ME565 Lecture 20: Numerical Solutions to PDEs Using FFT
ฝัง
- เผยแพร่เมื่อ 26 เม.ย. 2016
- ME565 Lecture 20
Engineering Mathematics at the University of Washington
Numerical Solutions to PDEs Using FFT
Notes: faculty.washington.edu/sbrunto...
Matlab code:
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Course Website: faculty.washington.edu/sbrunto...
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Grate lecture sires
Your lectures are awesome! I have to thank you very much :D
but I do have 2 questions I may have to ask
1. When we deal with DFT (and FFT) there are k as appeared in exp(-i*2*pi*k*n/N) and while your Keppa (=omega) is (2*pi/L)*k
From previous lectures, I thought k will run from k = [0, 1, 2, ..., N]. But in this lecture, your k = [-Nx/2: Nx/2 -1]. So, I can see and feel how it make sense that our k is now related to the space domain (x) in some ways. But I do not really mathimatically understand when and how it happens.
2. I can't help but notice that the way Fourier transform turns our PDE to ODE kind of force us to ignore any Boundary condition we may have in the process. In this lecture, we talk about an infinitely long rod. So, we can say that BC is far away and does not have any effect on our problems. But when we do it in Matlab, our cylinder does have a finite length.
For 2nd comment: by the definition, while doing FT to a derivative function, the implicit and handy assumption is the function is bounded (goes to zero at boundary). So in a sense it is not ignored but is presumed.
If there is a function with non-zero boundary condition, then the function can be turned into shifted version function with zero boundary condition. Basically the result is identical only the variable is shifted.
@@wowowowdog Thank you!
10/10 !!!!
My differential equation is of form:
EI*(d^4y/dx^4)+ky+m*(d^2y/dt^2)=Pδ(x-vt), −∞