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$$u_{tt}-16u_{xx}=0, \quad 0<x<3, \quad 0 < t < \infty,$$ $$u(x,0)=x(3-x), \quad u_t(x,0)=\cos(\pi x), \quad 0<x<3,$$ $$u(0,t)=u(3,t)=t, \quad 0 < t < \infty.$$

Determine $u(x,t)$ in terms of x and t for $(x,t)$ in regions 1, 2 , and 3 determined by the characteristics.

So I know for region 1 $u(x,t)$ can simply be found using d'Alembert's solution. What I am not sure of is for regions 2 and 3. Suppose $P: (x,t)$ is in region 2. You form a characteristic quadrilateral having one vertex on the line $x=0$ and two vertices on the piece of the characteristic from the origin bounding region 1. You can find this by $u(P)=u(A)+u(B)-u(C)$ where A is on x=0 and C and B are on the piece of the characteristic from the origin bounding region 1. I have how you find u(A), u(B), and u(C) in my notes, but I do not really understand it.

$\hskip1.5in$enter image description here

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You have two contradictory conditions for $u(x,0)$. – Christopher A. Wong Nov 8 '12 at 2:20
What are regions 1, 2, 3? – Pragabhava Nov 8 '12 at 2:22
@Pragabhava-Region 1 is the triangle that has its base on $t=0$, its left side is $x-4t=0$, and its right side is $x+4t=3$. Region 2 is the triangle that has a side on $x-4t=0$ (the left side of region 1), $x=0$, and $x+4t=3$. Region 3 has a side on $x+4t=3$ (the right side of region 1), $x=3$, and $x-4t=0$. I hope that makes sense. – Sprock Nov 8 '12 at 2:44
@Christopher A. Wong-that was a typo. It was supposed to be $u_t(x,0)=cos(\pi x)$. – Sprock Nov 8 '12 at 2:46
@Sprock Something like this? – Pragabhava Nov 8 '12 at 3:09
up vote 2 down vote accepted

For any parallelogram $ABCD$ in the $xt$-plane bounded by four characteristic lines, the sums of the values of $u$ at opposite vertices are equal, that is $$ u(A) + u(C) = u(B) + u(D) $$

Let $A = (x,t) \in \mbox{II}$, $B = (0,t_B)$, $C = (x_C,0)$ and $D = (x_D,t_D)$ as shown on the figure:

$\hskip1.5in$enter image description here

Then $u(x,t) = u(0,t_B) - u(x_C,0) + u(x_D,t_D)$. Now, it's easy to see that \begin{align} t_B &= t - \frac{x}{4}\\ x_C &= 4 t - x\\ t_D &= \frac{x}{4}\\ x_D &= 4t \end{align}

and then the solution on region II is $$ u(x,t) = u\big(0,t-\tfrac{x}{4}\big) - u\big(4t -x,0\big) + u_I\big(4t,\tfrac{x}{4}\big), $$ where $u_I$ is the d'Alambert solution in region I.

Can you do region III?

Moreover, can you solve for all time?

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How would the solution change if C and D were on the characteristic line $x-4t=0$? – Sprock Nov 8 '12 at 13:18
@Sprock Now that you know the solution on regions I and II, why dont you take the limit and see it yourself? – Pragabhava Nov 8 '12 at 16:25

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