I'm not great at maths and have hence been struggling on this for a while. I have been trying to find the numerical solution to the following, however I am not convinced it will actually reach steady state:
$$\frac{\partial V}{\partial t} = \frac{1}{r\rho}\left(\tau+r \frac{\partial\tau}{\partial r}\right),\quad \text{where}\quad \tau = P\left(\mu+b \sqrt{\dfrac{\rho_g}{P}}\frac{\partial V}{\partial r}\right)$$ Could anyone please advise if this will reach steady state over time, as so far in all of my attempts it has grown indefinitely.
For those who are interested it is the momentum equation in cylindrical coordinates, which I am trying to propagate to steady state in MATLAB.
Any help is much appreciated, thanks.
$$\frac{A}{r} + \frac{B}{r}\frac{\partial V}{\partial r} + B \frac{\partial^2 V}{\partial r^2} = \frac{\partial V}{\partial t}$$
Where $A$ and $B$ are some constants. We will have a look at the steady state solution of this equation, namely $\frac{\partial V}{\partial t} = 0$
$$\frac{A}{r} + \frac{B}{r}\frac{\partial V}{\partial r} + B \frac{\partial^2 V}{\partial r^2} = 0$$
By use of black magic or otherwise, we will find the partial solution
$$V = V_{hom} + V_{part} = V_{hom} -\frac{A}{B}r$$
Finally we will solve the homogeneous equation
$$\frac{B}{r}\frac{\partial V_{hom}}{\partial r} + B \frac{\partial^2 V_{hom}}{\partial r^2} = 0$$
Assuming $B$ and $r$ non-zero, we will simplify the equation as
$$\frac{\partial}{\partial r} \biggl (r \frac{\partial V_{hom}}{\partial r} \biggr) = 0$$
Integrating this equation two times will give us
$$V_{hom} = K_1 \log r + K_2$$
To the best of my understanding we have to set $K_1 = 0$, otherwise our solution approaches $-\infty$ at the origin. So the total steady state solution is
$$V = K_2 - \frac{A}{B}r$$
for some constant $K_2$.
Edit: Thanks to suggestion of @Semiclassical, there is an additional point to consider. Given the above steady state solution for $V$, its derivative
$$\frac{dV}{dr} = -\frac{A}{B} = const$$
Is constant and hence
$$\tau = \tau(\frac{dV}{dr}) = \tau(const) = const$$
This means that $\frac{d\tau}{dr} = 0$. When we plug in both results into the very original equation, we find that
$$\frac{\partial V}{\partial t} = \frac{\tau}{r\rho} = 0$$
This last equation cannot be satisfied for all $r$ by selecting $K_2$.
Hence we are forced to conclude that there are no steady state solutions to this equation, regardless of the boundary condition (that is, assuming that infinite boundary conditions are unphysical).