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Math Help - Tensor Calculus: total differential

  1. #1
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    Tensor Calculus: total differential

    I am having a problem understand a particular result in tensor calculus. It is quite a simple result. I shall set the scene.

    It is possible to define a double contraction of two second order tensors A, B, which is given by

    A:B=trace(A^T B)

    Now we come onto the idea of taking the gradient of a scalar-valued function of a second order tensor. Consider \phi(A) to be a scalar-valued function of second order tensors, A. It is a non-linear function, and we aim to approximate \phi at A by a linear function. Taylor expanding \phi at A apparently yields

    \phi(A+dA)=\phi(A)+d\phi + o(dA)

    where d \phi is given by the double contraction:

    d\phi = \frac{\partial \phi(A)}{\partial A} : dA

    Now I have tried and tried to understand how this follows from some form of Taylor expansion, but I'm not getting anywhere. Can someone point me in the right direction?

    Thanks,

    Lind
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  2. #2
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    Re: Tensor Calculus: total differential

    The Taylor's expansion of a function, f(x), around x= x_0 is f(x_0)+ f'(x_0)(x- x_0)+ terms of order [tex]x^2[/itex] and higher. In particular, if we write dx= x- x_0 that becomes f(x_0)+ f'(x_0)dx+ higher order terms in dA.

    That looks like just what you have.
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  3. #3
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    Re: Tensor Calculus: total differential

    HallsofIvy, thank you for your reply.

    Thanks for clearing up the step about Taylor series expansion. My main question is, why does this term  \frac{\partial \phi(A)}{\partial A} dA  end up being the double contraction given. I don't understand how these two things are related.

    Thanks,

    Lind
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