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I am trying to solve the following nonlinear ODE for a function $P(x)$:

$$\left(1-x^2\right)P''(x)+k(k+1)P(x)=cP(x)^\frac{k-1}{k+1}.$$

Here, $k$ and $c$ are arbitrary parameters. By rescaling $P(x)$, one can without loss of generality set $c=1$, so really this equation is only parameterized by $k$.

I wonder if there exists a closed form solution for this equation, at least for integer $k$? Although it looks very simple, my efforts to solve it have so far been unsuccessful. Nonetheless, I have been able to find some very simple solutions: for instance,

  • when $k=1$, the equation can be integrated, resulting in

$$P(x)=\frac{1}{2}+a\left(1-x^2\right)+b\left[x+\left(1-x^2\right)\mathrm{arctanh}\,x\right],$$

  • when $k=-2$, by inspection I found

$$P(x)=\sqrt{2}\frac{4x}{3+x^2},$$

  • when $k=-3$, by inspection I found

$$P(x)=120\frac{1-x^2}{\left(5-x^2\right)^2}.$$

These solutions look simple enough that I'm hopeful that a general solution exists. Perhaps there even exists some standard transformation to linearize such equations?

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  • $\begingroup$ Oh, I should add that the $c=0$ case is of course trivial: the equation becomes linear and the general solution is given by a superposition of two 2F1 hypergeometric functions. I am interested in the $c\neq0$ case. $\endgroup$ – Alex Lupsasca Jul 1 '17 at 10:44
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Using contact transformations, Lie showed that second order equations of this type can be transformed to the trivial equation $\frac{d^2 P^\prime}{dx^{\prime 2}} = 0$.

Edit: an algorithm for this is described in Kumei and Bluman When Nonlinear Differential Equations are Equivalent to Linear Differential Equations, SIAM J. Math. 42(5) (1982) 1157-1173.

Alternatively, a more general technique is described in Bluman and Dridi New Solutions for ordinary differential equations, J. Symb. Comp. 47 (2012) pp. 76-88. It uses a combination of invertible and non-invertible mappings. They treat several examples in detail.

Edit(2): The paper by N. Euler, Transformation Properties of $\ddot{x}+f_1(t)\dot{x}+f_2(t)x+f_3(t)x^n=0$ J. Nonlin. Math. Phys. 4(3-4) (1997) pp. 310-337 provides the conditions for an equation of this type to be integrable, as well as the conditions for it to have Lie symmetries. He also considers non-point transformations $X=F(x,t)$, $dT = G(x,t)dt$ to transform the equation to $\frac{d^2X}{dT^2}+k_1\frac{dX}{dT}+k_2X^p=0$ for constants $k_1$, $k_2$ and $p\in \mathscr{Q}$. I've tried substituting the corresponding values for your equation into his conditions, but the results underline your point about the lack of symmetries.

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  • $\begingroup$ Thank you for your answer, Phil. Correct me if I'm wrong, but isn't finding the contact transformation in question just as hard as solving the equation directly? $\endgroup$ – Alex Lupsasca Jul 3 '17 at 12:38
  • $\begingroup$ I am looking at the paper by Bluman and Dridi now. Once I understand how their Euler operator (18) acts, I'll apply their method to my equation. $\endgroup$ – Alex Lupsasca Jul 3 '17 at 12:39
  • $\begingroup$ @AlexLupsasca It may be useful to look at Dridi's thesis as well - tel.archives-ouvertes.fr/tel-00264288. $\endgroup$ – Phil Harmsworth Jul 4 '17 at 7:28
  • $\begingroup$ Hi Phil, I applied the method described in the paper by Bluman and Dridi. Unfortunately, it only produces the trivial solution to the linear equation that is obtained by setting c=0. This is because this restricts the equation to a subspace of solution space with enhanced symmetry--the situation is exactly analogous to the example they consider starting in the introduction. But thanks for mentioning the paper---their method is good to know! $\endgroup$ – Alex Lupsasca Jul 9 '17 at 10:56
  • $\begingroup$ @AlexLupsasca Very disappointing. I've added a reference to my original answer on a linearisation procedure. $\endgroup$ – Phil Harmsworth Jul 10 '17 at 11:51
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For $k=0$, according to Maple the general solution is

$$ \text{arctanh}(x) \pm \int^{P(x)/\sqrt{x^2-1}} \dfrac{du}{\sqrt{u^2 - 2 c \ln(u) + b}} = a $$

Another family of solutions (for all $k$) is $$ P(x) = \left( {\frac {k \left( k+1 \right) }{c}} \right) ^{-(k+1)/2}{x}^{k+1 } $$

For $k = -3$, there are some other closed-form solutions, e.g.

$$ P(x) = \frac{8}{c(1-x^2)} $$ and $$ P(x) = \frac{24 (21 x^2-5)}{c (3x^2+5)^2}$$

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  • $\begingroup$ Maxima happily confirmed the closed form solutions for $k=-3$, but didn't like the family of solutions for all $k$. $\endgroup$ – Phil Harmsworth Jul 2 '17 at 9:48
  • $\begingroup$ There are some problems with branches when $x < 0$, but it should be OK for $x > 0$. $\endgroup$ – Robert Israel Jul 2 '17 at 21:41
  • $\begingroup$ Thank you Robert! Unfortunately, the general family of solutions you found is not useful for me, because I am really trying to solve another (much nastier) equation for a function $y(x)$. I simplified the equation by a transformation from $y(x)$ to $P(x)$ but the general solution you found goes to a constant under the inverse transformation :( $\endgroup$ – Alex Lupsasca Jul 3 '17 at 12:42
  • $\begingroup$ That being said, I am curious how you found the other solutions? Perhaps the method can be generalized to other values of $k$? That would be almost as good as a general solution. I just tried to input the equation into Maple but I can't get it to spit out any of the answers you wrote, except for the $k=0$ one; is there a special argument that's required? Thank you once again! $\endgroup$ – Alex Lupsasca Jul 3 '17 at 12:44
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    $\begingroup$ What I did was guess that the solution should be a rational function with numerator and denominator of particular degrees, substitute that into the differential equation, and solve for the coefficients. $\endgroup$ – Robert Israel Jul 3 '17 at 19:08

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