Paper 3, Section II, C

Vector Calculus
Part IA, 2016

What is a conservative vector field on Rn\mathbb{R}^{n} ?

State Green's theorem in the plane R2\mathbb{R}^{2}.

(a) Consider a smooth vector field V=(P(x,y),Q(x,y))\mathbf{V}=(P(x, y), Q(x, y)) defined on all of R2\mathbb{R}^{2} which satisfies

QxPy=0\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0

By considering

F(x,y)=0xP(x,0)dx+0yQ(x,y)dyF(x, y)=\int_{0}^{x} P\left(x^{\prime}, 0\right) d x^{\prime}+\int_{0}^{y} Q\left(x, y^{\prime}\right) d y^{\prime}

or otherwise, show that V\mathbf{V} is conservative.

(b) Now let V=(1+cos(2πx+2πy),2+cos(2πx+2πy))\mathbf{V}=(1+\cos (2 \pi x+2 \pi y), 2+\cos (2 \pi x+2 \pi y)). Show that there exists a smooth function F(x,y)F(x, y) such that V=F\mathbf{V}=\nabla F.

Calculate CVdx\int_{C} \mathbf{V} \cdot d \mathbf{x}, where CC is a smooth curve running from (0,0)(0,0) to (m,n)Z2(m, n) \in \mathbb{Z}^{2}. Deduce that there does not exist a smooth function F(x,y)F(x, y) which satisfies V=F\mathbf{V}=\nabla F and which is, in addition, periodic with period 1 in each coordinate direction, i.e. F(x,y)=F(x+1,y)=F(x,y+1)F(x, y)=F(x+1, y)=F(x, y+1).