Find the complete integral of the equation (∂z/∂x_1)(∂z/∂x_2)(∂z/∂x_3)=z³x_1x_2x_3?

Jun 19, 2018

One solution is:

$z \left(\boldsymbol{x}\right) = \text{ const } \cdot \setminus {e}^{\frac{1}{2} \left({C}_{1} {x}_{1}^{2} + {C}_{2} {x}_{2}^{2} + {C}_{3} {x}_{3}^{2}\right)}$

Explanation:

Symmetry suggests that separation of variables should result in a solution:

So let $z \left(\boldsymbol{x}\right) = {X}_{1} \left({x}_{1}\right) \setminus {X}_{2} \left({x}_{2}\right) \setminus {X}_{3} \left({x}_{3}\right)$

$\therefore {\partial}_{{x}_{1}} \setminus z = {X}_{1}^{'} \setminus {X}_{2} \setminus {X}_{3}$

....with corresponding results for ${\partial}_{{x}_{2}} \setminus z$ and ${\partial}_{{x}_{3}} \setminus z$.

• X_1^' X_2^' X_3^' (X_1 X_2 X_3)^2 = (X_1 X_2 X_3)^3 \ x_1 x_2 x_3

• ${X}_{1}^{'} \setminus {X}_{2}^{'} \setminus {X}_{3}^{'} = {x}_{1} {X}_{1} \setminus {x}_{2} {X}_{2} \setminus {x}_{3} {X}_{3}$

• (X_1^')/(X_1 \ x_1) = ( x_2 X_2)/(X_2^') * (x_3 X_3)/(X_3^'

This is the same statement as:

• $f \left({x}_{1}\right) = \frac{1}{g \left({x}_{2}\right)} \cdot \frac{1}{h \left({x}_{3}\right)} q \quad \forall {x}_{i}$

These must be constant functions, or at very least they can be, so make an assumption or draw a conclusion:

$\left\{\begin{matrix}{f}_{1} = {C}_{1} \\ {g}_{2} = {C}_{2} \\ {h}_{3} = {C}_{3}\end{matrix}\right.$

• $\text{ where } {C}_{1} = \frac{1}{{C}_{2} {C}_{3}}$

It follows that:

$\frac{d {X}_{1}}{{X}_{1}} = {C}_{1} \setminus {x}_{1} \setminus d {x}_{1}$

Integrating:

$\ln {X}_{1} = {C}_{1} \setminus {x}_{1}^{2} / 2 + {D}_{1}$

• ${X}_{1} = {d}_{1} {e}^{{C}_{1} \setminus {x}_{1}^{2} / 2}$

Pattern-matching:

• $\left\{\begin{matrix}{X}_{2} = {d}_{2} {e}^{{C}_{2} \setminus {x}_{2}^{2} / 2} \\ {X}_{3} = {d}_{3} {e}^{{C}_{3} \setminus {x}_{3}^{2} / 2}\end{matrix}\right.$

$\textcolor{b l u e}{z \left(\boldsymbol{x}\right) = d \setminus {e}^{\frac{1}{2} \left({C}_{1} {x}_{1}^{2} + {C}_{2} {x}_{2}^{2} + {C}_{3} {x}_{3}^{2}\right)}} q \quad d = \textcolor{g r e e n}{{d}_{1} {d}_{2} {d}_{3}}$

Validating this possible solution:

• ${\partial}_{{x}_{1}} z = {C}_{1} {x}_{1} \setminus {d}_{1} \setminus {e}^{\frac{1}{2} \left({C}_{1} {x}_{1}^{2} + {C}_{2} {x}_{2}^{2} + {C}_{3} {x}_{3}^{2}\right)}$

• ${\partial}_{{x}_{2}} z = {C}_{2} {x}_{2} \setminus {d}_{2} \setminus {e}^{\frac{1}{2} \left({C}_{1} {x}_{1}^{2} + {C}_{2} {x}_{2}^{2} + {C}_{3} {x}_{3}^{2}\right)}$

• ${\partial}_{{x}_{3}} z = {C}_{3} {x}_{3} \setminus {d}_{3} \setminus {e}^{\frac{1}{2} \left({C}_{1} {x}_{1}^{2} + {C}_{2} {x}_{2}^{2} + {C}_{3} {x}_{3}^{2}\right)}$

${\partial}_{{x}_{1}} z \setminus {\partial}_{{x}_{2}} \setminus {\partial}_{{x}_{3}} z = \setminus {C}_{1} {C}_{2} {C}_{3} \setminus {x}_{1} {x}_{2} {x}_{3} \setminus {d}^{3} \setminus {e}^{\frac{3}{2} \left({c}_{1} {x}_{1}^{2} + {c}_{2} {x}_{2}^{2} + {c}_{3} {x}_{3}^{2}\right)}$

And because, by separating the variables, a specific conclusion was drawn:

• ${C}_{1} = \frac{1}{{C}_{2} {C}_{3}}$

$= {x}_{1} {x}_{2} {x}_{3} \setminus {\underbrace{{d}^{3} \setminus {e}^{\frac{3}{2} \left({c}_{1} {x}_{1}^{2} + {c}_{2} {x}_{2}^{2} + {c}_{3} {x}_{3}^{2}\right)}}}_{= {z}^{3}}$

$= {z}^{3} \setminus {x}_{1} {x}_{2} {x}_{3}$

So, at least one solution exists. Whether or not it is the exact solution depends upon the boundary values.