A direct proof of the Inside -Outside Theorem. Obtain this formula using Green\'

Question

A direct proof of the Inside -Outside Theorem. Obtain this formula using Green's Theorem only, without mention of the Identities. We have Ohm a Jordan domain with boundary curve (or curves) given by z = z(s) = (x(s), y(s)) as usual (are length parametrization). Also, N(z(s)) is the outward normal vector to Ohm at the point z(s), and u = u(z) epsilon (Ohm^+). Verify that (partial differential u| partial differential n)(z(s)) = v u(s)) middot N (z(s)). Verify N(z(s)) = (y'(s), -x' (s)). Use (a) to write (partial differential u/partial differential n)(z(s)) ds in the form p dx + q dy. Apply Green's Theorem to conclude the Inside-Outside Theorem: integral_Ohm partial differential u/partial differential n ds = integral_Ohm integral Delta u dx dy. Please answer question9. All the steps and calculations must be shown with formula. An incomplete answer will result in a thumb down and bad rating for you. A good answer will result in a thumb up and I would really appreciate it. Thanks

Explanation / Answer

Solution: Given that z= z(s) = (x(s),y(s)), N(z(s)) is outward

normal at z(s), and u=u(z)

Here x'(s) = (dx/ds)(s), y'(s) = (dy/ds)(s)

(a) grad u(z(s)). N(z(s))= <ux,uy>.<y',-x'> = y' ux-x'uy = (partial u)/(partial n)(z(s))

(b) We know that unit tangent vector at z(s)=(x(s),y(s))

is given by T(s) = z'(s) = <x'(s), y'(s)> =<dx/ds,dy/ds>.

Since N(s).T(s) = <y'(s),-x'(s)>. <x'(s), y'(s)> = -x'(s)y'(s) + x'(s)y'(s)

=0, so N(s) is normal to T(s).

Hence N(s) = <y'(s),-x'(s)>.

c) (partial u)/(partial n)(z(s))ds = grad u(z(s)). N(z(s))ds= <ux,uy>.<y'(s),-x'(s)> ds

= (uxy'(s) - uyx'(s))ds = uxy'(s)ds - uyx'(s)ds = uxdy - uydx (as x'(s)ds=dx, y'(s)ds=dy)

So (partial u)/(partial n)(z =(s))ds = uxdy - uydx= pdx +qdy,

where p = - uy and q =ux

d) By putting p=1 and q =u in Green's theorem I,

we get the result.

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