Cinch it

As $F(x, y, z) = a(x+y+z) + a'(x+y+z)$ and $F(x, y, z) = \nabla^2 F(x, y, z)$ , we have $a(x+y+z) + a'(x+y+z) = 3a''(x+y+z) + 3a^{[3]}(x+y+z) \Rightarrow \\ 3a^{[3]}(u) + 3a''(u) - a'(u) - a(u) = \hspace{0.25 em} 0$

The characteristic equation for this is $3x^3+3x^2-x-1 = 0$ , which factors into $(x-1)(3x^2-1) = 0$ . Thus, the solutions are $x = \pm \frac{1}{\sqrt{3}}, 1$ . Thus, the form of $a(u)$ is $a(u) = c_1 e^{\frac{1}{\sqrt{3}}u} + c_2 e^{-\frac{1}{\sqrt{3}}u} + c_3 e^{-u}$ . We know that $a(2u) = 2a(u) \cdot a'(u)$ . (As a point of interest, this is satisfied by $a(u) = \sinh(u)$ .) Thus, $c_1 e^{\frac{2 u}{\sqrt{3}}}+c_2 e^{-\frac{2 u}{\sqrt{3}}}+c_3 e^{-2 u} = 2 \left(\frac{c_1 e^{\frac{u}{\sqrt{3}}}}{\sqrt{3}}-c_3 e^{-u}-\frac{c_2 e^{-\frac{u}{\sqrt{3}}}}{\sqrt{3}}\right) \left(c_1 e^{\frac{u}{\sqrt{3}}}+c_2 e^{-\frac{u}{\sqrt{3}}}+c_3 e^{-u}\right) = \\ \frac{2 c_1^2 e^{\frac{2 u}{\sqrt{3}}}}{\sqrt{3}}-\frac{2 c_2^2 e^{-\frac{2 u}{\sqrt{3}}}}{\sqrt{3}} + \left(\frac{2}{\sqrt{3}}-2\right)c_1 c_3 e^{\frac{(\sqrt{3}-3)u}{3}} + \left(-\frac{2}{\sqrt{3}}-2\right)c_2 c_3 e^{\frac{-(\sqrt{3}+3)u}{3}} + 2{c_3}^2 e^{-2u}$

All of this looks far much worse than it is. We don't actually have to deal with evaluating these monsters. We know that $e^{px}$ and $e^{qx}$ are not proportional for all $x$ , so the only way the above equation can be satisfied is if the coefficients are equal. This gives the following system:

$\frac{2 c_1^2}{\sqrt{3}}=c_1 \\ -\frac{2 c_2^2}{\sqrt{3}}=c_2 \\ \frac{2 c_1 c_3}{\sqrt{3}}-2 c_1 c_3=0 \\ -\frac{2 c_2 c_3}{\sqrt{3}}-2 c_2 c_3=0 \\ -2 c_3^2=c_3$

The solutions of the above system in the format $(c_1, c_2, c_3)$ are $(0, 0, 0); (0, 0, -\frac{1}{2}); (0, -\frac{\sqrt{3}}{2}, 0); (\frac{\sqrt{3}}{2}, 0, 0); \text{ and } (\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}, 0)$ . Plugging these back into $a(u)$ and plugging in $u = 0$ gives the set of values $\left\{-\frac{1}{2},0,-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}\right\}$ , the mean of which is $-\frac{1}{8}$ . Thus, $a+b = \boxed{9}$ .