PDE27

From Exampleproblems

$u_{tt} - c^2 u_{xx}=2t=f(x,t)\,$

$u(x,0)=x^2=g(x)\,$
$u_t(x,0)=1=h(x)\,$

In general, combining Duhammel's principle and D'Alembert's formula, the solution is:

$u(x,t)=\frac{1}{2}\left[g(x+ct)+g(x-ct)\right]+\frac{1}{2c}\int_{x-ct}^{x+ct}h(\xi)d\xi + \frac{1}{2c}\int_0^t\int_{x-c(t-s)}^{x+c(t-s)} f(\xi,s)\,d\xi\,ds\,$ .

In this particular case,

$u(x,t) = \frac{1}{2}\left[(x-ct)^2+(x+ct)^2\right]+\frac{1}{2c}\int_{x-ct}^{x+ct} d\xi + \frac{1}{2c}\int_0^t\int_{x-c(t-s)}^{x+c(t-s)} 2s \,d\xi\,ds\,$