0
$\begingroup$

I'm trying to integrate the following term in the z-direction

$\displaystyle\int_b^a u\frac{\partial v}{\partial y} dz$

where the variable dependencies are

$a(x,y,t)$, $b(x,y,t)$, $u(x,y,z,t)$, $v(x,y,z,t)$

I'm not really sure how to apply Leibniz rule here, given the product $ u\dfrac{\partial v}{\partial y}$.

How do you make use of Leibniz rule when terms like $ u\dfrac{\partial v}{\partial y}$ are involved? Can you still pull the differential out in front some how?

$\endgroup$
0
$\begingroup$

What seems to be possible is to write the term $u \frac{\partial v}{\partial y}$ as result of the product rule, i.e. $$ u \frac{\partial v}{\partial y} = \frac{\partial( uv)}{\partial y} - v\frac{\partial{u}}{\partial y} $$ and then apply the Leibniz integral rule to the first term on the right: $$ \int_a^b \frac{\partial (uv)}{\partial y}\;dz = \frac{\partial}{\partial y}\left(\int_a^b (uv) \;dz\right) + \frac{\partial a}{\partial y} (uv)|_a - \frac{\partial b}{\partial y} (uv)|_b $$

$\endgroup$
  • $\begingroup$ I'm aware of this approach, but my question still stands. In the case you presented, there is still a $-v\frac{\partial u}{\partial y}$. Does Leibniz rule some how apply to this term? I know if I assume incompressible, I can remove this term, but I don't want to make that assumption at this point. $\endgroup$ – ThatsRightJack Feb 14 '17 at 21:33
  • $\begingroup$ Also, I think the last two terms in your solution shouldn't be integrals, but rather $(uv)|_b$ and $(uv)|_a$, meaning those terms evaluated at the integral bounds. $\endgroup$ – ThatsRightJack Feb 14 '17 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.