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Fluid-Dynamics · Material-Derivative

Material derivative of T = x^2 + y^2 with velocity [-y, x, 0]

\frac{D}{Dt}\left[x^{2} + y^{2}\right],\quad \mathbf{u} = \left\langle - y,\ x,\ 0 \right\rangle

Step by step

  1. T = x^{2} + y^{2},\quad \mathbf{u} = \left\langle - y,\ x,\ 0 \right\rangle,\quad \frac{D}{Dt} = \frac{\partial}{\partial t} + - y\,\frac{\partial }{\partial x} + x\,\frac{\partial }{\partial y} + 0\,\frac{\partial }{\partial z}

    The material derivative follows a fluid particle: the change at a fixed point, plus the change from being carried along by the flow.

  2. \frac{\partial T}{\partial t} = 0

    The local rate of change: differentiate with respect to t at a fixed position.

  3. \frac{\partial T}{\partial x} = 2 x

    Differentiate with respect to x, holding t and the other coordinates constant.

  4. \frac{\partial T}{\partial y} = 2 y

    Differentiate with respect to y, holding t and the other coordinates constant.

  5. \frac{\partial T}{\partial z} = 0

    Differentiate with respect to z, holding t and the other coordinates constant.

  6. \mathbf{u}\cdot\nabla T = \left(- y\right)\left(2 x\right) + \left(x\right)\left(2 y\right) + \left(0\right)\left(0\right) = 0

    The convective part: each velocity component times the matching slope.

  7. \frac{DT}{Dt} = 0 + \left(0\right) = 0

    Add the local and convective parts.

Reveal the answer
\frac{DT}{Dt} = 0