Problem 3

[Peishan Wang]

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

[Victor Ivrii]

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]

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]

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 it--press "Reload" in your web browser.

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

[Shida Wu]

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]

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]

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]

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]

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]

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.

[Qitan Cui]

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

[Qitan Cui]

part 1 solution

[Qitan Cui]

part 3

[Qitan Cui]

part 4

[Peishan Wang]

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