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Messages - Chi Ma

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1
Final Exam / Re: FE-7
« on: December 19, 2015, 10:57:57 AM »
$u$ can be written as a sum of 3 homogeneous harmonic polynomials.
\begin{equation}
u = -\frac{1}{5}(z^3-3zx^2) -\frac{1}{5}(z^3-3zy^2) + \frac{2}{5}z
\end{equation}

2
HA10 / Re: HA10-P2
« on: December 08, 2015, 01:44:06 AM »
Let $L = \frac{1}{\sqrt{2gu}} \sqrt{1+u^{\prime\,2}}$. Hamiltonian $H=u' L_{u'}-L=constant$ implies

\begin{equation} u' =  \sqrt{\frac{2A-u}{u}} \label{u} \end{equation} for some constant $A$. Reparameterize  $u = A - A\cos\theta$ to obtain the solution to ($\ref{u}$).
\begin{equation} x =  A (\theta - \sin \theta) + C \end{equation}

Condition $u(0)=0$ implies that $C=0$. The solution is a parametric cycloid:
\begin{equation} x =  A (\theta - \sin \theta) \end{equation}
\begin{equation} u =  A (1 - \cos \theta) \end{equation}

3
Web Bonus = Oct / Re: Web bonus problem : Week 3 (#4)
« on: December 06, 2015, 12:46:20 AM »
We look for solutions such that $u_{tt}-u_{xx}=u(1-2u^2)=0$.
$u_{tt}-u_{xx}=0$ implies that $u(x,t) = f(x \pm t)$ for some function $f$.
$u(1-2u^2)=0$ implies that either $u = 0$ or $u = \pm \frac{1}{\sqrt 2}$.

A kink may be described by
\begin{equation}
u(x,t) = \left\{\begin{array}{21}
 &\pm \frac{1}{\sqrt 2} \qquad & x \ge t\\
 & 0 & x < t \end{array} \right.
\end{equation}

A soliton may be described by
\begin{equation}
u(x,t) = \left\{\begin{array}{21}
 &\pm \frac{1}{\sqrt 2} \qquad & x = t\\
 & 0 & x \neq t \end{array} \right.
\end{equation}

4
Web Bonus = Oct / Re: Web bonus problem : Week 5 (#1)
« on: November 30, 2015, 07:38:44 PM »
Find a self–similar solution with finite $\int_{-\infty}^\infty u\,dx$ to the following pde $(\ref{f})$.
\begin{equation} u_{t} = (u u_{x})_{x}  \qquad -\infty< x <\infty, \quad t>0 \label{f} \end{equation}
Let $v(x,t) \equiv u_{\alpha,\beta,\gamma}(x,t) = \gamma u(\alpha x,\beta t)$. $v(x,t)$ satisfies $(\ref{f})$ if $\beta=\gamma\alpha^2$. Finite $\int_{-\infty}^\infty u\,dx$ implies $ \gamma=\alpha$ as $\alpha$ is assumed to be positive. Therefore, $\beta= \alpha^3$. Choose $\alpha = t^{-\frac{1}{3}}$.

\begin{equation}
 v(x,t) = t^{-\frac{1}{3}} u(t^{-\frac{1}{3}} x, 1) \equiv t^{-\frac{1}{3}} \phi (t^{-\frac{1}{3}} x) \label{v}
\end{equation}

Substitute the partial derivatives of $v(x,t)$ into $(\ref{f})$ and define $\xi \equiv t^{-\frac{1}{3}} x $.
\begin{align}
v_{t} &= (v v_{x})_ {x} \\
 -\frac{1}{3} t^{-\frac{4}{3}} \bigl( \xi\phi(\xi)\bigr)' &= t^{-\frac{4}{3}} \bigl(\phi(\xi) \phi’(\xi)\bigr)' \\
\phi(\xi) \bigl( \phi’(\xi) + \frac{1}{3} \xi \bigr) &= 0
\end{align}
Hence, either $\phi(\xi)=0$ or $\phi’(\xi) + \frac{1}{3} \xi=0$. We use the condition that $\int_{-\infty}^\infty v\,dx$ is finite to obtain a solution. In particular,
\begin{equation}
v(x,t) =\left\{\begin{aligned} -\frac{x^2}{6t}& & |x| \le C t^{\frac{1}{3}}  \\ 0& & |x| > C t^{\frac{1}{3}} \end{aligned}\right.
\end{equation}
for some $C>0$. Note that $\int_{-\infty}^\infty v\,dx = - \bigl( \frac{C}{3} \bigr) ^ {3}$ is a finite constant.

5
Web Bonus = Nov / Re: Web Bonus Problem to Week 9 (#1)
« on: November 30, 2015, 12:38:05 AM »
Substitute the partial derivatives of $u(x,y)=X(x)Y(y)$ into the pde and divide by $X^{(2)}Y^{(2)}$.

\begin{equation}
\frac{\frac{X^{(4)}}{X^{(2)}}}{\frac{Y^{(2)}}{Y}} + 2 + \frac{\frac{Y^{(4)}}{Y^{(2)}}}{\frac{X^{(2)}}{X}}=0
\end{equation}

One set of solutions can be obtained by assuming the following:
\begin{equation}
\frac{X^{(4)}}{X^{(2)}} = \frac{X^{(2)}}{X} = \omega^2 \qquad \frac{Y^{(4)}}{Y^{(2)}} = \frac{Y^{(2)}}{Y} = -\omega^2
\end{equation}

where $\omega > 0$. In this case, the solution is as follows:
\begin{equation}
u(x,y)=X(x)Y(y)=(A\cosh\omega x + B\sinh\omega x) (C\cos\omega y + D\sin\omega y)
\end{equation}

Similarly, another solution is as follows:
\begin{equation}
u(x,y)=X(x)Y(y)=(A\cos\omega x + B\sin\omega x) (C\cosh\omega y + D\sinh\omega y)
\end{equation}

6
HA7 / Re: HA7-P6
« on: November 15, 2015, 11:44:54 PM »
It's a convolution. We are applying Theorem 4a of Section 5.2.

7
HA6 / Quiz 4 Solution
« on: October 31, 2015, 01:42:46 PM »
Let $y = x - L$. The problem can be written as follows.

\begin{equation}
u_{tt} + Ku_{yyyy} = 0
\end{equation}

\begin{equation}
u(-L,t) = u_y(-L,t) = u(L,t) = u_y(L,t) = 0 \label{BC}
\end{equation}

Let $u(y,t) = Y(y)T(t)$ and apply separation of  variables.
\begin{equation}
\frac{Y^{(4)}(y)}{Y(y)} = \omega^4 \label{y}
\end{equation}

The general solution to (\ref{y}) is
\begin{equation}
Y(y) = A\cosh(\omega y) + B\sinh(\omega y) + C\cos(\omega y) + D\sin(\omega y)
\end{equation}

Boundary conditions (\ref{BC}) imply

\begin{align}
Y(-L) &= A\cosh(\omega L) - B\sinh(\omega L) + C\cos(\omega L) - D\sin(\omega L) = 0\\
Y(L) &= A\cosh(\omega L) + B\sinh(\omega L) + C\cos(\omega L) + D\sin(\omega L) = 0 \\
Y'(-L) &= -A\sinh(\omega L) + B\cosh(\omega L) + C\sin(\omega L) + D\cos(\omega L) = 0 \\
Y'(L) &= A\sinh(\omega L) + B\cosh(\omega L) - C\sin(\omega L) + D\cos(\omega L) = 0
\end{align}

This $4\times4$ system is equivalent to two $2\times2$ systems:

\begin{equation}
 \left[\begin{array}{cc} \sinh(\omega L) & -\sin(\omega L) \\ \cosh(\omega L) & \cos(\omega L) \end{array} \right] \left[\begin{array}{cc} A \\ C \end{array} \right] \equiv \Delta_2 \left[\begin{array}{cc} A \\ C \end{array} \right] = 0
\end{equation}

\begin{equation}
 \left[\begin{array}{cc} \sinh(\omega L) & \sin(\omega L) \\ \cosh(\omega L) & \cos(\omega L) \end{array} \right] \left[\begin{array}{cc} B \\ D \end{array} \right] \equiv \Delta_1 \left[\begin{array}{cc} B \\ D \end{array} \right] = 0
\end{equation}

For any given $\omega \in {\rm I\!R}$, $\Delta_1=\Delta_2=0$ iff $\omega = 0$. Since $0$ is not an eigenvalue, every eigenvalue corresponds to either an even or an odd eigenfunction. For odd eigenfunctions, the coefficients $[B \space D]'$ belong to the null space of $\Delta_1$. For even eigenfunctions, the coefficients $[A \space C]'$ belong to the null space of $\Delta_2$. Because both of these 2 null spaces are 1-dimensional, eigenvalues are simple. All eigenfunctions corresponding to the same eigenvalue are proportional.

A general odd eigenfunction can be written as follows:

\begin{equation}
Y(y) = B\sin(\omega_{odd} L)\sinh(\omega_{odd} y) - B\sinh(\omega_{odd} L) \sin(\omega_{odd} y)
\end{equation}

where $\omega_{odd}$ is a solution to $\tanh(\omega L) = \tan(\omega L)$.


A general even eigenfunction can be written as follows:

\begin{equation}
Y(y) = A\cos(\omega_{even} L)\cosh(\omega_{even} y) - A\cosh(\omega_{even} L) \cos(\omega_{even} y)
\end{equation}

where $\omega_{even}$ is a solution to $\tanh(\omega L) = - \tan(\omega L)$.

8
Web Bonus = Oct / Re: Web Bonus Problem to Week 7 (#2)
« on: October 29, 2015, 12:36:47 AM »
Rewrite the problem as follows:
\begin{align} &\frac{u_{xx}}{u} = - (H + \lambda) \qquad & |x|\le L \\
&\frac{u_{xx}}{u} = - \lambda \qquad & |x| > L
\end{align}

For $\lambda \in (−H,0)$, $−\lambda$ is positive and $-(H + \lambda)$ is negative. The solution is in exponential form outside the rectangular well and sinusoidal within the well. All eigenvalues therefore belong to the interval $−H < \lambda < 0$.

Consider even and odd eigenfunctions separately. Even eigenfunctions take the following form:
\begin{align} &u = A \cos\omega x \qquad & |x|\le L \\
&u = Be^{-\gamma |x|} \qquad & |x|> L
\end{align}
where $\omega^2 = H + \lambda$ and $\gamma^2 = -\lambda$.

Continuity and continuous first derivative at $L$ imply
\begin{align} A \cos\omega L &= Be^{-\gamma L} \\
A \omega \sin\omega L &= \gamma Be^{-\gamma L}
\end{align}

The corresponding eigenvalues are solutions to the following:
\begin{equation}\sqrt{H + \lambda} \tan(\sqrt{H + \lambda} L) = \sqrt{-\lambda}  \qquad \qquad −H < \lambda < 0 \end{equation}


Similarly, odd eigenfunctions take the following form:
\begin{align} &u = A \sin\omega x \qquad &|x|\le L \\
&u = Be^{-\gamma x} \qquad &x > L \\
&u = -Be^{\gamma x} \qquad &x < -L
\end{align}

Continuity and continuous first derivative at $L$ imply
\begin{equation} \omega \cot\omega L = -\gamma \end{equation}

The corresponding eigenvalues are solutions to the following:
\begin{equation}\sqrt{H + \lambda} \cot(\sqrt{H + \lambda} L )= -\sqrt{-\lambda}  \qquad \qquad −H < \lambda < 0 \end{equation}

9
HA3 / Re: HA3-P6
« on: October 08, 2015, 10:54:57 AM »
Thanks Prof. I have revised the solution.

10
HA3 / Re: HA3-P6
« on: October 08, 2015, 09:16:21 AM »
What happened to the Jacobian? Do we need to multiply the integral by $2c$?

11
HA3 / Re: HA3-P6
« on: October 07, 2015, 12:19:57 AM »
Transform the problem into the first quadrant of the characteristic coordinates $(\xi,\eta)$.
\begin{align}
-4c^2\tilde{u}_{\xi \eta} &= \tilde{f}(\xi, \eta) && \xi > 0, \eta > 0 \label{a}  \\
\tilde{u}|_{\xi=0} &= \tilde{g} \left( -\frac{\eta}{2c} \right) && \eta > 0  \\
\tilde{u}|_{\eta=0} &= \tilde{h} \left( \frac{\xi}{2c} \right)   && \xi > 0
\end{align}
The solution to $(\ref{a})$ is as follows.
\begin{align}
\tilde{u}(\xi,\eta)= -\frac{1}{4c^2} \int_0 ^\xi
\int_0 ^\eta \tilde{f}(\xi',\eta' )\,d\eta' d\xi' + \psi(\eta) + \phi(\xi)
\end{align}
The domain of dependence is a rectangle defined as $\tilde{R}(\xi,\eta) = \{ (\xi',\eta') \vert 0< \xi' < \xi,\, 0< \eta' < \eta\}$.

Assume $\psi(0) = \phi(0) = \frac{1}{2}g(0) = \frac{1}{2}h(0)$.
\begin{align}
\phi(\xi) &= h \left( \frac{\xi}{2c} \right) - \frac{1}{2}h(0)\\
\psi(\eta) &= g \left( -\frac{\eta}{2c} \right) - \frac{1}{2}g(0)
\end{align}

Translate back to the $(x,t)$ coordinates. Use the hint provided and the fact that the Jacobian is equal to $2c$.
\begin{align}
u(x,t) = -\frac{1}{2c}\iint _{R(x,t)} f(x',t')\,dx'dt' + h \left( \frac{x+ct}{2c} \right) + g \left( -\frac{x-ct}{2c} \right) - h(0)
\end{align}
where $R(x,t)=\{ (x',t'):\, 0< x'-ct' < x-ct,\, 0< x'+ct' < x+ct\}$.

12
HA3 / Re: HA3-P5
« on: October 06, 2015, 11:02:06 PM »
This is what I got for c:

\begin{align} u(x,t) = \frac{1}{2c^2}[F(x+ct)+F(x-ct)-2F(x)] \end{align}

For d, apply integration by part, which is why we need the third derivative $f(x)=F'''(x)$.

\begin{align}
u(x,t) &= \frac{1}{2c}\int^t_0[t'F''(x+c(t-t'))-t'F''(x-c(t-t'))]dt' \\
&= -\frac{1}{2c^2} \left\{  \int^t_0 t'dF'(x+c(t-t')) + \int^t_0 t'dF'(x-c(t-t')) \right\} \\
&= -\frac{1}{c^2}tF'(x) + \frac{1}{2c^3} [F(x+ct)+F(x-ct)]
\end{align}

13
HA2 / Re: HA2-P5
« on: September 30, 2015, 08:11:18 AM »
I think so. You probably will have 2 equations for the positive and negative square roots.

14
HA2 / Re: HA2-P4
« on: September 29, 2015, 01:27:15 AM »
$$u_x+3u_y=xy\\
\frac{dx}{1} = \frac{dy}{3} = \frac{du}{xy}\\
3x-y = C\\
du=xydx=x(3x-C)dx=(3x^2-Cx)dx\\
u = x^3 - \frac{C}{2}x^2 + \phi(3x-y)\\
u = x^3 - \frac{3x-y}{2}x^2 + \phi(3x-y)\\
u|_{x=0}=\phi(-y)=0\\
u = x^3 - \frac{3x-y}{2}x^2=-\frac{1}{2}x^3+\frac{1}{2}x^2y
$$

15
HA2 / Re: HA2-P5
« on: September 29, 2015, 12:58:57 AM »
Problem 5a
$$yu_x-xu_y=x\\
\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x}\\
x^2+y^2 = C\\
u = -y + \phi(x^2+y^2)$$

Problem 5b
$$yu_x-xu_y=x^2\\
\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x^2}\\
x^2+y^2 = C\\
\left\{\begin{array}{2}
x = r\cos\theta\\
y = r\sin\theta\\
\end{array}\right.\\
du = -xdy = -r^2\cos^2\theta d\theta\\
u = - \frac{r^2}{2} \left(\theta + \frac{1}{2}\sin2\theta\right) + \psi(r)$$

Problem 5c. I think there is a typo in equation (14). The coefficient on $u_x$ should be $y$. Yes indeed. V.I.
$$yu_x+xu_y=x\\
\frac{dx}{y} = \frac{dy}{x} = \frac{du}{x}\\
x^2-y^2 = C\\
u = y + \zeta(x^2-y^2)$$

Problem 5d
$$yu_x+xu_y=x^2\\
\frac{dx}{y} = \frac{dy}{x} = \frac{du}{x^2}\\
x^2-y^2 = C,\qquad -\infty < C < \infty$$

For $0 < C < \infty$, let $C = r^2$.
$$\left\{\begin{array}{2}x = r\sec\theta\\
y = r\tan\theta\\
\end{array}\right.\\
du = xdy = r^2\sec^3\theta d\theta\\
u =  \frac{r^2}{2} \left(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|\right) + \psi(r)$$

For $-\infty < C < 0$, let $C = -r^2$.
$$\left\{\begin{array}{2}
x = r\tan\theta\\
y = r\sec\theta\\
\end{array}\right.\\
du = xdy = r^2\tan^2\theta\sec\theta d\theta = r^2\left(\sec^3\theta - \sec\theta\right)d\theta\\
u =  \frac{r^2}{2} \left(\sec\theta \tan\theta - \ln|\sec\theta+\tan\theta|\right) + \phi(r)$$

At ${x = \pm y}$, $u$ is undefined and discontinuous.


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