Recent content by LPerrott

The proper way to think about it would be that the natural numbers are "finer" than the even numbers or the even numbers are "coarser" than the natural numbers ... I'm I correct in my thinking?

You don't use divergence theorem on the integral above. You need to first multiply by a test function, v, then integrate by parts. When you integrate by parts, the laplacian term will actually decompose into two terms. One will be gradient of u times gradient of v and the other will involve...

product rule and chain rule

that is correct

You don't evaluate any integrals ... and secondly what you have listed is not the bilinear form. The bilinear form is an operator on two functions B(u,v)= integral (_____) ... you don't actually evaluate anything. All you need to show is that B(u,v)=B(v,u) and B is coercive .. try reading up...

Look up Cantor diagonalization argument. That should help you. Proceed by contradiction. First, realize that the functions we are speaking about really yield a sequence of 1's and zero's. If the set of all these functions were countable, then we could put them into an matrix of infinite size...

In order to get the bilinear form you multiply the equation by a test function (in this case v), then you integrate over the domain Omega, then you integrate by parts (Note: you always integrate by parts when formulating a weak statement for a PDE .. the entire point is to transfer derivatives...

A quick an easy way to show that the inverse of A+I exists is to use the determinant. Find the determinant of A using the equation A^2011=0 and then find the determinant of A+I ... what does this tell you?

I told you exactly how to do it. When integrating a probability density function you need to integrate from -infty to infty, but since you have absolute value of x this is a problem, therefore you you just integrate 2 times the integral from 0 to infty. k=1/2 As far as the next part of...

I can tell you without even checking integration that is incorrect. k=-1 violates the first property of a prob density function ... mainly that f(x)>=0. Notice you have an absolute value of x in the function. This might be what's confusing you. Try solving the following equation for k 2k...

No. You need to look up the properties of a probability density function. Simply integrating the prob dens function isn't enough to find the value of k. If I want k = __ , then I need to start off with an equation integral of (___) = ___

What conditions does a function need to satisfy in order to be a probability density function? Hint: There are two of them, one is that f(x)>=0, but it is the the second condition that will help you find your answer.

From my understanding, the way the second equation is manifests itself onto the pressure field is through the Leray Projection. The Leray Projection, using Hodge orthogonal decomposition, projects the Sobolev space onto the space of divergence free functions (satisfying the second equation)...