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Fluid-Dynamics · Material-Derivative

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

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

Step by step

  1. T = t + x^{2} y,\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} = 1

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

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

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

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

    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 y\right) + \left(x\right)\left(x^{2}\right) + \left(0\right)\left(0\right) = x^{3} + 2 x y^{2}

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

  7. \frac{DT}{Dt} = 1 + \left(x^{3} + 2 x y^{2}\right) = x^{3} + 2 x y^{2} + 1

    Add the local and convective parts.

Reveal the answer
\frac{DT}{Dt} = x^{3} + 2 x y^{2} + 1