Fluid dynamics, flow of a 2D jet from a narrow slit

Question

I'm working through some exam problems, and I came across this one - the solution of which baffles me considerably.

A two-dimensional jet emerges from a narrow slit in a wall into fluid which is at rest. If the jet is thin, so that velocity $\vec u = (u, v)$ varies much more rapidly across the jet than along it, the fluid equation becomes:
$u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y} = \nu \frac{\partial^2 u}{\partial y^2}$
where constant $\nu$ is the viscosity coefficient. The bounday conditions are that the velocity and its derivatives tend to zero as we leave the jet (that is as $|y|\rightarrow\infty$) and $\frac{\partial u}{\partial y}$ at $y = 0$, as the motion is symmetrical about the x-axis.

The first question involves integrating across the jet to show that $\int u^2 dy$ is $x$ independent. So you end up with three integrals (subscripts denoting derivatives),

1) $\int u u_x dy$
2) $\int v u_y dy$
3) $\nu\int u_{yy} dy$

Somehow, for 1), you can write $\int u u_x dy = \frac{1}{2}\partial_x\int u^2 dy$ - is this an identity?

Secondly for 2) the solution states that: $\int v u_y dy = -\int v_y u dy$ from which you use the incompressibility condition to give $\int u_x u dy = \frac{1}{2}\partial_x\int u^2 dy$. Where does $vu_y=-v_yu$ come from, and again, is there an identity used in the final step?

And thirdly, this is a smaller issue but again, one that confused me a bit, the second question supposes that the streamfunction is self-similar and takes the form: $\psi = x^a f(\eta)$, $\eta = yx^b$. Which is fine, sub in for $u^2 = \psi_y^2$ and equate the power of the factor of $x$ that comes out to be zero. However, in the solution, it comes out as $2a = -b$ and I don't see how that works.

Suppose you take: $\eta = yx^b$, differentiate to get $\frac{d}{d\eta}\eta = \frac{dy}{d\eta}x^b$ so that $dy = d\eta x^{-b}$. When that's substituted into the integral, the factor that comes out is $x^{2a-b}$.

I have noticed some errors in the solutions, so just checking it's me, not them!

I'm sure this is a simple problem once you see the trick, thanks very much!

Answer 1

Is $\int u u_x dy = \frac{1}{2}\int \partial_x (u^2) dy = \frac{1}{2}\partial_x\int u^2 dy$ an identity?

Yes, the order of differentiation in the $x$-direction and integration in the $y$-direction can be exchanged under some mild technical assumptions.

Where does $vu_y= -v_y u$ come from?

Nowhere, it is not true.

Is there an identity to use in the final step?

Yes, you need the given information that the boundary terms at $y=\pm\infty$ vanish, so that you can integrate by part, e.g.,

$$\int (v u_y + v_y u ) dy = \int \partial_y(v u) dy = \left[ v u\right]^{y=\infty}_{y=-\infty}=0,$$

and

$$\int u_{yy} dy = \int \partial_y(u_y) dy = \left[u_y\right]^{y=\infty}_{y=-\infty}=0.$$

Piecing together the various parts yields

$$ \partial_x\int u^2 dy = 2 \int uu_x dy= \int u(u_x -v_y) dy = \int (u u_x + v u_y)dy=\nu\int u_{yy} dy = 0. $$

The second question supposes that the streamfunction is self-similar and takes the form: $\psi = x^a f(\eta)$, $\eta = yx^b$. Which is fine, sub in for $u^2 = \psi_y^2$ and equate the power of the factor of $x$ that comes out to be zero. However, in the solution, it comes out as $2a=−b$ and I don't see how that works.

You are almost there. Just be careful with the algebra.

$$\psi = x^af(yx^b),$$

so

$$ u = \psi_y = x^{a+b}f^{\prime}(yx^b),$$

and therefore

$$ \underbrace{\int u^2 dy}_{\mathrm{independent~of~} x} = x^{2a+b} \int f'(yx^b)^2 d(yx^b) = x^{2a+b} \underbrace{\int f^{\prime}(\eta)^2 d\eta}_{\mathrm{independent~of~} x}. $$