Let $f(t)$ be a function of class ${C^2}({R})$, which satisfies the ODE $4t \cdot f''(t) + 6f'(t) - f(t) = 0$. Use the chain rule to show that the function $u(x,y,z) = f({x^2} + {y^2} + {z^2})$ satisfies the PDE ${u_{xx}} + {u_{yy}} + {u_{zz}} = u$.
If i solve the ODE and substitute ${x^2} + {y^2} + {z^2}$ and then calculate ${u_{xx}},\,\,\,{u_{yy}},\,\,\,{u_{zz}}$ i get quite complex and very large expression. I have a feeling this problem should be solved in a different way, much easier way. Can someone give me a hint or show me how it should be done?
This is an example of problems where actually solving the equation leads to a complicated calculations.
All you need here is to use the fact that $f$ satisfies the equation $4tf''(t)+6f'(t)-f(t)=0$ (call it $(E)$).
By the Chain Rule, we have $u_x=2xf'(x^2+y^2+z^2)$ so $u_{xx}=2f'(x^2+y^2+z^2)+4x^2f''(x^2+y^2+z^2)$. By symmetry, $u_{yy}=2f'(x^2+y^2+z^2)+4y^2f''(x^2+y^2+z^2)$ and $u_{zz}=2f'(x^2+y^2+z^2)+4z^2f''(x^2+y^2+z^2)$.
so
$$ \begin{array}{rcl} u_{xx}+u_{yy}+u_{zz} & = & 2f'+4x^2f''+2f'+4y^2f''+2f'+4z^2f'' \\ & = & 4(x^2+y^2+z^2)f''(x^2+y^2+z^2)+6f'(x^2+y^2+z^2) \\ & = & 4tf''(t)+6f'(t) \end{array} $$
where $t=x^2+y^2+z^2$. But $f$ satisfies $(E)$ so
$4tf''(t)+6f'(t)=f(t)$
Therefore, $u_{xx}+u_{yy}+u_{zz}=u$.