The Student Room Group
johnbarkwith
Does anyone know how to show if a function of 2 variables is in L-2 Space (i.e. square integrable). I have only ever done this for functions of one variable and am not sure how the theory transfers over....


I have honestly no idea, perhaps its something to do with double integrals?
Reply 2
basically i need to know what the integration is with respect to....
Reply 3
L2L^2 usually means square integrable functions defined on the real line, i.e. measurable functions whose square is Lebesgue integrable, not functions that are integrable over the plane which are written L1(R2)L^1 (R^2). Are you sure of what is being asked?
Reply 4
i have been asked to show that the function u(r,θ)=r2/3sin((2θ+π)/3) u(r,\theta)=r^{2/3} sin((2\theta + \pi) /3) defined on the pi shaped domain 0r10\leq r \leq 1 , π/2θπ -\pi /2 \leq \theta \leq \pi is in H1(Ω)H^1(\Omega) but not in H2(Ω)H^2(\Omega). where the H's are Sobolev spaces. So i think i need to show that the first derivatives of u are square integrable but the second derivatives are not. This is where I became stuck, as the first and second derivatives are functions of two variables...

If you know how I could go about showing this I would be really gratefull, have been trying for ages I have been given the hint that I need to work in polar coordinates and use dxdy=rdrdθdxdy=rdrd\theta...
Reply 5
johnbarkwith
i have been asked to show that the function u(r,θ)=r2/3sin((2θ+π)/3) u(r,\theta)=r^{2/3} sin((2\theta + \pi) /3) defined on the pi shaped domain 0r10\leq r \leq 1 , π/2θπ -\pi /2 \leq \theta \leq \pi is in H1(Ω)H^1(\Omega) but not in H2(Ω)H^2(\Omega). where the H's are Sobolev spaces. So i think i need to show that the first derivatives of u are square integrable but the second derivatives are not. This is where I became stuck, as the first and second derivatives are functions of two variables...

If you know how I could go about showing this I would be really gratefull, have been trying for ages I have been given the hint that I need to work in polar coordinates and use dxdy=rdrdθdxdy=rdrd\theta...


Have you done any double integrals? The relation dxdy=rdrdθdxdy=rdrd\theta just comes from the Jacobian for polar co-ordinates. The actual integrals you need to consider in this example are actually pretty easy.

You may find

http://www.maths.ox.ac.uk/filemanager/active?fid=7161

some help - look at chapter 9.
Reply 6
Thanks for your help, i think i know how to compute the double integrals.. are you implying that to show whether the functions of two variables (i.e. the first and second derivatives of u) are square integrable i need to integrate there square with respect to dxdy and show that this is finite???
Reply 7
johnbarkwith
Thanks for your help, i think i know how to compute the double integrals.. are you implying that to show whether the functions of two variables (i.e. the first and second derivatives of u) are square integrable i need to integrate their square with respect to dxdy and show that this is finite???


Yes finite for the squares of both first partial derivatives, and infinite for the square of some second partial derivative.
Reply 8
I have calculated all of the first and second derivatives to be of the form arbsin(2θ+π/3)ar^b sin(2\theta + \pi/3) or arbcos(2θ+π/3)ar^b cos (2 \theta + \pi /3) for some constants a and b. However none of these have infinite inegrals over the domain when squared. can you see where I am going wrong????
Reply 9
sorry i was looking over the obvious... one of the second derivatives sqaured, multiplied by the r from the jacobian due to the change of coordinates, gives a minus power of r when integrated wrt r. when i then sub. in the limits of integration in particular the zero. I get 1/0 which is infinite...


thanks alot for your help,probably wouldnt have got it otherwise

Latest