A direct proof of the Inside Outside Theorem Obtain this for

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

Solution

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))= <u_x,u_y>.<y\',-x\'> = y\' u_x-x\'u_y = (\\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= <u_x,u_y>.<y\'(s),-x\'(s)> ds

= (u_xy\'(s) - u_yx\'(s))ds = u_xy\'(s)ds - u_yx\'(s)ds = u_xdy - u_ydx (as x\'(s)ds=dx, y\'(s)ds=dy)

So (\\partial u)/(\\partial n)(z =(s))ds = u_xdy - u_ydx= pdx +qdy,

where p = - u_y and q =u_x

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

we get the result.