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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.

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Let's see if we can solve this exactly.

Inverting the function gives us:

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

$$t'(y)=f(y)$$

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

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

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

$$f^2=\frac{y^2}{1+Cy^2+y^4}$$

$$f=\frac{y}{\sqrt{1+2C_1y^2+y^4}}$$

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

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:

$$t'(1)=\frac{1}{\sqrt{2(C_1+1)}}=1$$

$$C_1=-\frac{1}{2}$$

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

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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