# Finding characteristic time required to reach steady state in the pump equation

Consider the pump equation $$\rho \frac{\partial w}{\partial t} = K' + \frac{\mu}{r} \frac{\partial }{\partial r} \left(r \frac{\partial w}{\partial r} \right)$$

Subject to $$w(r=R,t)=0$$, $$w(r=0,t)$$ is finite, $$w(r,t=0) = \frac{K}{4\mu}(R^2 - r^2)$$.

Solving this equation by finding the steady state solution, and using $$w = \bar{w} + w_{ss}$$, I managed to find the solution for the above equation.

It looks like $$w(r,t) = \sum_{n=1}^{\infty} a_n J_0(j_{0,n} r/R) e^{-\frac{\lambda _n}{ \rho} t} + \frac{K'}{4\mu} (R^2-r^2)$$ where $$j_{0,n}$$ are the zeros of the Bessel $$J_0$$, and $$\lambda _n = \frac{\mu}{R^2} j_{0,n} ^2$$.

I have been asked to examine the centerline velocity $$r=0$$ and to determine the approximate time $$\tau _s$$ it would take to achieve the new steady state.

What does this mean? What is $$\tau _s$$? When I plug in $$r=0$$, the $$J_0$$ terms in the summation just go to $$1$$, so the expression for $$w(0,t)$$ is $$w(r,t) = \sum_{n=1}^{\infty} a_n e^{-\frac{\lambda _n}{ \rho} t} + \frac{K'}{4\mu} (R^2-r^2)$$

How do I go about figuring out $$\tau _s$$? Any advice would be appreciated.