Looking for a rectifying chart

[mihyaeru]

given task
Find a rectifying chart at (0,0) of the vector field
X: R^2 -> TR^2 , (x,y) -> X(x,y)
with X(x,y):= (cos^2(x+y)+sin^2(x-y))d/dx + (cos^2(x+y)-sin^2(x-y))d/dy .

attempt of solution
We know that there exists a rectifying chart since X(0,0)=(1,1).
A transversal direction would be (1,-1).

Therefore we get the differential equations:
dx/dt= cos^2(x+y)+sin^2(x-y) (1)
and
dy/dt= cos^2(x+y)-sin^2(x-y) (2)

Next we have to find theire solutions. Change of variables and integration lead me to the general solution of (1):

t+const = sec(2y)*tan^-1(sin(x-y)*sec(x+y))

=> tan(t*sin(2y))+const = sin(x-y)/sin(x+y)
where I got stuck... //Recall we have to solve this equation for x and (2) for y.

Maybe I should change my coordinate system at the very beginning, such that x+y=a and b=x-y? What do you think? Any ideas?

Kind regards,
mihyaeru

 

[jvicens]

Dear mihyaeru,

Thank you for sharing your attempt at solving this problem. It seems like you are on the right track, but as you mentioned, you have reached a point where you are stuck. Let's take a closer look at your equations and see if we can find a solution.

First, let's change the variables as you suggested, such that x+y=a and b=x-y. This will give us the following equations:

dx/dt= cos^2(a)+sin^2(b) (1)
dy/dt= cos^2(a)-sin^2(b) (2)

Now, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to rewrite these equations as:

dx/dt= 1 (1a)
dy/dt= cos(2a) (2a)

Integrating both equations with respect to t, we get:

x= t + const (1b)
y= 1/2sin(2a)t + const (2b)

Now, we can use these equations to eliminate t and obtain a solution for x and y in terms of a and b. Substituting (1b) into (1a), we get:

dx/dt= 1 = cos^2(a)+sin^2(b)
=> t= cos^2(a)+sin^2(b) - const

Similarly, substituting (2b) into (2a), we get:

dy/dt= cos(2a) = cos^2(a)-sin^2(b)
=> t= 1/2sin(2a) + const

Since both t's are equal, we can equate the right sides of these equations and solve for a and b:

cos^2(a)+sin^2(b) - const = 1/2sin(2a) + const
=> cos^2(a)-1/2sin(2a) + const = 0

This is a quadratic equation in terms of cos(a) with solutions:

cos(a)= 1/4 +/- sqrt(1/16-const)
=> a= arccos(1/4 +/- sqrt(1/16-const))

Now, we can use this solution for a in (1b) and (2b) to get the solutions for x and y:

x= t + const =