- #1
ralqs
- 99
- 1
D'Alembert's solution to the wave equation is
[tex]u(x,t) = \frac{1}{2}(\phi(x+ct) + \phi(x-ct)) + \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(\xi)d\xi[/tex] where [itex]\phi(x) = u(x,0)[/itex] and [itex]\psi(x) = u_t (x,0)[/itex]. I'm trying to understand this intuitively. The first term I get: a function like f = 0 (x/=0), = a (x=0) will "break into two functions" and become f = a/2 (x = +/- ct), = 0 (x /= +/- ct). But I can't see how the integral term comes about. Does anyone here have a good physical intuition about this? Thanks.
[tex]u(x,t) = \frac{1}{2}(\phi(x+ct) + \phi(x-ct)) + \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(\xi)d\xi[/tex] where [itex]\phi(x) = u(x,0)[/itex] and [itex]\psi(x) = u_t (x,0)[/itex]. I'm trying to understand this intuitively. The first term I get: a function like f = 0 (x/=0), = a (x=0) will "break into two functions" and become f = a/2 (x = +/- ct), = 0 (x /= +/- ct). But I can't see how the integral term comes about. Does anyone here have a good physical intuition about this? Thanks.