### Author Topic: Problem4  (Read 11535 times)

#### Vitaly Shemet

• Jr. Member
•  • Posts: 10
• Karma: 1 ##### 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 ##### 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

• Full Member
•   • Posts: 17
• Karma: 2 ##### 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

• Jr. Member
•  • Posts: 10
• Karma: 1 ##### 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

• Full Member
•   • Posts: 17
• Karma: 2 ##### 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

• Jr. Member
•  • Posts: 11
• Karma: 0 ##### 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 ##### 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 ##### 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 ##### Re: Problem4
« Reply #8 on: October 01, 2012, 08:59:59 PM »
Part (a):

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

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

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

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

#### Djirar

• Full Member
•   • Posts: 24
• Karma: 8 ##### Re: Problem4
« Reply #9 on: October 01, 2012, 09:00:18 PM »
problem 4

#### Rouhollah Ramezani

• Jr. Member
•  • Posts: 13
• Karma: 5 ##### 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

• Jr. Member
•  • Posts: 13
• Karma: 1 ##### Re: Problem4
« Reply #11 on: October 01, 2012, 09:20:49 PM »
My solution on Problem 4. (2 parts)    part1

#### Qitan Cui

• Jr. Member
•  • Posts: 13
• Karma: 1 ##### Re: Problem4
« Reply #12 on: October 01, 2012, 09:21:24 PM »
part 2

#### Victor Ivrii ##### Re: Problem4
« Reply #13 on: October 02, 2012, 07:20:48 AM »
Guys, when someone posts a complete and correct solution it is over! There is no need to post another solution unless you can point out flaws in the original one and unless you do it on forum nobody (me including) is not going to look further.

There may be an exception (a properly typed solution is better than the best scan) but the general rule is simple: the ultimate goal of the post is a usefulness for other students. Well, in HA1 I disregarded an abdominal quality scans (snapshots) but it was because they were useless