10.3.2.1. Planar jet
From the \(y\)-component of the momentum, \(P + \overline{v'^2} = P_0(x)\), and 1) if the fluctuation becomes negligible in the outer flow and 2) the outer flow is irrotational with unifrom velocity, then pressure is uniform in the whole domain, \(P_0(x) = P_0\), and Prandtl equations from turbulent flows become
\[\begin{split}\begin{cases}
u \partial_x u + v \partial_y u + \partial_y \left( \overline{u' v'} \right) = 0 \\
\partial_x u + \partial_y v = 0 \ ,
\end{cases}\end{split}\]
Self-similar solution. Looking for a self-similar solution
\[\begin{split}\begin{aligned}
u(x,y) & = U(x) f(\eta(x,y)) \\
\overline{u'v'}(x,y) & = U^2(x) g(\eta(x,y)) \ ,
\end{aligned}\end{split}\]
with \(\eta(x,y) = \frac{y}{\delta(x)}\). Through the incompressibility constraint, the \(y\)-component of the velocity reads
\[\begin{split}\begin{aligned}
v(x,y) - \underbrace{v(x,0)}_{=0}
& = \int_{t = 0}^{y} \partial_y v(x,t) \, dt = \\
& = - \int_{t=0}^{y} \partial_x u(x,t) \, dt = \\
& = \int_{t=0}^{y} \left\{ -U'(x) f(\eta(x,t)) + U(x) \dfrac{df}{d \eta}(\eta(x,t)) \frac{t \delta'(x)}{\delta^2(x)} \right\} \, dt = \\
& = \int_{\chi=0}^{\eta} \left\{ -U'(x) f(\eta(x,\delta(x) \chi)) \delta(x) + U(x) \dfrac{d f}{d \eta}(\eta(x, \delta(x) \chi)) \chi \delta'(x) \right\} \, d \chi \ ,
\end{aligned}\end{split}\]
being
\[\partial_x u(x,y) = U'(x) f(\eta(x,y)) - U(x) f'(\eta(x,y)) \frac{y \delta'(x)}{\delta^2(x)} \ ,\]
and having transformed the variable \(\chi = \frac{t}{\delta}\), and the extremes of integration accordingly.
Mass, momentum and energy fluxes across normal planes. Momentum flux reads
\[Q(x) = \int_{y=0}^{+\infty} u^2(x,y) \, dy \ .\]
Its derivative w.r.t. \(x\) reads
\[\begin{split}\begin{aligned}
\dfrac{d}{dx} Q(x)
& = \dfrac{d}{dx} \int_{y=-\infty}^{+\infty} u^2(x,y) \, dy = \\
& = \int_{y=-\infty}^{+\infty} \partial_x u^2 \, dy = \\
& = - \int_{y=-\infty}^{+\infty} \partial_y \left\{ uv + \overline{u'v'} \right\} \, dy = 0 \ ,
\end{aligned}\end{split}\]
if the flow is at rest at infinity. Using the self-similar assumption, this derivative reads
\[0 = \dfrac{d Q}{dx} = \dfrac{d}{dx} \left[ U^2(x) \delta(x) \int_{\eta=-\infty}^{+\infty} f^2(\eta) \, d \eta \right] \ ,\]
so that it follows
\[0 = \dfrac{d}{dx} \left( U^2(x) \delta(x) \right) = 2 U U' \delta + U^2 \delta' \ ,\]
or
\[U' = - \dfrac{\delta'}{2\delta} U \ .\]
Mass flux.
\[M(x) = \int_{y=-\infty}^{+\infty} u(x,y) \, dy \ , \]
and, if the self-similarity assumption holds, its derivative reads
\[\dfrac{d }{dx}M(x) = \dfrac{d}{dx} \left( U(x) \delta(x) \right) \int_{\eta=-\infty}^{+\infty} f(\eta) \, d\eta \ .\]
Energy flux.
\[E(x) = \int_{y=-\infty}^{+\infty} \dfrac{1}{2} u^3(x,y) \, dy \ , \]
and, if the self-similarity assumption holds, its derivative reads
\[\dfrac{d }{dx}E(x) = \dfrac{d}{dx} \left( \dfrac{1}{2} U^3(x) \delta(x) \right) \int_{\eta=-\infty}^{+\infty} f(\eta) \, d\eta \ .\]
Solving for the self-similar solution. Inserting the expressions of the velocity components and the turbulent stress into the \(x\)-component of the momentum equation,
\[\begin{split}\begin{aligned}
0
& = ( U f ) \left( U' f - U f' \eta \frac{\delta'}{\delta} \right) + \int_{\chi = 0}^{\eta} \left\{ -U' f \delta + U f' \chi \delta' \right\} \, d\chi \, U f' \frac{1}{\delta} + U^2 g' \frac{1}{\delta} = \\
& = U U' f^2 - U^2 f f' \eta \frac{\delta'}{\delta} - U U' f' \int_{\chi=0}^{\eta} f d \chi + U^2 f' \frac{\delta'}{\delta} \left(\left. \chi f \right|_{0}^{\eta} - \int_{\chi=0}^{\eta} f \, d \chi \right)+ U^2 g' \frac{1}{\delta} = \\
& = U U' f^2 - U U' f' \int_{\chi=0}^{\eta} f d \chi - U^2 f' \frac{\delta'}{\delta} \int_{\chi=0}^{\eta} f \, d \chi + U^2 g' \frac{1}{\delta}\ ,
\end{aligned}\end{split}\]
and introducing the relation between the derivative of \(U(x)\) and \(\delta(x)\), and dividing by \(\frac{U^2}{\delta}\)
\[\frac{\delta'(x)}{2} \left[ f^2(\eta) + f'(\eta) \int_{\chi=0}^{\eta} f(\chi) d \chi \right] = g'(\eta) \ .\]
For the existence of a self-similar solution, no coefficient can explicitly depend on \(x\) or \(y\), so the derivative of the boundary layer thickness must be constant
\[\delta'(x) = S \ ,\]
and thus a linear spreading is found,
\[\delta(x) - \delta(x_0) = S ( x - x_0 ) \ .\]
As the momentum flux is constant in \(x\), it follows that — setting the origin of the coordinates so that \(\delta(x) = S x\) — the maximum velocity (velocity on the symmetry line of the jet) decreases as
\[U(x) \propto \delta(x)^{-\frac{1}{2}} \propto x^{-\frac{1}{2}} \ .\]
Mass flux goes with \(M(x) \sim x^{\frac{1}{2}}\) (increases, entrainment), while the energy flux goes with \(E(x) \sim x^{-\frac{1}{2}}\) (decreases due to dissipation).
Solution with constant turbulent viscosity as closure model. With the assumption
\[- \overline{u'v'} = \nu_T \partial_y u \ ,\]
using self-similarity ansatz
\[-U^2 g = \nu_T U f' \frac{1}{\delta} \ ,\]
and with the additional assumption (non-physical, just to get a problem that can be solved analytically. An assumption more, an assumption less…),
\[\hat{\nu}_T := \frac{\nu_T}{\delta(x) U(x)} = \text{const.} \ ,\]
the two non-dimensional functions \(f(\eta)\) and \(g(\eta)\) are related as
\[g(\eta) = - \hat{\nu}_T f'(\eta) \ .\]
Thus, the solution of the equation
\[\frac{S}{2} \left[ f^2(\eta) + f'(\eta) \int_{\chi=0}^{\eta} f(\chi) d \chi \right] + \hat{nu}_T f''(\eta) = 0 \ ,\]
with the proper boundary conditions provides the non-dimensional velocity profile. Introducing the function \(F(\eta) = \int_{\chi=0}^{\eta} f(\chi) d \chi\),
\[\begin{split}\begin{aligned}
0
& = \frac{S}{2} \left( F'^{2}(\eta) + F''(\eta) F(\eta) \right) + \hat{\nu}T F'''(\eta) = \\
& = \frac{S}{2} \left( F(\eta) F'(\eta) \right)' + \hat{\nu}_T F'''(\eta) \ .
\end{aligned}\end{split}\]
Integrating once, and exploiting the boundary conditions (todo…)
\[\begin{split}\begin{aligned}
0
& = \frac{S}{2} F(\eta) F'(\eta) + \hat{\nu}_T F''(\eta) = \\
& = \frac{S}{4} \left( F^2(\eta) \right)' + \hat{\nu}_T F''(\eta) \ ,
\end{aligned}\end{split}\]
and integrating once again with the boundary condition \(F'(0) = 1\),
\[\frac{S}{4} F^2(\eta) = \hat{\nu}_T \left( 1 - F'(\eta) \right) \ ,\]
or
\[F'(\eta) = 1 - \frac{S}{4 \hat{\nu}_T} F^2(\eta) \ .\]
Integrating one last time,
\[F(\eta) = \frac{1}{\alpha} \text{tanh} \left( \alpha \eta \right) \ ,\]
with \(\alpha = \sqrt{\frac{S}{4 \hat{\nu}_T}}\).
