Continuum mechanics/Leibniz formula

< Continuum mechanics

The integral


       F(t) = \int_{a(t)}^{b(t)} f(x, t)~\text{dx}

is a function of the parameter t. Show that the derivative of F is given by


       \cfrac{dF}{dt} = \cfrac{d}{dt}\left( \int_{a(t)}^{b(t)} f(x, t)~\text{dx}
           \right) =  \int_{a(t)}^{b(t)} \frac{\partial f(x, t)}{\partial t}~\text{dx} +
             f[b(t),t]~\frac{\partial b(t)}{\partial t} - f[a(t),t]~\frac{\partial a(t)}{\partial t}~.

This relation is also known as the Leibniz rule.

Proof:

We have,


        \cfrac{dF}{dt} = \lim_{\Delta t\rightarrow 0}
           \cfrac{F(t + \Delta t) - F(t)}{\Delta t} ~.

Now,


  \begin{align}
   \cfrac{F(t + \Delta t) - F(t)}{\Delta t}  & = 
     \cfrac{1}{\Delta t} \left[
       \int_{a(t+\Delta t)}^{b(t+\Delta t)} f(x, t+\Delta t)~\text{dx} - 
       \int_{a(t)}^{b(t)} f(x, t)~\text{dx}\right] \\
     & \equiv 
     \cfrac{1}{\Delta t} \left[
       \int_{a+\Delta a}^{b+\Delta b} f(x, t+\Delta t)~\text{dx} - 
       \int_{a}^{b} f(x, t)~\text{dx}\right] \\
     & = 
     \cfrac{1}{\Delta t} \left[
       -\int_{a}^{a+\Delta a} f(x, t+\Delta t)~\text{dx} + 
       \int_{a}^{b+\Delta b} f(x, t+\Delta t)~\text{dx} - 
       \int_{a}^{b} f(x, t)~\text{dx}\right] \\
     & = 
     \cfrac{1}{\Delta t} \left[
       -\int_{a}^{a+\Delta a} f(x, t+\Delta t)~\text{dx} + 
       \int_{a}^{b} f(x, t+\Delta t)~\text{dx} + 
       \int_{b}^{b+\Delta b} f(x, t+\Delta t)~\text{dx} - 
       \int_{a}^{b} f(x, t)~\text{dx}\right] \\
     & = 
       \int_{a}^{b} \cfrac{f(x, t+\Delta t) - f(x,t)}{\Delta t}~\text{dx} + 
       \cfrac{1}{\Delta t}\int_{b}^{b+\Delta b} f(x, t+\Delta t)~\text{dx} - 
       \cfrac{1}{\Delta t}\int_{a}^{a+\Delta a} f(x, t+\Delta t)~\text{dx} ~.
  \end{align}

Since f(x,t) is essentially constant over the infinitesimal intervals a < x < a+\Delta a and b < x < b+\Delta b, we may write


   \cfrac{F(t + \Delta t) - F(t)}{\Delta t}  \approx
       \int_{a}^{b} \cfrac{f(x, t+\Delta t) - f(x,t)}{\Delta t}~\text{dx} + 
       f(b, t+\Delta t)~\cfrac{\Delta b}{\Delta t} - 
       f(a, t+\Delta t)~\cfrac{\Delta a}{\Delta t}~.

Taking the limit as \Delta t\rightarrow 0, we get


   \lim_{\Delta t \rightarrow 0} \left[\cfrac{F(t + \Delta t) - F(t)}{\Delta t}\right]  = 
       \lim_{\Delta t \rightarrow 0}\left[
       \int_{a}^{b} \cfrac{f(x, t+\Delta t) - f(x,t)}{\Delta t}~\text{dx}\right] 
       + \lim_{\Delta t \rightarrow 0}\left[f(b, t+\Delta t)~\cfrac{\Delta b}{\Delta t}\right] - 
         \lim_{\Delta t \rightarrow 0}\left[f(a, t+\Delta t)~\cfrac{\Delta a}{\Delta t}\right]

or,


   {
   \cfrac{dF(t)}{dt} = 
         \int_{a(t)}^{b(t)} \frac{\partial f(x, t)}{\partial t}~\text{dx} +
         f[b(t),t]~\frac{\partial b(t)}{\partial t} - f[a(t),t]~\frac{\partial a(t)}{\partial t}~.
   } \qquad\qquad\qquad\square


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