You need to specify a boundary condition along the bottom edge as well.

Maple is good in dealing with symbolic expressions involving scalars.  It can also handle vectors and matrices of explicitly specified sizes and contents.  But it does not have a facility to deal with purely symbolic vectors and matrices which you need in this instance.

I hope that such facilities will be added some time in the future, but I wouldn't expect it soon.

There may be a problem if the running command or proc refers to the previous calculations through their labels.

@Carl Love Oh, okay, and I have no idea regarding the validity of what Maple is showing as solutions.

You refer to an ODE but I see no ODEs in what you have shown.

Why don't you just upload a complete, self-contained worksheet like others do in this forum?  Look for a big fat green arrow in the dialog box where you reply to this request. Click on it to upload your worksheet.

All that and more can be done in Maple, but you need to supply the values of the coefficients, ca, cb, Ka, Kb, Kw and Va .  The optimal approach may vary, depending on what those coefficients are.

@WA573 In your worksheet, the expression for Phi is defined with a plut/minus sign.  I chose the plus sign.  Then, to obtain the level curves labeled h=−1,0.1 in the image that you have posted, the right-hand side will have to be h+2.

If you choose the minus sign, you will obtain the curves in the image when the right-hand side is h.

I have attached a modified worksheet with the minus sign for Phi, and the right-hand side set to h.

Download: mw2.mw

@Preben Alsholm Yes, I see the issue; I had checked the solution by hand and I had assumed that Maple will do the same thing, but as you have noted, (−1)^(1/3) is not −1 in Maple and one needs to watch out for that.

Your suggestion of rewriting the original ODE as ode := diff(v(t),t) = -2*(v(t)^2)^(1/3) addresses and resolves that issue perfectly.  As to the alternative suggestion, yes, that's also perfect, and in looking at that, I see that in offering my solution I have misread the requested initial condition v(0)=5 as v(0)=0.

How familiar are you with Maple?  Do you know how to plot any function at all?

@Earl That's a very nice demo.  As to your comment re "standard plot commands", I suppose you wish to replace your use of odeplot() with something else.  If you want, you may replace it with spacecurve(), or tubeplot().  Here is one possibility:

DispGeo := tubeplot(eval(convert(x(u, v(u)), list), dsol1), u = Pi/4 .. (3*Pi)/4, radius = 0.02, color = red, style = surface);

@Ahmed111 As Carl has pointed out, the trouble in Maple 18 may be due to "with(Physics)".  But that package is not needed in Linearize.mw.   Remove and try it again.

@Carl Love I misunderstood the question.  Then perhaps this is what is wanted:

Download: Lineaize.mw

@Earl The opening paragraph of the Wikipedia page that you referred to in your original message, begins with:

      In geometry, a geodesic is commonly a curve representing
      in some sense the shortest path between two points in a surface.

Note the qualifier "in some sense" there; it is not making an absolute statement that a geodesic is the shortest path!

Technically, a geodesic curve on a surface is defined to be the generalization, in a very precise technical sense, of a straight line in the plane. For instance, a great circle on a sphere is a geodesic; it's the closest thing to a straight line that lies on a sphere. There is no appeal here to "two points" or "shortest distance".

The shortest length property of a geodesic is a consequence (i.e., it is proved as a theorem) of the "straight line" definition alluded to above. It holds under certain circumstances but it is not always valid without qualifications.

Have a look at the picture of the green cone in my previous message. The shortest path between the two orange points is the obvious little arc (not drawn) that connects those points. However, look at the red curve which also connects those two points. Imagine this as a real physical cone and the red curve as a (loose) rope that winds around the cone as shown, and goes through those two points. What would happen if you pulled that rope real tight? You will be minimizing the rope's length! The rope will take the shape as drawn. So yes, that shape is also a "shortest distance" between those two points. It is the shortest among all curves that wind around the cone once. It is shortest because if you try to pull any harder, the rope will break; you cannot make it any shorter.

This should clarify my statement that there may be multiple geodesics between two curves on a surface.  In that connection, have a look at the red and cyan geodesics on the animated torus in my earlier reply.

@Earl Okay, here it is: A geodesic on a cone

Download worksheet: geodesic-on-cone.mw

@Earl Yes, there is an idea there which I had not thought about, and which can be useful in some situatios.   Specifically, if the surface is parametrized such that geodesics are single-valued functions of one of the surface's parameters, then we may characterize those geodesics as solutions of boundary values problems of a 2nd order ODE obtained through the Euler-Lagrange equations. This works particularly well with toruses as in the illustrated answer which I am going to post next.  It should work in the case of some other interesting surfaces but I haven't tried.