Diffusion with non-constant diffusivity
(Problem #9 on page 37)
Consider the diffusion-like equation
$$u_t = k(t)u_xx.$$
Note that this is very much like our basic diffusion equation but with a diffusivity coefficient that can vary with time. Suppose that we substitute $t$ with
$$t \to \tau = \int_0^t k(\eta) \, d\eta.$$
Show that the resulting equation is a standard diffusion equation with constant diffusivity.
Comments
Consider the diffusion-like equation
$$u_t = k(t)u_{xx}.$$
We are going to substitute $u(x,\tau)$ for $u(x,t)$ on both sides of the equation. We are allowed to do this because the function $u$ remains unchanged; it is simply rewritten in terms of a different variable.
Note that $\tau$ is not a function of x, so $u_{xx}(x,\tau)$ = $u_{xx}(x,t)$.
To evaluate $u_t(x, \tau)$, we apply the chain rule and obtain $u_{\tau}(x,\tau)\frac{\partial \tau}{\partial t}$, which by the definition of $\tau$ equals $u_{\tau}(x,\tau)\frac{\partial }{\partial t}(K(t) - K(0))$, where $K$ is the antiderivative of $k$. This gives us a final value of
$$u_t = u_{\tau}k(t).$$
Now our original equation becomes
$$u_{\tau}k(t) = k(t)u_{xx}.$$
Factoring out $k(t)$ from both sides, we get
$$u_{\tau} = u_{xx}.$$
That looks much easier to work with!
Start with the equation:
$$u_t = k(t)u_{xx} .$$
Substitute $\tau = \int_0^t k(\eta) d\eta$ in for $t$. Solving the integral gives:
$$\tau = K(t) - K(0).$$
After the substitution, $u(x,t)$ becomes $u(x,\tau).$ Since the function to find $\tau$ does not contain any $x$ terms, $u_{xx}$ is unaffected by the substitution:
$$u_{xx}(x,t) = u_{xx}(x,\tau).$$
Next, we need to relate $u_t$ to $u_\tau.$ We can do this by using the chain rule.
$$\frac{d}{dt}u(x,\tau) = u_x*\frac{d}{dt}u(x,\tau) + u_\tau*\frac{d}{dt}u(x,\tau) = 0 + u_\tau*\frac{d}{dt}[K(t)-K(0)]$$
Therefore $u_t(x,\tau) = u_\tau *k(t)$, since $k(t)$ is equal to $\frac{d}{dt}K(t)$
Replacing $u_t$ in the original equation with $u_\tau$ gives:
$$u_\tau*k(t) = k(t)u_{xx}.$$
Dividing both sides by $k(t)$ gives a final equation of $u_\tau = u_{xx}.$
This matches the standard diffusion equation, $u_t = Du_{xx}$, where $D = 1$ and is constant.