I’ve read a solution to a question in the Applied Mathematics book , but it seems incorrect to me . So here is the question :
Non-dimensionalize and find the leading order solution to the IVP:
$ m y’’ + a y’ + ky^3 = 0$ where $y(0)=0, my’(0)=I$ where $ m \ll 1$ and all other constants are $O(1)$
So they rescale as $t_c = t/(a/k)$ and $y_c = y/(I/a)$, the scale $a/k$ does not make it dimensionless so it seems incorrect to me ?
I took $t_c = t/T$ where $T = a^3/(kI^2)= O(1)$ ( and $t_c$ is dimensionless, as you can check as $a$ is $MT^-1$ and $k$ is $ML^{-2}T^{-2}$ and $ I $ is $MLT^{-1}$).
With this choice I rescale and find :
$ \epsilon y_c’’ + y_c’ + y_c^3 = 0$
with $ \epsilon = m/(aT) \ll 1 $
Note: $T$ was chosen so that the coefficient in front of $y_c^3$ is $1$
So for the outer solution you get to solve $ y_o’ + y_o^3=0$ which gives $ y_o = (C+2t_c)^{-1/2}$
For the inner part we set $t_2 = t_c/\epsilon$ and $Y(t_2)=y_c(t_c)$ to get $ Y’’ + Y’ + \epsilon Y^3 = 0$
Which gives $Y(t_2) = 1-e^{-t_2}$ with the initial condition $Y(0)=0, Y’(0) = 1 $
The matching gives $ Y(\infty)=1=y_o(0)=C^{-1/2}$
So I get the composite solution as :
$ y(t) = I/a\left((1+2ta\epsilon/m)^{-1/2}-e^{-at/m}\right)$
(where I used $ \epsilon=m/(aT)$ to simplify)
Here is the book solution :
Note : in some Lecture where you have $y$ instead of $y^3$ the solution appears to be $ y(t) = I/a \left( e^{-t/(a/k)}-e^{-(t/(a/k)/\epsilon}\right)$ where they scale here indeed with $T=a/k, \epsilon=mk/a^2$, which seems to be the solution provided here as $ u(\tau)=u(t/(a/k))=y(t)/(I/a)$