Vector fields, flow lines, and domains
Objective: Why can two distinct integral curves not cross at a regular point of a C¹ field?
A vector field assigns a direction and magnitude to every admissible point; its domain may contain holes or singularities that later control global conclusions.
Start from the definition: A vector field on D⊆Rⁿ is a map F:D→Rⁿ; an integral curve r(t) satisfies r′(t)=F(r(t)). The result to establish is: If F is continuously differentiable near a point, a unique local integral curve passes through that point.
The exact worked question is: Find the integral curves of F(x,y)=(x,−y) through (a,b). Compare it with this failure boundary: For F(x,y)=(−y,x)/(x²+y²), the origin is excluded; drawing an arrow there or treating the domain as all of R² erases the singularity.
If F is continuously differentiable near a point, a unique local integral curve passes through that point.
Rewrite r′=F(r) as a first-order autonomous differential system with the initial point specified.
Continuous differentiability makes F locally Lipschitz, so nearby velocity arrows cannot separate solutions arbitrarily fast.
The local existence-and-uniqueness theorem then supplies one trajectory through the point on a sufficiently short interval.
Find the integral curves of F(x,y)=(x,−y) through (a,b).
- The system is x′=x and y′=−y.
- Solving separately gives x=Ceᵗ and y=De⁻ᵗ.
- The initial point sets C=a and D=b, so xy=ab along each trajectory.
Result: The integral curve is r(t)=(aeᵗ,be⁻ᵗ), with hyperbolic trajectories except on the axes.