Lcl us find an expression for lhc slcady lcmpcraturcs T

Chapter 0, Problem 10.2

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Lcl us find an expression for lhc slcady lcmpcraturcs T (x. y) in a lhin semi-infinilc plate y :::: 0 whose faces arc insulated and whose edge y = 0 is kcpl al temperalurc zero excepl for lhe segmem - I < x < I. where il is kept al lcmpcrnlure unity (Fig. l 55 ). The function T(.r. y) is lo be bounded: this condition is natural if we consider lhc given plate as lhe limiting case of lhc plalc 0 ::: y ~ yo \vhose upper c:A. \B' ;\ T=O IJ T=I T= 0 /) .\ T=O II FIGURE 155 u =log_: -_I (~ .: 0. - :'.:.. < U1 - lh -:: >:r) ~ : - I r~ 2 - 1 . 368 ( I) (2) APPLICATIO!\S OF CONl" 0). when lxl < J. when Ix I > I : Cll.'\P. 10 also, IT(x. y) I < M \\-'here J\t/ is some positive constant. This is a Dirichlet problem (Scc.116Horthcuppcrhalfplancy '.:'.'.: O.Ourmcthodofsolu1ion\villbctoobtainancw Dirichlet problem for a region in the 1tF plane. Thal region \Viii be the image of !he half plane under a transformation w = f (:.)chat is analycic in the domain y > 0 and conformal along !he bound:.u-y y = Oexcept at the points( I. 0). where f (:.)is undefined. It will be a simple mancr to discover a bounded hannonic function satisfying 1he new problem. The cwo theorems in Chap. 9 will then be applied to transform the solution of the problem in the 11 v plane into a solucion of the original problem in the x y plane. Specifically. a harmonic function of 11 and v will be transfonned into a hannonic funccion of x and y. and the boundary condicions in the 11 t~ plane will be preserved on corresponding portions of !he bound;uy in the xy plane. There should be no confusion if we use the same symbol T to denote the di ffcrent cempernturc funciions in the cwo planes. Let us write J = r 1 cxp(i01) and :. +I = r2 cxpU02). where 0 ~ fh_ ~ ;rr (k = J. 2). The trnnsfonnacion (3) is defined on the upper half plane y ::::: 0. except for chc cwo points:. = I. since 0 ~ 81 - 02 ~ ;rr when y '.:'.'.: 0. (See Fig. 155.) Now the value of the logarichm is the p1incipal value when 0 ~ H1 - H2 ~ ;rr. and we recall from Example 3 in Sec. J 02 that the upper half plane y > 0 is chen mapped onto the horizontal strip 0 < r < n in the u: plane. As already noted in that example. 1he mapping is shov.,.n \Vilh corresponding boundary points in Fig. 19 of Appendix 2. Indeed. it was that figure which suggested lrnnsformation (3) here. The segment of the x axis beiween:. = - I and:.= I. where 111 -02 = ;rr, is mapped onto 1he upper edge of the strip: and the res! of the x axis. where 01 - 112 = 0. is mapped onto the lower edge. The required analyticity and confonnality conditions arc evidently satisfied by transfonnation (3). A bounded hannonic function of 11 and v that is zero on the edge v = 0 of the strip and unity on the edge v = ;rr is clearly ( 4) I T = -r: ;rr it is harmonic since ii is the imaginary componen1 of 1hc entire function (I /;rr )w. Changing to x and y coordinates by means of the equation (5) 1 u1 =In -_-- :.-11 + i arg (:.-J) -_-- . .. + l .. +I SEC. 120 A RELATED PROilLEM 369 \\'C find lhal [ (::-1)(;.+I>] [x2 +y2 -l+i2yl l' - aro - aro . - ' ::: (:. +- I)(::+- I) - ' ::: (.r + I )2 + r2 . or r = an::lan ( ) 2.y) ) . x- + y- - I The range of lhc arclangcnl runclion here is from 0 lo if since arg -- = H1 - A2 ( ::-1) :. + I and 0 :S 81 - 82 :S if. Expression (4) now takes lhc fonn (6) I ( 2r ) T = - arclan ) ) if x- + y- - 1 (0 :S arclan t :s if). Since the function (4) is harmonic in lhc slripO < L~ < if and since transfonnalion (3)isanalylicinlhchalfplancy > 0.wcmayapplylhclhcorcminScc.116loconcludc lhal lhe function (6) is harmonic in lhal hair plane. The boundary conditions for the two harmonic runclions are the same on corresponding parts of lhc bound