EDIT4:
- start of context
Apologies about context. I thought that it is an all too well known reference for re-counter on the topic of differential equations.
In the classical dynamic solution of Newtonian planetary orbits we have equilibrium equations along the radius and perpendicular to it. Integration of the perpendicular/circumferential components is seen as conservation of angular momentum or invariance of swept areal velocity and also as Kepler's second law in a differential form.
However, the radial dynamic equilibrium equations for radius or polar angle incorporating inverse squared law of gravitation are seldom solved directly with respect to time. But spatial conic orbits are obtained by elimination of time variable in the polar form eliminating the need for time dependence.
That is as:
$$ p/r(\theta) = 1 - \epsilon \cos(\theta). $$
The requirement of expressing $ r,\theta $ with respect to time and making it available in published literature remained or remains for a long time.. or so I think.
May be there exist advanced functions which came into existence after Newton's time which describe $ r(t),\theta(t) $ but usually unknown or not mentioned in text-books or no luck even after extensive googling. I still cannot understand avoidance of giving a label or name to these basic function solutions of non-linear ODEs in the Two Body Problem.
I seek an answer here for its indication due to lack of knowledge in this context. Changed labels of $ r,\theta $ and constants to focus on mathematics of differential equations.
- end of context
What substitution/s lead to solution of non-linear ODEs :
$$ \dfrac{d^2 u}{d t^2} +\dfrac{a}{u^2}-\dfrac{b}{u^3} =0 $$
$$ \dfrac {dv }{dt}= c/ u^2 $$
u, v are functions of $t$ and $ a, b , c $ are constants. Several trials involving log function did not lead somewhere.
EDIT1:
The uncoupled equation is involved with inverse functions in Mathematica. So I Include another coupled dependent variable $v$ along with $u$ with easy boundary conditions that gets some relief.
DSolve[{U''[t]+a/U[t]^2-b/U[t]^3==0,U[0]==ri,U'[0]==0,V'[t] U[t]^2==c,V[0]==0},{U,V},t]
EDIT2:
Changed the title to be more interested in answers than history.
EDIT3:
Its numerical solution is given here with code
:
mu=2.;h=1;ri=1.5;tmax=4.5;
NDSolve[{TH'[t] R[t]^2==h,R''[t]+mu/R[t]^2-h^2/R[t]^3==0,R'[0]==0,R[0]==ri,TH[0]==0},{TH,R},{t,0,tmax}];
{r[u_],th[u_]}={R[u],TH[u]}/.First[%];
Plot[{r[t],th[t]},{t,0,tmax}]
ParametricPlot[{r[t] Cos[th[t]],r[t] Sin[th[t]]},{t,0,tmax},PlotStyle->{Thick,Red},GridLines->Automatic,PlotLabel->NewtonianPlanetaryOrbit]
ParametricPlot[{1/r[t],Cos[th[t]]},{t,0,tmax},PlotStyle->{Thick,Purple},GridLines->Automatic,PlotLabel->" 1/r vs cos \[Theta]"]
Plot[th'[t],{t,0,tmax},PlotStyle->{Thick,Magenta},PlotRange->All,GridLines->Automatic,PlotLabel->\[Omega]]
Plot[th'[t]r[t]^2,{t,0,tmax},PlotStyle->{Blue},GridLines->Automatic,PlotLabel->" \[Omega] r^2"]
ParametricPlot[{th[t],th'[t]},{t,0,tmax},PlotStyle->{Thick,Purple},GridLines->Automatic,PlotLabel->" \[Theta] vs \[Omega]",PlotRange->All]
p=0.5
ecc=1-p/ri
Clear[mu,h,b,a,ri];
DSolve[{U''[t]+a/U[t]^2-b/U[t]^3==0,U[0]==ri,U'[0]==0,V'[t] U[t]^2==c,V[0]==0},{U,V},t]
NEXTnext NO next
DSolve[{U''[t]+-a/U[t]^3+b/U[t]^2==0,X[0]==ri,U'[0]==0},U,t]
some Mma outputs added
Let $\dfrac{du}{dt}=y(u)$, and we have $\dfrac{d^2u}{dt^2}=\dfrac{dy}{dt}=y\dfrac{dy}{du}$. This leads to $y\dfrac{dy}{du} = -\dfrac{a}{u^2}+\dfrac{b}{u^3}$. Hence, $$\frac{y^2}{2} = \frac{a}{u} - \frac{b}{2u^2}+\frac{C_1}{2}\Rightarrow y=\sqrt{\frac{2a}{u}-\frac{b}{u^2}+C_1}.$$ Further, we can get:$$\frac{udu}{\sqrt{C_1u^2+2au-b}}=dt.$$ When $a,b$ are given, and $C_1$ is determined by initial conditions, then we can go on to find out $F(u)=t+C_2$. If we are lucky enough, $u=F^{-1}(t+C_2)$ will be the final answer.
The following section is about the physics problem you mentioned, which is famous in classical mechanics. The basic equations are $$\ddot r-r\dot\theta^2=-\frac{a}{r^2}\\r^2\dot\theta=b\space(\text{const})$$ And we have $\ddot r+\dfrac{a}{r^2}-\dfrac{b}{r^3}=0$.
Our discussion above cannot give an explicit solution of $r(t)$ and $\theta(t)$, except for some special cases. This problem was first, as far as I know, considered comprehensively by Kepler. He reduced this problem to his famous equation.