# Unfamiliar directional derivative notation.

• Jul 20th 2010, 02:24 AM
scorpion007
Unfamiliar directional derivative notation.
In the following equation, we have
$u=u(x,y),~v=v(x,y)$ are scalar functions (solutions to a PDE).

http://img231.imageshack.us/img231/8171/math32.png

The directional derivative I'm familiar with is the following:

The dir. deriv of $u(x,y)$ in the unit direction $\hat{v}$ is

$\nabla{u}\cdot \hat{v}=u_x v_1 + u_y v_2$.

So if our vector $\vec{v} = (\cos(\theta), \sin(\theta))$ then I get the second step. But what does $u_\sigma\vert_\theta$ mean? And how did they get $\displaystyle\frac{dx}{d\sigma}$?
• Jul 20th 2010, 03:06 AM
HallsofIvy
$u_\sigma|\theta$ means "the derivative of u with respect to arc length along the direction $\theta$".

For example, suppose the path is just the straight line y= x. It's length, between $(x_0, x_0)$ and $(x_0+ x, x_0+ x)$ is, of course, $x\sqrt{2}$. That is, $\sigma= x\sqrt{2}$ and $x= \sigma/\sqrt{2}$.

If $u(x, y)= x^2+ 2y$, Then, along the path y= x, $u(x)= x^2+ 2= \sigma^2/2+ \sigma/\sqrt{2}$. It's derivative is $u_\sigma|\theta= \sigma+ \frac{1}{\sqrt{2}}$.

Of course, with y= x as the path, $\theta= \pi/4$ and with u(x,y)= x^2+ 2y, we have $\frac{\partial u}{\partial x}= 2x$ and [tex]\frac{\partial u}{\partial y}= 2.

$\frac{\partial u}{\partial x}cos(\theta)+ \frac{\partial u}{\partial y}sin(\theta)= 2x\frac{1}{\sqrt{2}}+ 2\frac{1}{\sqrt{2}}$ which, because $x= y= \sigma/\sqrt{2}$ is the same as $\frac{\partial u}{\partial x}\frac{\partial x}{\partial \sigma}+ \frac{\partial u}{\partial y}\frac{\partial y}{\partial \sigma}$.
• Jul 20th 2010, 03:40 AM
scorpion007
Did you mean u(x)=x^2+2x, not x^2 + 2 ? Then we get $\sigma^2/2+2\sigma/\sqrt{2}$ and the derivative is sigma + \sqrt{2}, not sigma + 1/sqrt{2}?