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Math Help - Unfamiliar directional derivative notation.

  1. #1
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    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).



    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}?
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  2. #2
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    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 (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}.
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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 \sigma^2/2+2\sigma/\sqrt{2} and the derivative is sigma + \sqrt{2}, not sigma + 1/sqrt{2}?
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