Curvilinear CoordinatesWatch

#1
I am attaching a question and the solution I have written.

There is one point that I am unhappy about so I have fudge it sort of speak.

The region is transformed from the triangle to the "trapezium" in the second page but I cannot justify analytically why the region "finishes" at u=0.

For instance why can not be a straight line (or indeed a curve) which joins (0,1) to (1,0).

(note there is also a singularity at x = 0 in the transformation equations)

Any suggestions?

jacobians_and_curvilinear_coodinates.pdf
IMG_0016_NEW.pdf
0
4 years ago
#2
I get the following:

If then:

So in terms of limits:

Does that help?
2
#3
(Original post by atsruser)
I get the following:

If then:

So in terms of limits:

Does that help?
I am not in a "fit state" at present, but I think this does it.
I will look at it tomorrow.
many thanks.

(PS will rep you as soon as I can rep you again)
0
#4
(Original post by atsruser)
I get the following:

If then:

So in terms of limits:

Does that help?
I am ashamed but I have done something stupid (blame the vino).

In the middle of the 1st page I have re-aranged incorrectly something that not even a GCSE student would have done.
0
4 years ago
#5
(Original post by TeeEm)
I am ashamed but I have done something stupid (blame the vino).

In the middle of the 1st page I have re-aranged incorrectly something that not even a GCSE student would have done.
Having thought about this, the transformation is a bit odd since the x-axis is apparently mapped to both the lines and ; it's not even a good function, if what I wrote above is correct.

However, if we look at the original equations then if (i.e. for points on the x-axis) we have:

except at the origin
which isn't identically zero on the x-axis

So I'm not sure what's going on. I think I've confused myself.
0
#6
(Original post by atsruser)
Having thought about this, the transformation is a bit odd since the x-axis is apparently mapped to both the lines and ; it's not even a good function, if what I wrote above is correct.

However, if we look at the original equations then if (i.e. for points on the x-axis) we have:

except at the origin
which isn't identically zero on the x-axis

So I'm not sure what's going on. I think I've confused myself.
I am aware of the singularity but I am happy to ignore it since I know the integral converges.

My problem was that I could not map the boundary correctly due to my stupidity.

When you rearranged x=x(u,v) and y=y(u,v) then the penny dropped that my rearrangements (although I eliminated differently) was not the same!

If you see my rearrangement for x = f(u,v) in the middle of the first page, you will laugh...

Anyway, the problem is sorted

0
4 years ago
#7
(Original post by TeeEm)
I am aware of the singularity but I am happy to ignore it since I know the integral converges.
My concern wasn't for the singularity though, but rather for the fact that 1) the mapping doesn't seem to be a function (it's multivalued on y=0, apparently) and 2) that I seem to get contradictory results depending on whether I play around with (x,y) or (u,v).

Did you come up with that transform yourself or it is a standard thing?
0
#8
(Original post by atsruser)
My concern wasn't for the singularity though, but rather for the fact that 1) the mapping doesn't seem to be a function (it's multivalued on y=0, apparently) and 2) that I seem to get contradictory results depending on whether I play around with (x,y) or (u,v).

Did you come up with that transform yourself or it is a standard thing?
I made the question backwards, i.e. I started with ex over the trapezium and picked 2 simple transformation equations at random to create an "unrecognisable" integrand,( where naturally I know the transformation that will unlock it.)

the integral has to exist because two different computer algebra systems will integrate it over the triangle to the answer I got.
0
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