@deniscr Instead of posting an image of your code, please post the code itself.  One cannot execute an image.

@amcmaple I have already said that plotting a phase portrait for a function is not a meaningful concept.  One plots a phase portrait for a differential equation, not a function.

What you have shown appears to be the solution of a differential equation.  What you need to show is the differential equation itself.

Additionally, it will help if you would explain why you are interesterd in this question since, as you say, you don't know what a phase portrait is.

A  "phase portrait for a function" is not a meaningful concept.  Perhaps you mean a phase portrait for a differential equation?  If so, then tell us what differential equation you have in mind because the details of the answer will depend on that.

Your references to the plot of dX/dt versus X indicate that the differential equation must be of the second order.  But because of the missing information, the discussion has been diverted to plotting solution curves (not phase portraits) of first order equations. Be more specific if you want good answers.

@spalinowy

@spalinowy Where do you see a nonlinearity?  The expression
              
is linear in u(t). That is what is meant by linearization.

@MargoMaples You say x_n+1 =x_n- (f(x_n)*d) / (f(x_n+d)-f(x_n)) and "d is a really small number".

Your problem of two intersecting circles leads to a nonlinear system of two equations in two unknowns. Consequently, the variable x_n in that iteration scheme would be a point in R².

This raises the question: What do you mean by x_n+d where x_n is in R² and d is a number?

Do you see the problem?  You cannot add a point and a number!

That's why I am asking for either a precise description of iteration scheme that you wish to apply, or an easily accessible reference where that description can be found.

Or... ask the person who gave you this problem for clarification.

@MargoMaples What is d?

It will help if you either describe precisely the algorithm that you have in mind, or refer to a web page that describes it.  Offering various permutations of the words Newton, Raphson, Modified, Secant is not a good substitute.

@spalinowy I have already shown how to linearize about any ψ₀(t).  In particular we see that if ψ₀(t)≡ 0, then the linearized expression is just zero. This is analogous to linearizing the function f(x) = x*sin(x) at x=0 — you just get zero.

I know the meanings of the "Newton method", "modified Newton method", "secant method", but I don't know "modified Newton secant method".  What is it?  References?

@Joe Riel And here is a cruder way:

@brian bovril You essentially have the answer in the first four lines of your spline.mw.  Just change the first line from Y := [3,10,2,7,7] to Y := [3,10,2,7,7,42]. That is, the answer "42" is not arrived at by accident; it is built in!

The result may be cast into Horner's form through convert(f, horner).

@rallezet I can't make sense of what you are asking. Don't be afraid of using more words to explain what you have in mind.

I think you had better explain more clearly what it is that you are asking.

@Samir Khan I tried displaying your globe on 4 different computers, all running 64-bit Linux:
1. Thinkpad X301
2. Thinkpad X301
3. Thinkpad X1
4. ThinkStation C30

The X1 and C30 display the globe  perfectly.   The two identical X301 machines both display the globe with the artifacts that I have shown.  I suppose that this is due to the limitations of the graphics cards in X301.

What I really like about your software is how responsive the globe is to the mouse motion—there is no resistance to the motion at all.  In contrast, the globe produced by the previously noted plot3d() method with the image=… option is somewhat difficult to rotate; it appears that the globe is recalculated/repainted with each incremental rotation.  So all else being equal, I prefer your globe.