## 6996 Reputation

16 years, 214 days

## Initial feedback...

You can expect better feedback in this forum if you upload your worksheet rather than what you have posted.  But just looking at the first few lines of your code, I see that you have strange stuff such as

`ibvc:=u(x, 0)=e^(K*x)/(1+exp^(0.5*K*x))^2,v(x, 0)=1/(1+exp^(0.5*K*x));`

What is e^(K*x)?   What is exp^(0.5*K*x))?  Think!

## Solving on a triangle...

@mmcdara Yes, with u(x,0)=0 it's possible to solve the problem in Maple by Fourier series.

Here is the outline of the idea.

We reflect the domain about the line y=0 and thus obtain a rectangle.  We solve Laplace's equation on the rectangle by taking for boundary values on the new edges the negatives of the values of the opposite pages.  Then the solution will be zero automatically along the line y=0 due to symmetry.

A detail to be taken care of is that the rectangle's edges are not parallel to the coordinate axes.  To apply the Fourier series, we rotate the rectangle by 45 degrees to bring its edges into alignment with the coordinate axes.  The PDE does not change since the Laplacian operator is rotationally invariant.

To avoid the rotation step, it would be much easier to rotate the original domain.  That is, instead of the triangle shown in the OP's drawing, we take a triangle formed by the coordinate axes and the line x+y=1 (or ax+by=c for some constants a, b, c).  Then the reflection of the triangle across the hypotenuse will produce a rectangle in which we may apply the standard Fourier series in the straightforward way.

Even then, there is an extra snag that needs to be taken care of.  The Fourier series method requires homogeneous boundary conditions, that is, u=0, along the four edges of the rectangle.  But our boundary data is not zero along the edges.  To take care of that, we introduce a function φ(x,y)  defined over the rectangle so that φ matches the given boundary conditions along the rectangle's edges.  If we let w(x,y) = u(x,y) - φ(x,y),  then w will be zero over the rectangle's edges, and therefore Fourier series may be applied to calculate it.  We note, however, since u satisfies the Laplace's equation, w satisfies the Poisson equation wxx + wyy = −φxxφyy. which may be solved by Fourier series just as easily.

Finally, there remains the question of how one constructs the function φ(x,y).  That is the subject of a topic that I posted here a few years ago.

There is some work involved in implementing the steps outlined above.  I haven't bothered to do the work considering that this approach applies to the very limited case when the domain is a right triangle, the boundary condition on the triangle's hypotenuse is zero, and also the fact that the OP has specified no such boundary condition.  In general such problems are best solved through the method of finite elements which is not available in Maple.

## Rectangular domain is okay...

@Oliveira Sure, you get a Fourier series representation of the solution.  That works on a rectangle but your original question was about solving over a triangle.

## Explanation...

@bstuan Let f(x) be your integrand and let m be the minimum value of (x+2)/(2x^3+1) on the interval (1,2). Since f(x) > m/ln(x), the integral of f(x) is more than the integral of m/ln(x) on (1,2). But the integral of m/ln(x) on (1,2) is infinity, and therefore so is the integral of f(x).

## Hint...

Hint: Examine the convergence of the integral of 1/ln(x) over the interval (1,2).  That should be good enough to reach a conclusion about your original problem.

## Missing boundary condition...

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

## Not in Maple...

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 co...

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 ...

@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.

## Need numbers...

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.

## h or h+2...

@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.

## Corrections...

@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?  D...

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

## tubeplot...

@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);` ﻿