I am really interested in knowing more about the Navier-Stokes Equation. One question: Assuming we can eliminate its nonlinear term, how do we even solve the equation? I provided the equation here in the y-direction for reference (Navier-Stokes Equation).
$$ \rho\left(\frac{\partial v}{\partial t}+u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+w \frac{\partial v}{\partial z}\right)=-\frac{\partial p}{\partial y}+f_{y}+\mu\left(\frac{\partial^{2} v}{\partial x^{2}}+\frac{\partial^{2} v}{\partial y^{2}}+\frac{\partial^{2} v}{\partial z^{2}}\right) $$
Thank you!
First off the Navier-Stokes equation are a set of non linear partial differential equation, which relate the pressure and velocity of a fluid in space and time. Without going into great detail the equations ask if one can determine the distribution of the velocity and pressure of a fluid as a function of time assuming we know the velocity at some fixed point in time (the starting point usually). In the absence of gravitational forces the problem is stated as followed:
Let $\mu>0$ and $u_0:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a divergence free vector field. Then there exist a smooth velocity field $u:\left[0,\infty\right)\times\mathbb{R}^n\rightarrow \mathbb{R}^n$ and pressure field $p:\mathbb{R}^n\rightarrow \mathbb{R}$ satisfying the equations \begin{eqnarray} %mass \dfrac{\partial }{\partial t}u +\left(u\cdot \bigtriangledown \right)u - \mu \bigtriangleup u + \bigtriangledown p & = & 0,\\ \\ \\ \nabla \cdot u & = & 0,\\ \\ \\ u\left(x,0\right) & = & u_0\left(x\right)\\ \end{eqnarray}
along with the bounded energy condition \begin{equation}\label{energy} \int_{\mathbb{R}^n}^{}{|u(x,t)|^2dx}<\epsilon,~~~~ \forall \epsilon > 0. \end{equation} and initial condition constraint \begin{equation}\label{EJEEJ} \forall \alpha \in \{1,2,\cdots, n\} ~~~\exists C_{\alpha}>0~~~~ s.t ~~~~ \forall ~~K>0 ~~|\partial^{\alpha} u_0(x)| \le \dfrac{C_\alpha}{(1+|x|)^k} \end{equation}
The scalar $\mu = \dfrac{\nu}{\rho_0}$ is called the kinetic 'viscosity' (a quantity that measures the friction of a fluid), $\bigtriangleup$ is the Laplacian space variable $\bigtriangleup = \sum_{i=1}^{n}{\dfrac{\partial^2}{\partial x_j^2}}$ with respect to the coordinates $\textbf{e}_1,\textbf{e}_2,\cdots, \textbf{e}_n$, and $\bigtriangledown$ denotes the gradient operation with respect to the same basis vectors. These equations are to be solved for a the velocity field $u(x,t)$ and pressure field $p(x,t)$ as time progresses.
Most non linear problems are very difficult to solve (just like this one). In general, the theory of PDE's are in fact very "rough" and in many cases there are different theory which correspond to different problem. To learn PDE well you need to learn the methods people use to solve them.
Even this however, will not lead you to the answer to the Naiver-Stokes equations. It is very likely new techniques are needed to solve these type of problems. Commonly, people focus on the vorticity equation in 'attempt' to obtain information concerning the behavior of the velocity $$\dfrac{d\omega}{dt}+(u\cdot \bigtriangledown) \omega - (\omega\cdot \bigtriangledown) u - \mu \bigtriangleup \omega = 0,~~~\omega : = \bigtriangledown\times u$$ (which is derived by taking the curl of the first equation mentioned in this post and noting the curl of the gradient of pressure vanishes). For example, if $\omega(x,0) = 0$ then $\omega(x,t) = 0$ for all time $t>0$ because the derivative of $\omega$ (as a function of time would be 0 at that point).
Assuming one day you can solve for the vorticity say $\omega_i(x,t) = \phi_i(x,t)$ for a family of function then you can substitute the result that you found $\phi_i(x,t)$ back into the vorticity equation and then try to solve the velocity as a function of the $\phi_i's$ because the vorticity is a relation between the velocity $u$ and the rotation $\omega$. Because the $u$ is independent of the curl of $u$ it would make since that you can solve this equation independent of $u$ (that is solve $\omega(x,t)$ for $t>0$). If you can do that then you can find the unique solution (assuming that after you accomplish the answer you find will be unique and you that you can verify it is a solution).