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APM3462012 => APM346 Math => Home Assignment 2 => Topic started by: Peishan Wang on September 26, 2012, 04:28:16 AM

Part (a) and (b), part (c) and (d) the same questions? Thanks

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)

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

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
It was fixed sometime ago but your computer could has cashed itpress "Reload" in your web browser.
There was an error in both variants: it claimed to be "Home Assignment 1" in the titlenow it fixed.

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?

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 (http://www.math.toronto.edu/courses/apm346h1/20129/L8.html) and note that it could be either odd or even

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?

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)

Part (a):
$$
\begin{equation*}
x > 2t > 0 \\
u(x, t) = \frac{1}{4} \int_{x2t}^{x+2t} dx' \\
= \frac{1}{4}(x + 2t  x + 2t) \\
= t. \\
\end{equation*}
$$
$$
\begin{equation*}
0<x<2t \\
u(x, t) = \frac{1}{4} \int_{2t  x}^{2t + x}dx' \\
= \frac{1}{4}(2t + x  2t + x) \\
= \frac{x}{2}. \\
\end{equation*}
$$
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):
$$
\begin{equation*}
x > 2t > 0 \\
u(x, t) = \frac{1}{4} \int_{x2t}^{x+2t} x' dx' \\
= \frac{1}{8}[(x+2t)^2  (x2t)^2] \\
= xt. \\
\end{equation*}
$$
$$
\begin{equation*}
0 < x < 2t \\
u(x,t) = \frac{1}{4} \int_{2t  x}^{2t+x} x'dx' \\
= \frac{1}{8}[(2t+x)^2  (2t  x)^2] \\
= xt. \\
\end{equation*}
$$
Part (d):
$x > 2t > 0$ as in (c): $u(x,t) = xt$
$$
\begin{equation*}
0 < x < 2t \\
u(x,t) = \frac{1}{4} \int_{0}^{2tx}x'dx' + \frac{1}{4} \int_{0}^{2tx} x'dx' \\
= \frac{1}{8}[(2tx)^2 + (2t + x)^2] \\
= t^2 + \frac{x^2}{4}. \\
\end{equation*}
$$

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
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}1\mathrm{d}s+\int_{x2t}^01\mathrm{d}s\Bigr]$$
$$= \frac{1}{2}x$$
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
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}1\mathrm{d}s+\int_{x2t}^01\mathrm{d}s\Bigr]$$
$$= t$$
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
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}x\mathrm{d}s+\int_{x2t}^0x\mathrm{d}s\Bigr]$$
$$= \frac{1}{8}\Bigl[(x+2t)^2+(x2t)^2\Bigr]$$
$$=xt$$
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
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}x\mathrm{d}s+\int_{x2t}^0x\mathrm{d}s\Bigr]$$
$$= \frac{1}{8}\Bigl[(x+2t)^2(x2t)^2\Bigr]$$
$$=\frac{1}{4}x^2+t^2$$
edit: No need to say all solutions are for $x<2t$.
edit: Fixed integral limits.

Please see my solution on the attachment. 5 parts in total.

part 1 solution

part 3

part 4

Should we consider solutions on different intervals (i.e. x>2t and 0<x<2t) since they differ in part (a) and part (d)?

part 5 complete

Should we consider solutions on different intervals (i.e. x>2t and 0<x<2t) since they differ in part (a) and part (d)?
To be impeccable, Yes. (but I guess solution where $x>2t$ is too trivial/standard to be of any particular interest here.)

Again MZ solutions are perfect and it is a cutoff. Only one remark: in (b) we have even initial functions and continuation must be even due to Neumann, in (c) we have odd initial functions and continuation must be odd due to Dirichlet. So in fact in these parts the answer is obviously the same for $x>t$ and $x<t$.
RR: you obviously put the wrong limits in some of the integrals. Not that it matters for credits  MZ got them
The rest is a flood, sorry. BTW, one can attach several docs to the post (OK, as configured currently it is up to 4 of the total size 192K). Also pdf is not the best possible format for a forum as forum does not display it: most of the browsers either cannot display pdf files or can display pdf files as standalone but not embedded into html page. *** png or jpg are better suited for a forum

Hi, I understand how to derive the solutions for Problem 3, but now I have a question. For example, in 3.a, the solution is
u=x/2 for 0<x<2t, but u(0,x)=x/2 which does not equal 0, the given initial condition. Also, du/dt(0,x)=0 which does not equal 1, the given initial condition. I know however that these initial conditions are satisfied for u=t as x>2t. Could you please explain this?

Hi, I understand how to derive the solutions for Problem 3, but now I have a question. For example, in 3.a, the solution is
u=x/2 for 0<x<2t, but u(0,x)=x/2 which does not equal 0, the given initial condition. Also, du/dt(0,x)=0 which does not equal 1, the given initial condition. I know however that these initial conditions are satisfied for u=t as x>2t. Could you please explain this?
Domain $ 0<x<2t$ does not approach line $t=0$ (except a single point) where initial condition must be satisfied.