Higher order derivatives?

[KLS2111]

The problem is verify that the function u= 1/(x^2 +y^2 +z^2) satisfies the eqn. u xx+u yy +u zz= 0

Here is what I have done, but I must have messed up the dervatives somewhere:
ux= -.5(x^2 + y^2 +z^2)^(-3/2) (2x +y^2 + z^2)
uy= -.5(x^2 + y^2 +z^2)^(-3/2) (2y +x^2 + z^2)
uz= -.5(x^2 + y^2 +z^2)^(-3/2) (2x +y^2 + x^2)

uxx= .75(x^2 +y^2 +z ^2)^(-5/2) (2x+y^2+z^2) * 2x+(x^2+y^2+z^2)^(-3/2) * (2+y^2+z^2)
uyy= .75(x^2 +y^2 +z ^2)^(-5/2) (2y+x^2+z^2) * 2y+(x^2+y^2+z^2)^(-3/2) * (2+x^2+z^2)
uxx= .75(x^2 +y^2 +z ^2)^(-5/2) (2x+y^2+x^2) * 2z+(x^2+y^2+z^2)^(-3/2) * (2+y^2+x^2)

Needless to say, I am not getting zero when adding them together..Any help on where I went wrong would be greatly appreciated!

 

[soroban]

Hello, KLS2111!

I assume that there is a square root in the function . . . You left it out.

\(\displaystyle \text{Verify that the function: }\:u \:=\:\frac{1}{\sqrt{x^2 +y^2 +z^2}}\,\text{ satisfies the equation: }\: u_{xx} + u_{yy} + u_{zz} \:=\: 0\)

You did mess up the derivatives . . .

We have: .\(\displaystyle u \;=\;(x^2+y^2+z^2)^{-\frac{1}{2}}\)


. . \(\displaystyle \begin{array}{cccccccccc}u_x &=& -\frac{1}{2}(x^2 + y^2 +z^2)^{-\frac{3}{2}}(2x) &=& -x(x^2+y^2+z^2)^{-\frac{3}{2}} \\ \\[-3mm] u_y &= & -\frac{1}{2}(x^2 + y^2 +z^2)^{-\frac{3}{2}}(2y) &=& -y(x^2+y^2+z^2)^{-\frac{3}{2}} \\ \\[-3mm] u_z &=& -\frac{1}{2}(x^2 + y^2 +z^2)^{-\frac{3}{2}}(2z) &=& -x(x^2+y^2+z^2)^{-\frac{3}{2}} \end{array}\)

. . \(\displaystyle \begin{array}{ccccccc}u_{xx} &=& -1\cdot(x^2+y^2+z^2)^{-\frac{3}{2}} - x(\text{-}\frac{3}{2})(x^2+y^2+z^2)^{-\frac{5}{2}}(2x) &=& -(x^2+y^2+z^2)^{-\frac{3}{2}} + 3x^2(x^2+y^2+z^2)^{-\frac{5}{2}} \\ u_{yy} &=& -1\cdot(x^2+y^2+z^2)^{-\frac{3}{2}} - y(\text{-}\frac{3}{2})(x^2+y^2+z^2)^{-\frac{5}{2}}(2y) &=& -(x^2+y^2+z^2)^{-\frac{3}{2}} + 3y^2(x^2+y^2+z^2)^{-\frac{5}{2}} \\ u_{zz} &=& -1\cdot(x^2+y^2+z^2)^{-\frac{3}{2}} - z(\text{-}\frac{3}{2})(x^2+y^2+z^2)^{-\frac{5}{2}}(2z) &=& -(x^2+y^2+z^2)^{-\frac{3}{2}} + 3z^2(x^2+y^2+z^2)^{-\frac{5}{2}} \\ \end{array}\)


\(\displaystyle \text{Then: }\:u_{xx} + u_{yy} + u_{zz} \;=\;\begin{Bmatrix}-(x^2+y^2+z^2)^{-\frac{3}{2}} + 3x^2(x^2+y^2+z^2)^{-\frac{5}{2}} \\ -(x^2+y^2+z^2)^{-\frac{3}{2}} + 3y^2(x^2+y^2+z^2)^{-\frac{5}{2}} \\ -(x^2+y^2+z^2)^{-\frac{3}{2}} + 3x^2(x^2+y^2+z^2)^{-\frac{5}{2}} \\ \end{Bmatrix}\)

. . . . . . . . . . . . . . . . \(\displaystyle =\;-3(x^2+y^2+z^2)^{-\frac{3}{2}} \;+\; 3x^2(x^2+y^2+z^2)^{-\frac{5}{2}} \;+\; 3y^2(x^2+y^2+z^2)^{-\frac{5}{2}} \;+\; 3z^2(x^2+y^2+z^2)^{-\frac{5}{2}}\)

. . . . . . . . . . . . . . . . \(\displaystyle =\;-3(x^2+y^2+z^2)^{-\frac{3}{2}} + (3x^2+3y^2+3z^2)(x^2+y^2+z^2)^{-\frac{5}{2}}\)

. . . . . . . . . . . . . . . . \(\displaystyle =\;-3(x^2+y^2+z^2)^{-\frac{3}{2}} + 3(x^2+ y^2 + z^2)(x^2+y^2+z^2)^{-\frac{5}{2}}\)

. . . . . . . . . . . . . . . . \(\displaystyle =\;-3(x^2+y^2+z^2)^{-\frac{3}{2}} + 3(x^2+y^2+z^2)^{-\frac{3}{2}}\)

. . . . . . . . . . . . . . . . \(\displaystyle = \qquad 0\)

 

[KLS2111]

Oops, you are right I did leave out the square root. Thank you for catching that! I see what I did wrong now. At some point in class we were told to write the variables that we weren't taking the derivative with respect to, however since theses would then be constants, their derivatives would be 0. Thank you for your help!