Author Topic: Web bonus problem -- Week 2  (Read 610 times)

Victor Ivrii

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Web bonus problem -- Week 2
« on: January 12, 2018, 09:03:13 AM »
Find solution $u=u(x,t)$ and describe domain, where it is uniquely defined
\begin{align}
&u_{tt}-u_{xx}=0,
\label{A}\\[2pt]
&u|_{t=x^2/2}= x^3,
\label{B}\\[2pt]
&u_t|_{t=x^2/2}= {\color{blue}{2}}x.
\label{C}
\end{align}
Correction: I replace $x$ by $2x$ in (\ref{C})
« Last Edit: January 13, 2018, 03:55:31 AM by Victor Ivrii »

Jaisen Kuhle

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Re: Web bonus problem -- Week 2
« Reply #1 on: January 12, 2018, 02:52:56 PM »
I just want to check if I did it right before writing out details:

u(t,x) = 3xt + xt2 for all (t,x)

Sheng Gao

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Re: Web bonus problem -- Week 2
« Reply #2 on: January 12, 2018, 07:47:21 PM »
This is my solution, not sure true or not. If it is correct, I will post more details. $u(x,t)=\sqrt2[(x+t)^\frac{3}{2}+(x-t)^\frac{3}{2}]+\frac{\sqrt2}{3}[(x+t)^\frac{1}{2}-(x-t)^\frac{1}{2}]$
« Last Edit: January 12, 2018, 08:16:17 PM by Sheng Gao »

Victor Ivrii

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Re: Web bonus problem -- Week 2
« Reply #3 on: January 12, 2018, 08:33:40 PM »
Jaisen: definitely does not satisfy even equation.

Sheng: satisfies equation but not conditions (I checked solution against old conditions).

I made an error. Please ignore old initial conditions. You need to post calculations, not just answers.

Ziyuan Wang

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Re: Web bonus problem -- Week 2
« Reply #4 on: January 14, 2018, 01:56:43 PM »
May I ask how do we make a function of x+x2/2 equals (x3+x2)/2. I have been thinking for a long time and still can't figure it out.

Victor Ivrii

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Re: Web bonus problem -- Week 2
« Reply #5 on: January 14, 2018, 02:00:06 PM »
Hint: What is the general solution of (\ref{A})?

Ziyuan Wang

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Re: Web bonus problem -- Week 2
« Reply #6 on: January 14, 2018, 05:00:30 PM »
Is it u(x,t)=C1(x+t)+C2(x-t). Then I plug in t=x2/2 for u and ut and I get C1(x+x2/2)= (x3+x2)/2 and C2(x-x2/2) = (x3-x2)/2 which is the question i asked before. Can you tell me a bit more? I still don't get it.

Jaisen Kuhle

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Re: Web bonus problem -- Week 2
« Reply #7 on: January 14, 2018, 06:08:52 PM »
Hint: What is the general solution of (\ref{A})?

For (1) it is $$ u(t,x)= \phi(x+t) + \psi(x-t)$$

For the Cauchy Problem it is:

$$u(t,x)=\frac{1}{2}\bigl[g(x+t)+g(x-t)\bigr]+
\frac{1}{2}\int_{x-t}^{x+t} h(y)\,dy$$
« Last Edit: January 14, 2018, 06:15:23 PM by Jaisen Kuhle »

Ziyuan Wang

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Re: Web bonus problem -- Week 2
« Reply #8 on: January 14, 2018, 06:28:20 PM »
Hint: What is the general solution of (\ref{A})?

For (1) it is $$ u(t,x)= \phi(x+t) + \psi(x-t)$$

For the Cauchy Problem it is:

$$u(t,x)=\frac{1}{2}\bigl[g(x+t)+g(x-t)\bigr]+
\frac{1}{2}\int_{x-t}^{x+t} h(y)\,dy$$

I thought sol'n to Cauchy problem only works in IVP with t=0.

Victor Ivrii

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Re: Web bonus problem -- Week 2
« Reply #9 on: January 14, 2018, 07:24:10 PM »
Ziyuan
What are $C_1$ and $C_2$ in your "solution"? Please, read Sect 2.3.


Jaisen
D'Alembert formula as you wrote works only in the case with the data at $t=0$ and could be modified for the case with the data at $t=t_0$ (which is constant).

You need to plug this general solution into initial conditions and then find $\phi$ and $\psi$ from there

Ioana Nedelcu

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Re: Web bonus problem -- Week 2
« Reply #10 on: January 14, 2018, 09:20:00 PM »
General solution for wave equation:
$$u(t,x)= \phi(x+t) + \psi(x-t)$$
For the first initial condition: $u|_{t=x^2/2}= x^3$ 
$$ x^3 = \phi(x+x^2/2) + \psi(x-x^2/2)\tag{A}$$
Second condition: $u_t|_{t=x^2/2}$
$$ 2x = \phi'(x+x^2/2) - \psi'(x-x^2/2)\tag{B}$$
$$ \implies x^2 = \phi(x+x^2/2) - \psi(x-x^2/2)\tag{C}$$
Combining the two equations,
$$ \phi(x+x^2/2) = (x^3 + x^2) /2 $$
$$ \psi(x+x^2/2) = (-x^2 + x^3) /2 $$

But now I'm not sure how to change the functions to get their argument to $x + t$ and $x - t$ to find $u$ in terms of x and t
« Last Edit: January 14, 2018, 10:39:14 PM by Victor Ivrii »

Jaisen Kuhle

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Re: Web bonus problem -- Week 2
« Reply #11 on: January 14, 2018, 10:39:07 PM »
Second condition: $u_t|_{t=x^2/2}$

$$ 2x = \phi'(x+x^2/2) - \psi'(x-x^2/2) $$
$$ \implies x^2 = \phi(x+x^2/2) - \psi(x-x^2/2) $$



I'm not following your implication here. It seems we have:

$$u_{x^2/2} = \phi_{x^2/2} - \psy_{x^2/2) = 2x$$

So don't we need to integrate with respect to $x^2/2$, in which case:

$$\phi(x+x^2/2) - \psi(x+x^2/2) = x^3 + C$$ 

Also, is this C truly constant?

Edit: Not sure what's wrong with my script for it not to display.
« Last Edit: January 14, 2018, 10:42:19 PM by Jaisen Kuhle »

Victor Ivrii

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Re: Web bonus problem -- Week 2
« Reply #12 on: January 14, 2018, 10:42:20 PM »
Ioana
Equation (A), (B) (I put tags here) are correct, but (C) does not follow from (B). Explain why.

Hint. We cannot integrate (B) but .....

Ioana Nedelcu

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Re: Web bonus problem -- Week 2
« Reply #13 on: January 15, 2018, 12:06:43 AM »
Thank you! I'm pretty sure my integration was wrong to begin with - I was confused with the functions' arguments and what the constants of integration should be (like Jaisen said)

If we can't integrate B, maybe we can differentiate A and solve for the functions this way?

Victor Ivrii

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Re: Web bonus problem -- Week 2
« Reply #14 on: January 15, 2018, 12:32:21 AM »
I'm pretty sure my integration was wrong to begin with - I was confused with the functions' arguments...
If we can't integrate (B), maybe we can differentiate (A)?
Yes, indeed. $\phi '(...)$ is a derivative with respect to ... , not $x$. And if you have ideas try them, rather than waiting my approval..