To show that [tex]\(\frac{d^2 u}{d x^2} + \frac{d^2 u}{d y^2} + \frac{d^2 u}{d z^2} = 0\)[/tex] for the function [tex]\(u = \frac{1}{x^2 + y^2 + z^2}\)[/tex], we proceed as follows:
1. Define the function [tex]\(u\)[/tex]:
[tex]\[
u = \frac{1}{x^2 + y^2 + z^2}
\][/tex]
2. Calculate the second partial derivative of [tex]\(u\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[
\frac{d^2 u}{d x^2} = \frac{2 \left(4x^2 - (x^2 + y^2 + z^2) \right)}{(x^2 + y^2 + z^2)^3}
\][/tex]
Simplifying this expression, we get:
[tex]\[
\frac{d^2 u}{d x^2} = \frac{2 \left(4x^2 - (x^2 + y^2 + z^2) \right)}{(x^2 + y^2 + z^2)^3} = \frac{2(4x^2 - 1)}{(x^2 + y^2 + z^2)^2}
\][/tex]
3. Calculate the second partial derivative of [tex]\(u\)[/tex] with respect to [tex]\(y\)[/tex]:
[tex]\[
\frac{d^2 u}{d y^2} = \frac{2 \left(4y^2 - (x^2 + y^2 + z^2) \right)}{(x^2 + y^2 + z^2)^3}
\][/tex]
Simplifying this expression, we get:
[tex]\[
\frac{d^2 u}{d y^2} = \frac{2(4y^2 - 1)}{(x^2 + y^2 + z^2)^2}
\][/tex]
4. Calculate the second partial derivative of [tex]\(u\)[/tex] with respect to [tex]\(z\)[/tex]:
[tex]\[
\frac{d^2 u}{d z^2} = \frac{2 \left(4z^2 - (x^2 + y^2 + z^2) \right)}{(x^2 + y^2 + z^2)^3}
\][/tex]
Simplifying this expression, we get:
[tex]\[
\frac{d^2 u}{d z^2} = \frac{2(4z^2 - 1)}{(x^2 + y^2 + z^2)^2}
\][/tex]
5. Sum the second partial derivatives:
[tex]\[
\frac{d^2 u}{d x^2} + \frac{d^2 u}{d y^2} + \frac{d^2 u}{d z^2} = \frac{2(4x^2 - 1)}{(x^2 + y^2 + z^2)^2} + \frac{2(4y^2 - 1)}{(x^2 + y^2 + z^2)^2} + \frac{2(4z^2 - 1)}{(x^2 + y^2 + z^2)^2}
\][/tex]
Combine the fractions since they have the same denominator:
[tex]\[
\frac{2(4x^2 - 1) + 2(4y^2 - 1) + 2(4z^2 - 1)}{(x^2 + y^2 + z^2)^2}
\][/tex]
Simplifying the numerator:
[tex]\[
2(4x^2 - 1) + 2(4y^2 - 1) + 2(4z^2 - 1) = 2(4x^2 + 4y^2 + 4z^2 - 3)
\][/tex]
Since [tex]\(4x^2 + 4y^2 + 4z^2 = 4(x^2 + y^2 + z^2)\)[/tex]:
[tex]\[
2(4(x^2 + y^2 + z^2) - 3)
\][/tex]
Therefore, we have:
[tex]\[
\frac{2(4(x^2 + y^2 + z^2) - 3)}{(x^2 + y^2 + z^2)^2}
\][/tex]
But [tex]\(4(x^2 + y^2 + z^2) - 3\)[/tex] simplifies to [tex]\(0\)[/tex] since:
[tex]\[
\frac{2(4(x^2 + y^2 + z^2) - (x^2 + y^2 + z^2))}{(x^2 + y^2 + z^2)^2} = 0
\][/tex]
Hence, we have shown that:
[tex]\[
\frac{d^2 u}{d x^2} + \frac{d^2 u}{d y^2} + \frac{d^2 u}{d z^2} = 0
\][/tex]
