# Free surface of inviscid fluid flow

The following problem seems like it should have a definite solution, but I've been thinking about it for months and haven't got anywhere. It might not be a well-posed problem, but if it isn't I'd like to understand why.

An incompressible, inviscid fluid of density $\rho$ flows continuously (in a steady state) as shown in the following diagram: We know the height of the fluid (and hence its pressure) at points $x_1$ and $x_2$, but we don't know the velocity of the fluid or its height at any other value for $x$. The top of the fluid is a free surface, i.e. it's determined by the properties of the flow rather than being specified as part of the problem. I've drawn it as slightly concave but I've no idea if that's right.

Let us assume that the velocity profile at $x_1$ is vertical (i.e. velocity does not vary with height above point $x_1$). Because the fluid is inviscid it seems to me that the constant vertical velocity profile should be maintained as the fluid travels to the right. So if we were to dye a vertical line of the fluid a particular colour, it would remain a vertical line as it travelled to the right, because the pressure differential across the line is constant with depth. If this is correct it means we can think of the velocity component in the $x$-direction, $v_x$, as a function of $x$ rather than $x$ and $y$.

Because the flow is incompressible we know that $h(x)v_x(x)$ must constant over space, and this is the value I want to solve for (although it might not have a unique value - in that case I just want to know the function $h(x)$). If we need to we can also assume we know the initial and final velocities, $v_x(x_1)=v_1$ and $v_x(x_2)=v_2$.

It seems like Bernoulli's equation should have some relevance here. That would certainly be the case if the fluid were confined to a pipe instead of having a free surface. (In this case the pressure difference would be independent of the difference in height, so we'd need to know that as well.) But every time I try to solve this problem using the Bernoulli equation I get into a terrible mess. I'm really not sure of the best way to approach this problem, so any insight anyone can offer would be much appreciated.

• I'm pretty sure that your "vertical line remains vertical" isn't necessarily true... – genneth Dec 19 '11 at 16:31
• I agree with @genneth. The condition $\frac{\partial v_x}{\partial y} = 0$ does not follow from zero viscidity. This could be an additional assumption since there is not enough data. But IMO approximation of potential flow is a better assumption for this case. – Maksim Zholudev Dec 19 '11 at 17:59
• I've edited the question to make it clear that I'm assuming that the initial velocity profile is vertical - I think inviscidity should imply that this constant vertical profile should be maintained across space. I'm not 100% sure though - I'll have to see if I can think of an argument justifying that. – Nathaniel Dec 21 '11 at 12:39
• (1) I agree with the fellows above me on this point - inviscidity does not imply $\partial_y v_x=0$. (2) if at $x_1$ the velocity is independent of $y$, then because it is horizontal at $y=0$, it must also be horizontal for $y=h(x_1)$, and therefore it $\partial_x h(x_1)=0$, and the tangent to the free surface will be horizontal on the left side. (3) The problem is not defined if you don't specify the full profile of $\vec v(y)$ at both edges of the problem. (4) This is a very nice question! I'm setting a bounty. – yohBS Dec 21 '11 at 20:01

The fluid is incompressible and has no sources inside. This means that the continuity (mass conservation) equation is $$\text{div}\,\vec{v} = 0. \qquad (1)$$ Now we follow the standard procedure and represent $\vec{v}$ as follows: $$\vec{v} = \text{rot}\,\vec{A}.$$ The divergence of any curl is zero so equation (1) is satisfied by any smooth vector field $\vec{A}(\vec{r})$.

For 2-dimensional flow we can assume $$\vec{A} = \Bigl(0, 0, \psi(x,y)\Bigr)$$ so that $$v_x = \frac{\partial \psi}{\partial y}; \quad v_y = -\frac{\partial \psi}{\partial x}. \qquad (2)$$

In fluid dynamics $\psi(x,y)$ is called stream function because the lines of constant $\psi$ are the streamlines.

We have two known stream lines: $$y = h(x)$$ and $$y = 0.$$ Let's select the stream function as follows: $$\psi(x,y) = C\frac{y}{h(x)}. (3)$$ For the upper streamline we have $\psi=C$ and for the lower line $\psi=0$. This is a strong assumption and the main point of the solution. The selection of $\psi$ is not definite here. Formula (3) is intuitive, it gives streamlines that are similar to $h(x)$ but coming more straight while approaching to the bottom.

Now we can use (2) and (3) to find $\vec{v}$: $$\vec{v}(x,y) = \left(\frac{C}{h(x)},\; Cy\frac{h'(x)}{h^2(x)}\right) \qquad (4)$$ where $C$ is some constant determined by the boundary conditions.

The velocity field depends on the unknown function $h(x)$.

### Finding $h(x)$

Function $h(x)$ can be found by applying the Bernoulli equation to the top streamline. Bernoulli equation for incompressible fluid is $$\frac{v^2\bigl(x, y(x)\bigr)}{2} + \frac{p\bigl(x, y(x)\bigr)}{\rho} + gy(x) = \text{const}$$ where
$y(x)$ is the streamline,
$p(x,y)$ is the pressure,
$\rho$ is the density of the fluid,
$g$ is the gravitational acceleration.

The upper streamline $y(x)=h(x)$ is in the equilibrium with the atmosphere air. This means that the pressure of the fluid is equal to the atmosphere pressure: $$p\bigl(x, y(x)\bigr) = p_0.$$ So $$\frac{v^2\bigl(x, h(x)\bigr)}{2} + gh(x) = \text{const} - \frac{p_0}{\rho} = D \qquad (5)$$

Substitution of (4) into (5) gives the differential equation for $h(x)$: $$\frac{C^2}{2h^2}\left(1 + h'^2\right) + gh = D$$ or $$\frac{dh}{dx} = \sqrt{\frac{2h^2}{C^2}(D-gh) - 1}$$ $$h(x_1) = h_1$$ This can be solved numerically if we know $C$ and $D$.

### Finding $C$ and $D$

The parameters $C$ and $D$ are determined by the boundary conditions. If we know the velocity at the point $(x_1, h_1)$ then
from (4): $$C = h_1 v_x(x_1, h_1)$$ and from (5): $$D = \frac{v^2(x_1, h_1)}{2} + gh_1$$

### Conclusion

There are two weak points in this solution:

1. the intuitive assumption (3);
2. the undefined constants $C$ and $D$.

Some boundary conditions can violate (3) and/or make calculation of $C$ and $D$ very difficult.

## Alternative

There is another way to select the stream function.

If we suppose the flow to be potential the velocity field will have the following form: $$\vec{v} = \nabla \varphi$$ where $\varphi(x,y)$ is the potential of the velocity vector field.

Then in addition to (2) we will have: $$v_x = \frac{\partial \varphi}{\partial x}; \quad v_y = \frac{\partial \varphi}{\partial y}. \qquad (6)$$

Now we can introduce the complex potential of the flow: $$W(x+iy) = \varphi(x,y) + i \psi(x,y)$$ The formulas (2) and (6) together are exactly the Cauchy-Riemann conditions for the function $W(z)$. This means that $W(z)$ describes some conformal map.

If we find a conformal map $W(z)$ that turns some rectangle into the blue area in the picture in the question for any $h(x)$, then we find a potential flow (flow with zero vorticity) that solves the problem. Some manipulations will still be required to find $h(x)$.

In fact any $W(z)$ always turns 2-dimensional potential flow with $$\varphi(x,y) = x$$ $$\psi(x,y) = y$$ and $$\vec{v} = (v_x, 0)$$ into something more interesting and still fitting the hydrodynamics equations. This works only for potential flows that are not always a good approximation.

Finding of $W(z)$ in this case is a mathematical problem and perhaps should be discussed somewhere else.

• Great, thanks very much - this gives me a lot of general insight into how to solve this type of problem. I think the weak points you mention aren't too bad for my purposes (and surely something like them will apply to any solution). In particular, I think I'm right in saying that your choice of $\psi$ is the only one that obeys my assumption of an initial velocity profile that's constant with height, so it can be justified that way. It certainly seems to be the only choice that maintains a constant velocity profile anyway. – Nathaniel Dec 22 '11 at 15:02
• @Nathaniel, this solution is not the only possible. I have added some remarks concerning potential flow approximation to the post. Potential flow has zero vorticity and a line that is initially vertical will not remain vertical in the case we consider. – Maksim Zholudev Dec 22 '11 at 15:52