I want to solve (numerically) the geodesics of the upper half plane and plot. The results are quite known (i) straight lines parallel to yy-axis and (ii) semicircles centered on the xx-axis. Now the problem is how to solve the coupled ODEs. I tried with NDSolve but the results are not coming. Can you please write a code and help me.

Here are the coupled ODEs:

¨x=2˙x˙yy¨y=−˙x2+˙y2y\begin{align*}

\ddot{x} &= \frac{2\dot{x}\dot{y}}{y}\\

\ddot{y} &= \frac{-\dot{x}^2+\dot{y}^2}{y}

\end{align*}

Thanking you in advance.

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dear chris can you please explain why the t in the NDSolve runs for 0 to 1 and then in parametric plot from 1 to 10. Also If I can put range of t 1 to 10 in NDSolve then two curves are coming ??? can you please explain

– ricci1729

May 25 ’13 at 10:01

thank you very much 🙂

– ricci1729

May 25 ’13 at 10:05

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2 Answers

2

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If you use NDSolve as follows

sol=

NDSolve[{ x”[t] == 2 x'[t] y'[t]/y[t],

y”[t] == (y'[t]^2 – x'[t]^2)/y[t],

x[0] == 1, y[0] == 1,

x'[0] == 1, y'[0] == 1}, {x[t], y[t]},

{t, 0, 8}];

Then

ParametricPlot[{x[t], y[t]} /. sol, {t, 0, 8}]

yields

Note that if/when the solution is driven towards ‘y=0’ you will run into problems.

You might also want to have a look at EquationTrekker

<< EquationTrekker` dear chris; that's very true . y=0 is the singularity that one can immediately see from the ODE's. But my question is, if you change the limit of t from 0 to 9 you see two curves. but why is that is not clear to me??? – ricci1729 May 25 '13 at 10:18 its a bit strange... – chris May 25 '13 at 10:28 I think you can solve the equations analytically. The change of variables y[t] -> Exp[z[t]] changes your system into

x”[t] == 2 x'[t] z'[t]

Exp[2 z[t]] z”[t] == – x'[t]^2

The first equation is easy and you get

z[t] == 1/2 Log[x'[t]] + a

with a a constant; the second equation, assuming x'[t]!=0 gives

1/2 Exp[2 a] x”[t]/x'[t] == -x[t] + b

with b a constant. This can be handled by Mathematica :

solx[t_] = x[t] /. First@

DSolve[1/2 Exp[2 a] x”[t]/x'[t] == -x[t] + b, x[t], t] /. {C[1] -> c, C[2] -> d}

from which we can retrieve the solution for z[t] :

solz[t_] = 1/2 Log[D[solx[t], t]] + a

In order to check against @chris let’s solve for the constants {a,b,c,d} by using the initial conditions :

const = FindRoot[

{solx[0] == 1, Derivative[1][solx][0] == 1,

solz[0] == 0, Derivative[1][solz][0] == 1},

{{a, 1}, {b, 2}, {c, 1}, {d, 1}}] //Chop ;

ParametricPlot[{solx[t] /. const, Exp[solz[t] /. const]}, {t, 0, 5}]

thank you very much but i know it is exactly solvable model. the question is how to generate this graph itself from the geodesic equation by NDSolve.

– ricci1729

May 28 ’13 at 8:00