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

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!

-

## 1 Answer

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

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

-
Out of curiosity, do you know anywhere I can find a proof for the identity (googling only throws up trig)? With the second equation, the solution given is: $\int vu_y dy = -\int v_y u dy = \int u_x u dy = \frac{1}{2}\partial_x \int u^2 dy$ So I assumed there was some relationship between the first two integrands. The viscosity integral is shown to be zero by your method in the 'official' markscheme. No doubt the same argument can be said for the 2nd integral, but they don't seem to have worked it out that way. And yes, I see my mistake with the last one! Thanks muchly! – Josh May 17 '11 at 16:27