1
Test 2 / Re: Solutions to TT2
« on: November 24, 2015, 04:13:56 PM »
Professor, in problem 4, is equation (4.9) possibly wrong? There was no $\pi$ anywhere, same for the result equation (4.10). I think you forgot $\frac{2}{\pi}$
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2
HA7 / Re: HA7-P3
« on: November 12, 2015, 11:03:39 AM »
Thanks Emily that helped a lot. Still hope prof can clarify this. I wish I can simply drop the coefficient.
3
HA7 / Re: HA7-P3
« on: November 12, 2015, 01:22:25 AM »
Professor, I was wondering if the coefficient before the integration of FT and IFT is important? Can we simply assume $1$? I have seen different forms as $ \frac{1}{2\pi} $ and $ \frac{1}{ \sqrt{2\pi}} $, when should we use which? It seems if we don't use the same coefficient we can't prove problem 1. Also for this question part (a), answer $ \frac{\pi}{\alpha}e^{-\alpha|x|} $ makes sense as well.
5
Test 1 / Re: TT1-P3
« on: October 21, 2015, 10:08:44 PM »
For $ 0 < x < 2t$:
\begin{equation} \phi (2t) + \psi (-2t) = e^{-2t} \end{equation}
Let $ x = -2t$
\begin{equation} \phi (-x) + \psi (x) = e^{x} \end{equation}
\begin{equation} \psi (x) = -\phi (-x) + e^{x} \end{equation}
Here $ \phi(-x) = -e^x$. So
\begin{equation} \psi (x) = e^x + e^{x} = 2e^x \ \ \ \ \ \ x < 0 \end{equation}
\begin{equation} u(x,t) = \phi (x+2t) + \psi (x-2t) = -e^{-x-2t} + 2e^{x-2t} \end{equation}
Please correct me if I'm wrong
\begin{equation} \phi (2t) + \psi (-2t) = e^{-2t} \end{equation}
Let $ x = -2t$
\begin{equation} \phi (-x) + \psi (x) = e^{x} \end{equation}
\begin{equation} \psi (x) = -\phi (-x) + e^{x} \end{equation}
Here $ \phi(-x) = -e^x$. So
\begin{equation} \psi (x) = e^x + e^{x} = 2e^x \ \ \ \ \ \ x < 0 \end{equation}
\begin{equation} u(x,t) = \phi (x+2t) + \psi (x-2t) = -e^{-x-2t} + 2e^{x-2t} \end{equation}
Please correct me if I'm wrong
6
Web Bonus = Oct / Re: Web bonus problem : Week 4 (#4)
« on: October 21, 2015, 10:02:55 PM »
a)
\begin{equation} \frac{\partial E(t)}{\partial t} = \int_0^\infty (u_tu_{tt} +c^2u_xu_{xt}) dx + au(0)u_t(0) \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = \int_0^\infty (c^2u_tu_{xx} +c^2u_xu_{xt}) dx + au(0)u_t(0) \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = c^2\int_0^\infty (u_tu_x)_x dx + au(0)u_t(0) \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = u_t(au - c^2u_x)|_{x=0} \end{equation}
We need
\begin{equation} \frac{\alpha_1}{\alpha_0} = \frac{a}{-c^2} \ \ \ \rightarrow \ \ \ a = - \frac{\alpha_1}{\alpha_0}c^2 \end{equation}
b)
\begin{equation} \frac{\partial E(t)}{\partial t} = \int_0^l (u_tu_{tt} +c^2u_xu_{xt}) dx + auu_t|_{x=0} + buu_t|_{x=l} \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} =u_t( c^2u_x +bu)|_{x=l} + u_t(au - c^2u_x)|_{x=0} \end{equation}
Thus we need
\begin{equation} -\frac{\beta_1}{\beta_0} = \frac{b}{c^2} \ \ \ \rightarrow \ \ \ b = - \frac{\beta_1}{\beta_0}c^2 \end{equation}
\begin{equation} -\frac{\beta_1}{\beta_0} = -\frac{a}{c^2} \ \ \ \rightarrow \ \ \ a = \frac{\beta_1}{\beta_0}c^2 \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = \int_0^\infty (u_tu_{tt} +c^2u_xu_{xt}) dx + au(0)u_t(0) \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = \int_0^\infty (c^2u_tu_{xx} +c^2u_xu_{xt}) dx + au(0)u_t(0) \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = c^2\int_0^\infty (u_tu_x)_x dx + au(0)u_t(0) \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = u_t(au - c^2u_x)|_{x=0} \end{equation}
We need
\begin{equation} \frac{\alpha_1}{\alpha_0} = \frac{a}{-c^2} \ \ \ \rightarrow \ \ \ a = - \frac{\alpha_1}{\alpha_0}c^2 \end{equation}
b)
\begin{equation} \frac{\partial E(t)}{\partial t} = \int_0^l (u_tu_{tt} +c^2u_xu_{xt}) dx + auu_t|_{x=0} + buu_t|_{x=l} \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} =u_t( c^2u_x +bu)|_{x=l} + u_t(au - c^2u_x)|_{x=0} \end{equation}
Thus we need
\begin{equation} -\frac{\beta_1}{\beta_0} = \frac{b}{c^2} \ \ \ \rightarrow \ \ \ b = - \frac{\beta_1}{\beta_0}c^2 \end{equation}
\begin{equation} -\frac{\beta_1}{\beta_0} = -\frac{a}{c^2} \ \ \ \rightarrow \ \ \ a = \frac{\beta_1}{\beta_0}c^2 \end{equation}
7
Web Bonus = Oct / Re: Web bonus problem : Week 4 (#2)
« on: October 21, 2015, 09:38:54 PM »
To prove energy conservation law, we need to show $\partial E(t)/ \partial t = 0$
\begin{equation} \frac{\partial E(t)}{\partial t} = \frac{1}{2} \int_0^\infty (2u_tu_{tt} +2c^2u_xu_{xt}+f(u)u_t) dx \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = \frac{1}{2} \int_0^\infty (2u_t(u_{tt}+f(u)) +2c^2u_xu_{xt}) dx \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = \frac{1}{2} \int_0^\infty (2c^2u_tu_{xx} +2c^2u_xu_{xt}) dx \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = c^2 \int_0^\infty \partial_x(u_tu_x) dx \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = c^2 ( u_tu_x|_{x=\infty} - u_tu_x|_{x=0} ) \end{equation}
For Dirichlet condition, $ u|_{x=0} = 0 \Rightarrow $u_x|_{x=0} = 0 $u_t|_{x=0} = 0$. Incorrect! You meant not $u_x$ but ?
We also know $u$ vanishes at $\infty$, thus $\partial E(t) / \partial t = 0 $
For Newmann condition, $ u_x|_{x=0} = 0 $. We also know $u$ vanishes at $\infty$, thus $\partial E(t) / \partial t = 0 $
Sorry don't know how to strike through an equation.
\begin{equation} \frac{\partial E(t)}{\partial t} = \frac{1}{2} \int_0^\infty (2u_tu_{tt} +2c^2u_xu_{xt}+f(u)u_t) dx \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = \frac{1}{2} \int_0^\infty (2u_t(u_{tt}+f(u)) +2c^2u_xu_{xt}) dx \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = \frac{1}{2} \int_0^\infty (2c^2u_tu_{xx} +2c^2u_xu_{xt}) dx \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = c^2 \int_0^\infty \partial_x(u_tu_x) dx \end{equation}
\begin{equation} \frac{\partial E(t)}{\partial t} = c^2 ( u_tu_x|_{x=\infty} - u_tu_x|_{x=0} ) \end{equation}
For Dirichlet condition, $ u|_{x=0} = 0 \Rightarrow $
We also know $u$ vanishes at $\infty$, thus $\partial E(t) / \partial t = 0 $
For Newmann condition, $ u_x|_{x=0} = 0 $. We also know $u$ vanishes at $\infty$, thus $\partial E(t) / \partial t = 0 $
Sorry don't know how to strike through an equation.
8
Test 1 / Re: TT1-P3
« on: October 21, 2015, 09:23:59 PM »
\begin{equation} u(x,t) = \phi (x+2t) + \psi (x-2t) \end{equation}
where $\phi (x) = -e^{-x}$ and $\psi (x) = 2e^{-x} $ when $ x > 2t$. When $ 0< x < 2t$, $ \psi (x) = 2e^x$
So for $ x > 2t$: \begin{equation} u(x,t) = -e^{-x-2t} + 2e^{2t-x} \end{equation}
For $ 0 < x < 2t$: \begin{equation} u(x,t) = -e^{-x-2t} + 2e^{x-2t} \end{equation}
where $\phi (x) = -e^{-x}$ and $\psi (x) = 2e^{-x} $ when $ x > 2t$. When $ 0< x < 2t$, $ \psi (x) = 2e^x$
So for $ x > 2t$: \begin{equation} u(x,t) = -e^{-x-2t} + 2e^{2t-x} \end{equation}
For $ 0 < x < 2t$: \begin{equation} u(x,t) = -e^{-x-2t} + 2e^{x-2t} \end{equation}
9
HA5 / Re: HA5-P7
« on: October 20, 2015, 11:20:28 PM »
Can professor explain if Emily's assumption about $u$ decay at infinity is right? If not, how do we do? If so can we justify?
10
HA5 / Re: HA5-P4
« on: October 20, 2015, 11:13:33 PM »
Yeming is right. I will post the result here. Correct me if I'm wrong.
Since the question asks a Dirichlet condition for the convection, in problem 3(c) we have shown the transformation $u(x,t) = U(x-ct,t)$ is not appropriate since $u(0,t) =U(-ct,t) =0$ is not a boundary condition for $U$.
Thus we use transformation $u(x,t) = v(x,t)e^{\alpha x + \beta t}$. Note in this case we have
\begin{equation}
u(0,t) = v(0,t)e^{\beta t} = 0 \rightarrow v(0,t) = 0
\end{equation}
This is a valid boundary condition for v.
Now we figure out how $g(x)$ changed here.
\begin{equation}
u(x,0) = v(x,0)e^{\alpha x} = g(x) = e^{-\epsilon |x|} \rightarrow v(x,0) = e^{-\epsilon |x| - \alpha x}
\end{equation}
Where $\alpha = \frac{c}{2k}$
Then general solution of Dirichlet problem gives
\begin{equation}
u(x,t) = \int_0^\infty (G(x,y,t) - G(x,-y,t))e^{-(\epsilon + \frac{c}{2k}) y}dy
\end{equation}
We can take $y$ out from $|y|$ since its all positive.
Do the same thing Catch has been doing. The final answer is
\begin{equation}
u(x,t) = \frac{e^{c^2-2cx}}{2}(1+erf(\frac{x-ct}{\sqrt{4kt}})) - \frac{e^{c^2+2cx}}{2}(1+erf(\frac{x+ct}{\sqrt{4kt}}))
\end{equation}
Since the question asks a Dirichlet condition for the convection, in problem 3(c) we have shown the transformation $u(x,t) = U(x-ct,t)$ is not appropriate since $u(0,t) =U(-ct,t) =0$ is not a boundary condition for $U$.
Thus we use transformation $u(x,t) = v(x,t)e^{\alpha x + \beta t}$. Note in this case we have
\begin{equation}
u(0,t) = v(0,t)e^{\beta t} = 0 \rightarrow v(0,t) = 0
\end{equation}
This is a valid boundary condition for v.
Now we figure out how $g(x)$ changed here.
\begin{equation}
u(x,0) = v(x,0)e^{\alpha x} = g(x) = e^{-\epsilon |x|} \rightarrow v(x,0) = e^{-\epsilon |x| - \alpha x}
\end{equation}
Where $\alpha = \frac{c}{2k}$
Then general solution of Dirichlet problem gives
\begin{equation}
u(x,t) = \int_0^\infty (G(x,y,t) - G(x,-y,t))e^{-(\epsilon + \frac{c}{2k}) y}dy
\end{equation}
We can take $y$ out from $|y|$ since its all positive.
Do the same thing Catch has been doing. The final answer is
\begin{equation}
u(x,t) = \frac{e^{c^2-2cx}}{2}(1+erf(\frac{x-ct}{\sqrt{4kt}})) - \frac{e^{c^2+2cx}}{2}(1+erf(\frac{x+ct}{\sqrt{4kt}}))
\end{equation}
11
HA5 / Re: HA5-P6
« on: October 19, 2015, 11:05:51 AM »
Maximum principle tells us that the maximum point will be on either t=0, x= lower limit (in this case, -2), x = higher limit (in this case 2). But from Yunheng's answer we can see the maximum point is indeed $(x,t) = (-1,1)$, not what the principle asserts.
The failure of the principle rooted in the possible negative value of $x$. A crucial step of the proof needs
\begin{equation}\label{eq:1}
v_t - kv_{xx} <0
\end{equation}
and the for the imaginary inner max point $(x_0,t_0)$,
\begin{equation} \label{eq:2}
v_t(x_0,t_0) - kv_{xx}(x_0,t_0) \ge 0
\end{equation}
to arrive at a contradiction. Where $v(x,t) = u(x,t) + \epsilon x^2$.
We can see in this example $k$ is changed to $x$, which is not a fixed positive value anymore, it has a chance of getting negative to fail both of these two equations. (Of course in the specific example t=1 is on the upper boundary, the second equation is proved differently than an inner point, but we nevertheless will arrive at (\ref{eq:2}) for contradiction purpose. Furthermore possible failure in (\ref{eq:1}) suffices.)
The failure of the principle rooted in the possible negative value of $x$. A crucial step of the proof needs
\begin{equation}\label{eq:1}
v_t - kv_{xx} <0
\end{equation}
and the for the imaginary inner max point $(x_0,t_0)$,
\begin{equation} \label{eq:2}
v_t(x_0,t_0) - kv_{xx}(x_0,t_0) \ge 0
\end{equation}
to arrive at a contradiction. Where $v(x,t) = u(x,t) + \epsilon x^2$.
We can see in this example $k$ is changed to $x$, which is not a fixed positive value anymore, it has a chance of getting negative to fail both of these two equations. (Of course in the specific example t=1 is on the upper boundary, the second equation is proved differently than an inner point, but we nevertheless will arrive at (\ref{eq:2}) for contradiction purpose. Furthermore possible failure in (\ref{eq:1}) suffices.)
12
HA5 / Re: HA5-P3
« on: October 18, 2015, 03:10:07 PM »(e)I will work on this part in a little while. So far I think I have gotten all of the same solutions as Rong Wei.Added my solution below.
For the case of the half-line and Dirichlet boundary condition, we will have the solution: \begin{equation}
u(x,t) = \frac{e^{\alpha{}x + \beta{}t}}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/4kt} - e^{-(x+y)^2/4kt}]g(y)dy \end{equation}
In the case of Neumann boundary conditions, we cannot use a similar method.
I don't think the solution is right here, unless we assume $v(x,0) = g(x)$. But usually we use $u(x,0) = g(x)$, then in this case \begin{equation}u(x,0) = v(x,0)e^{\alpha x} = g(x) \rightarrow v(x,0) = g(x)e^{-\alpha x} \end{equation}
Dirichlet condition transforms to:
\begin{equation} u(0,t) = v(0,t)e^{\beta t} = 0 \rightarrow v(0,t) = 0 \end{equation}
Thus we need to solve \begin{equation}v_t = kv_{xx} \end{equation}
\begin{equation} v(x,0) = g(x)e^{-\alpha x} \end{equation}
\begin{equation} v(0,t) = 0 \end{equation}
The answer is then the general result:
\begin{equation}
v(x,t) = \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/4kt} - e^{-(x+y)^2/4kt}]g(y)e^{-\alpha y}dy \end{equation}
Then
\begin{equation}
u(x,t) = \frac{e^{\frac{c}{2k}x -\frac{c^2}{4k}t}}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/4kt} - e^{-(x+y)^2/4kt}]g(y)e^{-\frac{c}{2k} y}dy \end{equation}
13
HA5 / Re: HA5-P2
« on: October 18, 2015, 11:40:19 AM »
The core is to invent a initial function that satisfies the boundary conditions as well as defined on the whole line.
\begin{equation} \end{equation}
a) need to think of an initial function, consider:
$$
f(x) = \left\{\begin{aligned}
&-g(-x) && -L<x<0 \\
&0 && x=...-2L, -L, 0, L, 2L ... \\
&g(x) &&0<x<L \\
&extended\ to\ be\ 2L-periodic \\
\end{aligned}
\right.$$
This function satisfy Dirichlet boundary conditions on the whole line.
Thus our solution is now:
\begin{equation} v(x,t) = \int_{-\infty}^{\infty} G(x,y,t)f(y)dy \end{equation}
Depends on the relative position of (x,t), the integral can be valued piecewisely.
b) by similar fashion, consider:
$$
f(x) = \left\{\begin{aligned}
&g(-x) && -L<x\le 0 \\
&g(x) &&0\le x<L \\
&extended\ to\ be\ 2L-periodic \\
\end{aligned}
\right.$$
This function is even, thus its derivative is an odd function, which satisfies Neumann condition.
\begin{equation} v(x,t) = \int_{-\infty}^{\infty} G(x,y,t)f(y)dy \end{equation} can be valued based on (x,t)'s position.
c) the condition ask for a function f(x) that has odd continuation at 0 + 2kL, but even continuation at L+2kL. Thus its period is 4L. consider:
$$
f(x) = \left\{\begin{aligned}
&-g(x-2L) && -2L<x\le -L\\
&-g(-x) && -L\le x<0 \\
&0 && x=...-4L,-2L , 0, 2L, 4L... \\
&g(x) &&0<x\le L \\
&g(2L-x) && L\le x<2L\\
&extended\ to\ be\ 4L-periodic \\
\end{aligned}
\right.$$
Solution \begin{equation} v(x,t) = \int_{-\infty}^{\infty} G(x,y,t)f(y)dy \end{equation} then can be valued based on (x,t)'s position.
\begin{equation} \end{equation}
a) need to think of an initial function, consider:
$$
f(x) = \left\{\begin{aligned}
&-g(-x) && -L<x<0 \\
&0 && x=...-2L, -L, 0, L, 2L ... \\
&g(x) &&0<x<L \\
&extended\ to\ be\ 2L-periodic \\
\end{aligned}
\right.$$
This function satisfy Dirichlet boundary conditions on the whole line.
Thus our solution is now:
\begin{equation} v(x,t) = \int_{-\infty}^{\infty} G(x,y,t)f(y)dy \end{equation}
Depends on the relative position of (x,t), the integral can be valued piecewisely.
b) by similar fashion, consider:
$$
f(x) = \left\{\begin{aligned}
&g(-x) && -L<x\le 0 \\
&g(x) &&0\le x<L \\
&extended\ to\ be\ 2L-periodic \\
\end{aligned}
\right.$$
This function is even, thus its derivative is an odd function, which satisfies Neumann condition.
\begin{equation} v(x,t) = \int_{-\infty}^{\infty} G(x,y,t)f(y)dy \end{equation} can be valued based on (x,t)'s position.
c) the condition ask for a function f(x) that has odd continuation at 0 + 2kL, but even continuation at L+2kL. Thus its period is 4L. consider:
$$
f(x) = \left\{\begin{aligned}
&-g(x-2L) && -2L<x\le -L\\
&-g(-x) && -L\le x<0 \\
&0 && x=...-4L,-2L , 0, 2L, 4L... \\
&g(x) &&0<x\le L \\
&g(2L-x) && L\le x<2L\\
&extended\ to\ be\ 4L-periodic \\
\end{aligned}
\right.$$
Solution \begin{equation} v(x,t) = \int_{-\infty}^{\infty} G(x,y,t)f(y)dy \end{equation} then can be valued based on (x,t)'s position.
14
HA4 / Re: HA4-P3
« on: October 15, 2015, 05:31:07 PM »My questions is about boundary condition. For the general form of 1D wave, there is no v present. We used boundary condition\begin{equation} u_{x=0} = 0 \end{equation}. Here you are asking v. I understand that we need conditions, but why here we need:\begin{equation} u_{x=vt} = something \end{equation} but not \begin{equation} u_{x=ct} = something \end{equation} or \begin{equation} u_{x=0} = something \end{equation}Thanks professor but I understand the process of doing this but not why we need initial condition Ux=vt here but not Ux=0, or similarly when with problems without v, why not use boundary condition Ux=ct?
This is completely incomprehensible. What do you really mean by this charade?
And what is v? where does it come from and how it relates to c here?
Hope that clarifies my questions. Sorry for the confusion.
15
HA4 / Re: HA4-P3
« on: October 15, 2015, 12:13:24 PM »Fei Fan Wu provided almost full explanations. Solution is given by $u=\phi(x+ct)+\psi(x-ct)$ and initial data provide us with $\phi(x)$ and $\psi(x)$ as $x>0$. So $\phi(x+ct)$ is defined as $x+ct>0$ and $\psi(x-ct)$ is defined as $x-ct>0$.
a) $v>c$. Then $x+ct > x-ct >0$ as $x>vt$ ($t>0$). completely defined!
b) $c>v> -c$. Then $x+ct$ as $x>vt$ but $x-ct$ could be less than $0$ so we need to find $\psi(x)$ for negative $x$. For this we need 1 boundary condition. Plug $u$ to it:
\begin{equation}
(\alpha + \beta c) \phi' ((v+c)t) + (\alpha - \beta c )\psi' ((v-c)t) =0,\qquad t>0.
\end{equation}
We can always find $\psi$ out of it except one special case which should be a mentioned.
c) $v<-c$. Then both $x-ct$ and $x+ct$ could be negative and we need to find $\phi(x)$ and $\psi(x)$ as $x<0$. We need two boundary conditions. Plug $u$ into boundary conditions:
\begin{align}
&\phi((v+c)t) + \phi'(v-c)t)=0,\\
$c\phi '( (v+c)t)-c\psi'((v-c)t)=0.
\end{align}
Integrating the second one (never mind, constant $C$ could be safely taken $0$) and comparing with the first one we find that $\phi(x)=\psi(x)=0$ as $x<0$.
QED
Thanks professor but I understand the process of doing this but not why we need initial condition Ux=vt here but not Ux=0, or similarly when with problems without v, why not use boundary condition Ux=ct?
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