- <fi(x,y) = 4 - ~Arg(z + 3) + *Arg(z + 1) - ~Arg(z-2).
<f!(x,y) = 4 - ~Arctan~ + ~Arctanrl:T - ~Arctant-:2.
(^7) ·"'(x '+',y) -- iArctan'Jr^1 -x•-y• '2y.
- "'(x y) = 5 - £l1Argi(l-z) = 5 -^20 Arctan l-x•-y•.
"P ' '1t' l+z 7r 2y - "'(x y) = 3 -^12 Argi- i•^2 = 3 - llArctan l- (x'+v•)2.
'+' ' ·Jr l+zlli 1r 4~y
ANSWERS 617Section 11 .3. Poisson's Integral Formula for the Upper Half-Plane:
page 443
(^1) • "'( 'I' x, y ) -- JL ln (x- l>2+i/ + x - I A ta -1L - ti.!A t ][ + 1
2 ,,. (x+l)-+y• ,,. re n.,_ 1 ,,. re anx+l.
- Both eY cos x and e- 11 cos x are haxmonic in the upper half-plane and satisfy
the boundary conditions. Also, Jim e-11 cos x = 0. It can be shown that the
y-oo
Poisson integral formula defines a bounded function in the upper half-plane;
therefore , the desired solu tion is <fi(x, y) = e-11 cos x. - Apply Leibniz's rule </!,,., + </! 1111 = ~ J~= U (t) [ (fl, + ~) (.x-tf•+y•] dt.
Th e term m 'bra ckte s m ' the m ' t egran d1s ' F:i?1 ffl (x-ty )"+v• + /iY'l' 11'" (x -t)'+y• =
2(3t•y:-6txy+3x'y-y3) + 2(-3t'y+6txy-3x•y+y3) = 0 Hence the integrand van-
((x- t)'+y2)3 ((.x-t)2+y^2 )3 ·
ishes and <fixx (x, y) +</! 1111 (x, y) = 0, which implies that <fi(x, y) is haxmonic.
(^7) • "'('I' - X ' Y ) - " 1£ Joo -oo (-x-tJ2+yU(t} dt (^2) - " 1£ Joo - oo (x+tU(t} )•+ydt 2 _ - 1£ ,,. J oo r-oo U(-t} (x-t) &-l+y)dt 2
Y. f"° U(- tJ dt 1£ f"° U(t } dt _ (x )
- ,,. - oo (x-t) +11• - " - oo (x-t)^2 +y^2 - </! ' Y ·
Section 11.5. Steady State Temperatures: page 4523. T (x, y) = 10 + ~ Arctan.,2~:L 1 - ~ Arctan.,2~:¥+1.
- T(x,y) = 100+^1 ~^0 Arctan(sinz+ 1 )-^1 ~^0 Arctan(sinz- 1).
- T(x,y) = ~ ln lz l.