Since x and t are independent variables, the only way a function of x and a function of t can be equal for all x and t is if they both equal a constant, say, -β2 (this curious choice will become clearer below!). Then equating each side of the equation to -β2 results in two equations, one in x and the other in t.
Our interest is in the x-dependent equation (i.e. the part that is independent of time). The equation in x is:
| 1 | d2ψ(x) | = -β2 | |
| ψ(x) | dx2 |
and hence we get
| d2ψ(x) | + &beta2ψ(x) = 0 |
| dx2 |
This is a time-independent classical wave equation. It is easy to solve, because all we have to do is take the β2sinψ(x) back over to the right hand side and then look for a function that, when we differentiate it twice, we get the same function again times -β2. One solution would be sinβx, and another (remember Euler again) would be eβi...
The solution ψ(x) = Asinβx is equivalent to the time-dependent solution in Exercise 3.5 if we set t = 0 (check this for yourself). Hence we can identify β with 2π/λ, and we can then rewrite the time-independent wave equation in the form:
| d2ψ(x) | + | 4π2 | &psi(x) = 0 |
| dx2 | λ2 |
(13.17)
