### Author Topic: Problem 3  (Read 11148 times)

#### Peishan Wang

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##### Problem 3
« on: September 26, 2012, 04:28:16 AM »
Part (a) and (b), part (c) and (d) the same questions? Thanks

#### Victor Ivrii

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##### Re: Problem 3
« Reply #1 on: September 26, 2012, 05:21:59 AM »
Part (a) and (b), part (c) and (d) the same questions? Thanks

There was misprint in (c) fixed almost immediately. Compare fourth lines in (a) vs (b) , (c) vs (d)

#### Benvenuto Triolo

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##### Re: Problem 3
« Reply #2 on: September 27, 2012, 11:16:02 AM »
The PDF file has the same fourth lines for a) and b), c) and d). But there is a difference if you go to the other link 'Home Assignment 2'.

There, the the fourth lines of a) and c) have the initial conditions as u, evaluated at x equals zero, is zero and the fourth lines of b) and d) are u differentiated with respect to x, where x equals equals zero, is zero.

Professor, could you fix this in the PDF file? I prefer the PDF file so I could work on the assignment without an internet connection.

Thank you

#### Victor Ivrii

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##### Re: Problem 3
« Reply #3 on: September 27, 2012, 01:41:03 PM »
Professor, could you fix this in the PDF file? I prefer the PDF file so I could work on the assignment without an internet connection.

Thank you

There was an error in both variants: it claimed to be "Home Assignment 1" in the title--now it fixed.

#### Shida Wu

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##### Re: Problem 3
« Reply #4 on: September 29, 2012, 02:50:24 PM »
What is meaning of "method of continuation"? I read the textbook and find it define the odd function to solve the problem, can I use that method?

#### Victor Ivrii

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##### Re: Problem 3
« Reply #5 on: September 29, 2012, 03:13:13 PM »
What is meaning of "method of continuation"? I read the textbook and find it define the odd function to solve the problem, can I use that method?

See also Lecture 8  and note that it could be either odd or even

#### Shida Wu

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##### Re: Problem 3
« Reply #6 on: September 29, 2012, 03:41:36 PM »
thanks, notes are more clear! However, my understanding for the problem is that I can define odd function to solve the Dirichlet boundary problem and define even function to solve Neumann boundary problem, isn't it?

#### Victor Ivrii

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##### Re: Problem 3
« Reply #7 on: September 29, 2012, 03:58:54 PM »
thanks, notes are more clear! However, my understanding for the problem is that I can define odd function to solve the Dirichlet boundary problem and define even function to solve Neumann boundary problem, isn't it?
You are right. However remember that method of continuation has certain preconditions (satisfied in this case)

#### Zarak Mahmud

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##### Re: Problem 3
« Reply #8 on: October 01, 2012, 09:00:43 PM »
Part (a):

$$\begin{equation*} x > 2t > 0 \\ u(x, t) = \frac{1}{4} \int_{x-2t}^{x+2t} dx' \\ = \frac{1}{4}(x + 2t - x + 2t) \\ = t. \\ \end{equation*}$$

$$\begin{equation*} 0<x<2t \\ u(x, t) = \frac{1}{4} \int_{2t - x}^{2t + x}dx' \\ = \frac{1}{4}(2t + x - 2t + x) \\ = \frac{x}{2}. \\ \end{equation*}$$

Part (b):
For $x > 2t > 0$ we obtain $u(x,t) = t$ as in (a).
$$\begin{equation*} 0<x<2t \\ u(x, t) = \frac{1}{4} \int_{0}^{2t - x} dx' + \frac{1}{4} \int_{0}{2t + x} dx' \\ = \frac{1}{4}(2t - x + 2t + x) \\ = t. \\ \end{equation*}$$

Part (c):
$$\begin{equation*} x > 2t > 0 \\ u(x, t) = \frac{1}{4} \int_{x-2t}^{x+2t} x' dx' \\ = \frac{1}{8}[(x+2t)^2 - (x-2t)^2] \\ = xt. \\ \end{equation*}$$

$$\begin{equation*} 0 < x < 2t \\ u(x,t) = \frac{1}{4} \int_{2t - x}^{2t+x} x'dx' \\ = \frac{1}{8}[(2t+x)^2 - (2t - x)^2] \\ = xt. \\ \end{equation*}$$

Part (d):
$x > 2t > 0$ as in (c): $u(x,t) = xt$
$$\begin{equation*} 0 < x < 2t \\ u(x,t) = \frac{1}{4} \int_{0}^{2t-x}x'dx' + \frac{1}{4} \int_{0}^{2t-x} x'dx' \\ = \frac{1}{8}[(2t-x)^2 + (2t + x)^2] \\ = t^2 + \frac{x^2}{4}. \\ \end{equation*}$$

#### Rouhollah Ramezani

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##### Re: Problem 3
« Reply #9 on: October 01, 2012, 09:01:11 PM »
a) Dirichlet boundary condition at $x=0$ means we need an odd continueation of $u|_{t=0}$ and $u_t|_{t=0}$; That is
$$u|_{t=0}=0,\qquad x<0$$
$$u_t|_{t=0}=-1,\qquad x<0$$
By application of d'Alembert's formula, solution to this problem is
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}1\mathrm{d}s+\int_{x-2t}^0-1\mathrm{d}s\Bigr]$$
$$= \frac{1}{2}x$$

b) Neumann boundary condition at $x=0$ means we need an even continueation of $u|_{t=0}$ and $u_t|_{t=0}$; That is
$$u|_{t=0}=0,\qquad x<0$$
$$u_t|_{t=0}=1,\qquad x<0$$
since both functions are even.
By application of d'Alembert's formula, solution to this problem is
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}1\mathrm{d}s+\int_{x-2t}^01\mathrm{d}s\Bigr]$$
$$= t$$

c) Dirichlet boundary condition at $x=0$ means we need an odd continueation of $u|_{t=0}$ and $u_t|_{t=0}$; That is
$$u|_{t=0}=0,\qquad x<0$$
$$u_t|_{t=0}=x,\qquad x<0$$
since both functions are odd.
By application of d'Alembert's formula, solution to this problem is
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}x\mathrm{d}s+\int_{x-2t}^0x\mathrm{d}s\Bigr]$$
$$= \frac{1}{8}\Bigl[(x+2t)^2+(x-2t)^2\Bigr]$$
$$=xt$$

d) Neumann boundary condition at $x=0$ means we need an even continueation of $u|_{t=0}$ and $u_t|_{t=0}$; That is
$$u|_{t=0}=0,\qquad x<0$$
$$u_t|_{t=0}=-x,\qquad x<0$$
By application of d'Alembert's formula, solution to this problem is
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}x\mathrm{d}s+\int_{x-2t}^0-x\mathrm{d}s\Bigr]$$
$$= \frac{1}{8}\Bigl[(x+2t)^2-(x-2t)^2\Bigr]$$
$$=\frac{1}{4}x^2+t^2$$

edit: No need to say all solutions are for $x<2t$.
edit: Fixed integral limits.
« Last Edit: October 02, 2012, 08:56:23 AM by Rouhollah Ramezani »

#### Qitan Cui

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##### Re: Problem 3
« Reply #10 on: October 01, 2012, 09:11:43 PM »
Please see my solution on the attachment. 5 parts in total.
« Last Edit: October 01, 2012, 09:15:12 PM by Qitan Cui »

#### Qitan Cui

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##### Re: Problem 3
« Reply #11 on: October 01, 2012, 09:12:21 PM »
part 1 solution

#### Qitan Cui

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##### Re: Problem 3
« Reply #12 on: October 01, 2012, 09:12:49 PM »
part 3

#### Qitan Cui

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##### Re: Problem 3
« Reply #13 on: October 01, 2012, 09:13:18 PM »
part 4

#### Peishan Wang

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##### Re: Problem 3
« Reply #14 on: October 01, 2012, 09:13:46 PM »
Should we consider solutions on different intervals (i.e. x>2t and 0<x<2t) since they differ in part (a) and part (d)?