Question

If \[{u^2} = \dfrac{1}{{\left( {{x^2} + {y^2} + {z^2}} \right)}}\], then show that \[\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {z^2}}} = 0\].

Answer 1

Hint: First, differentiate the given equation partially with respect to \[x\]. Again, differentiate the given equation partially with respect to \[x\] and calculate the second derivative. Follow these steps to calculate the second derivative of the given equation with respect to \[y\], and \[z\]. In the end, add the values of all three second-order derivatives and simplify the equation to reach the required answer.

Formula used:
\[\dfrac{\partial }{{\partial x}}\left( {\dfrac{1}{{{x^n}}}} \right) = - n{x^{ - n - 1}}\]
Product rule of differentiation: \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]

Complete step by step solution:
The given equation is \[{u^2} = \dfrac{1}{{\left( {{x^2} + {y^2} + {z^2}} \right)}}\].
To prove: \[\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {z^2}}} = 0\]

Take square root on the both sides of the above equation.
We get,
\[u = \dfrac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\]

Let’s differentiate the above equation partially with respect to the variable \[x\].
\[\dfrac{\partial }{{\partial x}}\left( u \right) = \dfrac{\partial }{{\partial x}}\left( {\dfrac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right)\]
Apply the exponent rule.
\[\dfrac{{\partial u}}{{\partial x}} = \dfrac{\partial }{{\partial x}}\left( {{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{ - \dfrac{1}{2}}}} \right)\]
Apply the chain rule and \[\dfrac{\partial }{{\partial x}}\left( {\dfrac{1}{{{x^n}}}} \right) = - n{x^{ - n - 1}}\].
\[\dfrac{{\partial u}}{{\partial x}} = \dfrac{{ - 1}}{{2{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}}\dfrac{\partial }{{\partial x}}\left( {{x^2} + {y^2} + {z^2}} \right)\]
\[ \Rightarrow \dfrac{{\partial u}}{{\partial x}} = \dfrac{{ - 1}}{2}\left( {\dfrac{1}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}}} \right)\left( {2x} \right)\]
\[ \Rightarrow \dfrac{{\partial u}}{{\partial x}} = - {\left( {{u^2}} \right)^{\dfrac{3}{2}}}\left( x \right)\]
\[ \Rightarrow \dfrac{{\partial u}}{{\partial x}} = - {u^3}x\]
Again, differentiate the above differential equation partially with respect to the variable \[x\].
\[\dfrac{\partial }{{\partial x}}\left( {\dfrac{{\partial u}}{{\partial x}}} \right) = \dfrac{\partial }{{\partial x}}\left( { - {u^3}x} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}u}}{{\partial {x^2}}} = - \dfrac{\partial }{{\partial x}}\left( {{u^3}x} \right)\]
Apply the product rule.
\[\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} = - \left( {{u^3}\dfrac{\partial }{{\partial x}}\left( x \right) + x\dfrac{\partial }{{\partial x}}\left( {{u^3}} \right)} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}u}}{{\partial {x^2}}} = - \left( {{u^3} + x\left( {3{u^2}} \right)\dfrac{{\partial u}}{{\partial x}}} \right)\]
Substitute \[\dfrac{{\partial u}}{{\partial x}} = - {u^3}x\] in the above equation.
\[\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} = - \left( {{u^3} + x\left( {3{u^2}} \right)\left( { - {u^3}x} \right)} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}u}}{{\partial {x^2}}} = - \left( {{u^3} - 3{x^2}{u^5}} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}u}}{{\partial {x^2}}} = 3{x^2}{u^5} - {u^3}\] \[.....\left( 1 \right)\]

Similarly, double differentiate the given equation with respect to the variables \[y\] and \[z\].
We get,
\[\dfrac{{{\partial ^2}u}}{{\partial {y^2}}} = 3{y^2}{u^5} - {u^3}\] \[.....\left( 2 \right)\]
And \[\dfrac{{{\partial ^2}u}}{{\partial {z^2}}} = 3{z^2}{u^5} - {u^3}\] \[.....\left( 3 \right)\]

Now add the equations \[\left( 1 \right)\], \[\left( 2 \right)\] and \[\left( 3 \right)\].
We get,
\[\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {z^2}}} = 3{x^2}{u^5} - {u^3} + 3{y^2}{u^5} - {u^3} + 3{z^2}{u^5} - {u^3}\]
\[ \Rightarrow \dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {z^2}}} = 3{u^5}\left( {{x^2} + {y^2} + {z^2}} \right) - 3{u^3}\]
Resubstitute the value \[\left( {{x^2} + {y^2} + {z^2}} \right) = {u^{ - 2}}\] in the above equation.
\[\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {z^2}}} = 3{u^5}\left( {{u^{ - 2}}} \right) - 3{u^3}\]
\[ \Rightarrow \dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {z^2}}} = 3{u^3} - 3{u^3}\]
\[ \Rightarrow \dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \dfrac{{{\partial ^2}u}}{{\partial {z^2}}} = 0\]
Hence, proved.

Note: Students often get confused and directly apply the product rule without using the chain rule of differentiation.
The chain rule is used to take derivatives of composites of functions and this happens by chaining together their derivatives.
Chain rule: If \[F\left( x \right) = f\left( {g\left( x \right)} \right)\], then \[F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\]