# Approximated solution to differential equation in the form $f(u)u'^2+(u-u_0)^2=k$

I'm trying to solve the following differential equation, that arises from conservation of energy in a physical problem. $R,k$ are constants. $$(1+R^4u^4)u'^2+(u-u_0)^2=k$$ Now, according to my book I should find "approximate solutions" around the point of equilibrium $u_0$. I started my reasoning with simpler equations and I have some questions:

1. Is there a general good method to solve equations of the form $1/2k_1u'^2+1/2k_2u^2=k_3$? What I usually do is to resort to the physical problem of an harmonic oscillator with mass $k_1$, elastic constant $k_2$ and energy $k_3$ to find a cosinusoidal solution. I can find the amplitude but not the phase difference. In alternative I take the derivative of both sides of the equation and solve the second order linear differential equation that arises. I can find the amplitude of the oscillations by substitution of $u=A\cos t$ in the original equation. Are there more direct approaches?

2. In the case of an equation of the form $1/2k_1u'^2+1/2k_2(u-u_0)^2=k_3$ I solve them either by inspection or by sobstituting $\xi = u-u_0,\ \xi'=u'$. Is that procedure correct?

3. Now the original problem $(1+R^4u^4)u'^2+(u-u_0)^2=k$. My book gives what I think is a wrong answer : $u_0+(\sqrt{k}/u_0)\cos(\sqrt{1+R^4u_0^2})$. How to obtain the correct solution? I thought to substitute $u^4 = u_0^4$ in the leftmost term. Am I allowed to do this or it would be too rough an approximation? If I approximate even to first order $u^4$ the differential equation becomes too difficult for me

Request for clarification: is $u_0$ a constant or a function of independent variable? If it's a constant, then it appears that $k$ must be zero; otherwise $u\equiv u_0$ is not a solution.

Re: 1. There is a general approach to autonomous ODE of 1st order. It works for the simpler equations in 1 and 2, but for the original one the computations are probably complicated.

Re: 2. The substitution $u=u_0+\xi$ is reasonable. The equality $\xi'=u'$ is correct as long as $u_0'=0$, see my request for clarification above.

Re: 3. Again, writing $u=u_0+\xi$ is a reasonable place to begin. If you are willing to neglect some "small" terms with $\xi$, the equation may become manageable.

More precisely, in the equation $(1+R^4 (u_0+\xi)^4)(\xi')^2+\xi^2=k$ we may decide to neglect all terms in which $\xi$ and its derivative appear to power higher than 2. This leaves us with $(1+R^4 u_0^4)(\xi')^2+\xi^2=k$, which you know how to solve. Yes, the solution in the book appears to be wrong. I expect a trigonometric function with argument $x/\sqrt{1+R^4 u_0^4}$ where $x$ is the independent variable.

• You are right. $u_0$ is a constant. Physically it is related to the energy of the system, so for "equilibrium solution" I intend the solution when $k$ is zero. I'm interested in the solutions where $k$ is small – Ralph Jul 15 '12 at 21:07
• @Ralph I expanded my answer. – user31373 Jul 15 '12 at 21:51

$$A(u) u'^2 +B(u)=k$$
$$A(u) u'^2 =k-B(u)$$
$$\frac{\sqrt{A(u)}}{\sqrt{k-B(u)}} u' =1$$
$$\frac{\sqrt{A(u)}}{\sqrt{k-B(u)}} du =dx$$