Closed curve and orthogonal curvilinear coordinate system

[traianus]

Hello, I have a "simple" problem for you guys. I am not expert in math and so try to be simple.

I explain the problem by starting with one example. The polar coordinate system has the following main property: with two parameters, rho and theta, each point is described as the intersection of a circle and a straight line. These two curves are orthogonal and so the curvilinear coordinate system is orthogonal.

Another example is if we use ellipses and hyperbolas: they are orthogonal. The transformation is simple:

x = c*cosh(a)* cos(b) (1)
y = c*sinh(a) * sin(b) (2)

where a>0, and 0<=b<=2pi

Now my question. Suppose that we like to use a set of curvilinear coordinate system (plane x-y only) which is orthogonal. Suppose that instead of ellipses or circles one of the type of curves is a closed path defined by (for example) the equation

x^8/A^8 + y^8/B^8 = 1

which is closed curve similar to an ellipse. Ho do I find a similar trasnformation (like equations (1) and (2)) with the property that the curvilinear coordinate system is also orthogonal? Please help me!

 

[Hurkyl]

Suppose that instead of ellipses or circles one of the type of curves is a closed path defined by (for example) the equation

x^8/A^8 + y^8/B^8 = 1

Did you mean the family of curves (parametrized by k > 0)

x^8/A^8 + y^8/B^8 = k

where A and B are fixed constants?

Ho do I find a similar trasnformation (like equations (1) and (2)) with the property that the curvilinear coordinate system is also orthogonal?

In the obvious way! Just look at what you want: you want functions of two variables a and b such that:

(1) The derivative WRT a is parallel to x^8/A^8 + y^8/B^8 = k
(2) The derivative WRT b is perpendicular to x^8/A^8 + y^8/B^8 = k

So you should be able to set up a system of partial differential equations. I guess it might not be very easy to solve, though.

 

[traianus]

Thank you. Can you show the system of partial differential equations that solves the problem? Actually in my case I like to have the coordinate system of the "hyper-ellipse"

x^(2n)/A^(2n) + y^(2n)/B^(2n) = k

where n is a positive integer.

 

[Hurkyl]

If C(k) is the loop defined by:

x^8 / A^8 + y^8 / B^8 = k

and (p, q) is a point on C(k)

then can you find a tangent vector to C(k) at (p, q)? What about a vector perpendicular to C(k) at (p, q)?

 

[traianus]

It is an implicit function. So I think that the tangent has equation

7*p^7/A^8*(x-p) + 7*q^7/B^8*(y-q) = 0 (1)

from (1) it follows that the vector perpendicular to the curve is

Vperp= q^7/B^8 * i - p^7/A^8*j (2)

where i,j are the unit vectors. The vector parallel to the curve is then

Vpar = p^7/A^8*i+ q^7/B^8*j (3)

Let me see if I get your point. Curvilinear coordinates means that there is a transformation of the type

x = f(a,b) (4)

y = g(a,b) (5)

Now I have to obtain the system. Please correct if am wrong. If I am not, can you write the system of equations?

 

[Hurkyl]

Your work is off. I think you just switched which is perpendicular and which is parallel.

You're right that it's an implicit equation: when you differentiate, you should get:

$$\frac{8 p^7}{A^8} \, dx + \frac{8 q^7}{B^8} \, dy = 0$$

is the equation for motion along your loop at the point (p, q).

Let me see if I get your point. Curvilinear coordinates means that there is a transformation of the type

x = f(a,b) (4)

y = g(a,b) (5)

Yep.

Since you wanted one of your parameters to go around the loop, that means when you partially differentiate, the result should be the tangent vector! i.e.

$\frac{\partial}{\partial a} \langle f(a, b), g(a, b) \rangle$ = the tangent vector at <f(a, b), g(a, b)>

and a similar condition for the other direction!

 

[HallsofIvy]

Finding "orthogonal trajectories" used to be a standard problem in ordinary d.e.s. If a family of curves is given by f(x,y)= c, then
$$\frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}y'= 0$$
so
$$y'= -\frac{\frac\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$

The "orthogonal trajectories" must satisfy the differential equation
$$y'= \frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}$$


In the case that $$\frac{x^{2n}}{A^{2n}}+ \frac{y^{2n}}{B^{2n}}= k$$ (you still haven't answered the question as to whether A and B are fixed constants while k determines the member of the family- actually, I suspect that there are some relations between A, B, k), assuming that A, B are constant while k varies over the family, the orthogonal trajectories satisfy
$$y'= \frac{A^{2n}}{B^{2n}}\frac{y^{2n-1}}{x^{2n-1}}$$

 

[traianus]

Thank you for you answers. A, B are constant, as n. So, what is the parameteric representation

y = f(a,b)

x = g(a,b)

in which when a = constant I have the curve

$$\frac{x^{2n}}{A^{2n}}+ \frac{y^{2n}}{B^{2n}}= k$$

and when b = constant I have a family of curves orthogonal to it (like it happens for polar coordinates in which the lines are perpendicular to the circles)?

 

[traianus]

A and B must be positive. If I move from a curve fto another one, A and B can change, but they can be function ONLY of the parameter a. Also, they have to be positive.

 

[traianus]

Nobody knwos?