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Messages - Yeming Wen

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HA7 / Re: HA7-P6
« on: November 06, 2015, 07:56:52 AM »
part b.
First, we make a partial fourier transform $u(x,y) \mapsto \hat{u}(k,y)$.
Then $\hat{u}_{kk} = -k^2 \hat{u}$.
Thus we have $$\hat{u}_{yy}-k^2\hat{u}=0$$
Solve this ODE, we have $$\hat{u}=A(k)e^{-|k|y}+B(k)e^{|k|y}$$
Since the domain goes to infinity and we don't want the solution to escape to infinity, then $B(k)=0$.
Thus, $$\hat{u}=A(k)e^{-|k|y}$$
Notice $$\hat{u}_y=-|k|\cdot A(k)e^{-|k|y}$$
So we plug in the boundary condition, we get $$\hat{u}_y=-|k|\cdot A(k)=\hat{f}(k) \implies A(k)=-\frac{\hat{f}(k)}{|k|}$$
Thus $$\hat{u} = -\frac{\hat{f}(k)}{|k|}e^{-|k|y}$$
And we write $u$ in a fourier integral form as $$u=\int^{\infty}_{-\infty} -\frac{\hat{f}(k)}{|k|}e^{-|k|y} e^{ikx} dk$$

HA7 / Re: HA7-P4
« on: November 05, 2015, 04:20:34 PM »
Solution for a(i) is attached. Please correct me if I am wrong.

Maybe the transform for $e^{-\frac{ax^2}{2}}$ should be $\sqrt{\frac{2\pi}{a}}e^{-\frac{k^2}{2a}}$ instead of $\sqrt{\frac{2\pi}{a}}e^{-\frac{ak^2}{2}}$?

HA6 / Re: hm6 Q1
« on: October 29, 2015, 12:00:25 PM »
I think the question is asking the case where $\lambda =0$.

HA6 / Re: HW6-P2
« on: October 28, 2015, 05:19:08 PM »
Just wondering, the boundary condition in this question is $X(0)=0$ and $X'(l)+\beta X(l)=0$?

Quiz 4 / Re: Will Quiz 4 be next week?
« on: October 22, 2015, 01:25:50 PM »
It says we will have quiz 4 on the official website, which covers HW6.

Test 1 / Re: TT1-P5
« on: October 21, 2015, 11:28:16 PM »
Notice we only need to integrate $$\frac{1}{\sqrt{4\pi kt}}\int^1_{-1} ye^{-\frac{(x-y)^2}{4kt}}dy$$
Then we change of variable $z=\frac{y-x}{\sqrt{4kt}}$.
So the integral becomes
\frac{1}{\sqrt{4\pi kt}}\int^a_b\sqrt{4kt}(x+\sqrt{4t}z)e^{-z^2}dz &= \frac{1}{\sqrt{\pi}}\int^a_b (x+2\sqrt{t}z)e^{-z^2}dz \\
&= \frac{x}{\sqrt{\pi}}\int^a_b e^{-z^2}dz+\frac{1}{\sqrt{\pi}}\int^a_b 2\sqrt{t}ze^{-z^2}dz\\
&= \frac{x}{2}[E(a)-E(b)] +\frac{2\sqrt{t}}{\sqrt{\pi}}(-\frac{1}{2}e^{-z^2}) \rvert^a_b\\
&= \frac{x}{2}[E(a)-E(b)] + \sqrt{\frac{t}{\pi}}[e^{-\frac{(1+x)^2}{4t}}-e^{-\frac{(1-x)^2}{4t}}]
where $$a=\frac{1-x}{2\sqrt{t}}$$ and $$b=\frac{-1-x}{2\sqrt{t}}$$ and $E$ is the error function.

HA5 / Re: HA5-P5
« on: October 20, 2015, 09:42:13 PM »
For part a, when we increase $T$, we expand our domain, so $M(T)$ should be at least non-decreasing.
Similarly, for part b, when we expand the domain, $m(T)$ is non-increasing.

HA5 / Re: HA5-P4
« on: October 20, 2015, 09:32:17 PM »
I think 4(c) is asking us to find the solution for convection heat equation, which it is in 3(e), not the ordinary heat equation.

HA4 / Re: HA4-P2
« on: October 15, 2015, 04:20:46 PM »
Suppose $u=\phi_1(x+c_1t)+\psi_1(x-c_1t)$ for $x>0$ and $u=\phi_2(x+c_2t)+\psi_2(x-c_2t)$ for $x<0$.
Also we can solve $\phi_1(x)=\phi(x)-\frac{1}{2}\phi(0)$, $\psi_1(x)=\frac{1}{2}\phi(0)$ for $x>0$.
And $\phi_2=\psi_2=0$ for $x<0$.
It follows that $u=\phi(x+c_1t)$ for $x>c_1t$ and $u=0$ for $x<-c_2t$.
Now we are interested in $\psi_1(x)$ when $x<0$ and $\phi_2(x)$ when $x>0$.
Notice from boundary condition, for $t>0$
Solve for $\psi_1$ and $\phi_2$, we have
$$\psi_1(x)=\frac{c_1\beta-ac_2}{ac_2+c_1\beta}\phi_1(-x) \enspace \text{for   } x<0$$
$$\phi_2(x)=\frac{2c_2\phi_1(\frac{c_1x}{c_2})}{ac_2+c_1\beta} \enspace \text{for   } x>0$$
Thus, for $-c_2t<x<0$,
For $0<x<c_1t$,

HA4 / Re: HA4-P5
« on: October 15, 2015, 12:11:06 PM »
Can we say $t=0\implies \zeta=\eta$?

HA4 / Re: HA4-P5
« on: October 15, 2015, 11:37:02 AM »
By the hint, we do a change of variable. Let $$x=\frac{1}{2}(\zeta+\eta)$$ $$t=\sinh(\frac{1}{2}(\zeta-\eta))$$
Then we compute $u_{\zeta}$ and $u_{\zeta\eta}$.
By chain rule,
$$u_\zeta=\frac{1}{2}u_x+u_t\cdot t_\zeta=\frac{1}{2}(u_x+u_t\cdot \cosh(\frac{1}{2}(\zeta-\eta)))$$
Similarly, $$u_{\zeta\eta}=\frac{1}{2}([u_x]_\eta+[u_t\cdot \cosh(\frac{1}{2}(\zeta-\eta)]_\eta)$$
$$[u_t\cdot \cosh(\frac{1}{2}(\zeta-\eta)]_\eta=[u_t]_\eta\cosh(\frac{1}{2}(\zeta-\eta))+u_t[\cosh(\frac{1}{2}(\zeta-\eta))]_\eta=\frac{1}{2}u_{tx}\cosh(\frac{1}{2}(\zeta-\eta))+u_{tt}[\cosh(\frac{1}{2}(\zeta-\eta))]^2-\frac{1}{2}u_t\sinh(\frac{1}{2}(\zeta-\eta))$$
Notice that $$[\cosh(\frac{1}{2}(\zeta-\eta))]^2=1+[\sinh(\frac{1}{2}(\zeta-\eta))]^2=t^2+1$$
We have $$u_{\zeta\eta}=\frac{1}{2}[\frac{1}{2}u_{xx}-\frac{1}{2}(t^2+1)u_{tt}-\frac{1}{2}tu_t]=\frac{1}{4}(u_{xx}-(t^2+1)u_{tt}-tu_t)$$
Combine with the original PDE, we have $u_{\zeta\eta}=0$. So $u=g(\eta)+f(\zeta)$.

MT / Re: MT problem 3
« on: October 30, 2014, 01:31:46 AM »
\begin{equation} x^3 y''' + 6x^2 y'' + 5x y' -5y = x^2 \ln x  \label{A} \end{equation}
First we notice that it is Euler equation. We let \begin{equation*} t = \ ln x \end{equation*}
Then Let D denote the operator, which is \begin{equation*} D = \frac{d}{dt} \end{equation*}
Plug in the equation (\ref{A}), we get
\begin{equation*}  D(D-1)(D-2)y + 6D(D-1)y + 5Dy -5y =  e^{2t} \end{equation*}
Expend it, we can have,
\begin{equation*}  D^3y +  3D^2y + Dy - 5y = e^{2t} \end{equation*}
which is,
\begin{equation}  y''' + 3y'' + y' - 5y = e^{-2t} \label{B}  \end{equation}
Now we have to solve (\ref{B}), which is a constant third order ODE.
First we solve the homogeneous one, which is
\begin{equation}  y''' + 3y'' + y' - 5y = 0  \label{C} \end{equation}
The characteristic equation is
\begin{equation*} r^{3} + 3r^2 + r - 5 = (r - 1)(r^2 + 4r + 5) = 0 \end{equation*}
So the solution to (\ref{C}) is
\begin{equation*} y = c_1e^{t} + e^{-2t}(c_2 \cos(t) + c_3 \sin(t)) \end{equation*}
Now it remains to find a particular solution to (\ref{B}).
Since 2 is not a root of the corresponding characteristic polynomial, then we can guess $y=Ae^{2t}$
Plug in (\ref{B}), we have $A = \frac{1}{17}$.
So a general solution to (\ref{B}) is
\begin{equation}  y = c_1e^{t} + e^{-2t}(c_2 \cos(t) + c_3 \sin(t)) + \frac{1}{17}e^{2t}  \label{D} \end{equation}
Substitute $t = \ lnx $
We have
\begin{equation}  y = C_1x + x^{-2}(C_2 \cos(\ ln x) + C_3 \sin(\ ln x)) + \frac{1}{17}x^2   \end{equation}

TT1 / Re: TT1-problem 2
« on: October 09, 2014, 11:13:46 AM »
Part a)
First we make the coefficient of $y''$ one, which results in
y''(x) - \frac{(2\ln x+3)}{x(\ln x+1)} y'(x) + \frac{(2\ln x+3)}{x^2(\ln x+1)}y(x)=0
By Abel' Thm,
\ln W=\int{\frac{(2\ln x+3)}{x(\ln x+1)}}dx\label{A}
Using change of variables to evaluate the integral,let
\mu=\ln x + 1
The the integral becomes
\int{\frac{2\mu+1}{\mu}}du =\int{2}du+\int{\frac{1}{\mu}}du \label{B}
We have \begin{equation*} \ln W=(2\mu+\ln \mu)+C_1   \end{equation*}
Plug in x back, which is \begin{equation} W= C_2x^2(\ln x + 1) \end{equation}

Part b)
To verify y = x is a solution. Just plug in, we have
\begin{equation*} x^3(\ln x+1)\cdot  0 -(2\ln x+3)x^2 \cdot 1 + (2\ln x+3) x^2  = 0 \end{equation*}
Then to find another independent solution, notice that $W > 0$ for $x > 1$
So we can just let $C_2 = 1$. And we know $W = \left| \begin{matrix} x & y \\1 &  y’ \end{matrix}\right|$
Now we have a new ode
\begin{equation}  y' - \frac{1}{x} y = x(\ln x + 1), \qquad x>1. \end{equation}
Solve the homogeneous ode first,
\begin{equation*} z' - \frac{1}{x}z = 0 \end{equation*}
We get $z=x$. Then we know $y=\mu(x)x$.

And $y'=\mu'x+\mu$, we plug in into (3). We have \begin{equation*} u'=(\ln x + 1) \end{equation*}
Then \begin{equation*} u=\int{\ln x +1}dx \end{equation*}
We use integral by parts to evaluate $\int{\ln x}dx$ first.
\begin{equation*} \int{\ln x} dx = x\ln x - \int{1}dx = x\ln x - 1 \end{equation*}
So we have \begin{equation*} \mu x=\ln x + x + C \end{equation*}
Finally, we have $y_2=(x\ln x+x+C)x$.

TT1 / Re: TT1-problem 4
« on: October 09, 2014, 09:28:21 AM »
Observe that the equation is an Euler equation, then we let
\begin{equation*} t = \log x
Then the equation becomes
 y'' - y' - 6y =10 e^{-2t}- 6
It is the same ODE from question 3 .
Use the particular solution from 3 i.e
y=-2te^{-2t} + 1
Plug in $x$ back, we have
\begin{equation*} y=\frac{-2\log x}{x^2} + 1. \end{equation*}

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