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 Exercise [10.14] 
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Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Exercise [10.14]
We have,

\nabla^{2}\alpha=0and

\beta=\int{\frac{\partial{\alpha}}{\partial{x}}}dy

Now,

\frac{\partial{\beta}}{\partial{y}}=\frac{\partial}{\partial{y}}\int{\frac{\partial{\alpha}}{\partial{x}}}dy=\frac{\partial{\alpha}}{\partial{x}}

And

\frac{\partial{\beta}}{\partial{x}}=\frac{\partial}{\partial{x}}\int{\frac{\partial{\alpha}}{\partial{x}}}dy=\int{\frac{\partial^{2}\alpha}{\partial{x^{2}}}}dy=-\int{\frac{\partial^{2}\alpha}{\partial{y^{2}}}}dy=-\frac{\partial}{\partial{y}}\int{\frac{\partial{\alpha}}{\partial{y}}}dy=-\frac{\partial{\alpha}}{\partial{y}}


14 Jul 2008, 11:29
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