Mechoption

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15 years, 282 days

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These are answers submitted by Mechoption

Yes - I knew I was brain farting and there had to be a way to do this !

Thanks !

Try the following:

(1) Define the vector as: [[-0.000163,-8.84352*10^(-13)+(I*1.71511)*10^(-12)],[0,1.49867*10^(-13)+(I*1.45355)*10^(-12)],[0.000163,1.15763*10^(-12)+(I*9.08261)*10^(-13)]];

(2) Define your equation as a*exp(-I*4375*t)+b*exp(I*4375*t)

(3) Define your variable as t

(4) Solve as CurveFitting[LeastSquares]((1),(3),curve=(2));

Done :)

See http://www.maplesoft.com/support/help/AddOns/view.aspx?path=Task/CurveFitting_LeastSquares

This looks great, thanks! but alas my pde training/knowledge is quite rusty.  I'm trying to figure out how you came up with that 'constant solution' ? My solution to the equation is:

 C(r, t) = _C1*_C3*exp(t*Di*_c[1])*sinh(sqrt(_c[1])*r)/(r*exp(t*lambda))+_C3*exp(t*Di*_c[1])*_C2*cosh(sqrt(_c[1])*r)/(r*exp(t*lambda))+Beta/lambda+Beta*_C2*sinh(sqrt(lambda/Di)*r)/(r*lambda*_C1)+Beta*_C3*cosh(sqrt(lambda/Di)*r)/(r*lambda*_C1)

Now, I see that the last three terms are independant of time, and therefore presumably make up your constant solution somehow?

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