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If I have a material derivative, lets say $$\frac{D \rho }{D t} = -c\rho$$ Is the solution in the lagrangian frame of reference $$\rho_L = \rho_{0L} \exp{(-ct)}$$ , or I is it wrong to treat a material derivative as a standard derivative?

Thanks for your time

Petros
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  • My understanding of $D/Dt$ is that it is the operator $$\frac{D}{Dt} = \frac{\partial}{\partial t} + \vec u \cdot \nabla,$$ where $\vec u$ is the movement of the reference frame. So unless $\vec u = 0$, I don't see how the two are equal. – Gregory Oct 26 '17 at 18:50
  • The material derivative is indeed a standard derivative; it's the derivative of the composition $\rho(t, \mathbf x(t))$ where $\mathbf x'$ is the flow velocity. Therefore, if I don't miss something, I think that you are correct about the solution in the Lagrangian frame (i.e. the frame that follows with the flow). – md2perpe Oct 26 '17 at 19:01

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