We're given that if the temperature $u(t,x)$ of a homogeneous bar satisfies the heat equation, we're to show that the associated heat flux $w(t,x)$ is another solution to the same equation.
I know that $$w(t,x) =-\kappa(x)\frac{\partial u}{\partial x}$$ and the solution for the heat equation is given by $$u(t,x)=\sum_{n=1}^\infty b_n \exp\left[-\frac{\gamma\ n^2\pi^2t} {l^2} \right] \sin(\frac{ n\pi x}{l})$$
My question is where do I begin with this problem? Do I just start like I would with solving the original heat equation and applying separation of variables to $w(t,x)$ like $$w(t,x) = T(t)X(x)$$ and proceeding from there? Or do I start at the solution of the heat equation and plug stuff in until things work out? Thanks.
Just found out it's actually way easier than either of those two suggestions. We can treat the thermal conductivity variable $\kappa(x)$ as a constant, so that we have the heat flux given by $$w(t,x) =-\kappa\frac{\partial u}{\partial x},$$
we first take the derivative of the heat equation with respect to $x$: $$\frac{\partial^2 u}{\partial x \partial t} = -\kappa\frac{\partial^3 u}{\partial x^3}$$
Then changing the order of partial derivatives and substituting gives us $$\frac{\partial u}{\partial t}\left(-\kappa\frac{\partial u}{\partial x}\right) = \left(-\kappa\frac{\partial u}{\partial x}\right) \frac{\partial^2 u}{\partial x^2}$$
Which is just $$\frac{\partial u}{\partial t}w(t,x) = w(t,x) \frac{\partial^2 u}{\partial x^2}$$