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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.

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