Verify that the function 𝑢=1√𝑥2+𝑦2+𝑧2 is a solution of the three-dimensional Laplace equation 𝑢𝑥𝑥+𝑢𝑦𝑦+𝑢𝑧𝑧=0.

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Verify a Solution to the Three Dimensional Laplace Equation

Question: Verify that the function $u=\frac{1}{\sqrt{x^2+y^2+z^2}}$ is a solution of the three dimensional Laplace equation $u_{xx}+u_{yy}+u_{zz}=0$ .

All you have to do is take the derivative twice of each variable and then add them together.

$u_{xx}+u_{yy}+u_{zz}=0$

$\Rightarrow \frac{\partial^2}{\partial x^2}\frac{1}{\sqrt{x^2+y^2+z^2}}+\frac{\partial^2}{\partial y^2}\frac{1}{\sqrt{x^2+y^2+z^2}}+\frac{\partial^2}{\partial z^2}\frac{1}{\sqrt{x^2+y^2+z^2}}=0$

Since all three derivatives will be identical in operation, we only need to compute one.

$u_x=\frac{\partial}{\partial x}\frac{1}{\sqrt{x^2+y^2+z^2}}=\frac{\partial}{\partial x}(x^2+y^2+z^2)^{-\frac12}$

$=-\frac12(x^2+y^2+z^2)^{-\frac12-1}\frac{\partial}{\partial x}(x^2+y^2+z^2)=-\frac12(x^2+y^2+z^2)^{-\frac32}(2x)$

$=-\frac{x}{(x^2+y^2+z^2)^{\frac32}}$

$u_{xx}=-\frac{\partial}{\partial x}\frac{x}{(x^2+y^2+z^2)^{\frac32}}$

$=-\frac{(x^2+y^2+z^2)^{\frac32}\frac{\partial}{\partial x}x-x\frac{\partial}{\partial x}(x^2+y^2+z^2)^{\frac32}}{\left[(x^2+y^2+z^2)^{\frac32}\right]^2}$

$=-\frac{(x^2+y^2+z^2)^{\frac32}-x\frac32(x^2+y^2+z^2)^{\frac32-1}2x}{\left[(x^2+y^2+z^2)^{\frac32}\right]^2}$

$=\frac{3x^2(x^2+y^2+z^2)^{\frac32-1}-(x^2+y^2+z^2)^{\frac32}}{\left[(x^2+y^2+z^2)^{\frac32}\right]^2}$

$=\frac{(x^2+y^2+z^2)^{\frac32}\left[3x^2(x^2+y^2+z^2)^{-1}-1\right]}{\left[(x^2+y^2+z^2)^{\frac32}\right]^2}$

$=\frac{(x^2+y^2+z^2)^{\frac32}\left[3x^2(x^2+y^2+z^2)^{-1}-1\right]}{(x^2+y^2+z^2)^{\frac32}(x^2+y^2+z^2)^{\frac32}}$

$=\frac{3x^2(x^2+y^2+z^2)^{-1}-1}{(x^2+y^2+z^2)^{\frac32}}$

$=\frac{\frac{3x^2}{x^2+y^2+z^2}-\frac{x^2+y^2+z^2}{x^2+y^2+z^2}}{(x^2+y^2+z^2)^{\frac32}}$

$=\frac{\frac{3x^2-(x^2+y^2+z^2)}{x^2+y^2+z^2}}{\frac{(x^2+y^2+z^2)^{\frac32}}{1}}$

$=\frac{3x^2-(x^2+y^2+z^2)}{(x^2+y^2+z^2)(x^2+y^2+z^2)^{\frac32}}$

$=\frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{\frac52}}$

Now, since

$u_{xx}=\frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{\frac52}}$

We can easily see $u_{yy},u_{zz}$

$u_{yy}=\frac{2y^2-x^2-z^2}{(x^2+y^2+z^2)^{\frac52}},\qquad u_{zz}=\frac{2z^2-x^2-y^2}{(x^2+y^2+z^2)^{\frac52}}$

Then,

$u_{xx}+u_{yy}+u_{zz}=0$

$\Rightarrow \frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{\frac52}}+\frac{2y^2-x^2-z^2}{(x^2+y^2+z^2)^{\frac52}}+\frac{2z^2-x^2-y^2}{(x^2+y^2+z^2)^{\frac52}}=0$

$\Rightarrow 2x^2-y^2-z^2+2y^2-x^2-z^2+2z^2-x^2-y^2=0$

$\Rightarrow 0=0$

Thus, $u=\frac{1}{\sqrt{x^2+y^2+z^2}}$ is a solution of the three dimensional Laplace equation $u_{xx}+u_{yy}+u_{zz}=0$ .

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Verify a Solution to the Three Dimensional Laplace Equation

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