# A solution of Euler's equation

1. Jul 6, 2010

### coverband

1. The problem statement, all variables and given/known data
The velocity vector for a flow is u = (xt, yt, -2zt). Given that the density is constant and that the body force is F = (0,0,-g) find the pressure, P(x,t) in the fluid which satisfies $$P = P_0(t)$$ at x = 0

2. Relevant equations
Euler's equation: $$\frac{Du}{Dt}=-\frac{1}{\rho}\nabla P + F$$

3. The attempt at a solution

$$\nabla P = \rho(-x-xt^2, -y - yt^2, 2z - 4zt^2 - g)$$
How do you get P from this. Back of the book gives
$$P = - \frac {1}{2} \rho (x^2+y^2)(1+t^2) + \rho z^2(1-2t^2)-\rho gz + P_0(t)$$
How did they get that?

2. Jul 6, 2010

### Fightfish

What do you understand by the del operator? You simply have to solve a set of three partial differential equations to reconstruct P.

3. Jul 6, 2010

### HallsofIvy

More specifically, you have
$$\frac{\partial P}{\partial x}= \rho(x- xt^2)= \rho(1- t^2)x$$
$$\frac{\partial P}{\partial y}= \rho(y- yt^2)= \rho(1- t^2)y$$
and
$$\frac{\partial P}{\partial z}= \rho(z- zt^2)= \rho(1- t^2)z$$

Which can be treated as three ordinary differential equations with 't' as a fixed parameter.

4. Jul 6, 2010

### hunt_mat

The arbitrary function of t comes into play as you're doing a partial integration, so when you take the gradient again, all functions of t get knocked out as you're doing partial differentiation.

5. Jul 6, 2010

### coverband

Thanks I was getting delta mixed up with nabla!