496 CHAPTER 18. DIFFERENTIAL FORMS

whether it requires an odd or even number of switches to obtain (1,2,3) . Thus ε123 =1,ε213 = −1 and so forth. Show that ∑k ε i jkε irs = δ jrδ ks− δ jsδ kr. We often agreeto add over a repeated index to avoid having to write the summation symbol. Thus∑k ε i jkε irs = ε i jkε irs. If you do this, avoid having the repeated index repeated morethan once in any term. Here δ i j is 1 if i = j and 0 if i ̸= j.

9. Let U be a bounded open set inRp and suppose u ∈C2 (U)∩C(U)

such that ∇2u≥ 0

in U. Then letting ∂U =U \U, it follows that

max{

u(x) : x ∈U}= max{u(x) : x ∈ ∂U} .

The symbol ∇2 is the Laplacian. Thus ∇

2u = ∑i uxixi . In terms of repeated indexsummation convention, ∇

2u = u,ii. Hint: Suppose this does not happen. Then thereexists x0 ∈U with

u(x0)> max{u(x) : x ∈ ∂U} .Since U is bounded, there exists ε > 0 such that

u(x0)> max{

u(x)+ ε |x|2 : x ∈ ∂U}.

Therefore, u(x)+ ε |x|2 also has its maximum in U because for ε small enough,

u(x0)+ ε |x0|2 > u(x0)> max{

u(x)+ ε |x|2 : x ∈ ∂U}

for all x ∈ ∂U . Now let x1 be the point in U where u(x)+ ε |x|2 achieves its max-imum. Now recall the second derivative test from single variable calculus. Explainwhy at a local maximum of f you must have ∇

2 f ≤ 0. Apply this to the functionx→ u(x)+ ε |x|2 at the point x1 and get a contradiction. This is called the weakmaximum principle.

10. Review the cross product from calculus. Show that in R3,(a×b)i = ε i jka jbk wheresummation is over repeated indices. Using the above reduction identity of Problem8, simplify (a×b)×c in terms of dot products. Then do the same for a×(b×c).

11. Show that ∇ · (∇×v) = 0. Now show∫

∂V ∇×v ·ndA = 0 where V is a region forwhich the divergence theorem holds and v is a C2 vector field. ∇×v is the curl of v.In the new notation, (∇×v)i = ε i jk∂ jvk where ∂ j is an operator which means to takethe partial derivative with respect to x j. It is understood here that the coordinates arerectangular coordinates. The first part of this is real easy if you remember the bigtheorem about equality of mixed partial derivatives.

12. Let U be a bounded open set in R2 which has a Lipschitz boundary so that thedivergence theorem holds. Let the axes be oriented in the usual way as in calculus.Let P(x,y) ,Q(x,y) be two smooth functions defined on Ū . What does the divergencetheorem say for the vector field (Q,−P)? If r : [a,b]→ R2 is Lipschitz with r (a) =r (b) and r is one to one with ∂U = r ([a,b]) and r′ (t) is in the direction of k×nso r′ (t)×k is in the direction of n, a statement about orientation. Thus for r (t) =(x(t) ,y(t)) ,

n= c

∣∣∣∣∣∣i j k

x′ (t) y′ (t) 00 0 1

∣∣∣∣∣∣= c(y′ (t)i− x′ (t)j

), so c =

1√y′ (t)2 + x′ (t)2