Vtgyjfy

I'm trying to prove that any function $u=u(x,y,z)$ of the form $$u(x,y,z)=frac{f(t+r)}{r}+frac{g(t-r)}{r},$$ with $r^2=x^2+y^2+z^2$, satisfies the differential equation $ u_{xx}+u_{yy}+u_{zz}=u_{tt}, $ but I'm stuck when relating the derivatives.

From what I get, $$u_{tt} = dfrac{1}{r}(f''+g''),$$ as $r$ does not depend on $t$.

Furthermore, $$ u_x = dfrac{r_x}{r^2}left( r(f'-g')-(f+g) right), $$ so, if I'm not mistaken,

$$u_{xx} = dfrac{x^2}{r^3} (f''+g'') + dfrac{r^2-3x^2}{r^4}(f'-g') + dfrac{x^2-r^2}{r^5}(f+g). $$

(By symmetry, $u_{yy}$ and $u_{zz}$ have the same form).

But I don't know how to relate this expression to the other second derivatives, so I think I'm missing something...

Any ideas are greatly appreciated.
Thanks in advance!

EDIT: Corrected a $r^2$ term in the expression for $u_x$. Also, thanks to @Klaramun's and @TedShifrin's suggestions, I managed to reduce the expression for $u_{xx}$, noting that $f_x=f_y=f_z=f',, g_x=g_y=g_z=g'$, so I fixed that a couple lines up.

First, consider the function
$$h(x,y,z)=frac{f(t+r)}{r} tag 1$$
and with $r^2=x^2+y^2+z^2quad;quad rfrac{partial r}{partial x}=x quad;quad frac{partial r}{partial x}=frac{x}{r}$
$$h_x=left(frac{f'(t+r)}{r}-frac{f(t+r)}{r^2} right)frac{partial r}{partial x}=xfrac{f'}{r^2}-xfrac{f}{r^3} $$
In interest of space, we write $f$ instead of $f(t+r)$ and $f'$ instead of $f'(t+r)$ .
$$h_{xx}=left(frac{f'}{r^2}-frac{f}{r^3}right)+left(xfrac{f''}{r^2}-xfrac{f'}{r^3} right)frac{x}{r}+left(-2xfrac{f'}{r^3}+3xfrac{f}{r^4} right)frac{x}{r}$$
$$h_{xx}=frac{f'}{r^2}-frac{f}{r^3}+x^2frac{f''}{r^3}-x^2frac{f'}{r^4} -2x^2frac{f'}{r^4}+3x^2frac{f}{r^5}$$
$$h_{xx}=x^2left(frac{f''}{r^3} -3frac{f'}{r^4}+3frac{f}{r^5}right)+frac{f'}{r^2}-frac{f}{r^3}$$
On the same manner, we get :
$$h_{yy}=y^2left(frac{f''}{r^3} -3frac{f'}{r^4}+3frac{f}{r^5}right)+frac{f'}{r^2}-frac{f}{r^3}$$
$$h_{zz}=z^2left(frac{f''}{r^3} -3frac{f'}{r^4}+3frac{f}{r^5}right)+frac{f'}{r^2}-frac{f}{r^3}$$
Adding the three equations
$$h_{xx}+h_{yy}+h_{zz}=(x^2+y^2+z^2)left(frac{f''}{r^3} -3frac{f'}{r^4}+3frac{f}{r^5}right)+3frac{f'}{r^2}-3frac{f}{r^3}$$
$$h_{xx}+h_{yy}+h_{zz}=r^2left(frac{f''}{r^3} -3frac{f'}{r^4}+3frac{f}{r^5}right)+3frac{f'}{r^2}-3frac{f}{r^3}$$
$$h_{xx}+h_{yy}+h_{zz}=frac{f''}{r}$$
From Eq.$(1)quad:quad h_t=frac{f'}{r}quad$and$quad h_{tt}=frac{f''}{r}quad$thus
$$h_{xx}+h_{yy}+h_{zz}=h_{tt}$$
This proves that
$$frac{f(t+r)}{r}quad text{is solution of}quad u_{xx}+u_{yy}+u_{zz}=u_{tt}$$
Second, on the same manner one prouves that
$$frac{g(t-r)}{r}quad text{is solution of}quad u_{xx}+u_{yy}+u_{zz}=u_{tt}$$
Any linear combination of solutions of the PDE $(1)$ is solution of the PDE because the PDE is linear. Thus
$$frac{f(t+r)}{r}+frac{g(t-r)}{r} quadtext{satisfies}quad u_{xx}+u_{yy}+u_{zz}=u_{tt}$$

Ok, so thanks to @Klaramun and @TedShifrin I managed to find the proper way to solve it. With the expression for $u_{xx}, , u_{yy}, , u_{zz}$ as added in the question, and noting that $r^2=x^2+y^2+z^2$, it's immediate that $$u_{xx}+u_{yy}+u_{zz} = dfrac{1}{r}(f''+g'') = u_{tt}.$$
Hope this helps somebody :)