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FDTD algorithm in Ruby
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include Math | |
require 'pp' | |
X = 10 | |
T = 4 | |
PI = 3.14 | |
dt = 1 | |
dx = 1 | |
# HW4 prob.2 | |
#V = 0.8 | |
#psi = [ | |
# [0, 0, Math.sin(PI/4), sin(2*PI/4), sin(3*PI/4), 0, 0, 0, 0, 0, 0], | |
# [0, 0, Math.sin((1-0.8)*PI/4), sin((2-0.8)*PI/4), sin((3-0.8)*PI/4), sin((4-0.8)*PI/4), sin((5-0.8)*PI/4), 0, 0, 0, 0], | |
#] | |
#def s(x,t); 0; end | |
# HW6 problem3 | |
V = 0.5 | |
psi = [ [0]*X ] | |
def s(x,t) | |
if x == 5 && t == 0 | |
1 | |
else | |
0 | |
end | |
end | |
(psi.size-1).upto(T-1) { |t| | |
psi[t+1] = [] | |
0.upto(X-1) { |x| | |
right = x == X-1 ? 0 : psi[t][x+1] | |
left = x == 0 ? 0 : psi[t][x-1] | |
old = t == 0 ? 0 : psi[t-1][x] | |
psi[t+1][x] = -old + 2*psi[t][x] + V*V*dt*dt/dx/dx * (left + right - 2*psi[t][x]) + dt*dt*s(x,t) | |
} | |
} | |
pp psi | |
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