ROUGH DRAFT authorea.com/112566
Main Data History
Export
Show Index Toggle 0 comments
  •  Quick Edit
  • Problem 6.6

    [6.6] Show that, if \(f\) is holomorphic and nonzero in \(G\), then \(ln|f(x,y)|\) is harmonic in G.

    Assume \(f=u+iv\) is holomorphic and nonzero in \(G\). Since \(f\) is holomorphic, its real and imaginary parts are harmonic and therefore infinitely differentiable by Proposition 6.4 and Corollary 6.9. The first derivative of \(ln|f(x,y)|\) with respect to \(x\) is \[\begin{aligned} \\ (ln|f(x,y)|)_x &= [ln(\sqrt{u^{2}+v^{2}})]_x\\ &= \frac{1}{2}[ln(u^{2}+v^{2})]_x\\ &=\frac{2uu_x+2vv_x}{2(u^{2}+v^{2})}\\ &=\frac{uu_x+vv_x}{u^{2}+v^{2}}\end{aligned}\] And the second derivative with respect to \(x\) is \[\begin{aligned} (ln|f(x,y)|)_{xx}&= (\frac{uu_x+vv_x}{u^{2}+v^{2}})_x\\ &= \frac{(uu_x + uu_{xx} +v_xv_x + vv_{xx})(u^{2}+v^{2})-(uu_x + vv_x)(2uu_x+2vv_x)}{(u^{2}+v^{2})^{2}}\end{aligned}\] Similarly the first and second derivatives of \(ln|f(x,y)|\) with respect to \(y\) are \[\begin{aligned} (ln|f(x,y)|)_y &= [ln(\sqrt{u^{2}+v^{2}})]_y\\ &= \frac{1}{2}[ln(u^{2}+v^{2})]_y\\ &=\frac{2uu_y+2vv_y}{2(u^{2}+v^{2})}\\ &=\frac{uu_y+vv_y}{u^{2}+v^{2}}\end{aligned}\] and \[\begin{aligned} (ln|f(x,y)|)_{yy}&= (\frac{uu_y+vv_y}{u^{2}+v^{2}})_y\\ &= \frac{(uu_y + uu_{yy} +v_yv_y + vv_{yy})(u^{2}+v^{2})-(uu_y + vv_y)(2uu_y+2vv_y)}{(u^{2}+v^{2})^{2}}\end{aligned}\] Since \(f\) is nonzero and \(u\) and \(v\) are harmonic in \(G\), the second partials of \(ln|f(x,y)|\) are defined on \(G\). Now consider the Laplace equation, \[\begin{aligned} (ln|f(x,y)|)_{xx}+(ln|f(x,y)|)_{yy} &=\frac{1}{(u^{2}+v^{2})^{2}}[(uu_x + uu_{xx} +v_xv_x + vv_{xx})(u^{2}+v^{2})-(uu_x + vv_x)(2uu_x+2vv_x)\\ &+ (uu_y + uu_{yy} +v_yv_y + vv_{yy})(u^{2}+v^{2})-(uu_y + vv_y)(2uu_y+2vv_y)]\end{aligned}\] We will show that the numerator expression will be equal to zero and thus the Laplace equation will equal zero.\[\begin{aligned} (uu_x + uu_{xx} +v_xv_x + vv_{xx})(u^{2}+v^{2}) &-(uu_x + vv_x)(2uu_x+2vv_x)+(uu_y + uu_{yy} +v_yv_y + vv_{yy})(u^{2}+v^{2})\\ &-(uu_y + vv_y)(2uu_y+2vv_y)\\ &=u^{2}u^{2}_x+u^{3}u_{xx}+u^{2}v^{2}_x+u^{2}vv_{xx}+v^{2}u^{2}_x+v^{2}uu_{xx}+v^{2}v^{2}_x+v^{3}v_{xx}\\ &-2u^{2}u^{2}_x-2uvv_xu_x-2uvu_xv_x-2v^{2}v^{2}_x+u^{2}u^{2}_y+u^{3}u_{yy}+u^{2}v^{2}_y+u^{2}vv_{yy}+\\ &v^{2}u^{2}_y+v^{2}uu_{yy}+v^{2}v^{2}_y+v^{3}v_{yy}-2u^{2}u^{2}_y-2uvv_yu_y-2uvu_yv_y-2v^{2}v^{2}_y\\ &-2u^{2}u^{2}_y-2uvv_yu_y-2uvu_yv_y-2v^{2}v^{2}_y\end{aligned}\] But since \(u\) and \(v\) are harmonic, \(u_{xx}+u_{yy}=0\) and \(v_{xx}+v_{yy}=0\). Using this fact and combining like terms gives us, \[\begin{aligned} -u^{2}u^{2}_x+u^{2}v^{2}_x+v^{2}u^{2}_x-v^{2}v^{2}_x-4uvu_xv_x-u^{2}u^{2}_y+u^{2}v^{2}_y+v^{2}u^{2}_y-v^{2}v^{2}_y-4uvu_yv_y\end{aligned}\] And now we can use the Cauchy-Riemann equations to get \[\begin{aligned} -u^{2}u^{2}_x+u^{2}v^{2}_x+v^{2}u^{2}_x-v^{2}v^{2}_x-4uvv_yv_x-u^{2}v^{2}_x+u^{2}u^{2}_x+v^{2}v^{2}_x-v^{2}u^{2}_x+4uv_xv_y\end{aligned}\] Which simplifies to 0 and thus \((ln|f(x,y)|)_{xx}+(ln|f(x,y)|)_{yy}=0\). Since \(ln|f(x,y)|\) has continuous second partials on G and the Laplace equation is satisfied, \(ln|f(x,y)|\) is harmonic on \(G\).

    [Someone else is editing this]

    You are editing this file