Error for perturbed solution vs Numeric

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member 428835
Hi PF!

I'm solving an PDE where the analytic solution is called ##F(x)## (unknown). To approximate the analytic solution I made a naive expansion in some small parameter ##\epsilon## such that ##F(x) = f_0(x)+\epsilon f_1(x)+O(\epsilon^2)##, where I know ##f_0(x)## and ##f_1(x)##. I then solved the PDE numerically, let's call that solution ##F_n##. Then the error ##(F_n - (f_0(x)+\epsilon f_1(x)))/F_n## should be ##O(\epsilon^2)##. However, when I let ##\epsilon=.9## and then ##\epsilon=.8## my error is still about ##0.15##. How can this be?

I should say I know the numeric and asymptotic solutions are correct.
 
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joshmccraney said:
Hi PF!

I'm solving an PDE where the analytic solution is called ##F(x)## (unknown). To approximate the analytic solution I made a naive expansion in some small parameter ##\epsilon## such that ##F(x) = f_0(x)+\epsilon f_1(x)+O(\epsilon^2)##, where I know ##f_0(x)## and ##f_1(x)##. I then solved the PDE numerically, let's call that solution ##F_n##. Then the error ##(F_n - (f_0(x)+\epsilon f_1(x)))/F_n## should be ##O(\epsilon^2)##. However, when I let ##\epsilon=.9## and then ##\epsilon=.8## my error is still about ##0.15##. How can this be?

I should say I know the numeric and asymptotic solutions are correct.
##\epsilon=.9## isn't really all that small. Do you get better results for, say, ##\epsilon=.1##?
 
Mark44 said:
##\epsilon=.9## isn't really all that small. Do you get better results for, say, ##\epsilon=.1##?
So for ##\epsilon=0.01## I'm getting an error of 0.165 but for ##\epsilon=0.2## the error is 0.14. How could this ever be possible?