Toronto Math Forum
APM3462012 => APM346 Math => Home Assignment X => Topic started by: Thomas Nutz 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
$$
E(t)=\frac{1}{2}\int_0^L (u_t^2+c^2u_x^2+\gamma u^2)dx
$$
is timeindependent, 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!

$\renewcommand{\Re}{\operatorname{Re}}$
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
$$
E(t)=\frac{1}{2}\int_0^L (u_t^2+c^2u_x^2+\gamma u^2)dx
$$
is timeindependent, where u satisfies b.c. and $u_{tt}c^2u_{xx}+\gamma u=0$.
Yes
The time independence of the integral means that
$$
\frac{\partial}{\partial t}u_tu^*_t+c^2\frac{\partial}{\partial t}u_xu^*_x+\gamma \frac{\partial}{\partial t}u u ^* =0
$$
Nowe 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
\begin{equation}
\mathcal{E} _t + \mathcal{P}_x = 2\Re u^*_t f
\end{equation}
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]

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))
$$
(u_t^*u_t)_t=u^*_tu_{tt}+u_tu_{tt}^*=(fig)(f'+ig')+(f+ig)(f'ig')=2(ff'+gg')=2Re(u_{tt}u_t^*)
$$

Thanks for your very quick response!
Is the 3 in your first point supposed to be a 2?
Yescorrected
I obtain
$$
(u_t^*u_t)_t=3Re(u_{tt}u_t^*)
$$
as I obtain (with u_t=f(t)+ig(t))
$$
(u_t^*u_t)_t=u^*_tu_{tt}+u_tu_{tt}^*=(fig)(f'+ig')+(f+ig)(f'ig')=2(ff'+gg')=2Re(u_{tt}u_t^*)
$$
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 complexconjugate to the first one.

Are we supposed to deal with the complex variables in Term test?or just the real valued like last year term test1?

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. Complexvalued functions of the real variables is a completely different animal

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 :)

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 :)
Definitely notread a Hint

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!

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

by sqr I mean "to the power of 2".

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 mathwriting.
After some simple edits :)
Click quote to see what I changed.

Just have edited the original post :)

Thanks, Zarak,
BTW, you missed "$$"
Now Levon you see how to write beautiful math :)
This is correct.

Thanks for your help Prof. Ivrii :)

will alpha: nonnegative and beta:nonpositive for part b) be a more accurate answer?

will alpha: nonnegative and beta:nonpositive for part b) be a more accurate answer?
Yes