(adsbygoogle = window.adsbygoogle || []).push({}); Verify that the function U = (x^2 + y^2 + z^2)^(-1/2) is a solution of the three-dimensional Laplace equation Uxx + Uyy + Uzz = 0.

First I solved for the partial derivative Uxx,

Ux

= 2x(-1/2)(x^2 + y^2 + z^2)^(-3/2)

= -x(x^2 + y^2 + z^2)^(-3/2)

Uxx

= -(x^2 + y^2 + z^2)^(-3/2) + -x(2x)(-3/2)(x^2 + y^2 + z^2)^(-5/2)

= 3(x^2)(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2)

From there I saw that for finding the partial derivative Uyy & Uzz I would just have the change the variable being squared in the beginning of the function. So,

Uxx =3(x^2)(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2)

Uyy =3(y^2)(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2)

Uzz =3(z^2)(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2)

Because, through my observation, everything should just stay the same. However, when added together I get,

Uxx + Uyy + Uzz

= 3(3x^2 + 3y^2 + 3z^2)[(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2)]

As you can see this ridiculously long function is not zero which leads to my question. Why didn't everything cancel itself out? Wasn't that suppose to happen, everything cancels itself out so I can say that Laplace's Rule works and it all equals to zero? I'm confused.

Again, thank you so much for reviewing my question and not being deterred at the sheer sight of my derivation of the partial derivatives and algebra.

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# Homework Help: Verify that the function U is a solution for Laplace Equation.

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