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I need a check on the following exercise:

Consider the following Cauchy problem: \begin{cases} y''(x)=y'(x)^2 - 2 \\ y(0)=0 \\ y'(0) = 1 \end{cases}

i) Show the solution is defined for all $x \in \mathbb{R}$

ii) Compute $\lim_{x \rightarrow +\infty} y'(x)$ and $\lim_{x \rightarrow +\infty} y(x)$


My attempt:

i) I recast everything to the first order, hence I define the vector function

$$F(t,y,y')=[y'^2-2,y']^T$$

I would like to prove sublinearity in order to show that the solution is globally defined:

$$||F(t,y,y')|| \leq h + k ||[y,y']||$$

Using the expression for $F$: $$y'^4 - 3 y'^2 +4$$ but I don't know how to find a sublinearity condition here: I should bound the latter expression with $y^2 + y'^2$

So, I notice that the function $F=[F_1,F_2]$ is such that $\partial_y F_1 = \partial_y F_2 = 0$ and $\partial_{y'}F_1 = 1$ and $\partial_{y'}F_2 = 2y'$. This means that $F$ il globally Lipschitz, so in principle, existence and uniqueness could be applied iteratively and define a solution for every $x \in \mathbb{R}$.

IS THERE A WAY TO SHOW IT WITH SUBLINEARITY?

ii) Here I note that, after the reduction to the first order, I have (call $y'=z$) the ODE $$z' = z^2-2$$ with $z(0)=1$. By existence and uniquess, and using the stationary solutions $y=\pm \sqrt{2}$, I have that $z$ starts from $1$ and then it decreases. The limit must exists, since the solution is defined on the whole $\mathbb{R}$ and it's monotone. Then $$\lim_{x \rightarrow \infty} z(x)= \lim_{x \rightarrow \infty} y'(x)= -\sqrt{2}$$

To compute $$\lim_{x \rightarrow \infty} y(x)$$ I note that $y'(x)=z(x)$, and if it would be finite , then $$\lim_{x \rightarrow \infty} y'(x) = 0$$ But this limit is precisely the one I have just copmuted, i.e. $-\sqrt{2}$, therefore this limit must be $+\infty$ or $-\infty$. Since $y'(x)=z(x)$ and $z(x)$ is monotonically decreasing, then this limit must be $-\infty$.


Is eveything okay?

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You did fine in ii), for the proof of global existence you only need to sharpen that argument in that no other solution can reach or cross the constant solutions in $z'=z^2-2$. That is, any solution starting in $[-\sqrt2,\sqrt2]$ remains bounded to that interval and thus exists for all times. Then use that $y'=z$ has a bounded right side, as you already did half-ways.


For this special form of the differential equation, a Riccati DE in $z=y'$, there is a particularly easy way to obtain the solution formula.

Consider $u(x)=\exp(-y(x))$. Then $u'(x)=-\exp(-y(x))y'(x)$ and $$ u''(x)=-e^{-y(x)}(y''(x)-y'(x)^2)=2u(x) $$ The solution of this now second order linear DE for $u$ with constant coefficient and initial condition $u(0)=\exp(-y_0)=1$, $u'(0)=-\exp(-y_0)y_0'=-1$ is $$ u(x)=\cosh(\sqrt2 x)-\frac1{\sqrt2}\sinh(\sqrt2 x). $$ Thus $$ y(x)=-\ln(u(x))=-\sqrt2x-\ln\left(\tfrac12(1-\tfrac1{\sqrt2})+\tfrac12(1+\tfrac1{\sqrt2})e^{-2\sqrt2 x}\right) $$ The asymptotic behavior can be read off this solution formula.

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  • $\begingroup$ thanks, but I don't have to find the analytic solution. Btw, I've just seen that the limits are the same. The problem is in showing global existence: how could I do ? $\endgroup$ – andereBen Aug 23 '20 at 7:03
  • $\begingroup$ I just wanted to show a particularly easy way to obtain the solution formula (that applies only to this special form of equation, Riccati in $z=y'$). // You did fine in ii), you only need to sharpen that in that no other solution can reach or cross the constant solutions in $z'=z^2-2$, that is, any solution starting in $[-\sqrt2,\sqrt2]$ remains bounded to that interval and thus exists for all times. Then use that $y'=z$ has a bounded right side. $\endgroup$ – Lutz Lehmann Aug 23 '20 at 11:06
  • $\begingroup$ Yes, so $|y(t)|= |\int_0^t z(s)ds| \leq t \sqrt{2}$ and hence we have sublinearity for $x$ ;) Many thanks Lutz $\endgroup$ – andereBen Aug 23 '20 at 11:30
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Let $z=y'$, the the ODE is $$z'=z^2-2 \implies -\int \frac{dz}{2-z^2}=x+C \implies -\frac{1}{\sqrt{2}} \tanh^{-1}\frac{z}{\sqrt{2}}=x+C$$ Since at $x=0, z=1$, so $C=-\frac{1}{\sqrt{2}}\coth^{-1} \sqrt{2}$. Next we get $$z=-\sqrt{2}\tanh[\sqrt{2}(x+C)]=\frac{dy}{dx}\implies y=\int -\sqrt{2}\tanh[\sqrt{2}(x+C)] dx+D~~~(1)$$ $$\implies y(x)=-\ln \cosh[\sqrt{2}(x+C)]+D.$$ Yse $y(0)=0$, $D=\ln \cosh \sqrt{2}C$ Finally, we have $$y(x)=\ln \left (\frac{\cosh \sqrt{2}C}{\cosh[\sqrt{2}(x+C)]}\right)~~~(2)$$ Domain of this solution is all $x \in \Re$. From (1), $y'(\infty)=-\sqrt{2}$ and from (2) $y(\infty)=-{\infty}$

See the fig. below for $y'(x)$ (red line) and $y(x)$ (blue line)enter image description here

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  • $\begingroup$ This is not a qualitative study, you just found the analytical solution $\endgroup$ – andereBen Aug 23 '20 at 7:04
  • $\begingroup$ @anderBen Normally, qualitative analysis is done if a problem is not solvable. But if it is luckily solvable one should start drawing the required inferences. It is said that estimates and numerical results are never solutions of a problem; the exact analytic formula is actually the one and complete. Cheers. $\endgroup$ – Z Ahmed Aug 23 '20 at 7:15
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    $\begingroup$ I totally agree with you, but the exercise does not allow to use such techniques. $\endgroup$ – andereBen Aug 23 '20 at 7:17

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