Question

Consider the following vector field: a(t) where a(t) is an arbitrary time dependent function. (a) Show that the origin is a hyperbolic trajectory. (b) Argue that the graph of y 2 is the global unstable manifold of the origin. What requirements must be made on the function a(t) in order that these conclusions are true?

Answer 1

(a) Since passing through origin , a= 0 , b = 0

Take the unit circle, (x - 0)² + (y-0)² = 1

i.e x² + y² = 1, when describing a sector of this circle , we can talk about the arc length r$\Theta$, and always obtain the same number since r = 1

Take unit length hyperbola x² - y² = 1, and consider area enclosed by an arc of this hyperbola, starting at point (1,0), if both end or arc are connected to origin (0,0)  

Relation to Euler's constant....

exp(x) =  $\sum_{n=0 }^{\infty }$ xⁿ/n! = 1 + x + x²/2! + x³/3! + x⁴/4! +.......

sin(x) = $\sum_{n=0 }^{\infty }$ (-1)ⁿ x²ⁿ⁺¹ / (2n+1)! = x - x³/3! + x⁵/5! - .......

cos(x) = $\sum_{n=0 }^{\infty }$    (-1)ⁿ x²ⁿ / (2n)! = 1 - x²/2! + x⁴ / 4! - .........

sinh(x) =  $\sum_{n=0 }^{\infty }$ x²ⁿ⁺¹ / (2n+1)! = x +  x³/3! + x⁵/5! + ..........

cosh(x) =  $\sum_{n=0 }^{\infty }$ x²ⁿ/2n! = 1 +  x²/2! + x⁴/4! + ..........

The trignometry function are closley relatedto exponential function, and thus to e . The circular trignometric function have these alternating signs, which are best explained as a purely imaginary argument to the exponential function, The hyperbolic function don't have this, so expressing via e. If we want to express everything via the exp function like this,

sin(x) = exp(ix) - exp(-ix) / 2i, sinh(x) = exp(x) - exp(-x) / 2

cos(x) =  exp(ix) + exp(-ix) / 2, cos(x) = exp(x) + exp(-x) / 2

(b) The unstable manifold consist of those p such that $\phi$t (p) $\rightarrow$ (0,0) as t $\rightarrow$ - $\infty$. The condition forces that c1 = 0 so that the unstable manifold is given by

x = c₂e^2t ,

y = 0

i.e it is the entire x - axis