Skip to content
Back to exploration
Calculus / Chinese

A counterexample to path independence

Charles队长 · Bilibili · 1:08

Open original
READ & KEEP

The explanation, unpacked.

Reviewed learning material · Video analysis · English
Read the full overview

The video exposes a hidden topological prerequisite in the theorem of path independence for line integrals. It starts with the standard condition (∂Q/∂x=∂P/∂y\partial Q/\partial x = \partial P/\partial y) and tests it against a rotational vector field undefined at the origin. Although the partial derivatives match perfectly, calculating integrals along two different semi-circular paths yields contradictory results (π\pi vs −π-\pi), and the closed loop integral is non-zero (2π2\pi). The video concludes by correcting the theorem statement: path independence holds only in simply connected domains. A visual comparison between a solid disk (simply connected) and an annulus (multiply connected) illustrates why the presence of a singularity ('hole') invalidates the naive application of the theorem.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Theorem Introduction & Counterexample Setup0:13Verifying Conditions & False Conclusion0:22Calculation Reveals Contradiction0:45Root Cause & Corrected Theorem

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The video opens with a common formulation of the theorem regarding path-independent line integrals: if the cross-partial derivatives are equal (∂Q∂x=∂P∂y\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}), then the integral depends only on endpoints. To test this, we examine a specific 2D vector field F⃗=(−yx2+y2,xx2+y2)\vec{F} = (\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}). The screen displays the flow lines of this field, showing arrows rotating counter-clockwise around the origin.

We proceed to verify the analytic conditions algebraically. Calculating the partial derivative of QQ with respect to xx gives y2−x2(x2+y2)2\frac{y^2-x^2}{(x^2+y^2)^2}. Similarly, the partial derivative of PP with respect to yy yields the exact same expression. Since they are identical, one might naively conclude based on the initial theorem that the line integral is path-independent.

Travel from (-1,0) to (1,0) along the upper and lower semicircles. The upper route is clockwise and gives -π; the lower route is counterclockwise and gives +π. Different results with identical endpoints establish path dependence. A complete counterclockwise turn gives 2π, also ruling out a conservative field.

Where did the logic fail? The issue lies in the domain definition. The vector field has a singularity at the origin (0,0)(0,0) where it is undefined, creating a 'hole' in the space. Thus, the correct version of the theorem requires the domain DD to be simply connected. The video ends by contrasting a green solid circle (simply connected, valid) with a red ring containing a hole (multiply connected, invalid), emphasizing that topology dictates the applicability of calculus theorems. A hole does not make every field path dependent; it means zero curl alone no longer guarantees path independence.

Knowledge cards

01

Path Independence Theorem (Complete Form)

On a simply connected open domain, C1 coefficients with Qx=PyQ_x=P_y give path independence. Simple connectivity is a sufficient domain condition for this general criterion, not a necessary condition for a particular field: a gradient field on an annulus can still be path independent.

If D is simply connected and ∂Q∂x=∂P∂y, then the line integral is path independent.\text{If } D \text{ is simply connected and } \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}, \text{ then the line integral is path independent.}
02

The Vortex Field Counterexample

Away from the origin this field satisfies the zero-curl condition Qx=PyQ_x=P_y, not the Cauchy–Riemann equations. Its punctured domain allows nonzero circulation, so it is not globally conservative.

F⃗=(−yx2+y2,xx2+y2)\vec{F} = \left( \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2} \right)
03

Algebraic Verification of Equality

Both components satisfy the Cauchy-Riemann-like condition for conservativeness wherever defined. This highlights that checking derivatives alone is insufficient without analyzing the continuity and connectivity of the underlying region.

∂Q∂x=y2−x2(x2+y2)2=∂P∂y\frac{\partial Q}{\partial x} = \frac{y^2 - x^2}{(x^2 + y^2)^2} = \frac{\partial P}{\partial y}
04

Zero curl and nonzero circulation

Although the curl is zero away from the origin, a counterclockwise loop enclosing it once has circulation 2π. Green’s theorem cannot be applied directly to the disk containing the undefined origin, so there is no contradiction with the theorem.

∮LF⃗⋅dr⃗=∫02π1 dt=2π≠0\oint_L \vec{F} \cdot d\vec{r} = \int_0^{2\pi} 1 \, dt = 2\pi \neq 0
05

Simply Connected vs Multiply Connected

A simply connected domain allows every closed curve to shrink to a point continuously. An annulus or punctured plane fails this test because curves encircling the hole cannot pass through the missing material. Green's Theorem relies heavily on this distinction.

Explore the knowledge in this video

Open video knowledge graph →

  • Definite integrals ApplicationAt 0:52
    Why this connection?

    The reviewed path-independence card gives a sufficient criterion for line integrals: on an open simply connected domain, continuously differentiable components satisfying Qx=PyQ_x=P_y yield path independence. The punctured-plane example instead has zero curl away from the origin but circulation 2π2\pi. Simple connectivity is sufficient for this general criterion, not necessary for every particular conservative field.

Questions this video answers

Find a method

↗
Meet the concept

↗
Meet the concept

↗
Understand why

↗
Understand why

↗
Understand why

↗
Meet the concept

↗
Find a method

↗