Recent content by crocomut

Excellent. I knew it didn't make sense and just needed confirmation, I got the integral from an obscure paper on a computational code and that confirms what I suspected - the integral is not written properly. Thanks.

Thanks for your reply, I apologize for the confusion but I don't think I was clear, the apostrophe was meant to indicate different letters for variables, not a derivative. With that in mind, I will rephrase my question and actually simplify the integrand since that is not what is important...

I have the following integral: \int_0^{f(x,y)}{f' \sin(y-f')df'} Now suppose that f(x,y) = x*y, my question is how do I write the integral in terms of x and y only? Can I do something like this? Since df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy we can obtain...

Thank you for your time, I think I was correct in the way I was thinking and just needed confirmation. Your answer was great.

This is not a homework question or anything like that, I am just trying to learn about the log-law of the wall in fluid dynamics. So in reality what I have to do is solve for u in the following expression (there are constants in there but I removed for clarity): ∂u/∂z = 1/z The way...

Yes, this is what I attempted and I got zu - ∫udz Does it make sense?

How do I compute the integral ∫z ∂u(z)/∂z dz Thanks.

In this textbook, how exactly are 2.2.19 and 2.3.21 equal? http://i.imgur.com/LrcXE.png

Your answer is exactly what I was thinking but, as you can see from http://i.imgur.com/VmbKS.jpg"in my hydrodynamics lecture, it is not what my professor claims. Hence the confusion.

Sorry, let me correct and ask again:For a solenoidal velocity field \nabla \cdot \mathbf{u} = 0 which means that \nabla is perpendicular to \mathbf{u} . Similarly, for an irrotational velocity field \nabla \times \mathbf{u} = \mathbf{0} which means that \nabla is parallel to \mathbf{u}...

For a solenoidal velocity field [ tex ] \nabla \cdot \mathbf{u} [ /tex ] which means that [ tex ] \nabla [/tex ] is perpendicular to [ tex ] \mathbf{u} [ /tex ]. Similarly, for an irrotational velocity field [ tex ] \nabla \times \mathbf{u} [ /tex ] which means that [ tex ] \nabla [/tex ] is...

Suppose we have a function u=f(x,y,z) If \frac{\partial u}{\partial x} = 0 then u is independent of x and is u=f(y,z) only. Correct?

Sorry, here it is: Gabriel Godin - Magnitude of Stokes' drift in coastal waters - Ocean Dynamics, 1995 - Springer

My question pertains to the following article: http://tinyurl.com/4uw9h2a I have attached the relevant section to this post. My question is whether Godin's assertion is correct or not - namely the sentence "Such a development ... additional terms" and the last sentence in the attachment...

Hi HallsofIvy, Thanks so much for your answer, I actually did it last night. Maybe you can have a look at my result and confirm: In the end I got that the diagonal entries of D are: di+1 = di\sqrt{b_i/c_{i+1}} The square root is why bici+1 > 0 Croco