A Nonlinear Oscillator,
Numerical Solution
Problem 3, on P.S. #3, 2002. Numerical check of perturbation
solution, problem not solved here.
Finds a numerical solution to:
y'' + y 
with y'(0)=0 and y(0)=1.
The period in the numerical solution is then compared to the prediction made from the perturbation expansion. The math for the
perturbation expansion is not shown here.
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The following line was changed to text. You may want to change the Format>Style to "Input" to execute the statement, and confirm that Mathematica cannot find an analytical solution:
DSolve[{y''[t]+y[t]+eps*y[t]^3==0,y[0]==1,y'[0]==0},y[t],t]
So we resort to a numerical solution for a specified value of eps:
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Plot the analytical solution for eps=0 in black and the numerical
solution for eps=0.5 in red:
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Here we make a more convenient interface to access the interpolating function that contains the numerical solution. Notice the replacement rule is inside a nested list, which is accessed by the [[1,1]]:
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Let's give it a try:
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Let's find t where y(t)=0 in the numerical solution:
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Now make a module to find the first t where y(t)=0 for a
specified value of ε.
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Try it:
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Make a table of ratio of the period of oscillation to that with ε=0.
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Compare the numerical result, which we assume to be exact, (green) with the perturbation result (red).
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Now compare with the perturbation result out to 
:
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