi \bigr) +\frac{1}{\sin^2(\phi)}\partial_{\theta}^2.

Plug $u=P(\rho)Y(\phi,\theta)$ into $\Delta u=0$:
P''(\rho)Y(\phi,\theta) + \frac{2}{\rho}P' (\rho)Y(\phi,\theta) + \frac{1}{\rho^2} P(\rho)\Lambda Y(\phi,\theta)=0

which could be rewritten as

\frac{\rho^2 P''(\rho) + \rho P' (\rho)}{P(\rho)}+
\frac{\Lambda Y(\phi,\theta)}{Y(\phi,\theta)}=0

and since the first term depends only on $\rho$ and the second only on$\phi, \theta$ we conclude that both are constant:
\begin{align}
\rho^2 P'' +2\rho P' = \lambda P,\\
\Lambda Y(\phi,\theta)=-\lambda Y(\phi,\theta).
\end{align}
The first equation is of Euler type and it has solutions $P:=\rho^l$ iff
$\lambda= l(l+1)$. However if we are considering ball, solution must be infinitely smooth in its center due to some general properties of Laplace equation and this is possible iff $l=0,1,2,\ldots$ and in this case $u$ must be a polynomial of $(x,y,z)$.Which in this case, u depends on $x^2+y^2+z^2$ and $z$

Then suppose that
\begin{align}
u= &z^4 + a (1-x^2-y^2-z^2) + bz^2 (1-x^2-y^2-z^2)  + c(1-x^2-y^2-z^2) ^2\\\\
=&z^4 + a (1-x^2-y^2-z^2) + b (z^2-z^2x^2-z^2y^2-z^4) + c(1-x^2-y^2-z^2) ^2

\end{align}
\begin{align}
\Delta u &= uxx+uyy+uzz\\
&=12 z^2  -6a + 2b( 1 -\rho^2 -7 z^2) -6c +20c\rho^2\\\text{Suppose $1-\rho^2-z^2=0$}\\
&=12z^2-6a+2b(-6z^2)-6c+20c(1-z^2)\\
&=12z^2-6a-12bz^2-6c+20c-20cz^2\\
&=(12-12b-20c)z^2-6a-6c=0
\end{align}
Then we have
\begin{align}
12-12b-20c=0\\
-6a-6c=0
\end{align}
Then $a = -c$, $b=1-\frac{5}{3}c$, and c is arbitrary.

5
##### Web Bonus Problems / Re: Web Bonus Problem 4
« on: March 11, 2015, 09:58:31 PM »

Now hint:
a) consider $u_s (x,t)= s^l u(s x, s^k t)$ and prove that if $u$ satisfies original problem then $u_s$ satisfies it for all $s>0$ iff $k=1$, $l=0$.
b) So, $u_s(x,t)= u(sx, st)$ satisfies it and we are interested in self-similar solution $u(x,t)=u(sx,st)$ for all $s>0$. Plugging $s=t^{-1}$ we arrive to $u(x,t)=v (xt^{-1})$ (with $v(y)=u(y, 1)$.
c) Plugging $u(x,t)=v (xt^{-1})$ into original equation we have an ODE. Which?
d) Find continuous solution of this ODE such that $v(y)=-1$ as $y<-1$ and $v(y)=1$ as $y>1$ (Think why).
e) Plug into $u$

a) $s^k = s$ because one derivative with respect to t is also one derivative with respect x. Thus $k = 1$;
By total energy rule, $\int_{-\infty}^{\infty} u_s dx = s^l\int _{-\infty}^{\infty}u dx$ Thus $s^l = 1\implies l=0$

b) $u_s= u(sx,st)$. Let $s= t^{-1}$ we have $u_s= u(t^{-1}x, 1) = v(t^{-1}x)$
c) plug in $v(t^{-1}x)$
\begin{align}
ut&=-t^{-2}xv'(t^{-1}x)\\
ux&=v'(t^{-1}x)t^{-1}\\
-t^{-2}xv'(t^{-1}x)&=v(t^{-1}x)(v'(t^{-1}x)t^{-1}\\
-t^{-1}xv'(t^{-1}x)&=v(t^{-1}x)(v'(t^{-1}x)\\
\end{align}
Let $t^{-1}x=\xi$ we get

-\xi v'(\xi)=v(\xi)v'(\xi)
\label{A}

$-\xi = v(\xi)$ is the ODE?

6
##### Web Bonus Problems / Re: Web Bonus Problem 4
« on: March 07, 2015, 07:57:40 PM »
No, this does not fly. To construct a solution you need to fill by characteristics the whole half-plane $t>0$. So far you covered only $x<-t$ where $u=-1$ and $x>t$ where $u=1$ leaving sector $-t<x<t$ empty.

You need to apply the method of self-similar solutions to find continuous $u(x,t)$  there.
Sorry professor I quoted my post accidentally, could you help me delete the previous post? -t<x<t sector has two values which is definitely not right but I don't know what do you mean apply the self-similar solution, can you give me more hints please?

7
##### Web Bonus Problems / Re: Web Bonus Problem 4
« on: March 06, 2015, 12:13:48 AM »
Characteristic equation:
\begin{align}
\frac{dt}{1} &=\ \frac{dx}{-u} = \frac{du}{0}\\
\frac{du}{dt}&=\ 0 \implies u=f(C)\\
x &=\ C - tu\\
C &=\ x+tu
\end{align}

Impose boundary condition: $t=0$ and $x = C$,
\begin{align}
u &=\ f(C) =f(x+tu)\\
u|_{t=0}&=f(x) =\left\{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\end{align}
then,

\end{align}

8
##### HA6 / Re: HA6 problem 1
« on: March 05, 2015, 10:15:48 PM »
a.
\begin{align}
\hat{f}(\omega) &=\ \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\alpha |x|}e^{-i\omega x}dx\\
&=\  \frac{1}{2\pi}[\int_{-\infty}^{0} e^{\alpha x -i\omega x} +\int_{0}^{\infty} e^{-\alpha x -i\omega x}]dx\\
&= \frac{1}{2\pi}[\int_{-\infty}^{0} (\cos (\omega x)-i\sin (\omega x))e^{\alpha x } +\int_{0}^{\infty} (\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+i\sin (\omega x))e^{-\alpha x } +\int_{0}^{\infty} (\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+i\sin (\omega x)+\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+\cos (\omega x))e^{-\alpha x}]dx\\
&= \frac{2}{2\pi}\int_{0}^{\infty}\cos(\omega x)e^{-\alpha x} dx\\
&= \frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}
\end{align}

b.
1)
Let $f(\omega) = e^{-\alpha |x|}$, and let
\begin{align}
g(x) &=\ e^{-\alpha |x|}\cos(\beta x)\\
&=\ e^{-\alpha |x|}(\frac{1}{2}e^{i\beta x}+\frac{1}{2}e^{-i\beta x})\: \mbox{ (By Trig identity)}\\
&=\ \frac{1}{2}e^{-\alpha |x|}e^{i\beta x}+\frac{1}{2}e^{-\alpha |x|}e^{-i\beta x}
\end{align}
Which satisfies the form $\hat{g}(\omega) = \frac{1}{2}\hat{f}(\omega - \beta) + \frac{1}{2}\hat{f}(\omega + \beta)$.

From part a we have found $\hat{f}\frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}$
Thus\begin{align}
\hat{g}(\omega) = \frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})
\end{align}

2)
Let $f(\omega) = e^{-\alpha |x|}$, and let
\begin{align}
g(x) &=\ e^{-\alpha |x|}\sin(\beta x)\\
&=\ e^{-\alpha |x|}(\frac{1}{2i}e^{i\beta x}-\frac{1}{2i}e^{-i\beta x})\: \mbox{ (By Trig identity)}\\
&=\ \frac{1}{2i}e^{-\alpha |x|}e^{i\beta x}-\frac{1}{2i}e^{-\alpha |x|}e^{-i\beta x}
\end{align}
Which satisfies the form $\hat{g}(\omega) = \frac{1}{2i}\hat{f}(\omega - \beta) - \frac{1}{2i}\hat{f}(\omega + \beta)$.

From part a we have found $\hat{f}\frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}$
Thus\begin{align}
\hat{g}(\omega) = \frac{1}{2i}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2)})-\frac{1}{2i}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})
\end{align}

c.
Let $f(\omega) = e^{-\alpha |x|}$, and let
$g(x) = xe^{-\alpha |x|}$
By the theorem $\hat{g}(\omega) = i\hat{f'}(\omega)$, We have
\begin{align}
\hat{g}(\omega) &=\ i \frac{d\hat{f}(\omega)}{d\omega}\\
&=\  i \frac{d}{d\omega }(\frac{\alpha }{\pi (\alpha ^2 +\omega ^2)})\\

&=\ i \frac{\alpha}{\pi}\frac{-2\omega}{(\alpha ^2 +\omega ^2)^2}\\
\end{align}

d.
Similar to part c),
Let $f(\omega) = e^{-\alpha |x|}cos(\beta x)$, and let $g(x) = xe^{-\alpha |x|}cos(\beta x)$

By the theorem $\hat{g}(\omega) = i\hat{f'}(\omega)$, where
$\hat{f}(\omega) = \frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})$

We have
\begin{align}
\hat{g}(\omega) &=\ i \frac{d\hat{f}(\omega)}{d\omega}\\

&=\ i \frac{d}{d\omega }\hat{f}(\omega)\\
& =\ i \frac{d}{d\omega }(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})\\
&=\ i\frac{\alpha}{\pi}[(\frac{-(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2})+\frac{-(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]\\
&=\ \frac{-\alpha i}{\pi}[\frac{\omega - \beta}{(\alpha^2 + (\omega - \beta)^2)^2}+\frac{\omega + \beta}{(\alpha^2 + (\omega + \beta)^2)^2}]
\end{align}

Similar to above, we can get $g(x) = xe^{-\alpha |x|}sin(\beta x)$ by applying part b.2 formula and use the method we used in part c, then we get:
$\hat{g}(\omega) = \frac{-\alpha }{\pi}[\frac{\omega - \beta}{(\alpha^2 + (\omega - \beta)^2)^2}-\frac{\omega + \beta}{(\alpha^2 + (\omega + \beta)^2)^2}]$

9
##### Test 1 / Re: TT1 Problem 2
« on: February 12, 2015, 10:23:59 PM »
I don't know how to attach a pdf properly such that we can see its preview without download it.

You cannot. So scan to png or jpg. V.I.

10
##### Test 1 / Re: TT1 Problem 5
« on: February 12, 2015, 10:00:28 PM »
Question 5

11
##### HA3 / Re: HA3 problem 4
« on: February 05, 2015, 09:09:54 PM »
a.$M(T)=$max $u(x,t)$, where $0\leq x\leq l, 0\leq t\leq T]$

We claim that M(T) is an non decreasing function of T.

Let $T_1 \leq T_2$ and the rectangle $R_1 = {0\leq x\leq l,0\leq t \leq T_1}$ and $R_2 = {0\leq x\leq l,0\leq t \leq T_2}$.

The bottom and lateral sides of $R_1$ contained in $R_2$. By the uniqueness property, the diffusion equation of u on $R_2$ is an extension of u of $R_1$ and thus$M(T1)\leq M(T2)$

b.$m(T)=$ min $u(x,t)$, where $0\leq x\leq l, 0\leq t\leq T]$

We claim that m(T) is an non increasing function of T.

Let $T_1 \leq T_2$ and the rectangle $R_1 = {0\leq x\leq l,0\leq t \leq T_1}$ and $R_2 = {0\leq x\leq l,0\leq t \leq T_2}$.

The bottom and lateral sides of $R_1$ contained in $R_2$. By the uniqueness property, the diffusion equation of u on $R_2$ is an extension of u of $R_1$ and thus  $m(T2)\leq m(T1)$

12
##### HA2 / Re: HA2 problem 1
« on: January 29, 2015, 10:35:13 PM »
b. When $v=2$, we need to impose $u|_{x=2t}=0, t>0$

First when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)

Then when $2t<x<3t$, $x+3t>0$, but $x-3t<0$.

\begin{align}
\phi(x) = \frac{1}{2}\cos(x)+\frac{1}{6}\int_0^x \! \sin(y) dx.\\
\phi(x) = \frac{1}{2}\cos(x) - \frac{1}{6}(\cos(x)-1)\\
\phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}\\
\end{align}
We need to impose the extra condition such that
\begin{align}
u = \phi(5t) + \psi(-t) = 0\\
\phi(5t) = -\psi(-t)\\
\psi(t) = \phi(5t)\\
\psi(x-3t) = \frac{1}{3}\cos(5x-15t) + \frac{1}{6}\\
\end{align}
Then we get $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x-15t)) + \frac{1}{3}$ for $2t<x<3t$.

c. When $v=-4$, we need to impose $u|_{x=-4t}=ux|_{x=-4t}=0, t>0$

Firstly, when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)

Then when $-4t<x<-3t$, $x+3t<0$ and $x-3t<0$.

\begin{align}
\phi(5t) + \psi(-t) = 0\\
\phi'(5t) + \psi'(-t) = 0\\
\end{align}
Then we get $\phi'(5t)=0$ and $\psi(-t) = 0 \implies u(x,t) = 0$ for $-4t<x<-3t$.
Lastly, -3t<x<3t, the situation is similar to part b) $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x-15t)) + \frac{1}{3}$ .

d. When $v=-3$, we need to impose $u|_{x=-3t}=0, t>0$

First, when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)

Then when $-3t<x<3t$, $x+3t>0$ and $x-3t<0$.

Then $\phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}$

We need to find $\psi(x)$ by imposing the condition:

\begin{align}
\phi(0) + \psi(-6t) = 0\\
0 + \psi(-6t) = 0\\
\psi(x-3t) = 0
\end{align}

Then we get $u(x,t) = \phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}$ for  $-3t<x<3t$.

13
##### Web Bonus Problems / Re: Web Bonus Problem 1
« on: January 25, 2015, 08:43:56 AM »
The general solution for $u$ is $u = f(x_0) = f(x-ut)$, now consider the first set of initial data.
When t = 0,
Case1. If $x < -a$, then we have $u = 1$
Case2. If  $x > a$, then we have $u = -1$
Case3. If $-a\le x \le a$, then we have $u =- \frac{x}{a}$.

u = -\frac{x-ut}{a}\rightarrow u = \frac{x}{t-a}

Then the solution looks like
\begin{align}
u(x,t)=&\left\{\begin{aligned}
1& && x<-a+t,\\
\frac{x}{t-a}& && -a+t\le x \le a-t,\\
-1& && x>a-t;
\end{aligned}\right.
\\
\end{align}

Thus as t increase, -a+t and a-t runs toward each other, the the solution graph is squeezing. The solution is define for $0<t<a$.

14
##### Web Bonus Problems / Re: Web Bonus Problem 1
« on: January 24, 2015, 12:42:21 AM »
I interpret this question in the following:

This is a quasilinear equation with coefficient u.
As we solved in HA1, the general solution for $u$ is $u = f(x_0) = f(x-ut)$, now consider the first set of initial data.
Case1. If $x_0 < -a$, then we have $u = -1$ and also $x_0 = x+t < -a$.
Case2. If  $x_0 > a$, then we have $u = 1$ and also $x_0 = x-t > a$.
Case3. If $-a\le x \le a$, then we have $u = \frac{x}{a}$.

x_0 = x - \frac{x}{a}t\rightarrow u = \frac{x}{a+t}

Then the solution looks like
\begin{align}
u(x,t)=&\left\{\begin{aligned}
-1& && x<-a-t,\\
\frac{x}{a+t}& && -a-t\le x \le a+t,\\
1& && x>a+t;
\end{aligned}\right.
\\
\end{align}

15
##### HA1 / Re: HA1 problem 4
« on: January 22, 2015, 11:59:45 PM »
c.
Suppose in general case, $yu x - 4xu y = f(x, y)$
Use polar angles, define

x=r\cos \theta \rightarrow x {\theta} = -r \sin\theta\\
y = 2r\sin \theta \rightarrow y{\theta} = 2r\cos\theta\\
yu x - 4xu y = -2u\theta.

Since trajectories are closed (elliptic shape) $\theta \in (0, 2\pi)$ periodically.
Let $-2u \theta = g(\theta, r)$ then $u(\theta, r) = \int_0^{2\pi}\! g(\theta, r) d\theta$ but g needs to be 0 over the period, i.e $\int_0^{2\pi}\! g(\theta, r) d\theta = 0$.

In part a)
y = -2u\theta\\
r\cos\theta = u
and integral over this period is 0. Thus solution for part a) is valid.

In part b)
x^2 = -2u\theta\\
-\frac{1}{2}r^2 \frac{1}{2}(\cos 2\theta +1) = u\theta\\
u = -\frac{1}{8}r^2\sin 2\theta - \frac{1}{4}r^2\theta = u\\
the integral over period is not zero. Thus the periodic trajectories would not work.

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