Author Topic: Problem 1  (Read 8590 times)

Laurie Deratnay

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Problem 1
« on: September 27, 2012, 10:06:01 AM »
Hi - In the pdf of home assignment 2 in problem 1 the inequalities throughout are different than the other version of the assignment (i.e. pdf version has 'greater than or equal to' and the other version in only '>').  Which one is correct? 
« Last Edit: September 30, 2012, 06:56:47 AM by Victor Ivrii »

Victor Ivrii

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Re: problem 1 typo?
« Reply #1 on: September 27, 2012, 11:10:32 AM »
Hi - In the pdf of home assignment 2 in problem 1 the inequalities throughout are different than the other version of the assignment (i.e. pdf version has 'greater than or equal to' and the other version in only '>').  Which one is correct?

Really does not matter, but I changed pdf to coincide

James McVittie

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Re: problem 1 typo?
« Reply #2 on: September 29, 2012, 08:13:56 AM »
Can we assume that for part (C) of Problem 1 that the Cauchy conditions are evenly reflected for x < 0?

Victor Ivrii

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Re: problem 1 typo?
« Reply #3 on: September 29, 2012, 09:04:56 AM »
Can we assume that for part (C) of Problem 1 that the Cauchy conditions are evenly reflected for x < 0?

Sure, you can but it will not be useful as your domain is $x>vt$ rather than $x>0$. Just use the general solution.

Peishan Wang

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Re: problem 1 typo?
« Reply #4 on: September 29, 2012, 03:08:00 PM »
Professor I have a question for part (c). Does the solution have to be continuous? For example I have f(x) on x>2t, g(x) on -2t<x<2t and h(x) on -3t<x<-2t. Should f(2t) = g(2t) and g(-2t)=h(-2t) (so the overall solution is continuous)?

My problem is that some of the f, g, h involve a constant K and I was wondering if I should use continuity to specify what K is.

Thanks a lot!

Victor Ivrii

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Re: problem 1 typo?
« Reply #5 on: September 29, 2012, 03:10:20 PM »
Professor I have a question for part (c). Does the solution have to be continuous? For example I have f(x) on x>2t, g(x) on -2t<x<2t and h(x) on -3t<x<-2t. Should f(2t) = g(2t) and g(-2t)=h(-2t) (so the overall solution is continuous)?

My problem is that some of the f, g, h involve a constant K and I was wondering if I should use continuity to specify what K is.

Thanks a lot!

You should be able to find constants from initial and boundary conditions. Solutions may be discontinuous along lines you indicated

Calvin Arnott

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Re: problem 1 typo?
« Reply #6 on: September 29, 2012, 04:57:27 PM »
You should be able to find constants from initial and boundary conditions. Solutions may be discontinuous along lines you indicated

Ah, excellent! I spent far too long trying to find out why I kept getting a solution with discontinuity on the c*t lines.

Victor Ivrii

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Re: problem 1 typo?
« Reply #7 on: September 29, 2012, 05:10:55 PM »
Consider this:
\begin{align*}
&u_{tt}-c^2u_{xx}=f(x,t)\qquad \text{as  } x>0,t>0,\\
&u|_{t=0}=g(x),\\
&u_t|_{t=0}=h(x),\\
&u|_{x=0}=p(t)
\end{align*}
has a continuous solution if and only if $p(0)=g(0)$ (compatibility condition) but with the Neumann BC solution would be always $C$ (albeit not necessarily $C^1$.

BTW  heat equation
\begin{align*}
&u_{t}-ku_{xx}=f(x,t)\qquad \text{as  } x>0,t>0,\\
&u|_{t=0}=g(x),\\
&u|_{x=0}=p(t)
\end{align*}
also has a continuous solution if and only if $p(0)=g(0)$ (compatibility condition) but the discontinuity stays in $(0,0)$ rather than propagating along characteristics as for wave equation.

Rouhollah Ramezani

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Re: Problem 1
« Reply #8 on: October 01, 2012, 09:00:03 PM »
A) The problem as is has a unique solution. No extra consitions are necessary. Since $x>3t$, we are confident that $x-2t$ is always positive. I.e. initial value functions are defined everywhere in domain of $u(t,x)$. Using d'Alembert's formula we write:
\begin{equation*}
u(t,x)=\frac{1}{2}\Bigl[e^{-(x+2t)}+e^{-(x-2t)}\Bigr]+\frac{1}{4}\int_{x-2t}^{x+2t}e^{-s}\mathrm{d}s
\end{equation*}
$$= \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$
Stated in this way, $u(t,x)$ is determined uniquely in its domain. OK


B) In this case, we need an extra boundary condition at $u_{|x=t}=0$ to find the unique solution:
For $x>2t$ general formula for $u$ is as part (A):
$$u(t,x) = \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$
For $t<x<2t$, story is different:
$$u(t,x)= \phi(x+2t)+\psi(x-2t)$$
where $\phi(x+2t)= \frac{1}{4}\bigl[e^{-(x+2t)}+1\bigr]$ is determined by initial conditions at $t=0$. To find $\psi(x-2t)$, we impose $u_{|x=t}=0$ to solution:
$$u_{|x=t}=\phi(3t)+\psi(-t)=0$$
$$\Rightarrow \psi(s)=-\phi(-3s)$$
$$=\frac{-1}{4}\bigl[e^{3s}+1\bigr]$$
Hence the general solution for $t<x<2t$ is:
$$u(t,x)= \frac{1}{4}\bigl[e^{-(x+2t)}+1\bigr]-\frac{1}{4}\Bigl[e^{3x-6t}+1\Bigr]$$
Note that we would not be able to determine $\psi$ if we did not have the extra condition  $u|_{x=t}$. Also note that $u_x|_{x=t}=\frac{1}{2}(e^{-3t}) \neq 0$. This means the problem would have been overdetermined, without any solution, if we considered boundary condition $u|_{x=t}=u_x|_{x=t}=0$.


C) In this case we need to impose the strongest boundary condition to get the unique solution:
Case $x>2t$ is identical to part (A) and (B):
$$u(t,x) = \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$
We find the solution in region $-3t<x<-2t$ by imposing boundary conditions $u|_{x=-3t}=u_x|_{x=-3t}=0$ to $u(t,x)= \phi(x+2t)+\psi(x-2t)$. This gives
$$\phi(-t)+\psi(-5t)=0$$
$$\phi'(-t)+\psi'(-5t)=0$$
Differentiating first equation and adding to the second we get $-4\psi'(-5t)=0$. Therefore $\psi(s)=C$, $\phi(s)=-C$ and $u(t,x)$ is identically zero.Note that we would not be able to find $\phi$ and $\psi$ uniquely, if we did not have both boundary conditions at $x=-3t$.

From continuety of $u$ in $t>0$, $x>-3t$, we conclude $u|_{x=-2t}=0$. This helps us to find solution for $-2t<x<2t$. Analogous to part (B) we write:
$$u(t,x)= \phi(x+2t)+\psi(x-2t)$$
Impose $u|_{x=-2t}=0$ to $u$ to get $\psi(-4t)=-\phi(0)=C$. Therefore solution here is:
$$u(t,x)=\frac{1}{4}e^{-(x+2t)}+C$$
By continuity at $x=2t$, we get $C=\frac{3}{4}$. General solution for part (C) can be explicitly formulated as
\begin{equation*}
   u(x,y)=
      \left\{\begin{aligned}[h]
         &\frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}, & x>2t\\
         &\frac{1}{4}e^{-(x+2t)}+\frac{3}{4}, & -2t<x<2t\\
         &0, & -3t<x<-2t\\
         \end{aligned}
      \right.
\end{equation*}
« Last Edit: October 07, 2012, 01:53:31 AM by Rouhollah Ramezani »

Victor Ivrii

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Re: Problem 1 -- not done yet!
« Reply #9 on: October 02, 2012, 06:56:08 AM »
Posted by: Rouhollah Ramezani
« on: October 01, 2012, 09:00:03 pm »

A) is correct

B) definitely contains an error which is easy to fix. Why I know about error? --  solution of RR  is not $0$ as $x=t$

C) Contains a logical error in the domain $\{−2t<x<2t\}$ (`middle sector`) which should be found and fixed. Note that the solution of RR there is not in the form $\phi(x+2t)+\psi(x-2t)$

RR deserves a credit but there will be also a credit to one who fixes it




So, for a wave equation with a propagation speed $c$ and moving boundary (with a speed $v$) there are three cases (we exclude exact equalities $c=\pm v$ ) -- interpret them as a piston in the cylinder:
  • $-c<v<c$ The piston moves with a subsonic speed: one condition as in the case of the staying wall
  • $v>c$ The piston moves in with a supersonic speed: no conditions => shock waves etc
  • $v<-c$ The piston moves out with a supersonic speed: two conditions.
3D analog: a plane moving in the air. If it is subsonic then everywhere on its surface one boundary condition should be given but for a supersonic flight no conditions on the front surface, one on the side surface and two on the rear (with $\vec{v}\cdot \vec{n} >c$, $-c< \vec{v}\cdot \vec{n} <c$ and $\vec{v}\cdot \vec{n} <-c$ respectively where $\vec{v}$ is the plane velocity and $\vec{n}$ is a unit outer normal at the given point to the plane surface. The real fun begins at transonic points where $\vec{v}\cdot \vec{n} =\pm c$).





PS MathJax is not a complete LaTeX and does not intend to be, so it commands like \bf do not work outside of math snippets (note \bf); MathJax has no idea about \newline as it is for text, not math. For formatting text use either html syntax (in plain html) or forum markup

PPS \bf is deprecated, use \mathbf instead

« Last Edit: October 02, 2012, 10:54:47 AM by Victor Ivrii »

Rouhollah Ramezani

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Re: Problem 1 -- not done yet!
« Reply #10 on: October 07, 2012, 02:54:17 AM »
Posted by: Rouhollah Ramezani
« on: October 01, 2012, 09:00:03 pm »

A) is correct

B) definitely contains an error which is easy to fix. Why I know about error? --  solution of RR  is not $0$ as $x=t$

C) Contains a logical error in the domain $\{−2t<x<2t\}$ (`middle sector`) which should be found and fixed. Note that the solution of RR there is not in the form $\phi(x+2t)+\psi(x-2t)$


B) is fixed now.
C) is also amended, but I probably failed to spot the "logical error" and it is still there.


PS MathJax is not a complete LaTeX and does not intend to be, so it commands like \bf do not work outside of math snippets (note \bf); MathJax has no idea about \newline as it is for text, not math. For formatting text use either html syntax (in plain html) or forum markup

Yes, I realized that after. And I found out from your other post that we can actually use an html editor+MathJax instead of LaTeX. right?

PPS \bf is deprecated, use \mathbf instead

Wilco.

Victor Ivrii

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Re: Problem 1 -- not done yet!
« Reply #11 on: October 07, 2012, 11:15:56 AM »
Posted by: Rouhollah Ramezani

C) Contains a logical error in the domain $\{−2t<x<2t\}$ (`middle sector`) which should be found and fixed. Note that the solution of RR there is not in the form $\phi(x+2t)+\psi(x-2t)$


C) is also amended, but I probably failed to spot the "logical error" and it is still there.

You presume that $u$ should be continuous, which is not the case. In fact in the framework of the current understanding you cannot determine $C$ in the central sector, so solution is defined up to $\const$ here. One needs to dig dipper in the notion of the weak solution.

Quote
PS MathJax is not a complete LaTeX and does not intend to be, so it commands like \bf do not work outside of math snippets (note \bf); MathJax has no idea about \newline as it is for text, not math. For formatting text use either html syntax (in plain html) or forum markup

Yes, I realized that after. And I found out from your other post that we can actually use an html editor+MathJax instead of LaTeX. right?


Not really: you do not use html syntax but a special SMF markdown which translates into html (so you cannot insert a raw html-- but Admin  can if needed.

Di Wang

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Re: Problem 1
« Reply #12 on: October 14, 2012, 10:23:39 PM »
A) The problem as is has a unique solution. No extra consitions are necessary. Since $x>3t$, we are confident that $x-2t$ is always positive. I.e. initial value functions are defined everywhere in domain of $u(t,x)$. Using d'Alembert's formula we write:
\begin{equation*}
u(t,x)=\frac{1}{2}\Bigl[e^{-(x+2t)}+e^{-(x-2t)}\Bigr]+\frac{1}{4}\int_{x-2t}^{x+2t}e^{-s}\mathrm{d}s
\end{equation*}
$$= \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$
Stated in this way, $u(t,x)$ is determined uniquely in its domain. OK


B) In this case, we need an extra boundary condition at $u_{|x=t}=0$ to find the unique solution:
For $x>2t$ general formula for $u$ is as part (A):
$$u(t,x) = \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$
For $t<x<2t$, story is different:
$$u(t,x)= \phi(x+2t)+\psi(x-2t)$$
where $\phi(x+2t)= \frac{1}{4}\bigl[e^{-(x+2t)}+1\bigr]$ is determined by initial conditions at $t=0$. To find $\psi(x-2t)$, we impose $u_{|x=t}=0$ to solution:
$$u_{|x=t}=\phi(3t)+\psi(-t)=0$$
$$\Rightarrow \psi(s)=-\phi(-3s)$$
$$=\frac{-1}{4}\bigl[e^{3s}+1\bigr]$$
Hence the general solution for $t<x<2t$ is:
$$u(t,x)= \frac{1}{4}\bigl[e^{-(x+2t)}+1\bigr]-\frac{1}{4}\Bigl[e^{3x-6t}+1\Bigr]$$
Note that we would not be able to determine $\psi$ if we did not have the extra condition  $u|_{x=t}$. Also note that $u_x|_{x=t}=\frac{1}{2}(e^{-3t}) \neq 0$. This means the problem would have been overdetermined, without any solution, if we considered boundary condition $u|_{x=t}=u_x|_{x=t}=0$.


C) In this case we need to impose the strongest boundary condition to get the unique solution:
Case $x>2t$ is identical to part (A) and (B):
$$u(t,x) = \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$
We find the solution in region $-3t<x<-2t$ by imposing boundary conditions $u|_{x=-3t}=u_x|_{x=-3t}=0$ to $u(t,x)= \phi(x+2t)+\psi(x-2t)$. This gives
$$\phi(-t)+\psi(-5t)=0$$
$$\phi'(-t)+\psi'(-5t)=0$$
Differentiating first equation and adding to the second we get $-4\psi'(-5t)=0$. Therefore $\psi(s)=C$, $\phi(s)=-C$ and $u(t,x)$ is identically zero.Note that we would not be able to find $\phi$ and $\psi$ uniquely, if we did not have both boundary conditions at $x=-3t$.

From continuety of $u$ in $t>0$, $x>-3t$, we conclude $u|_{x=-2t}=0$. This helps us to find solution for $-2t<x<2t$. Analogous to part (B) we write:
$$u(t,x)= \phi(x+2t)+\psi(x-2t)$$
Impose $u|_{x=-2t}=0$ to $u$ to get $\psi(-4t)=-\phi(0)=C$. Therefore solution here is:
$$u(t,x)=\frac{1}{4}e^{-(x+2t)}+C$$
By continuity at $x=2t$, we get $C=\frac{3}{4}$. General solution for part (C) can be explicitly formulated as
\begin{equation*}
   u(x,y)=
      \left\{\begin{aligned}[h]
         &\frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}, & x>2t\\
         &\frac{1}{4}e^{-(x+2t)}+\frac{3}{4}, & -2t<x<2t\\
         &0, & -3t<x<-2t\\
         \end{aligned}
      \right.
\end{equation*}

sorry, but I am still confused about part B, I understand when x<2t will fail the general solution of u. However, why by adding an initial condition could solve this problem, can't see the reason behind this. Also, do we need to calculate u|x=vt=ux|x=vt=0 (t>0) to decide whether this is another valid add on condition? Appreciate your help

Di Wang

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Re: Problem 1
« Reply #13 on: October 14, 2012, 11:31:13 PM »
same thing for part C, how would we decide condition c is the necessary initial condition for unique solution

Victor Ivrii

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Re: Problem 1
« Reply #14 on: October 15, 2012, 03:01:22 AM »
sorry, but I am still confused about part B, I understand when x<2t will fail the general solution of u. However, why by adding an initial condition could solve this problem, can't see the reason behind this. Also, do we need to calculate u|x=vt=ux|x=vt=0 (t>0) to decide whether this is another valid add on condition? Appreciate your help

I am not sure what the question is. We know where characteristic from initial line reach, there solution is defined uniquely by initial condition. Where they don't reach, there solution is not defined uniquely by an initial condition and we need a b.c.