Trigonometry! 4

$u^2$ can be evaluated to $a^2 + b^2+2\sqrt{a^2b^2+(a^2-b^2)^2\left(\frac{\sin^22\theta}{4}\right)}$

So, for the maximum value of $u^2$ , we take $\sin^{2} 2\theta = 1$ and $u^2$ becomes $2a^{2} + 2b^{2}$ .

For the minimum value of $u^{2}$ , we take $\sin ^{2} 2\theta = 0$ and $u^{2}$ becomes $(a+b)^{2}$ .

Hence the answer, $\boxed{(a-b)^{2}}$

By the definitions of trigonometric functions, we quickly see that the maximum will occur at $\theta=\frac{\pi}{4}$ and that the minimum will occur at $\theta=0$ or $\theta=\frac{\pi}{2}$ (in this case it doesn't matter because of symmetry). Thus evaluating $u^2$ at maximum we first get:

$u=\sqrt{a^{2}(\frac{\sqrt{2}}{2})^2 + b^{2}(\frac{\sqrt{2}}{2})^2} + \sqrt{a^{2}(\frac{\sqrt{2}}{2})^2 + b^{2}(\frac{\sqrt{2}}{2})^2}$

Which simplifies to:

$u=2\sqrt{\frac{a^2+b^2}{2}}$

Thus making $u^2 = 2(a^2+b^2)$ at maximum.

Evaluating $u$ at minimum ( $\theta=0$ chosen below) gives:

$u=\sqrt{a^{2}(1) + b^{2}(0)}+ \sqrt{a^{2}(0) + b^{2}(1)}$

Thus making $u$ equal to:

$u=a+b$

and $u^2$ equal to:

$u^2=(a+b)^2$ at its minimum.

Thus difference of maximum and minimum values of $u^2$ is:

$2(a^2+b^2) -(a+b)^2$ = $\boxed{(a-b)^2}$