VC3.38

From Exampleproblems

Note that in the xy-plane, $r^2=x^2+y^2\,$ so that

$\frac{\partial r}{\partial x}=\frac{x}{r}\,$ --(1)

Also,for xy-plane $\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\,$ --(2)

Since $u=\log r\,$ ,using (2),we have

$\nabla^2 u=\nabla^2\log r=[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}]\log r=\frac{\partial^2}{\partial x^2}(\log r)+\frac{\partial^2}{\partial x^2}(\log r)\,$ --(3)

Now, $\frac{\partial}{\partial}\log r=\frac{1}{r}\frac{\partial r}{\partial x}=\frac{1}{r}\frac{x}{r}\,$ ,by using(1)

Therefore, $\frac{\partial^2}{\partial x^2}\log r=\frac{\partial}{\partial x}[\frac{\partial}{\partial x}[\frac{x}{r^2}]=\frac{1}{r^2}-\frac{2x}{r^4}\,$ --(4)

Similarly, $\frac{\partial^2}{\partial y^2}\log r=\frac{1}{r^2}-\frac{2y^2}{r^4}\,$ --(5)

Adding (4) and (5), we have

$\frac{\partial^2}{\partial x^2}\log r+\frac{\partial^2}{\partial y^2}\log r=\frac{2}{r^2}-\frac{2}{r^4}(x^2+y^2)=\frac{2}{r^2}-\frac{2}{r^4}(r^2)=\frac{2}{r^2}-\frac{2}{r^2}=0\,$

or $\nabla^2 u=0\,$ by using (3)

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