PDE6.5

Transform the BVP into one with homogeneous boundary conditions.

$u_t=u_{xx} + w(x,t)\,$

$u_x(0,t) = \alpha(t)\,$
$u_x(1,t) = \beta(t)\,$
$u(x,0) = \phi(x)\,$

$0<x<1\,$
$t>0\,$

Let $v(x,t) = u(x,t) - \frac{\beta(t)}{2} x^2 - (x-\frac{x^2}{2})\alpha(t)\,$

$v_x = u_x - \beta(t) x - \alpha(t) + \alpha(t) x\,$

$v_{xx} = u_{xx} - \beta(t) + \alpha(t)\,$

$v_t = u_t - \frac{\beta'(t)}{2} x^2 - (x-\frac{x^2}{2})\alpha'(t)\,$

Now the new initial boundary value problem can be stated.

The new DE is $v_t = v_{xx} + \beta(t) - \alpha(t) - \frac{\beta'(t)}{2}x^2 - (x-\frac{x^2}{2})\alpha'(t)+w(x,t)\,$

The new IC is $v(x,0) = u(x,0) - \frac{\beta(0)}{2} x^2 - (x-\frac{x^2}{2})\alpha(0)\,$

The new BCs are

$v_x(0,t) = u_x(0,t) - \alpha(t) = 0\,$

$v_x(1,t) = u_x(1,t) - \beta(t) = 0\,$