## Variable initial condn NDSolve with earlier version

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In an earlier version 8 (ParametricNDSolve not available) I tried to integrate with different initial values $$(yi= 0.1, 0.4,0.8); (zi^2+yi^2=1)$$ radially on unit semi circle to sketch tractrices but not successful. Is there a workaround?

$$\sin \phi = y/a,\, \tan \phi= \frac{dy}{dz}$$ a = 1; zmax = 1; ri = 0.8; zi = -Sqrt[a^2 - ri^2]; ar = 0.5;
{zi, ri}
NDSolve[{R'[z]/(R[z] (1 + R'[z]^2)^0.5) == 1/a, R == ri},
R, {z, zi, zmax}];
r[t_] = R[t] /. First[%];
ps8 = Plot[{r[z], 0}, {z, zi, zmax}, PlotStyle -> {Red, Thick},
AspectRatio -> ar, GridLines -> Automatic];
semicirc =
ParametricPlot[a {Cos[t], Sin[t]}, {t, 0, Pi}, AxesOrigin -> {0, 0},
GridLines -> Automatic];
Show[{semicirc, ps1, ps4, ps8}, PlotRange -> All]


What is your version? – Alex Trounev – 2019-03-30T22:32:13.573

Mathematica Version 8 – Narasimham – 2019-03-30T22:44:45.587

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It would be better to portray both solutions. Using DSolve, we construct two functions (for each branch of the solution)

trac1[ri_, x_] :=
Block[{r = ri, a = 1, z = x},
s = DSolve[{R'[z]/(R[z] (1 + R'[z]^2)^(1/2)) == 1/a, R == ri},
R[z], z]; R1 = R[z] /. s[]; R1]

p1 = Plot[
Evaluate[Table[trac1[ri, x], {ri, {.2, .4, 0.8}}]], {x, -1, 1}];

trac2[ri_, x_] :=
Block[{r = ri, a = 1, z = x},
s = DSolve[{R'[z]/(R[z] (1 + R'[z]^2)^(1/2)) == 1/a, R == ri},
R[z], z]; R2 = R[z] /. s[]; R2]

p2 = Plot[
Evaluate[Table[trac2[ri, x], {ri, {.2, .4, 0.8}}]], {x, -1, 1}];

With[{a = 1},
semicirc =
ParametricPlot[a {Cos[t], Sin[t]}, {t, 0, Pi},
AxesOrigin -> {0, 0}, GridLines -> Automatic, PlotPoints -> 50]];

Show[{semicirc, p1, p2}] Thanks for separation of the plots, now I know how we can better vary a single boundary condition parameter. The tractrix through $(0,0.8)$ however has positive curvature, seems not alright to me.Will it change on later versions? I again plotted by repetitive computation as Shown at right..I am still trying to find under what conditions curves whose differential equations are given by constant $\sin \phi /r, \cos \phi/r$ cease to be orthogonal. – Narasimham – 2019-03-31T09:48:36.047

@Narasimham You're right. We cannot solve this problem using NDSolve[]. Therefore, I changed the code. See update. – Alex Trounev – 2019-03-31T12:20:00.700

Fine. What still bothers me is ...a trace is found for orthogonal trajectory even outside unit semi-circle earlier to boundary start point .. it seems the computation re-visits DE to somehow make it more comprehensive, a matter not bargained for a priori.!! Seems we are not in control of the DE.. – Narasimham – 2019-03-31T12:42:46.707

@Narasimham Can you explain what you want to get? – Alex Trounev – 2019-03-31T14:53:12.950

For example tractix should not pass through boundary points $( .8,-.6 ),( .4,-.9165 ), ( .1,,.9950 )$ but should actually start at these points. I want to know why it is not so in the simple ODE integration. – Narasimham – 2019-03-31T17:44:55.757

@Narasimham I do not understand what is the boundary point? – Alex Trounev – 2019-03-31T17:51:59.450

E.g., the point $( -0.8,0.6 )$ is initial point in the initiial value problem. Even though tractrix function is monotonous the locus moves on either side not just down. – Narasimham – 2019-04-01T05:02:11.190