# Could any one help me

1. Apr 27, 2008

### islamphysics

dose any one have a derivation for NS equations in the cylindrical coordinates starting from x-y coordinates

Or Even a derivation for the continuity equation

2. Apr 27, 2008

### HallsofIvy

Staff Emeritus
NS is "Navier-Stokes"?

In cylindrical coordinates, $r= (x^2+ y^2)^{1/2}$ and $\theta= arctan(y/x)$.

$$\frac{\partial r}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)= \frac{x}{\sqrt{x^2+ y^2}}= \frac{x}{r}$$
$$= \frac{r cos(\theta)}{r}= cos(\theta)$$

$$\frac{\partial r}{\partial y}= sin(\theta)$$

$$\frac{\partial \theta}{\partial x}= \frac{1}{1+ (y/x)^2}(-y/x^2)=\frac{-y}{x^2+ y^2}$$
$$= \frac{-r sin(\theta)}{r^2}= -\frac{sin(\theta)}{r}$$

$$\frac{\partial \theta}{\partial y}= \frac{1}{1+ (y/x)^2}(1/x)= \frac{x}{x^2+ y^2}$$
$$= \frac{r cos(\theta)}{r^2}= \frac{cos(\theta)}{r}$$

Using the chain rule
$$\frac{\partial \phi}{\partial x}= \frac{\partial \phi}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial \phi}{\partial \theta}\frac{\partial \theta}{\partial x}$$

You should be able to finish that, and then the second derivatives yourself. It's tedious but requires no deep mathematics.