# How to solve this relatively simple non-linear ODE?

I'm having troubles in simplifying a differential equation to find its solutions. Consider this ODE: $$\frac{1}{r} \, \frac{d}{dr} \Bigl( \frac{r}{B} \, \frac{d B}{dr} \Bigr) = k \, B^2, \tag{1}$$ where $$k$$ is just a real constant and $$B(r)$$ is the function to be found. I already know one solution of this equation (it was found by guessing, trial and errors): $$B(r) = \frac{\beta}{1 + \lambda r^2}, \tag{2}$$ where $$\beta$$ and $$\lambda = k \beta^2 / 4$$ are arbitrary constants. My goal is to transform (1) into an integral so I could find back the solution (2). I'm yet unable to find a change of variable that makes that equation solvable analyticaly. I tried $$u = 1/B$$, and $$B = e^{\phi(r)}$$ (and few other trials). Any idea how to solve (1)?

• You already found a one-parameter family of solutions (which is no small feat). What is left to do is to show that there are no other solutions. Two more remarks: (1) Your solution is for the right hand side $-kB^2$, not $+kB^2$. (b) Set $u(r) = \log B(r)$. Then $u$ is a solution of the Gelfand problem $\Delta u + ke^ {2u}= 0$ in 2 dimensions, using polar coordinates. But you probably knew this already :) Jun 2, 2020 at 19:15
• @HansEngler, your comment suggest that there is no procedure or change of variable to transform (1) into an elementary integral, is that right?
– Cham
Jun 2, 2020 at 19:44
• I am not aware of one. The linear version, essentially $\frac{1}{r}\frac{d}{dr}(r u(r)) + k u = 0$, is a Bessel equation with non-elementary solution. Jun 2, 2020 at 21:01
• I would be interested in any other analytical solutions to (1), which are not special cases of (2).
– Cham
Jun 3, 2020 at 0:41

Since the equation features only multiplication and division, it is worth checking whether it has some kind of homogeneous continuous group of symmetries. Indeed, it is straight-forward to check that the one-parameter Lie group of transformations \begin{align} & r \, \mapsto \, e^{s}r\\ & B \, \mapsto \, e^{-s}B \end{align} leaves the equation $$\frac{1}{r} \frac{d}{dr}\left(\frac{r}{B} \frac{dB}{dr}\right) \,=\, k\, B^2$$ invariant (unchanged) Furthermore, the quantity $$C = rB$$ is also invariant under the action of the one-parameter group of symmetries. The vector field that generates this group of transformation is $$r\,\frac{\partial }{\partial r} \, -\, B\,\frac{\partial }{\partial B}$$ Now, using the invariant quantity $$C$$, we can construct a change of variables (a local diffeomorphism) that rectifies the latter vector field. In other words, if we define the change of variables \begin{align} &s = \ln(r)\\ &C = rB \end{align} with inverse \begin{align} &r = e^s\\ &B = e^{-s}\,C \end{align} the vector field above transforms into $$r\,\frac{\partial }{\partial r} \, -\, B\,\frac{\partial }{\partial B} \, =\, \frac{\partial }{\partial s}$$ and so, in the new coordinates $$s, \,C$$ the original equation should be invariant under the simple translation-transformation \begin{align} &s \mapsto s + \tilde{s}\\ &C \mapsto C \end{align} which means that the transformed equation should not depend explicitly on the new $$s$$ variable. Indeed, since $$dr = e^s ds$$ we can write the equation in the new coordinates as follows: $$\frac{1}{r} \frac{d}{dr}\left(\frac{r}{B} \frac{dB}{dr}\right) \,=\, k\, B^2$$ $$\frac{1}{e^s} \, \frac{d}{e^sds}\left(\frac{e^s}{e^{-s}\,C} \frac{d}{e^s ds}\Big(e^{-s}\,C\Big)\right) \,=\, k\, \big(e^{-s}\,C\big)^2$$ $$e^{-2s} \, \frac{d}{ds}\left(\frac{e^s}{C} \frac{d}{ds}\Big(e^{-s}\,C\Big)\right) \,=\, k\, e^{-2s}\,C^2$$ $$\frac{d}{ds}\left(\frac{e^s}{C} \frac{d}{ds}\Big(e^{-s}\,C\Big)\right) \,=\, k\, C^2$$ $$\frac{d}{ds}\left(\frac{e^s}{C} \Big(e^{-s}\,\frac{dC}{ds} - e^{-s}\,C\Big)\right) \,=\, k\, C^2$$ $$\frac{d}{ds}\left(\frac{e^s}{C} \, e^{-s}\, \Big(\frac{dC}{ds} - C\Big)\right) \,=\, k\, C^2$$ $$\frac{d}{ds}\left(\frac{1}{C}\, \Big(\frac{dC}{ds} - C\Big)\right) \,=\, k\, C^2$$ $$\frac{d}{ds}\left(\frac{1}{C}\, \frac{dC}{ds} - 1\right) \,=\, k\, C^2$$ so now it is clear that the equation in the new variables is $$\frac{d}{ds}\left(\frac{1}{C}\, \frac{dC}{ds}\right) \,=\, k\, C^2$$ and does not depend on the variable $$s$$ explicitly. The next step is to introduce the extra variable $$P \, =\, \frac{1}{C}\, \frac{dC}{ds}$$ which turns the equation into $$\frac{dP}{ds} \, =\, k\,C^2$$ and as a result of this substitution, we obtain the following system of differential equations \begin{align} &\frac{dC}{ds} \, =\, C \, P\\ &\\ &\frac{dP}{ds} \, =\, k\,C^2 \end{align} We can try to find a first integral of the latter system as follows \begin{align} &\frac{dP}{dC} \, =\, \frac{k\,C^2}{C\,P} \end{align} \begin{align} &\frac{dP}{dC} \, =\, \frac{k\,C}{P}\\ \end{align} $$P\,dP \, =\, k\,C \,dC$$ so after integrating both sides $$P^2 \,=\, k\,C^2 \, +\, b$$ where $$b$$ is an arbitrary constant. Substituting back $$P \,=\, \frac{1}{C} \, \frac{dC}{ds}$$ we obtain the separable, integrable, differential equation $$\left(\frac{1}{C}\,\frac{dC}{ds}\right)^2 \, =\, k\,C^2 \, +\, b$$ $$\frac{1}{C}\,\frac{dC}{ds}\, =\, \pm \sqrt{\,k\,C^2 \, +\, b\,}$$ $$\frac{dC}{ds}\, =\, \pm\, C\,\sqrt{\,k\,C^2 \, +\, b\,}$$ The solution to the latter equation can be obtain by integrating $$\pm\int \, \frac{dC}{C\,\sqrt{\,k\,C^2 \, +\, b\,}} \, =\, \int ds$$ and, consequently, it can be written in implicit form as follows $$\mp\,\frac{1}{\sqrt{b}}\, \tanh^{-1}\left(\sqrt{\,1 \, + \,\frac{k}{b} \,C^2\,}\right) \, =\, s - a$$ where $$a$$ is another arbitrary constant. Rewrite it $$\sqrt{\,1 \, + \,\frac{k}{b} \,C^2\,} \, =\,\tanh\Big( \mp \sqrt{b}\,(s - a)\,\Big)$$ $$1 \, + \,\frac{k}{b} \,C^2\, =\,\tanh^2\Big( \mp \sqrt{b}\,(s - a)\,\Big)$$ $$\frac{k}{b} \,C^2\, =\,\tanh^2\Big( \mp \sqrt{b}\,(s - a)\,\Big) \, -\, 1$$ and solve for $$C$$ $$C\, =\,\pm\,\sqrt{\,\frac{b}{k}\,\tanh^2\Big( \mp \sqrt{b}\,(s - a)\,\Big) \, -\, \frac{b}{k}\,}$$ Finally, return to the original variables \begin{align} &s = \ln(r)\\ &C = rB \end{align} which leave us with the solution $$r\,B\, =\,\pm\,\sqrt{\,\frac{b}{k}\,\tanh^2\Big( \mp \sqrt{b}\,\big(\ln(r) - a\big)\,\Big) \, -\, \frac{b}{k}\,}$$ or explicitly $$B\, =\,\pm\,\frac{1}{r}\sqrt{\,\frac{b}{k}\,\tanh^2\Big( \mp \sqrt{b}\,\big(\ln(r) - a\big)\,\Big) \, -\, \frac{b}{k}\,}$$ After relabling the constants $$\alpha = \mp\,a \,\sqrt{b}$$ and $$\beta = \mp\,\sqrt{b}$$, one can write the final solution as follows $$B\, =\,\pm\,\frac{1}{r}\sqrt{\,\frac{\beta^2}{k}\,\tanh^2\Big( \beta\,\ln(r) - \alpha\,\Big) \, -\, \frac{\beta^2}{k}\,}$$ $$B\, =\,\pm\,\frac{\beta}{r}\sqrt{\,-\,\frac{1}{k \, \cosh^2\Big( \beta\,\ln(r) - \alpha\,\Big)} \,}$$ $$B\, =\,\pm\,\frac{\beta}{\sqrt{-\,k\,}\,}\,\frac{1}{\,r \, \cosh\Big( \beta\,\ln(r) - \alpha\,\Big)}$$ $$B\, =\,\pm\,\frac{\beta}{\sqrt{-\,k\,}\,}\,\frac{1}{\,r \, \cosh\Big( \beta\,\ln(r) - \alpha\,\Big)}$$ so maybe it simplifies to something like this $$B\, =\,\,\frac{\beta}{\sqrt{-\,k\,}\,}\, \frac{2 \, \lambda \,\, r^{\beta-1}}{\,1 \, + \, \lambda^2 \,\, r^{2\beta}}$$ where the constant $$\lambda = \pm e^{-\alpha}$$
• @Cham I know. After all, $\tanh$ is a fraction of exponents, but I did not feel like going further. I have focused mostly on the systematic method for solving the equation, as requested. Feb 1 at 18:57