Given the problem setup, we need to solve for the velocity $u(x)$ given the provided equations and assumptions.

Step 1: Substitute the equation of state into the Navier-Stokes equation

The given Navier-Stokes equation for an incompressible, inviscid fluid with no body forces is: $\rho \left( \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} \right) = -\frac{\partial P}{\partial x}$ Assuming steady flow, $\frac{\partial u}{\partial t} = 0$, the equation reduces to: $\rho u \frac{\partial u}{\partial x} = -\frac{\partial P}{\partial x}$ Given the equation of state $\rho = \frac{mP}{kT}$, substituting this into the above gives: $\frac{mP}{kT} u \frac{\partial u}{\partial x} = -\frac{\partial P}{\partial x}$

Step 2: Simplify using the given temperature distribution

We're given $T(x) = T_1 + (T_2 - T_1)\left(\frac{x}{L}\right)$, and $P = \frac{1}{2} \rho u^2$. Substituting the pressure expression $P = \frac{1}{2} \rho u^2$ into the equation gives: $\frac{m}{kT(x)} \left(\frac{1}{2} \rho u^2\right) u \frac{\partial u}{\partial x} = -\frac{\partial }{\partial x}\left(\frac{1}{2} \rho u^2\right)$ Expanding the derivatives: $\frac{m \rho u^3}{2kT(x)} \frac{\partial u}{\partial x} = -\frac{1}{2} \rho \left( 2u \frac{\partial u}{\partial x} \right)$ Simplifying: \frac{m u^3}{2kT(x)} = -u \quad \text{(since \rho cancels out)} Divide by $u$ (assuming $u \neq 0$): $\frac{m u^2}{2kT(x)} = -1$ $u^2 = -\frac{2kT(x)}{m}$ However, $u^2$ must be positive, so the equation implies the right-hand side should be negative: $u(x) = \sqrt{\frac{2kT(x)}{m}}$ Substitute $T(x) = T_1 + (T_2 - T_1)\left(\frac{x}{L}\right)$: $u(x) = \sqrt{\frac{2k\left( T_1 + \left(T_2 - T_1\right)\frac{x}{L} \right)}{m}}$

Final Answer

The velocity $u(x)$ is: $u(x) = \sqrt{\frac{2k}{m} \left( T_1 + \frac{T_2 - T_1}{L}x \right)}$

Would you like more details on any step? Here are some related questions:

1. How does the assumption of steady flow affect the Navier-Stokes equations?
2. What are the physical implications of the temperature distribution on the velocity profile?
3. How would the solution change if the fluid were viscous?
4. What other assumptions could simplify the Navier-Stokes equations?
5. How does the equation of state affect the dynamics of the flow?

Tip: When solving fluid dynamics problems, always double-check the physical feasibility of the derived equations, especially concerning quantities like pressure and velocity that must remain real and positive.