Exercise [10.13]

Sameed Zahoor · 02 Apr 2008, 08:59

Differentiate as follows:
$\frac{\partial\alpha}{\partial{x}}=\frac{\partial\beta}{\partial{y}}\Rightarrow\frac{\partial}{\partial{x}}(\frac{\partial\alpha}{\partial{x}})=\frac{\partial}{\partial{x}}(\frac{\partial\beta}{\partial{y}})$
$\frac{\partial\alpha}{\partial{y}}=-\frac{\partial\beta}{\partial{x}}\Rightarrow\frac{\partial}{\partial{y}}(\frac{\partial\alpha}{\partial{y}})=\frac{\partial}{\partial{y}}(-\frac{\partial\beta}{\partial{x}})$
Adding both these equations we have:
$\nabla^2\alpha=0$ (because $\frac{\partial^{2}\beta}{\partial{x}\partial{y}}=\frac{\partial^{2}\beta}{\partial{y}\partial{x}})$
Similarly we can prove:
$\nabla^2\beta=0$ by differentiating the first equation wrt

and the second wrt to

.

Exercise [10.13]

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