# Under what conditions is the material derivative of an integral not equal to the integral of the material derivative of a quantity?

I have always been curious, but the general form of the Helmholtz equation in fluid mechanics states that

$$\frac{D}{Dt}\int_{C}\vec{v}\cdot\vec{dl}$$ = 0

which is the statement that the circulation around a closed loop remains constant as it convects with the bulk flow.

An alternate representation of the same equation which is often written in these books is

$$\int_{C} \frac{D}{Dt} (\vec{v}\cdot\vec{dl})$$ = 0

I have noticed this equivalence being used in most fluid mechanics books without explaining how this can be done. I was curious under what conditions can the material derivative be brought inside the integral. Is there a specific scenario when the above equivalence breaks apart?

## 1 Answer

The material derivative is $$\frac{\mathrm D}{\mathrm Dt} = \frac{\partial}{\partial t} + \vec v\cdot\nabla$$ So your first expression, where the the expression being differentiated is an integral over space and therefore only a function of time, simplifies to $$\frac{\mathrm D}{\mathrm Dt}\oint \vec v\cdot\mathrm d\vec l = \frac{\partial}{\partial t}\oint\vec v\cdot\mathrm d\vec l = \oint\frac{\partial\vec v}{\partial t}\cdot\mathrm d\vec l$$ Your second expression expands to become $$\oint \frac{\mathrm D\vec v}{\mathrm Dt}\cdot\mathrm d\vec l = \oint\left(\frac{\partial\vec v}{\partial t} + \vec v\cdot\nabla\vec v\right)\cdot\mathrm d\vec l = \oint\frac{\partial\vec v}{\partial t}\cdot\mathrm d\vec l + \oint (\vec v\cdot\nabla)\vec v\cdot\mathrm d\vec l$$ The difference between the two is the integral $$\oint(\vec v\cdot\nabla)\vec v\cdot \mathrm d\vec l$$ So the expressions are the same when this is $$0$$. I don't know enough about fluid dynamics to tell you when this is the case but hopefully this helps.