How do I differentiate under the line integral along an open curve?

Question

Is there a way to perform the following differentiation under the line integral sign ?

$$ I = \frac{d}{dt}\left(\int_{\Gamma\left(t\right)}\vec{u}\cdot d\vec{x}\right) $$

provided the vector $ \vec{u}=\vec{u}\left(\vec{x},t\right)$ varies in space and also varies with the variable $ t $. And provided that $ \vec{u} $ may not be conservative.

The curve $\Gamma$ may also be allowed to depend on the variable $ t $ and may even be allowed to undergo translations and rotations. But, the most important thing about $\Gamma$ is that it be an open curve and not a closed one. This is because I know that the closed curve would be a special case of the 2-D Liebnitz rule.

Is there a general formula for this operation ? and, if so, how could one prove it?

Answer 1

This can be quite an uncomfortable thing to encounter the first time. There are a number of rules for this operation depending on the situation or how many dimensions you are working in. One of the rules which may be fitting is called the Reynolds transport theorem (seen as you've tagged fluid mechanics, I assume you mean this theorem) which is the same form as the integral you have above. This rule and more can be found here.

Post your progress and see if I or others can help further!

I should also mention some examples here. As you will see, it is a very powerful method. Good luck!

Answer 2

If the curve $\Gamma\left(t\right)$ is open then in 2-D, for example, the line integral may be written simply as a sum of two definite integrals.

$$L = \int_{x_{0}}^{x_{1}\left(t\right)}u_{x} dx+\int_{y_{0}}^{y_{1}\left(t\right)}u_{y} dy$$

If the initial point $(x_{0},y_{0})$ is fixed, the Leibniz integral rule requires that

$$\frac{d}{dt}\int_{\Gamma}\vec{u}.(dx,dy)=\frac{d}{dt}\int_{x_{0}}^{x_{1}\left(t\right)}u_{x} dx+\frac{d}{dt}\int_{y_{0}}^{y_{1}\left(t\right)}u_{y} dy\\ =\int_{x_{0}}^{x_{1}\left(t\right)}\frac{\partial u_{x}}{\partial t} dx+\int_{y_{0}}^{y_{1}\left(t\right)}\frac{\partial u_{y}}{\partial t} dy+ u_{x}\left(x_{1}\left(t\right),t\right)\frac{dx_{1}}{dt}+u_{y}\left(y_{1}\left(t\right),t\right)\frac{dy_{1}}{dt}$$

This then simplifies to

$$\frac{d}{dt}\int_{\Gamma\left(t\right)}\vec{u}\cdot d\vec{x}=\int_{\Gamma\left(t\right)}\frac{\partial \vec{u}}{\partial t}\cdot d\vec{x} +\vec{u}\left(\vec{x_{\Gamma}},t\right)\cdot \frac{d \vec{x_{\Gamma}}}{dt} $$

where I have defined $\vec{x_{\Gamma}}$ to the the end point on $\Gamma$.

under the same assumptions, is this equation true in 3-D?

Answer 3

It seems to be possible to consider the problem i a much more general form.

If the vector $\vec{v_{n}}$ is the normal velocity field of all points along the curve $\Gamma$ and if the vector $\vec{v_{t}}$ is the tangent velocity field along the curve $\Gamma$, then this argument concludes that the derivative under the line integral should be

$$\frac{d}{dt}\int_{\Gamma\left(t\right)}\vec{u}\cdot d\vec{x}=\left[\vec{u}\left(\vec{x_{2}},t\right)\cdot \vec{v_{t_2}} -\vec{u}\left(\vec{x_{1}},t\right)\cdot \vec{v_{t_1}}\right]+\int_{\Gamma\left(t\right)}\left[\frac{\partial \vec{u}}{\partial t}+\left(\vec{v_{n}}\cdot\nabla\right)\vec{u}\right]\cdot d\vec{x} $$ $$+\int_{\Gamma\left(t\right)}\vec{u}\cdot\frac{\partial \vec{v_{n}}}{\partial l}dl$$

where $\vec{x_{1}}$ and $\vec{x_{2}}$ are the coordinates of the initial and final points on $\Gamma$ respectively, while $\vec{v_{t_1}}$ and $\vec{v_{t_2}}$ are the corresponding tangential velocity components. The length along the curve $\Gamma$ i denoted by $l$.