# Interchange of Lagrangian/material derivative and volume integral

In hydrodynamics there are two basic approaches. The first is the Eulerian specification where the coordinate system is fixed. In that case, the partial time derivative and volume integral operators can be interchanged: $$\int_V\frac{\partial f}{\partial t}dV=\frac{\partial}{\partial t}\int_VfdV,$$ the volume being fixed in space. The second specification is the Lagrangian one where you follow one fluid "particle" and the relevant time derivative is the Lagrangian/material derivative: $$\frac{\partial f}{\partial t}+\mathbf{u}\cdot\nabla f\equiv\frac{Df}{Dt},$$ where $$\mathbf{u}$$ is the flow field. In that case, the Lagrangian derivative can be interchanged with the volume integral, but only if the volume is that of the fluid "particle": $$\int_{V_u}\frac{Df}{Dt}dV_u=\frac{D}{Dt}\int_{V_u}fdV_u.$$

Now I've seen in a paper the author using this latter interchange of operators, but for a fixed volume comprising of the entire system (actually the volume going up to infinity). Is this valid? How do you make that step? I couldn't find any info on this. Maybe you just sum all the individual volume integrals and do the interchange with all of them? And then what happens to the Lagrangian derivative?

• This is really more general than just fluids, if the integration bounds depend on time then you can't just switch them. If the volume is the entire control volume and it's fixed then it necessarily won't change in time. At least that is how I am interpreting it. Commented May 12, 2023 at 15:36
• In my case the volume is goes up to infinity, and so it encompasses the entire system, and thus doesn't vary on time. What confuses me, is what to do with the Lagrangian derivative. Part of it depends on time through $\mathbf{u}$, although it is zero at the boundaries of the volume. Commented May 12, 2023 at 15:43

I wonder if your Reynolds transport theorem is correct ?

The correct formula would be (with $$dm = \rho dV$$):

$$\frac{D}{Dt}\int_{V_u} f \rho dV=\int_{V_u}\frac{Df}{Dt} \rho dV$$ It can also be formulated as follows:

$$\frac{D}{Dt}\int_{V_u}fdV= \int_{V_u}(\frac{\partial f }{\partial t }+ \vec{\nabla }(f \vec{u})) dV$$

For example, for the mass $$M(t)$$ :

$$\frac{D M}{Dt}=\frac{D}{Dt}\int_{V_u} \rho dV= \int_{V_u}(\frac{\partial \rho }{\partial t }+ \vec{\nabla }(\rho \vec{u})) dV$$

With :

$$\frac{\partial \rho }{\partial t} + \vec{\nabla }(\rho \vec{u})=\frac{\partial \rho }{\partial t}+\rho \vec{\nabla }( \vec{u})+ \vec{u} \vec{\nabla }(\rho)=\frac{D \rho }{D t}+ \rho \vec{\nabla }( \vec{u})$$

Finally :

$$\frac{D}{Dt}\int_{V_u} \rho dV= \int_{V_u}(\frac{D \rho }{D t}+\rho \vec{\nabla }( \vec{u})) dV$$

Hope it can help and sorry for my poor english.