Here is my problem (made by the teacher, using in class the book Partial Differential Equations, by Walter Strauss):
Solve the following wave equation on the half-line
- $U_{tt}$ - $c^2$$U_{xx}$ = $0$, $x>0, t>0$
- $U(x,0)$ = $U_t$$(x,0)$ = $1$, $x>0$
- $U(0,t)$ = $0$, $t \ge 0$
And my math skills are a little rusty, so please correct me and tell me where I need to practice if I make a mistake.
What troubles me in solving the equation is the value of the boundary condition $1$ - I can't figure out a way to find eigensolutions because of it, and that there's only one initial condition for x. Here is the work I have now:
My solution will have the form:
$U(x,t) = X(x) * T(t)$
Separating the values and moving X, T around gets us to:
$\frac{T''(t)}{c^2T(t)} = \frac{X''(x)}{X(x)} = -\lambda$
Which gives us two characteristic equations:
- $X''(x) + \lambda X(x) = 0$
- $T''(t) + c^2\lambda T(t) = 0$
Staring with X, say $\lambda = -\beta^2 < 0$. Then:
$X(x) = Ae^{\beta x} + Be^{-\beta x}$
$X(0) = A + B = 0$, so $A = -B$ and $X(x) = A(e^{\beta x} - e^{-\beta x})$
But there's not a second condition, so I can't go any further and determine A.
Say $\lambda = 0$. Then:
$X(x) = A + Bx$, so $X(0) = A = 0$, and $X(x) = Bx$, but without a second condition, I can't determine B.
Say $\lambda = \beta^2 > 0$. Then:
$X(x) = Acos(\beta x) + Bsin(\beta x)$, so $X(0) = A = 0,$ and $X(x) = Bsin(\beta x)$, but no further condition to solve for B.
Now for the characteristic equation for T:
Say $c^2 \lambda = -\beta^2 < 0$. Then after solving the system of linear equations, you eventually get to the solution of:
$T(t) = \frac{\beta - 1}{2\beta}e^{\beta t} + \frac{1 -\beta}{2\beta}e^{\beta t}$
If $c^2 \lambda = 0$ we get the trivial solution, but if $c^2 \lambda = \beta^2 > 0$:
$T(t) = cos(\beta t) + \frac{1}{\beta}sin(\beta t)$
As in, I don't get a system of eigensolutions like I am hoping for and don't know what to do next...how do I keep moving with this problem?