Trying to solve the following differential equation:

$u' + \frac{1}{2} u^2 = -\frac{a}{cosh^2 (x)}-k$

Following the regular method I first solved the corresponding uniform ODE:

$ u' + \frac{1}{2} u^2 = 0 \\ -\frac{du}{u^2} = \frac{1}{2}dx \\ \frac{1}{u} = \frac{x+C}{2} \\ u = \frac{2}{x+C} $

Next, to find a particular solution of non-uniform equation I replace C with some function of independent variable: $C \rightarrow f(x)$

$ u(x) = \frac{2}{x+f} \\ u'(x) = - \frac{2(1+f')}{(x+f)^2} \\ - \frac{2(1+f')}{(x+f)^2} + \frac{2}{(x+f)^2} = -\frac{a}{cosh^2(x)}-k \\ \frac{-f'}{(x+f)^2} = -\frac{a}{cosh^2(x)} -k$

And I'm totally stuck. Any ideas?

EDIT: Originally the ODE was retrieved from following PDE:

$H_{xx} + H_{yy} + \frac{1}{4} H_y ^2 + \frac{1}{4} H_x ^2 + \cot(y) H_y -1 = 0$

Where the indices denote derivatives over corresponding variables. The PDE itself comes from more complicated PDE after the substitution $H(x,y) = \ln F(x,y)$ and some algebra.

I'll be more than happy if there's some more convenient method of solving it.

EDIT2: just noticed that PDE is not the correct one. Still, if there are any convenient methods of solving it, feel free to post.

  • $\begingroup$ From Wolfram|Alpha, I got some horrifying looking solution... wolframalpha.com/input/… $\endgroup$ – Karn Watcharasupat Dec 27 '17 at 18:02
  • $\begingroup$ Quite strange. When I fed the equation to Wolfram Mathematica, it just didn't give me any solution. $\endgroup$ – Tajimura Dec 27 '17 at 18:06
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    $\begingroup$ Meanwhile, I've discovered that I forgot a term in the equation. Will edit the question. $\endgroup$ – Tajimura Dec 27 '17 at 18:06
  • $\begingroup$ what are you using the DE for though? If you have the initial conditions the solutions might get simpler. $\endgroup$ – Karn Watcharasupat Dec 27 '17 at 18:07
  • $\begingroup$ @Karn, I've got it after separating the variables in a PDE. The PDE itself comes from Einstein equation in bispherical coordinates. $\endgroup$ – Tajimura Dec 27 '17 at 18:09

The equation is non-linear, so the whole mechanism with homogeneous and inhomogeneous solutions does not work.

You can set $u=2\dfrac{v'}{v}$, $u'=2\dfrac{v''v-v'^2}{v^2}$ so that $$ u'+\frac12u^2=2\frac{v''}{v}=f(x) $$ gives a linear ODE of second order $$ v''(x)-\frac12f(x)v(x)=0 $$ which still is usually not symbolically solvable but gives you more equations with named solutions to compare to. Also, a power series solution might be easier to compute in this form.

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  • $\begingroup$ Schrödinger! Great :) $\endgroup$ – Tajimura Dec 28 '17 at 8:29

$$u' + \frac{1}{2} u^2 = -\frac{a}{\cosh^2 (x)}-k$$ Let $u=2\frac{v'}{v} \quad\to\quad u'=2\frac{v''}{v}-2\frac{v'^2}{v^2}$

$2\frac{v''}{v}-2\frac{v'^2}{v^2} + \frac{1}{2} \left(2\frac{v'}{v}\right)^2 = -\frac{a}{\cosh^2 (x)}-k$

$$v'' +\left(\frac{a}{2\cosh^2 (x)}+\frac{k}{2}\right)v=0 \tag 1$$ This is a second order linear ODE of the kind : $$\quad v''+f(x)\:v=0\quad$$ where $f(x)$ is a known function.

It is well known that this kind of ODEs are not always analytically solvable. Some of them are solvable in terms of special functions. For example if $f(x)=\frac{A}{x^2}+\frac{B}{x}+C$ the solutions involve the Kummer and Whittaker functions. In case of $f(x)=Cx^p$ the solutions involve the Bessel functions. A few other examples can be found involving some standard special functions.

Unfortunately, the case $\quad f(x)=-\frac{a}{\cosh^2 (x)}-k\quad$ is not known, as far as I know. On my opinion, you cannot expect a solution of Eq.$(1)$ on closed form until a new family of special functions be created and standardized.

In such circumstances, other approaches such as series expansion or numerical calculus could provide approximate results.

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