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$y''(x)=x-y^2(x)$

I'm particularly interested in solutions when $x>0$.

I've performed asymptotic analysis and reached the conclusion that solutions must behave as $\pm\sqrt{x}$ when $x\rightarrow \infty$ but I don't know what to do next.

I've also tried looking for a solution of the form $y(x)=y_p(x)\pm\sqrt{x}$, but I got to an even more difficult differential equation for $y_p(x)$

I can also deduce the obvious thing that $y''>0$ when $x>y^2$ and $y''<0$ when $x<y^2$, so there are like three regions. My guess is that there will be solutions which tend to the parabola $y=\sqrt{x}$ in every region, but I'm really stuck.

Any insights would be very appreciated.


EDIT: When I say ''solve'', I mean that I want a feel on how possible solutions behave (e.g., are there any oscillating solutions?, if I have a solution of an IVP where $y^2(x_0)>x_0$, what will that solution do?...)

Also, I would be thrilled if there were solutions expressable as elementary functions.

Now that i think about it, a better phrasing than ''solve this ODE'' would be ''study [the behaviour of solutions of] this differential equation''

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  • $\begingroup$ What do you mean by "solve" this ODE? Probably there is no closed-form solution in terms of elementary functions. Your ODE wrote reminds me of the first-order ODE $y'(x) = y^2 - x$, which is not solvable using elementary functions. See Example 4.2 of math.uconn.edu/~kconrad/blurbs/analysis/contraction.pdf, which gives a reference to a page in Ritt's book on integration in finite terms. $\endgroup$
    – KCd
    Jun 21 '14 at 20:56
  • $\begingroup$ Sorry, I should've been more clear. I mean to find an explicit solution in case there is one (I would even like a series expansion if you can see properties of the function there), or if there isn't a simple way to write it down, at least to know the behaviour of solutions (e.g., are there oscillations?) $\endgroup$
    – Alubeixu
    Jun 21 '14 at 20:58
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    $\begingroup$ Well, please give us an initial condition: $y(1) = 1$? $y(0) = 5$? And you should clear up what you have in mind in the question itself (if you can you edit your question) so that people reading it know right away what you specifically intend. $\endgroup$
    – KCd
    Jun 21 '14 at 21:04
  • $\begingroup$ Notice that the solutions cannot oscillate, since they are convex functions. $\endgroup$ Jun 22 '14 at 0:21
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    $\begingroup$ Actually, except for the factors of $x$ and $y$, this is the first Painlevé equation.More precisely, a transformation $y=\alpha z, x=\gamma t$ with suitable $\alpha,\gamma$ leads to PI, i.e. $d^2z/dt^2=6z^2+t$. A lot of information about solutions can be found here. $\endgroup$
    – Helmut
    Oct 10 '18 at 8:41
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If you are interested in the behavior near infinity, you can provide very good approximations through the following method: $$ y=(x^{1/2}+A(x)),\qquad y''=(-\frac{1}{4}x^{-3/2}+A''(x))$$ $$ y^2 = x + 2x^{1/2}A(x) + A^2(x)$$ $$ y^2 + y'' = x + 2 x^{1/2} A(x) -\frac{1}{4}x^{-3/2}+A''(x)+A^2(x) \tag{1}$$ Now set $A(x)$ in order to cancel out the second and third term in the RHS: $$ A(x)=\frac{1}{8x^2}+B(x). $$ You get: $$ y^2 + y'' = x+ \left(2\sqrt{x}+\frac{1}{4x}\right) B(x)+\frac{3}{4x^4}+B''(x)+B^2(x),\tag{2}$$ hence by setting: $$ B(x) = - \frac{3}{8x^{9/2}+x^3}+C(x)$$ you get the quite close approximation:

$$ \sqrt{x}+\frac{1}{8x^2}-\frac{3}{8x^{9/2}+x^3}\leq y(x)\leq \sqrt{x}+\frac{1}{8x^2}\tag{3}$$

that can even refined through successive steps of the method, even if the bounds become more and more involved.

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  • $\begingroup$ How do you know the inequalities in (3) are satisfied, i.e., how do you know $0 \leq C(x) \leq 3/(8x^{9/2} + x^3)$ for large $x$ under suitable initial conditions? $\endgroup$
    – KCd
    Jun 21 '14 at 22:54
  • $\begingroup$ At each step, you add a term that is $O\left(\frac{1}{x^{f(n)}}\right)$ ($f(n)$ increasing), with alternate sign. You can also prove that the constants hidden in the $O$-notation decay pretty fast to zero, hence the approximation gets better and better at each step. $\endgroup$ Jun 21 '14 at 23:01
  • $\begingroup$ Maybe the simpler way to achieve this is to apply the differential operator $y\to y''+y^2-x$ to both sides of $(3)$. $\endgroup$ Jun 21 '14 at 23:02
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    $\begingroup$ Differential operators don't respect inequalities, so what does the fact that the operator applied to the right side of (3) is positive (equal to $49/(64x^4)$) tell us? Setting $g(x) = \sqrt{x} + 1/(8x^2)$, we have $g''+ g^2 - x > 0 = y'' + y^2 -x$, so $g''+ g^2 > y'' + y^2$. These are nonlinear expressions in $g$ and $y$. How does that inequality imply $g(x)\geq y(x)$ for large $x$? $\endgroup$
    – KCd
    Jun 21 '14 at 23:56
  • $\begingroup$ For big $x$, the term $y^2-g^2$ is dominant on the term $y''-g''$ (we know that $y$ behaves like $x^{1/2}$ AND that $y''$ behaves like $x^{-3/2}$, so $y^2-g^2$ is roughly $x^2$ times $y''-g''$), hence the positivity of the differantial operator implies $g^2>y^2$, or just $g>y$, for big $x$. $\endgroup$ Jun 22 '14 at 0:14

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