## Derivation

The 2-D Navier-Stokes equations are given as follows:

$$\begin{align}

\nabla\vec{V}&=0\\

\rho\frac{D\vec{V}}{Dt}&=-\nabla p+\mu\nabla^{2}\vec{V}

\end{align}$$

$$\begin{align}

\nabla\vec{V}&=0\\

\rho\frac{D\vec{V}}{Dt}&=-\nabla p+\mu\nabla^{2}\vec{V}

\end{align}$$

Letting \(V_{x}=U+u'\), \(V_{y}=V+v'\), and \(p=P+p'\) and performing a small disturbance analysis gives the small perturbation version of the Navier-Stokes Equations

$$\begin{align}\frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}&=0\\

\frac{\partial u'}{\partial t}+U\frac{\partial u'}{\partial x}+V\frac{\partial u'}{\partial y}+u'\frac{\partial U}{\partial x}+v'\frac{\partial U}{\partial y}&=-\frac{1}{\rho}\frac{\partial p'}{\partial x}+\frac{\mu}{\rho}\left(\frac{\partial^{2}u'}{\partial x^{2}}+\frac{\partial^{2}u'}{\partial y^{2}}\right)\\

\frac{\partial v'}{\partial t}+U\frac{\partial v'}{\partial x}+V\frac{\partial v'}{\partial y}+u'\frac{\partial V}{\partial x}+v'\frac{\partial V}{\partial y}&=-\frac{1}{\rho}\frac{\partial p'}{\partial y}+\frac{\mu}{\rho}\left(\frac{\partial^{2}v'}{\partial x^{2}}+\frac{\partial^{2}v'}{\partial y^{2}}\right)

\end{align}\tag{1}$$

$$\begin{align}\frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}&=0\\

\frac{\partial u'}{\partial t}+U\frac{\partial u'}{\partial x}+V\frac{\partial u'}{\partial y}+u'\frac{\partial U}{\partial x}+v'\frac{\partial U}{\partial y}&=-\frac{1}{\rho}\frac{\partial p'}{\partial x}+\frac{\mu}{\rho}\left(\frac{\partial^{2}u'}{\partial x^{2}}+\frac{\partial^{2}u'}{\partial y^{2}}\right)\\

\frac{\partial v'}{\partial t}+U\frac{\partial v'}{\partial x}+V\frac{\partial v'}{\partial y}+u'\frac{\partial V}{\partial x}+v'\frac{\partial V}{\partial y}&=-\frac{1}{\rho}\frac{\partial p'}{\partial y}+\frac{\mu}{\rho}\left(\frac{\partial^{2}v'}{\partial x^{2}}+\frac{\partial^{2}v'}{\partial y^{2}}\right)

\end{align}\tag{1}$$

Assuming parallel flow, where \(U\approx U(y)\) and \(V\approx0\), we can simplify this to the the following form of the Navier-Stokes equations.

$$\begin{align}\frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}&=0\\\frac{\partial u'}{\partial t}+U\frac{\partial u'}{\partial x}+v'\frac{\partial U}{\partial y}&=-\frac{1}{\rho}\frac{\partial p'}{\partial x}+\frac{\mu}{\rho}\left(\frac{\partial^{2}u'}{\partial x^{2}}+\frac{\partial^{2}u'}{\partial y^{2}}\right)\\\frac{\partial v'}{\partial t}+U\frac{\partial v'}{\partial x}&=-\frac{1}{\rho}\frac{\partial p'}{\partial y}+\frac{\mu}{\rho}\left(\frac{\partial^{2}v'}{\partial x^{2}}+\frac{\partial^{2}v'}{\partial y^{2}}\right)\end{align}\label{simp_NS}\tag{2}$$

In this analysis, disturbances are assumed to be Tollmien-Schlichting waves, with the general form as follows.

$$\begin{align}\psi&=\phi(y)e^{i(\alpha x-\omega t)}\\u'&=\frac{\partial\psi}{\partial y}=\frac{\partial\phi}{\partial y}e^{i(\alpha x-\omega t)}\\v'&=-\frac{\partial\psi}{\partial x}=-i\alpha\phi e^{i(\alpha x-\omega t)}\end{align}\tag{3}$$

The temporal and spatial derivatives are then calculated as follows.

$$\begin{align}\frac{\partial u'}{\partial t}&=-i\omega\frac{\partial\phi}{\partial y}e^{i(\alpha x-\omega t)}\\\frac{\partial u'}{\partial x}&=i\alpha\frac{\partial\phi}{\partial y}e^{i(\alpha x-\omega t)}\\\frac{\partial^{2}u'}{\partial x^{2}}&=-\alpha^{2}\frac{\partial\phi}{\partial y}e^{i(\alpha x-\omega t)}\\\frac{\partial u'}{\partial y}&=\frac{\partial^{2}\phi}{\partial y^{2}}e^{i(\alpha x-\omega t)}\\\frac{\partial^{2}u'}{\partial y^{2}}&=\frac{\partial^{3}\phi}{\partial y^{3}}e^{i(\alpha x-\omega t)}\\\frac{\partial v'}{\partial t}&=-\alpha\omega\phi e^{i(\alpha x-\omega t)}\\\frac{\partial v'}{\partial x}&=\alpha^{2}\phi e^{i(\alpha x-\omega t)}\\\frac{\partial^{2}v'}{\partial x^{2}}&=i\alpha^{3}\phi e^{i(\alpha x-\omega t)}\\\frac{\partial v'}{\partial y}&=-i\alpha\frac{\partial\phi}{\partial y}e^{i(\alpha x-\omega t)}\\\frac{\partial^{2}v'}{\partial y^{2}}&=-i\alpha\frac{\partial^{2}\phi}{\partial y^{2}}e^{i(\alpha x-\omega t)}\end{align}\tag{4}$$

We can then substitute each of these derivatives into Equation \(\ref{simp_NS}\) and we get the following relations.

$$\begin{align}e^{i(\alpha x-\omega t)}\left[i\alpha\frac{\partial\phi}{\partial y}-i\alpha\frac{\partial\phi}{\partial y}\right]&=0\\-\rho e^{i(\alpha x-\omega t)}\left[-i\omega\frac{\partial\phi}{\partial y}+i\alpha U\frac{\partial\phi}{\partial y}+-i\alpha\phi\frac{\partial U}{\partial y}-\frac{\mu}{\rho}\left(-\alpha^{2}\frac{\partial\phi}{\partial y}+\frac{\partial^{3}\phi}{\partial y^{3}}\right)\right]&=\frac{\partial p'}{\partial x}\\-\rho e^{i(\alpha x-\omega t)}\left[-\alpha\omega\phi+U\alpha^{2}\phi-\frac{\mu}{\rho}\left(i\alpha^{3}\phi-i\alpha\frac{\partial^{2}\phi}{\partial y^{2}}\right)\right]&=\frac{\partial p'}{\partial y}\end{align}$$

To eliminate the pressure fluctuation term, differentiate the x- and y-momentum equations by \(y\) and \(x\), respectively.

$$\begin{align}\frac{1}{-\rho e^{i(\alpha x-\omega t)}}\frac{\partial^{2}p'}{\partial x\partial y}&=-i\omega\frac{\partial^{2}\phi}{\partial y^{2}}+i\alpha\frac{\partial U}{\partial y}\frac{\partial\phi}{\partial y}+i\alpha U\frac{\partial^{2}\phi}{\partial y^{2}}-i\alpha\frac{\partial\phi}{\partial y}\frac{\partial U}{\partial y}\\&\quad-i\alpha\phi\frac{\partial^{2}U}{\partial y^{2}}+\frac{\mu}{\rho}\left(\alpha^{2}\frac{\partial^{2}\phi}{\partial y^{2}}-\frac{\partial^{4}\phi}{\partial y^{4}}\right)\\\frac{1}{-i\alpha\rho e^{i(\alpha x-\omega t)}}\frac{\partial^{2}p'}{\partial x\partial y}&=-\alpha\omega\phi+U\alpha^{2}\phi+\frac{\mu}{\rho}\left(-i\alpha^{3}\phi+i\alpha\frac{\partial^{2}\phi}{\partial y^{2}}\right)\end{align}\tag{5}$$

Equating the two momentum equations gives

$$-i\omega\frac{\partial^{2}\phi}{\partial y^{2}}+i\alpha U\frac{\partial^{2}\phi}{\partial y^{2}}-i\alpha\phi\frac{\partial^{2}U}{\partial y^{2}}+\frac{\mu}{\rho}\left(2\alpha^{2}\frac{\partial^{2}\phi}{\partial y^{2}}-\frac{\partial^{4}\phi}{\partial y^{4}}-\alpha^{4}\phi\right)+i\alpha^{2}\omega\phi-iU\alpha^{3}\phi=0$$

This simplifies to the Orr-Sommerfeld Equation.

$$\left(U-\frac{\omega}{\alpha}\right)\left(\frac{\partial^{2}\phi}{\partial y^{2}}-\alpha^{2}\phi\right)-\phi\frac{\partial^{2}U}{\partial y^{2}}+\frac{i\nu}{\alpha}\left(\frac{\partial^{4}\phi}{\partial y^{4}}-2\alpha^{2}\frac{\partial^{2}\phi}{\partial y^{2}}+\alpha^{4}\phi\right)=0\label{orrsommerfeld}\tag{6}$$

$$\begin{align}\frac{1}{-\rho e^{i(\alpha x-\omega t)}}\frac{\partial^{2}p'}{\partial x\partial y}&=-i\omega\frac{\partial^{2}\phi}{\partial y^{2}}+i\alpha\frac{\partial U}{\partial y}\frac{\partial\phi}{\partial y}+i\alpha U\frac{\partial^{2}\phi}{\partial y^{2}}-i\alpha\frac{\partial\phi}{\partial y}\frac{\partial U}{\partial y}\\&\quad-i\alpha\phi\frac{\partial^{2}U}{\partial y^{2}}+\frac{\mu}{\rho}\left(\alpha^{2}\frac{\partial^{2}\phi}{\partial y^{2}}-\frac{\partial^{4}\phi}{\partial y^{4}}\right)\\\frac{1}{-i\alpha\rho e^{i(\alpha x-\omega t)}}\frac{\partial^{2}p'}{\partial x\partial y}&=-\alpha\omega\phi+U\alpha^{2}\phi+\frac{\mu}{\rho}\left(-i\alpha^{3}\phi+i\alpha\frac{\partial^{2}\phi}{\partial y^{2}}\right)\end{align}\tag{5}$$

Equating the two momentum equations gives

$$-i\omega\frac{\partial^{2}\phi}{\partial y^{2}}+i\alpha U\frac{\partial^{2}\phi}{\partial y^{2}}-i\alpha\phi\frac{\partial^{2}U}{\partial y^{2}}+\frac{\mu}{\rho}\left(2\alpha^{2}\frac{\partial^{2}\phi}{\partial y^{2}}-\frac{\partial^{4}\phi}{\partial y^{4}}-\alpha^{4}\phi\right)+i\alpha^{2}\omega\phi-iU\alpha^{3}\phi=0$$

This simplifies to the Orr-Sommerfeld Equation.

$$\left(U-\frac{\omega}{\alpha}\right)\left(\frac{\partial^{2}\phi}{\partial y^{2}}-\alpha^{2}\phi\right)-\phi\frac{\partial^{2}U}{\partial y^{2}}+\frac{i\nu}{\alpha}\left(\frac{\partial^{4}\phi}{\partial y^{4}}-2\alpha^{2}\frac{\partial^{2}\phi}{\partial y^{2}}+\alpha^{4}\phi\right)=0\label{orrsommerfeld}\tag{6}$$

## Nondimensionalization

The Orr-Sommerfeld equation is nondimensionalized using the following nondimensional parameters,

$$\bar{U}=\frac{U}{U_{\infty}}\quad\xi=\frac{y}{\delta}\quad\bar{\phi}=\frac{\phi}{U_{\infty}\delta}\quad\bar{c}=\frac{c}{U_{\infty}}\quad\bar{\alpha}=\alpha\delta\quad Re_{\delta}=\frac{U_{\infty}\delta}{\nu}\tag{7}$$

where \(c=\frac{\omega}{\alpha}\) and \(\delta\) is the boundary layer thickness. Substituting these into Equation \(\ref{orrsommerfeld}\) gives

$$\begin{align}0&=\left(\bar{U}U_{\infty}-\bar{c}U_{\infty}\right)\left(\frac{1}{\delta^{2}}\frac{\partial^{2}}{\partial\xi^{2}}\left(\bar{\phi}U_{\infty}\delta\right)-\left(\frac{\bar{\alpha}}{\delta}\right)^{2}\left(\bar{\phi}U_{\infty}\delta\right)\right)-\frac{\bar{\phi}U_{\infty}\delta}{\delta^{2}}\frac{\partial^{2}}{\partial\xi^{2}}\left(\bar{U}U_{\infty}\right)\\&\quad+\frac{i\nu\delta}{\bar{\alpha}}\left(\frac{1}{\delta^{4}}\frac{\partial^{4}}{\partial\xi^{4}}\left(\bar{\phi}U_{\infty}\delta\right)-\frac{2\bar{\alpha}^{2}}{\delta^{4}}\frac{\partial^{2}}{\partial\xi^{2}}\left(\bar{\phi}U_{\infty}\delta\right)+\frac{\bar{\alpha}^{4}}{\delta^{4}}\left(\bar{\phi}U_{\infty}\delta\right)\right)\end{align}\tag{8}$$

Applying the chain rule for each partial derivative gives

$$\begin{align}0&=U_{\infty}\left(\bar{U}-\bar{c}\right)\left(\frac{U_{\infty}}{\delta}\frac{\partial^{2}\bar{\phi}}{\partial\xi^{2}}-\frac{\bar{\alpha}^{2}U_{\infty}}{\delta}\bar{\phi}\right)-\frac{U_{\infty}^{2}}{\delta}\frac{\partial^{2}\bar{U}}{\partial\xi^{2}}\bar{\phi}\\&\quad+\frac{i\nu\delta}{\bar{\alpha}}\left(\frac{U_{\infty}}{\delta^{3}}\frac{\partial^{4}\bar{\phi}}{\partial\xi^{4}}-\frac{2\bar{\alpha}^{2}U_{\infty}}{\delta^{3}}\frac{\partial^{2}\bar{\phi}}{\partial\xi^{2}}+\frac{\bar{\alpha}^{4}U_{\infty}}{\delta^{3}}\bar{\phi}\right)\end{align}\tag{9}$$

Finally, factor out \(\frac{U_{\infty}^{2}}{\delta}\) and substitute for \(Re_{\delta}\) to get

$$\left(\bar{U}-\bar{c}\right)\left(\frac{\partial^{2}\bar{\phi}}{\partial\xi^{2}}-\bar{\alpha}\bar{\phi}\right)-\frac{\partial^{2}\bar{U}}{\partial\xi^{2}}\bar{\phi}+\frac{i}{\bar{\alpha}Re_{\delta}}\left(\frac{\partial^{4}\bar{\phi}}{\partial\xi^{4}}-2\bar{\alpha}^{2}\frac{\partial^{2}\bar{\phi}}{\partial\xi^{2}}+\bar{\alpha}^{4}\bar{\phi}\right)=0\tag{10}$$

For convenience, derivatives with respect to the station coordinate \(\xi\) are hereafter denoted with prime notation. This gives the final nondimensional form of the Orr-Sommerfeld equation:

$$\left(\bar{U}-\bar{c}\right)\left(\bar{\phi}''-\bar{\alpha}\bar{\phi}\right)-\bar{U}''\bar{\phi}+\frac{i}{\bar{\alpha}Re_{\delta}}\left(\bar{\phi}''''-2\bar{\alpha}^{2}\bar{\phi}''+\bar{\alpha}^{4}\bar{\phi}\right)=0\tag{11}$$

Hello Christopher, thanks a lot. This will help me a lot in my exam.

ReplyDeleteThanks a ton, Christopher.

ReplyDeleteHi Christopher. I really don't know how u^\prime.\dfrac{\partial u^\prime}{\parital x} and v^\prime.\dfrac{\partial u^\prime}{\parital x} are gone at the 2nd equation, in (1)? I'll very happy if you explain it to me. Thank you so much.

ReplyDeleteit is due to small disturbance assumption, meaning that the problem is linearized, so any products of primes are neglected (small term squared).

DeleteThanks

ReplyDeleteReally thank you Christopher from Japan!

ReplyDeleteHello Christopher, I'd like to ask whether this is for 3D or 2D? And where does the gravity term ( gravity x density) from the navier stokes equation disappear? This is really easy to understand, thank you!

ReplyDeleteit's 2D and the body force (gravity) is neglected, such as in flow along straight pipe without any elevation change.

DeleteThanks a lot this help for examination because complete calculation available but I follow viscous fluid book by white where we observe many difficulties about calculation

ReplyDeleteThis is a good one Christopher. Pls how do we derive the fluctuating x and y momentum equation?

ReplyDeleteWhy we eliminate pressure term???

ReplyDeleteThanks Chris, amazing work. Only thing: I noticed a typo in equations 10 and 11, where your alpha_bar term in the second parenthesis goes from a power to 2 to a power of 1.

ReplyDelete