# If (u(y),0,0) are the velocity components of an incompressible Newtonian fluid flow due to a pressure gradient in the X-direction, then u(y) is a. (a) linear function of y. (b) quadratic function of y. (c) cubic function of y. (d) constant.

Determining a component of velocity vector of a fluid flow explicitly
If (u(y),0,0) are the velocity components of an incompressible Newtonian fluid flow due to a pressure gradient in the X-direction, then u(y) is a
(a) linear function of y.
(c) cubic function of y.
(d) constant.
I started with Euler dynamical equation as follows:
$\frac{D\stackrel{\to }{q}}{Dt}=\stackrel{\to }{F}-\frac{1}{\rho }\left(\mathrm{\nabla }p\right)$
where $\stackrel{\to }{q}=\left(u\left(y\right),0,0\right)$ and $\stackrel{\to }{F}$ is the external force, $\mathrm{\nabla }p$ being pressure gradient. Now as per question we have along X-direction that
$\frac{\mathrm{\partial }u\left(y\right)}{\mathrm{\partial }t}+u\left(y\right)\frac{\mathrm{\partial }u\left(y\right)}{\mathrm{\partial }x}+0.\frac{\mathrm{\partial }u\left(y\right)}{\mathrm{\partial }y}+0.\frac{\mathrm{\partial }u\left(y\right)}{\mathrm{\partial }z}=0-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{\mathrm{\partial }u\left(y\right)}{\mathrm{\partial }t}=-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}u\left(y\right)=-\int \left(\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }x}\right)dt+C$
C being arbitrary constant. But from here I am unable to conclude whether this is linear or quadratic or cubic or constant. Is my thinking correct here?
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Step 1
Neglecting body force terms other than gravity which can be absorbed into the pressure field, the x-component of velocity satisfies
$\begin{array}{}\text{(*)}& \frac{\mathrm{\partial }u}{\mathrm{\partial }t}+u\frac{\mathrm{\partial }u}{\mathrm{\partial }x}+v\frac{\mathrm{\partial }u}{\mathrm{\partial }y}+w\frac{\mathrm{\partial }u}{\mathrm{\partial }z}=-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }x}+\frac{\mu }{\rho }{\mathrm{\nabla }}^{2}u,\end{array}$
where $\mu$ is the viscosity.
The flow is steady, fully developed (u depends only on y), and unidirectional. Consequently the partial derivatives of u with respect to t, x, and z vanish and $v=w=0$. Thus, (*) reduces to
$\begin{array}{}\text{(**)}& \frac{{d}^{2}u}{d{y}^{2}}=\frac{1}{\mu }\frac{\mathrm{\partial }p}{\mathrm{\partial }x}=\frac{G}{\mu }\end{array}$
The pressure gradient G is independent of y. This follows from the y-component of the Navier-Stokes equations which reduces to
$\frac{\mathrm{\partial }p}{\mathrm{\partial }y}=0$
Step 2
Integrating both sides of (**) twice with respect to y we get the quadratic function
$u\left(y\right)=\frac{G}{2\mu }{y}^{2}+{C}_{1}y+{C}_{2},$
where ${C}_{1}$ and ${C}_{2}$ are integration constants.