### Author Topic: Problem4  (Read 22181 times)

#### Vitaly Shemet

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##### Problem4
« on: September 30, 2012, 11:33:01 AM »
Are we allowed in part "a" to use initial wave equation for equality proof? ([d^2u/dt^2-d^2u/dx^2=0] in de/dt=dp/dx and de/dx=dp/dt (partials) )
« Last Edit: September 30, 2012, 12:10:56 PM by VitalyShemet »

#### Victor Ivrii

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##### Re: Problem4
« Reply #1 on: September 30, 2012, 12:13:59 PM »
Are we allowed in part "a" to use initial wave equation for equality proof? ([d^2u/dt^2-d^2u/dx^2=0] in de/dt=dp/dx and de/dx=dp/dt (partials) )

Yes, you are allowed

#### Bowei Xiao

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##### Re: Problem4
« Reply #2 on: September 30, 2012, 01:38:47 PM »
And I believe that's the only condition u have  ,Are u sure u are not gonna use it?

#### Vitaly Shemet

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##### Re: Problem4
« Reply #3 on: September 30, 2012, 02:04:47 PM »
Quote
And I believe that's the only condition u have  ,
Not only. u(x,t) must be continuous and have partials up to 3rd degree

#### Bowei Xiao

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##### Re: Problem4
« Reply #4 on: September 30, 2012, 02:07:28 PM »
well..yes,that's true, I always assume the Utt exisits otherwise the equation doesn't make sense though

#### Shu Wang

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##### Re: Problem4
« Reply #5 on: October 01, 2012, 12:14:16 AM »
hi y'all, a quick question,
would it be appropriate to assume x & t are independent variables in this question? as like, they are presumably not correlated in any function of each other.

Also, what does it mean by rho = T = 1 ? (is T the the kinetic energy or something?)
« Last Edit: October 01, 2012, 12:42:25 AM by Shu Wang »

#### Zarak Mahmud

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##### Re: Problem4
« Reply #6 on: October 01, 2012, 01:20:04 PM »
hi y'all, a quick question,
would it be appropriate to assume x & t are independent variables in this question? as like, they are presumably not correlated in any function of each other.

Also, what does it mean by rho = T = 1 ? (is T the the kinetic energy or something?)

When modeling a physical string, $T$ represents the tension force and $\rho$ is the mass density. I think they are just asking us to consider $c = 1$ for this problem, since $c = \sqrt{\frac{T}{\rho}}$.

#### Victor Ivrii

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##### Re: Problem4
« Reply #7 on: October 01, 2012, 02:58:36 PM »
hi y'all, a quick question,
would it be appropriate to assume x & t are independent variables in this question? as like, they are presumably not correlated in any function of each other.

Also, what does it mean by rho = T = 1 ? (is T the the kinetic energy or something?)

When modeling a physical string, $T$ represents the tension force and $\rho$ is the mass density. I think they are just asking us to consider $c = 1$ for this problem, since $c = \sqrt{\frac{T}{\rho}}$.

$\rho$ is the linear mass density. On the first question: yes, $x,t$ are independent variables ($t$ is a time and $x$ is a spatial coordinate)

#### Zarak Mahmud

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##### Re: Problem4
« Reply #8 on: October 01, 2012, 08:59:59 PM »
Part (a):

$$\begin{equation*} \rho \frac{\partial^2u}{\partial t^2} - T \frac{\partial^2u}{\partial x^2} = 0 \\ \frac{1}{c^2} \frac{\partial^2u}{\partial t^2} - \frac{\partial^2u}{\partial x^2} = 0 \\ c = \sqrt{\frac{T}{\rho}} = 1 \\ \end{equation*}$$

$$\frac{\partial e}{\partial t} = \frac{1}{2}\left(u_t \frac{\partial u_t}{\partial t} + \frac{\partial u_t}{\partial t} u_t + \frac{\partial u_x}{\partial t}u_x + u_x\frac{\partial u_x}{\partial t}\right) \\ =u_t \frac{\partial u_t}{\partial t} + u_x\frac{\partial u_x}{\partial t} \\ =u_tu_{tt} + u_xu_{xt}$$
$$\frac{\partial p}{\partial x} = \frac{\partial}{\partial x}\left(u_tu_x \right) \\ =\frac{\partial u_t}{\partial x}u_x + u_t\frac{\partial u_x}{\partial x} \\ = u_{tx}u_x + u_tu_{xx} \\$$
$$\frac{\partial e}{\partial t} = \frac{\partial p}{\partial t} \\ u_tu_{tt} + u_xu_{xt} = u_{tx}u_x + u_tu_{xx} \\ u_tu_{tt} = u_tu_{xx} \\ u_t(u_{tt} - u_{xx}) = 0 \\$$
$u_{tt} - u_{xx} = 0$ because this is precisely the wave equation with $c = 1$.

$$\frac{\partial p}{\partial t} = \frac{\partial}{\partial t}\left(u_tu_x \right) \\ =u_{tt}u_{xt} + u_tu_{xt} \\$$
$$\frac{\partial e}{\partial x} = \frac{1}{2}\frac{\partial}{\partial x}(u_t^2 + u_x^2) \\ \frac{1}{2}(u_{tx}u_t + u_tu_{tx} + u_{xx}u_x + u_xu_{xx}) \\ = u_{tx}u_t + u_{xx}u_x \\$$
$$\frac{\partial p}{\partial t} = \frac{\partial e}{\partial x} \\ u_{tt}u_x + u_tu_{xt} = u_{tx}u_t + u_{xx}u_x \\ u_x(u_{tt} - u_{xx}) = 0 \\$$

Part (b):
$$\frac{\partial^2 e}{\partial t^2} = \frac{\partial}{\partial t}\frac{\partial e}{\partial t} = \frac{\partial}{\partial t}\frac{\partial p}{\partial x} = \frac{\partial}{\partial x}\frac{\partial p}{\partial t} = \frac{\partial}{\partial x}\frac{\partial e}{\partial x} = \frac{\partial^2 e}{\partial x^2} \\ e_{tt} - e_{xx} = 0 \\$$
$$\frac{\partial^2 p}{\partial t^2} = \frac{\partial}{\partial t}\frac{\partial p}{\partial t} = \frac{\partial}{\partial t}\frac{\partial e}{\partial x} = \frac{\partial}{\partial x}\frac{\partial e}{\partial t} = \frac{\partial}{\partial x}\frac{\partial p}{\partial x} = \frac{\partial^2 p}{\partial x^2} \\ p_{tt} - p_{xx} = 0 \\$$

#### Djirar

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##### Re: Problem4
« Reply #9 on: October 01, 2012, 09:00:18 PM »
problem 4

#### Rouhollah Ramezani

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##### Re: Problem4
« Reply #10 on: October 01, 2012, 09:01:58 PM »
a) Both Proofs are straightforward:
$$e_t=u_tu_{tt}+u_xu_{xt}$$
$$=u_tu_{xx}+u_xu_{tx}$$
$$=\frac{\partial u_tu_x}{\partial x}$$
$$=p_x$$
Above we used definition of $p$ and $e$ as well as the original PDE $u_{tt}=u_{xx}$.

$$p_t=u_{tt}u_x+u_tu_{xt}$$
$$=u_{xx}u_x+u_tu_{tx}$$
$$=\frac{\partial}{\partial x}\frac{1}{2}(u_t^2+u_x^2)$$
$$=e_x$$
Again, we used definition of $p$ and $e$ as well as the original PDE $u_{tt}=u_{xx}$.

b) Direct result of differentiating the identities of part (a) is
$$e_{tt}=p_{xt}$$
$$=p_{tx}=e_{xx}$$
$$\rightarrow e \text{ satisfies the PDE.}$$
And
$$p_{xx}=e_{tx}$$
$$=e_{xt}=p_{tt}$$
$$\rightarrow p \text{ satisfies the PDE.}$$

#### Qitan Cui

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##### Re: Problem4
« Reply #11 on: October 01, 2012, 09:20:49 PM »
My solution on Problem 4. (2 parts)    part1

#### Qitan Cui

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##### Re: Problem4
« Reply #12 on: October 01, 2012, 09:21:24 PM »
part 2