User:OldTimeNESter/sandbox

I'm working my way through a textbook on PDEs, and I'm stuck on the Fourier sine transform. Here is the PDE to transform:

$$ \begin{align} u_t & = \alpha^2 u_{xx} \\ u_x(0,t) & = 0 \\ u(x,0) & = H(1 - x) \\ \end{align} $$

where $H(x)$ is the Heaviside step function. So, a standard heat equation with an interesting initial condition. For the transformed equation, I get:

$$ U'(t) = \alpha^2 \lbrack -\omega^2U(t) + \frac{2C\omega}{\pi} \rbrack $$

Note that I substituted $C$ for the boundary condition, since the derivative of a constant is zero. When I try and transform the initial condition, I get the following integrals:

$$ F_s [u(x,0)] = \begin{cases} \frac{2}{\pi} \int_{0}^{0}\ (1) sin(\omega t)\, dt, & x = 0 \\ \frac{2}{\pi} \int_{0}^{\infty}\ (0) sin(\omega t)\, dt, & x > 0 \\ \end{cases} $$

Both of these evaluate to zero, so I thought the initial condition $u(x,0)$ was zero; however, that gives me the following ODE:

$$ \begin{align} U'(t) + \omega^2\alpha^2U(t) & = \frac{2C\omega\alpha^2}{\pi} \\ U(0) & = 0 \\ \end{align} $$

The homogeneous solution is $U(t) = A\exp^{-\omega^2\alpha^2t}$, but plugging in the initial condition leaves $A=0$ , which I don't think is right.