By H. Begehr

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**Additional info for Complex Analytic Methods for Partial Diff. Eqns. - An Intro. Text**

**Example text**

Then 9(C, z) log ll - z( l Applying on { ICI = p = 1) 87- 8p=pa_=CaC+a-=2Re(y( we get area z( 9((, z) = 2Re C C (log 11 - z(1 - log IC - zI) a C -Re 1(z( +C- z - )7=-Re\C-i+ C-z}C 1 _ Iz12 IC-z12 Therefore in D any harmonic function being continuous in D may be represented by _ 2 U(Z) = 2n 1 u(()-( zl2darg( . 1(1=' Complex Analytic Methods for Partial Differential Equations 32 This is the POISSON formula with the POISSON kernel S+z 1 - Iz12 (-z _ 1 - r2 1-2rcos(19-V)+r2' IC-z12 which for an arbitrary circle IC - al < p is _ 2 _ r2 (z _ iB z=re ,rp-e z2a=p2-2rpcos(99-gyp)+r2' z=a+re"°,,=a+pe' Red Solving the CAUCHY-RIEMANN equations with a given harmonic function the conju- gate harmonic function can locally be calculated.

Forgiven A, (P E C°(8D; ll" ), 0 < a < 1, with A(() # 0 on OD find an analytic function w in D such that Re{a(()w(())= ,p((), (E 8D . Remark. With A = p + iv, w = u + iv we have Re{aw} = leu + vv = cp on 8D . While the DIRICHLET problem, coinciding with the RIEMANN-HILBEItT problem if A = 1 just prescribes the real part of the analytic function on the boundary in the RIEMANN-HILBERT boundary condition a linear combination of the real and imaginary parts of the function looked for is given. Although this problem is more general Complex Analytic Methods for Partial Differential Equations 46 the solution will be found by reducing it to the DIRICHLET problem.

This follows from the argument principle, in the case of an unbounded domain D applied to D n {jzI < R} for sufficiently large R and then passing with R to infinity. Let r be a smooth simply closed curve and G, g E Riemann problem. C°(r; 0), 0 < a < 1, with G(() # 0 on r. Find analytic functions 0+ in D+ and in D' such that 0+(() = G(()O (() + g((), (E r . Remark. This problem is sometimes called the problem of linear conjugacy. Theorem 14. For 0 < rc the homogeneous RIEMANN boundary value problem (g = 0) has K + 1 linearly independent solutions 00 (z) = zke'+(s), z E D+ , 0