### Author Topic: problem 3  (Read 14495 times)

#### Thomas Nutz

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##### problem 3
« on: October 13, 2012, 06:11:36 PM »
Dear all,

I don't know what to do with problem 3. We are asked to find conditions on the three parameters $\alpha$, $\beta$ and $\gamma$ s.t. the integral
$$E(t)=\frac{1}{2}\int_0^L (|u_t|^2+c^2|u_x|^2+\gamma |u|^2)dx$$

is time-independent, where u satisfies b.c. and $u_{tt}-c^2u_{xx}+\gamma u=0$.

The time independence of the integral means that
$$\frac{\partial}{\partial t}u_tu^*_t+c^2\frac{\partial}{\partial t}u_xu^*_x+\gamma \frac{\partial}{\partial t}u u ^* =0$$

but I can`t find $u$, as there is this $u$ term in the wave equation, and the boundary conditions do not help me with this equation neither. Any hints? Thanks a lot!

#### Victor Ivrii

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##### Re: problem 3
« Reply #1 on: October 13, 2012, 06:29:14 PM »
$\renewcommand{\Re}{\operatorname{Re}}$
Dear all,

I don't know what to do with problem 3. We are asked to find conditions on the three parameters $\alpha$, $\beta$ and $\gamma$ s.t. the integral
$$E(t)=\frac{1}{2}\int_0^L (|u_t|^2+c^2|u_x|^2+\gamma |u|^2)dx$$

is time-independent, where u satisfies b.c. and $u_{tt}-c^2u_{xx}+\gamma u=0$.
Yes
Quote
The time independence of the integral means that
$$\frac{\partial}{\partial t}u_tu^*_t+c^2\frac{\partial}{\partial t}u_xu^*_x+\gamma \frac{\partial}{\partial t}u u ^* =0$$
No-we are looking for time independence of integral in the spacial limits, not of the integrand (the latter would be much stronger requirement).

• You can differentiate products. Actually you do not need double each term: f.e. $(u_t u^*_t)_t = 2\Re(u_{tt}u_{t}^* )$.
• In the second term integrate by parts by $x$ to transform $u_x u^*_{tx}$ into $\pm u_{xx}u^*_t$ plus boundary terms.

Equivalently

\mathcal{E} _t + \mathcal{P}_x =  2\Re u^*_t f

where $\mathcal{E}=|u_t|^2+c^2 |u_x|^2 + \gamma |u|^2$ and you need to find expression for $\mathcal{P}$, and $f=0$ if equation is fulfilled.

• You need to assemble integral terms and prove that they together are $0$.
• Finally, you need to consider boundary terms and using boundary conditions find when they are $0$ (to prove (a)). Similarly for (b) but derivative must be -- what?
[/list]
« Last Edit: October 13, 2012, 08:15:47 PM by Victor Ivrii »

#### Thomas Nutz

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##### Re: problem 3
« Reply #2 on: October 13, 2012, 07:46:50 PM »
Thanks for your very quick response!
Is the 3 in your first point supposed to be a 2? I obtain
$$(u_t^*u_t)_t=3Re(u_{tt}u_t^*)$$
as I obtain (with u_t=f(t)+ig(t))
$$(u_t^*u_t)_t=u^*_tu_{tt}+u_tu_{tt}^*=(f-ig)(f'+ig')+(f+ig)(f'-ig')=2(ff'+gg')=2Re(u_{tt}u_t^*)$$

#### Victor Ivrii

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##### Re: problem 3
« Reply #3 on: October 13, 2012, 08:20:26 PM »
Thanks for your very quick response!
Is the 3 in your first point supposed to be a 2?
Yes--corrected
Quote
I obtain
$$(u_t^*u_t)_t=3Re(u_{tt}u_t^*)$$
as I obtain (with u_t=f(t)+ig(t))
$$(u_t^*u_t)_t=u^*_tu_{tt}+u_tu_{tt}^*=(f-ig)(f'+ig')+(f+ig)(f'-ig')=2(ff'+gg')=2Re(u_{tt}u_t^*)$$

Too complicated: just use that $2Re (v)=v+v^*$ and then $(u_tu^*_t)_t=u_{tt}u^*_t +u^*_{tt}u_t =2\Ree(u_{tt}u^*_t )$ as the second term is complex-conjugate to the first one.

#### Bowei Xiao

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##### Re: problem 3
« Reply #4 on: October 14, 2012, 05:59:23 PM »
Are we supposed to deal with the complex variables in Term test?or just the real valued like last year term test1?

#### Victor Ivrii

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##### Re: problem 3
« Reply #5 on: October 14, 2012, 06:22:11 PM »
Are we supposed to deal with the complex variables in Term test?or just the real valued like last year term test1?

We do not consider complex variables. Complex-valued functions of the real variables is a completely different animal

#### Levon Avanesyan

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##### Re: problem 3
« Reply #6 on: October 14, 2012, 07:42:16 PM »
Here are the answers I got.
a) $alpha$ = $beta$
b) $beta$ < $alpha$
Are they correct Prof. Ivrii?

P.S. Dont post the solution because a) dont know how to type math, b) dont have a scanner  c) it is written in a very messy style
« Last Edit: October 14, 2012, 07:54:51 PM by Levon Avanesyan »

#### Victor Ivrii

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##### Re: problem 3
« Reply #7 on: October 14, 2012, 09:22:49 PM »
Here are the answers I got.
a) $alpha$ = $beta$
b) $beta$ < $alpha$
Are they correct Prof. Ivrii?

P.S. Dont post the solution because a) dont know how to type math, b) dont have a scanner  c) it is written in a very messy style

#### Levon Avanesyan

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##### Re: problem 3
« Reply #8 on: October 15, 2012, 12:40:31 AM »
So, after some more time spent on this problem as a result of integration I get
$$\Re(\beta(|u_t|^2))(x=L) - \Re(\alpha(|u_t|^2))(x=0).$$
Does this imply that the answers should be

a) $\Re(\alpha)=0, \Re(\beta)=0$
b) $\Re(\alpha)>0, \Re(\beta)<0$

P.S. Zarak, Thanks for editing!
« Last Edit: October 15, 2012, 12:58:26 PM by Levon Avanesyan »

#### Victor Ivrii

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##### Re: problem 3
« Reply #9 on: October 15, 2012, 03:04:16 AM »
So, after some more time spent on this problem as a result of integration I get
Re(alpha*sqr(|u_t|))(x=0) + Re(beta*sqr(|u_t|))(x=L).
Does this imply that the answers should be
a)Re(alpha)=0, Re(beta)=0
b)Re(alpha)<0, Re(beta)<0

P.S. I am sorry for my weird math-writing.

Now you closer to the truth but I not sure what "sqr" means and if it is a $\sqrt{\ \ \ }$ then it cannot be there

#### Levon Avanesyan

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##### Re: problem 3
« Reply #10 on: October 15, 2012, 12:44:45 PM »
by sqr I mean "to the power of 2".

#### Zarak Mahmud

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##### Re: problem 3
« Reply #11 on: October 15, 2012, 12:53:35 PM »
So, after some more time spent on this problem as a result of integration I get
$$\Re(\alpha(|u_t|^2))(x=0) + \Re(\beta(|u_t|^2))(x=L).$$
Does this imply that the answers should be

a) $\Re(\alpha)=0, \Re(\beta)=0$
b) $\Re(\alpha)>0, \Re(\beta)<0$

P.S. I am sorry for my weird math-writing.

After some simple edits
Click quote to see what I changed.
« Last Edit: October 15, 2012, 12:56:05 PM by Zarak Mahmud »

#### Levon Avanesyan

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##### Re: problem 3
« Reply #12 on: October 15, 2012, 12:59:39 PM »
Just have edited the original post

#### Victor Ivrii

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##### Re: problem 3
« Reply #13 on: October 15, 2012, 01:04:39 PM »
Thanks, Zarak,
BTW, you missed "$-$"
Now Levon you see how to write beautiful math

This is correct.

#### Levon Avanesyan

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##### Re: problem 3
« Reply #14 on: October 15, 2012, 01:12:07 PM »
Thanks for your help Prof. Ivrii
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