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Solve the initial value problem: $$ u_{tt} - u_{xx} = 0 \\ u(x,0) = 0 \\ u_t(x,0) = \begin{cases} \cos \pi (x-1) & \text{if } 1 < x < 2 \\ 0 & \text{otherwise} \end{cases} $$ Find and draw the solution at $t=3$ for

(a). $x\in (-\infty, \infty)$.

(b). $x\in (0, \infty)$ with $u_x(0,t) = 0$.

(c). $x\in (-\infty, 3)$ with $u(3,t) = 0$.

(d). $x\in (0,3)$ with $u_x(0,t)=u_x(3,t)=0$.

For (a) since we are in the entire real line, we can just use d'Alembert's solution and obtain $u = \sin(\pi(x-3t))-\sin(\pi(x+3t))$, but how do we do it for the other cases that we have semi infinite domain?

For (d) we can use separation of variables, correct?

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  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. I have edited your question to reflect this principle. $\endgroup$ – Brian Apr 9 at 3:38
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You should only ask questions about a single problem. This is not the place for a homework dump.

With that being said, I'm going to solve the first problem. First, define the primitive of $u_t(x,0)=g(x)$ as

$$ G(s) = \int g(s)\ ds = \begin{cases} \dfrac{1}{\pi}\sin\big(\pi(s-1)\big), & 1 < s < 2 \\ 0, & \text{otherwise} \end{cases} $$

Then the solution given by d'Alembert's formula is

$$ u(x,t) = \frac{G(x+t) - G(x-t)}{2} $$

where you have to check case by case for both $x+t$ and $x-t$. Below is a graph ($x$ vs. $t$) of the 4 characteristic lines going through $x=1$ and $x=2$.

enter image description here

The numbered regions are as follows:

\begin{array}{rrr} \text{I}: && x - t < 1, && x + t < 1 \\ \text{II}: && x - t < 1, && 1 < x + t < 2 \\ \text{III}: && x - 1 < 1, && x + t > 2 \\ \text{IV}: && 1 < x - t < 2, && 1 < x + t < 2 \\ \text{V}: && 1 < x - t < 2, && x + t > 2 \\ \text{VI}: && x - t > 2, && x + t > 2 \\ \end{array} Then, you can simplify the solution

$$ u(x,t) = \begin{cases} 0, && (x,t) \in \text{I, III, VI} \\ \dfrac{1}{2\pi}\sin\big(\pi(x+t-1)\big), && (x,t) \in \text{II} \\ \dfrac{1}{2\pi}\Big[\sin\big(\pi(x+t-1)\big) - \sin\big(\pi(x-t-1)\big) \Big], && (x,t) \in \text{IV} \\ -\dfrac{1}{2\pi}\sin\big(\pi(x-t-1)\big), && (x,t) \in \text{V} \end{cases} $$

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  • $\begingroup$ can you give a hint in to how to approach part b) ? $\endgroup$ – Mikey Spivak Apr 11 at 3:08
  • $\begingroup$ @MikeySpivak I commented on your other question. (b) is an even extension (of the above free-space solution) across $x=0$. (c) is an odd extension across $x=3$. For (d) separation of variables will do the trick. $\endgroup$ – Dylan Apr 11 at 6:11
  • $\begingroup$ can you explain a little bit more? $\endgroup$ – Mikey Spivak Apr 16 at 3:28
  • $\begingroup$ @MikeySpivak The solution for (b) is given by $v(x,t) = u(x,t) + u(-x,t)$ where $u(x,t)$ is the solution in (a). Similarly, the solution is (c) is given by $v(x,t) = u(x,t) - u(6-x,t)$ $\endgroup$ – Dylan Apr 16 at 17:35
  • $\begingroup$ how do you get part (c)? $\endgroup$ – Mikey Spivak Apr 16 at 18:40

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