MathJax example
01. If \(u\) and \(v\) are harmonic functions show that \(\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)+i\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\) is analytic.
Ans:
Let \(P=\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)\) and \(Q=\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\).
Now we have to show that \(P+iQ\) is analytic, i.e., we have to show that \(P\) and \(Q\) will satisfies
CR-equations
\(\frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)-\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\)
\(\Longrightarrow \frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}=\frac{\partial^2 u}{\partial x \partial y}-\frac{\partial^2 v}{\partial x^2}-\frac{\partial^2 u}{\partial y \partial x}-\frac{\partial^2 v}{\partial y^2}=-\left(\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}\right)=0\)
[because \(\frac{\partial^2 v}{\partial x \partial y}=\frac{\partial^2 v}{\partial y \partial x}\) and since \(u\) and \(v\) are harmonic, \(u\) and \(v\) satisfies
Laplace equations]. Hence \(\frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}=0\)\(\Longrightarrow \frac{\partial P}{\partial x}=\frac{\partial Q}{\partial y}\)
Similarly, consider
\(\frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}=\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)+\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\)
\(\Longrightarrow \frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}=\frac{\partial^2 u}{\partial y^2}-\frac{\partial^2 v}{\partial y \partial x}+\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 v}{\partial x \partial y}=\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)=0\)
[because \(\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial^2 u}{\partial y \partial x}\) and since \(u\) and \(v\) are harmonic, \(u\) and \(v\) satisfies
Laplace equations]. Hence \(\frac{\partial Q}{\partial x}+\frac{\partial Q}{\partial y}=0\)
\(\Longrightarrow \frac{\partial Q}{\partial x}=-\frac{\partial Q}{\partial y}\)
Hence \(P+iQ\) is analytic.
02. If \(f(z)\) is analytic, show that \(\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)|f(z)|^2=4|f'(z)|^2\) is analytic.
Ans:
Let \(f(z)=u+iv\) be analytic. \(\Longrightarrow |f(z)|=\sqrt{u^2+v^2}\Longrightarrow |f(z)|^2=u^2+v^2=\phi\)<\br>
Hence, we need to show that \(\phi_{xx}+\phi_{yy}=4|f'(z)|^2\).<\br>
$$\phi=u^2+v^2\Longrightarrow \phi_x=2uu_x+2vv_x \Longrightarrow \phi_{xx}=2[uu_{xx}+u_x^2+vv_{xx}+v_x^2]\rightarrow (1)$$
$$\phi=u^2+v^2\Longrightarrow \phi_y=2uu_y+2vv_y \Longrightarrow \phi_{yy}=2[uu_{yy}+u_y^2+vv_{yy}+v_y^2]\rightarrow (2)$$
Now add (1) and (2)
$$\phi_{xx}+\phi_{yy}=2[u(u_{xx}+u_{yy})+v(v_{xx}+v_{yy})+u_x^2+v_x^2+u_y^2+v_y^2]\rightarrow (3)$$
Since \(f(z)\) is analytic, \(u\) and \(v\) are harmonic \(\Longrightarrow u_{xx}+u_{yy}=0,v_{xx}+v_{yy}=0\), also using
CR-equations in (3), we get.
$$\phi_{xx}+\phi_{yy}=2[2u_x^2+2v_x^2]=4[u_x^2+v_x^2]\rightarrow (4)$$
But \(f'(z)=u_x+iv_x \Longrightarrow |f'(z)|=u_x^2+v_x^2\), hence we can conclude that
$$\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)|f(z)|^2=4|f'(z)|^2$$
03. If \(f(z)\) is a regular analytic, show that \(\left(\frac{\partial}{\partial x}|f(z)|\right)^2+\left(\frac{\partial}{\partial y}|f(z)|\right)^2=|f'(z)|^2\)
Ans:
Let \(f(z)=u+iv\) be analytic. \(\Longrightarrow |f(z)|=\sqrt{u^2+v^2}=\phi\Longrightarrow |f(z)|^2=u^2+v^2=\phi^2\)<\br>
Consider \(\phi^2=u^2+v^2\)
$$\Longrightarrow 2\phi \phi_x=2uu_x+2vv_x\Longrightarrow \phi \phi_x=uu_x+vv_x \rightarrow (1)$$
$$\text{similarly} 2\phi \phi_y=2uu_y+2vv_y\Longrightarrow \phi \phi_y=uu_y+vv_y\rightarrow (2)$$
Squaring and adding both (1) and (2), we will get
$$\phi^2 (\phi_x^2+\phi_y^2)=(uu_x+vv_x)^2+(uu_y+vv_y)^2$$
$$=(u^2 u_x^2+v^2 v_x^2+2uvu_x v_x)+(u^2u_y^2+v^2v_y^2+2uvu_y v_y)$$
Since \(f(z)\) is analytic, \(u\) and \(v\) satisfies
CR-equations.
$$\Longrightarrow \phi^2 (\phi_x^2+\phi_y^2)=u^2(u_x^2+v_x^2)+v^2(u_x^2+v_x^2)=(u^2+v^2)(u_x^2+v_x^2)=\phi^2 (u_x^2+v_x^2)$$
Hence, we get
$$\phi_x^2+\phi_y^2=u_x^2+v_x^2$$
Also we have \(f(z)=u+iv\Longrightarrow f'(z)=u_x+iv_x \Longrightarrow |f'(z)|=\sqrt{u_x^2+v_x^2}\), hence above equation will become
$$\phi_x^2+\phi_y^2=|f'(z)|^2$$
No comments:
Post a Comment