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Thread: Unfamiliar directional derivative notation.

  1. #1
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    Unfamiliar directional derivative notation.

    In the following equation, we have
    $\displaystyle u=u(x,y),~v=v(x,y)$ are scalar functions (solutions to a PDE).



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

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

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

    So if our vector $\displaystyle \vec{v} = (\cos(\theta), \sin(\theta))$ then I get the second step. But what does $\displaystyle u_\sigma\vert_\theta$ mean? And how did they get $\displaystyle \displaystyle\frac{dx}{d\sigma}$?
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  2. #2
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    $\displaystyle u_\sigma|\theta$ means "the derivative of u with respect to arc length along the direction [itex]\theta[/itex]".

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

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

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

    $\displaystyle \frac{\partial u}{\partial x}cos(\theta)+ \frac{\partial u}{\partial y}sin(\theta)= 2x\frac{1}{\sqrt{2}}+ 2\frac{1}{\sqrt{2}}$ which, because $\displaystyle x= y= \sigma/\sqrt{2}$ is the same as $\displaystyle \frac{\partial u}{\partial x}\frac{\partial x}{\partial \sigma}+ \frac{\partial u}{\partial y}\frac{\partial y}{\partial \sigma}$.
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  3. #3
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    Did you mean u(x)=x^2+2x, not x^2 + 2 ? Then we get $\displaystyle \sigma^2/2+2\sigma/\sqrt{2}$ and the derivative is sigma + \sqrt{2}, not sigma + 1/sqrt{2}?
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