I am trying to determine the asymptotic behavior as $t \to \infty$ of the solution to the IVP:

$$y''(t) - y(t) + \frac{1}{[y(t)]^3} = 0\\y(0) = 1;\, y'(0) = 1$$

Without the nonlinearity, we have solutions of the form $e^{\pm t}$. I think the nonlinear solution has similar behavior, but I'm not sure how to show this or determine the precise asymptotic form.


Let's see if we can solve this exactly.

Inverting the function gives us:

$$-\frac{t''}{t'^3} - y + \frac{1}{y^3} = 0 \\ t''+\left(y-\frac{1}{y^3} \right) t'^3=0$$


$$f'=\left(\frac{1}{y^3}-y \right) f^3$$

$$-\frac{1}{2 f^2}=-\frac{1}{2 y^2}-\frac{y^2}{2}+C$$




$$t(y)= \int \frac{y ~dy}{\sqrt{1+2C_1y^2+y^4}}=\frac{1}{2} \int \frac{du}{\sqrt{1+2C_1 u+u^2}}=\frac{1}{2} \log \left(C_1+u+\sqrt{1+2C_1 u+u^2} \right)+C_2$$

So we have:

$$t(y)=\frac{1}{2} \log \left(C_1+y^2+\sqrt{1+2C_1 y^2+y^4} \right)+C_2$$

Substituting the first condition gives us:

$$C_2=-\frac{1}{2} \log \left(C_1+1+\sqrt{2(C_1+1)) } \right)$$

For the second condition we find:



$$C_2=-\frac{1}{2} \log \frac{3}{2}$$

So we get finally the exact implicit solution:

$$t(y)=\frac{1}{2} \log \left(y^2-\frac12+\sqrt{y^4-y^2+1} \right)-\frac{1}{2} \log \frac{3}{2}$$

We can find $y(t)$ by solving the above equation, which would reduce to a quadratic one:

$$y^2-\frac12+\sqrt{y^4-y^2+1}=\frac{3}{2} e^{2t}$$

$$\sqrt{y^4-y^2+1}=\frac{3}{2} e^{2t}+\frac12 -y^2$$

$$y^4-y^2+1=y^4-\left(1+3 e^{2t} \right)y^2+\left(\frac{1}{2}+\frac{3}{2} e^{2t} \right)^2 $$

$$3 e^{2t} y^2=\left(\frac{1}{2}+\frac{3}{2} e^{2t} \right)^2 -1$$

$$y(t)= \frac{e^{-t}}{2\sqrt{3}} \sqrt{\left(1+3 e^{2t} \right)^2 -4}$$

$$y(t)= \frac{e^{-t}}{2} \sqrt{3 e^{4t}+2e^{2t} -1}$$

This is the exact solution, which can be checked by direct substitution. From this we can find the asymptotic.

For $t \to +\infty$ we have:

$$y(t) \asymp \frac{\sqrt{3} }{2}e^{t}$$

The asymptotic gives a very good approximation for $t>2$, see the plot (blue is the exact solution, orange the asymptotic):

enter image description here


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