Navier Stokes solution doesnt make sense to me

Question

I tried solving the Navier Stokes equation for a Newtonian fluid in a pipe.

$\frac{du}{dt}-\frac{\sigma}{\rho}\frac{du}{dx}=g$ with initial condition $u(0,t)=2$ which means that the flow velocity at $x=0$ is $2$.

$\frac{dt}{dx} = -\frac{\rho}{\sigma}\rightarrow t =-\frac{\rho}{\sigma}x+c\rightarrow c = t+\frac{\rho}{\sigma}x $

We the introduce two ned variable $\xi = c , \eta=t$ to put the PDE into canonical form:

$\xi_{t} = 1 ,\xi_{x} =\frac{\rho}{\sigma} $ and

$\eta_{t}=1,\eta_{x}=0$

$u_{x} = \xi_{x}u_{\xi}+\eta_{x}u_{\eta} = \frac{\rho}{\sigma}u_{\xi}$

$u_{t} =\xi_{t}u_{\xi}+\eta_{t}u_{\eta} = u_{\xi}+u_{\eta}$

Substituting into the original equation we get:

$u_{\eta}=g\rightarrow u(\xi,\eta) = \eta g+f(\xi)$ But $\eta = t$ and $\xi = t+\frac{\rho}{\sigma}x$ so we get $u(x,t) = gt+f(t+\frac{\rho}{\sigma}x)$. By applying the condition $u(0,t)=2\rightarrow gt+f(t)=2\rightarrow f(t) = 2-gt \rightarrow f(t+\frac{\rho}{\sigma}x) = 2-gt-g\frac{\rho}{\sigma}x$ and by substituting $f(t+\frac{\rho}{\sigma}x) \rightarrow u(x,t) = 2-g\frac{\rho}{\sigma}x$ which doesnt make sense since it can take negative values and predict a negative flow velocity.What am I doing wrong?

Answer 1

According to Wikipedia in the Navier Stokes equation the Cauchy stress tensor $\sigma = \tau-p$ where $\tau$ describes viscocity and $p$ describes the volumetric pressure

Since we can make the meaningful assumption that when $|p|>0$ fluid will flow in the pipe $|p|>|\tau|\rightarrow \tau-p<0$ so in my equation the term $-g\frac{\rho}{\sigma}>0$ so $2-g\frac{\rho}{\sigma}x$ cannot be $<0$(for $x>0$)