Toronto Math Forum
APM346-2012 => APM346 Math => Home Assignment X => Topic started by: Thomas Nutz on October 13, 2012, 06:11:36 PM
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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!
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$\renewcommand{\Re}{\operatorname{Re}}$
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
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
\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]
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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))
$$
(u_t^*u_t)_t=u^*_tu_{tt}+u_tu_{tt}^*=(f-ig)(f'+ig')+(f+ig)(f'-ig')=2(ff'+gg')=2Re(u_{tt}u_t^*)
$$
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Thanks for your very quick response!
Is the 3 in your first point supposed to be a 2?
Yes--corrected
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.
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Are we supposed to deal with the complex variables in Term test?or just the real valued like last year term test1?
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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
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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 :)
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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 not--read a Hint
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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!
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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
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by sqr I mean "to the power of 2".
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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.
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Just have edited the original post :)
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Thanks, Zarak,
BTW, you missed "$-$"
Now Levon you see how to write beautiful math :)
This is correct.
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Thanks for your help Prof. Ivrii :)
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will alpha: non-negative and beta:non-positive for part b) be a more accurate answer?
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will alpha: non-negative and beta:non-positive for part b) be a more accurate answer?
Yes