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Problem of the Week #12 - June 18th, 2012

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW!! Here's this week's problem.

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Problem: We say that a function $f(x,y)$ is harmonic if it satisfies the Laplace equation $\dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 0$. Suppose that $\phi(x,y)$ and $\psi(x,y)$ are harmonic functions. Let $u$ and $v$ be functions defined as follows:
\[u(x,y) = \phi_x\phi_y+\psi_x\psi_y\quad\text{and}\quad v(x,y) = \tfrac{1}{2}(\phi_x^2+\psi_x^2-\phi_y^2 - \psi_y^2).\]
Show that $u(x,y)$ and $v(x,y)$ satisfy the Cauchy-Riemann equations

\[\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\quad \text{and}\quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.\]

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by hmmm16 and Sudharaka.

Here's Sudharaka's solution:

\[u(x,y) = \phi_x\phi_y+\psi_x\psi_y\]

\[\Rightarrow\frac{\partial}{\partial x}u(x,y)=\phi_{xx}\phi_y+\phi_{yx}\phi_x+\psi_{xx}\psi_y+\psi_{x}\psi_{yx}~~~~~~~~~~(1)\]


\[v(x,y) = \tfrac{1}{2}(\phi_x^2+\psi_x^2-\phi_y^2 - \psi_y^2)\]


\[\Rightarrow\frac{\partial}{\partial y}v(x,y) = \phi_{x}\phi_{xy}+\psi_{x}\psi_{xy}-\phi_{y}\phi_{yy}-\psi_{y}\psi_{yy}~~~~~~~~~~~(2)\]


We shall assume that \(\phi\) and \(\psi\) have commutative second partial derivatives. Then,


\[\phi_{xy}=\phi_{yx}\mbox{ and }\psi_{xy}=\psi_{yx}\]


By (2),


\[\frac{\partial}{\partial y}v(x,y) = \phi_{x}\phi_{yx}+\psi_{x}\psi_{yx}-\phi_{y}\phi_{yy}-\psi_{y}\psi_{yy}~~~~~~~~~~~~~~(3)\]


Since \(\phi\) and \(\psi\) are harmonic functions,


\[\phi_{xx}=-\phi_{yy}\mbox{ and }\psi_{xx}=-\psi_{yy}\]


By (3),


\[\frac{\partial}{\partial y}v(x,y) = \phi_{x}\phi_{yx}+\psi_{x}\psi_{yx}+\phi_{y}\phi_{xx}+\psi_{y}\psi_{xx}~~~~~~~~~~~~(4)\]


By (1) and (4),


\[\frac{\partial}{\partial x}u(x,y)=\frac{\partial}{\partial y}v(x,y)\]


Similarly,


\[\frac{\partial}{\partial y}u(x,y)=\phi_{x}\phi_{yy}+\phi_{xy}\phi_y+\psi_{x}\psi_{yy}+\psi_{xy}\psi_{y}~~~~~~~~~~(5)\]


\[-\frac{\partial}{\partial x}v(x,y) = -\phi_{x}\phi_{xx}-\psi_{x}\psi_{xx}+\phi_{y}\phi_{yx}+\psi_{y}\psi_{yx}\]


By our previous assumption,


\[-\frac{\partial}{\partial x}v(x,y) = -\phi_{x}\phi_{xx}-\psi_{x}\psi_{xx}+\phi_{y}\phi_{xy}+\psi_{y}\psi_{xy}~~~~~~~~~~~~~(6)\]


Since \(\phi\) and \(\psi\) are harmonic functions,


\[\phi_{xx}=-\phi_{yy}\mbox{ and }\psi_{xx}=-\psi_{yy}\]


By (6),


\[-\frac{\partial}{\partial x}v(x,y) = \phi_{x}\phi_{yy}+\psi_{x}\psi_{yy}+\phi_{y}\phi_{xy}+\psi_{y}\psi_{xy}~~~~~~~~~~~(7)\]


By (5) and (7),


\[\frac{\partial}{\partial y}u(x,y)=-\frac{\partial}{\partial x}v(x,y)\]


Q.E.D.
 
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