122Chapter 7The Fourier Transform and its ApplicationsSolutions to Exercises 7.51.To solve the Dirichlet problem in the upper half-plane with the given boundaryfunction, we use formula (5). The solution is given byu(x,y)=yπ∞-∞f(s)(x-s)2+y2ds=50yπ1-1ds(x-s)2+y2=50πtan-11 +xy+ tan-11-xy,where we have used Example 1 to evaluate the definite integral.5.Appealing to (4) in Section 7.5, withy=y1, y2,y1+y2, we findF(Py1)(w) =e-y1|w|,F(Py2)(w) =e-y2|w|,F(Py1+y2)(w) =e-(y1+y2)|w|.HenceF(Py1)(w)· F(Py2)(w) =e-y1|w|e-y2|w|=e-(y1+y2)|w|=F(Py1+y2)(w).ButF(Py1)(w)· F(Py2)(w) =F(Py1*Py2)(w),HenceF(Py1+y2)(w) =F(Py1*Py2)(w);and soPy1+y2=Py1*Py2.9.Modify the solution of Example 1(a) to obtain that, in the present case, thesolution isu(x,y) =T0πtan-1a+xy+ tan-1a-xy.To find the isotherms, we must determine the points (x,y) such thatu(x,y) =T.As in the solution of Example 1(b), these points satisfyx2+y-acot(πTT0)2=acsc(πTT0)2.Hence the points belong to the arc in the upper half-plane of the circle with center(0,acot(πTT0)) and radiusacsc(πTT0). The isotherm corresponding toT=T02is thearc of the circlex2+y-acot(π2)2=acsc(π2)2,orx2+y2=a2.Thus the isotherm in this case is the upper semi-circle of radius
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