PDEFT1

From Exampleproblems

$u_t=ku_{xx}\,$

$u(0,t) = 0\,$
$u(x,0) = f(x)\,$
$t>0,\,\,0<x<\infty\,$

Since the boundary condition makes the solution equal to zero at $x = 0$ , the Fourier sine transform is most appropriate.

$U(x,t) = \int_0^\infty u(y,t) \sin(xy)\,dy\,$

Take the partial wrt to $t$ and use the DE in the transform integral.

${\partial\over\partial t}U(x,t) = \int_0^\infty u_t(y,t) \sin(xy) \,dy = k \int_0^\infty u_{yy}(y,t) \sin(xy)\,dy\,$

Integrate the integral on the right twice by parts.

$= k\left([u_y(y,t)\sin(xy)]_0^\infty - x\int_0^\infty u_y \cos(xy)\,dy\right)\,$

The first term evaluated at 0 is 0 because of the sine function, and evaluated at infinity is 0 because we assume that there is no heat flow past an infinite boundary.

$= -kx \left([u(y,t) \cos(xy)]_0^\infty + x\int_0^\infty u(y,t) \sin(xy) \,dy\right)\,$

The boundary condition at 0 gives $u (0, t) = 0$ and again we assume that there is no heat at infinity. Otherwise, there would be heat flow past that point.

$= -kx^2\int_0^\infty u(y,t) \sin(xy)\,dy\,$