VC3.30

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$\nabla^2 u=[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}]u\,$

= $[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}]\,$

Now, $\frac{\partial u}{\partial x}=\frac{\partial}{\partial x}(x^2-y^2+4z)=2x\,$ implies $\frac{\partial^2 u}{\partial x^2}=2\,$

$\frac{\partial u}{\partial y}=\frac{\partial}{\partial y}(x^2-y^2+4z)=-2y\,$ implies $\frac{\partial^2 u}{\partial y^2}=-2\,$

$\frac{\partial u}{\partial z}=\frac{\partial}{\partial z}(x^2-y^2+4z)=4\,$ implies $\frac{\partial^2 u}{\partial x^2}=0\,$

Therefore, $\nabla^2 u=2-2+0=0\,$

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