# Prove Bernoulli Function is Constant on Streamline

I have an incompressible, inviscid fluid, under the influence of gravity, with a velocity potential:

$$\mathbf{u} = (-\cos(x)\sin(y), \sin(x)\cos(y), 0)$$

Using Euler's equations,

$$\mathbf{u} \cdot (\mathbf{\nabla} \cdot \mathbf{u}) = \frac{-1}{\rho} \mathbf{\nabla}p - g\mathbf{\hat{z}}$$

that is,

$$\frac{\partial p}{\partial x} = \frac{\rho}{2} \sin(2x)$$ $$\frac{\partial p}{\partial y} = \frac{\rho}{2} \sin(2y)$$ $$\frac{\partial p}{\partial z} = -\rho gz$$

the pressure, $p$, is then:

$$p = \frac{-\rho}{4}(\cos(2x) + \cos(2y) + 4gz)$$

and the streamlines are given by:

$$\frac{dx}{-\cos(x)\sin(y)} = \frac{dy}{\sin(x)\cos(y)}$$

$$\int \frac{-\sin(x)}{\cos(x)}dx = \int \frac{\sin(y)}{\cos(y)}dy$$

$$\implies \ln|\cos(x)| = \ln|\sec(x)| + c$$

$$\implies \cos(x)\cos(y) = A = \phi(x,y)$$

Where $A$ is constant. I can't show that the Bernoulli equation is constant along the streamlines.

$$\frac{1}{\rho}p + \frac{\mathbf{u}^2}{2} + gz$$ $$= \frac{-1}{4}(\cos(2x) + \cos(2y) + 4gz) + \frac{(-\cos(x)\sin(y))^2 + (\sin(x)\cos(y))^2}{2} + gz$$ $$= \frac{1}{2}(\sin^2(x) - \cos^2(x)\cos(2y))$$

Which is not of the form $\phi(x,y)$. Am I missing some trigonometric identity or have I made a mistake in my earlier calculations?

$$\sin^2(x) - \cos^{2}(x)\cos(2y) = \sin^2(x) - \cos^2(x)(2\cos^2(y)-1)$$ $$= 1 - 2\cos^2(x)\cos^2(y) = 1 - 2\phi^2$$
• No, you get $|\cos x|=|\sec x|\cdot e^c$, or $|\cos x\cos y|=A$. – Rory Daulton Feb 13 '16 at 13:09