- #1
Dessert
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- 0
dx/dt = η(t)
dy/dt = ζ(t)
where
<η(t)>=<ζ(t)>=0
<η(t)η(t')> = 2Dδ(t-t')
<ζ(t)ζ(t')> = 2Dδ(t-t')
If <η(t)ζ(t')> = 0, we have the standard 2-D diffusion equation and the analytical solution is known.
If <η(t)ζ(t')> = 2Dδ(t-t'), or η(t) = ζ(t), we can transform it into a 1-D problem and the analytical solution is also known.
What if <η(t)ζ(t')> = 2Dcδ(t-t') where 0<c<1 which is correlation of the two noises? We can still write down a 2-D diffusion equation, but is the analytical solution known?
dy/dt = ζ(t)
where
<η(t)>=<ζ(t)>=0
<η(t)η(t')> = 2Dδ(t-t')
<ζ(t)ζ(t')> = 2Dδ(t-t')
If <η(t)ζ(t')> = 0, we have the standard 2-D diffusion equation and the analytical solution is known.
If <η(t)ζ(t')> = 2Dδ(t-t'), or η(t) = ζ(t), we can transform it into a 1-D problem and the analytical solution is also known.
What if <η(t)ζ(t')> = 2Dcδ(t-t') where 0<c<1 which is correlation of the two noises? We can still write down a 2-D diffusion equation, but is the analytical solution known?
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