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358 Advanced Calculus, Fifth EditionWe now let 6 + 0. As in Section 5.11, the left-hand side approaches div F, evaluatedat (XO, yo); the right-hand side approaches -2np(xo, yo), as in Section 4.3[Eq. (4.49)]. Therefore for every point (x, y) interior to E,Since F = grad U, we have alsodiv F = -2nj~(x, y). (5.178)div grad U = V ~U = -2np(x, y) (5.179)jat every such point. Equation (5.179) is called the Poisson equation.Remark. The replacement of div grad U by V ~U in (5.179) assumes that the potentialU has continuous second partial derivatives at (x, y). This can be shown tobe valid if, for example, p(x, y) has continuous first derivatives. (See Chapter 5 ofthe book by Sternberg and Smith listed at the end of the chapter.)The first two identities of Green are given in Problems 11 and 12 following Section45.7. The third identity is the equationHere u is assumed to have continuous first partial derivatives in a domain containingthe closed region E bounded by the curve C and to be harmonic inside C. The point(x, y) is interior to C, n is the exterior normal on C, and r is as above the distancefrom a general point (c, r]) on C to (x, y).We remark that the termis a logarithmic potential of a single layer on C and the termis the logarithmic potential of a double layer on C. Hence the identity asserts that aharmonic function can be represented as the sum of two such logarithmic potentials.To prove the identity, one starts with the second identity of Green:I(Problem 12(a) following Section 5.7). We apply this first to our given region E,taking v = log l/r, where r is as above the distance from (c, r]) to (x, y), but take
Chapter 5 Vector Integral Calculus 359(x, JJ) exterior to E. Then v is harmonic as a function of (c, g) in E, so that (5.181)becomesau 1 alog -V'U dc dg =SS t an r anEC- log - - u - logIf, in addition, u is harmonic in E, then (5.182) becomes(5.182)This is valid for (x, y) outside of E.Now for (x, y) interior to E we draw a small circle Cg of radius 6, with centerat (x, y), and let Eg be the region obtained from E by deleting the interior of thiscircle from E (Fig. 5.45). We again take v = log 1 / r and can apply Eq. (5.181) tothe region Eg, since (x, y) is exterior to this region. We obtainOn Cs the normal is exterior to Eg, SO that a/an = -a/&.Thus in polar coordinatesr, 8, with center at (x, y), so that ds = -6 dB on Cs, traced clockwise,rFigure 5.45Proof of third Green identity.
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3.6 Combined Operations 183*3.7 Cur
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