One, (20 points) Solve the equation:
$$
\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=\sqrt{a}(a)
$$
$0)$.
Discuss the solution of this equation for the value of the positive number $a$.


Given $\sqrt{x-\sqrt{2 x-1}}=\frac{|x-1|}{\sqrt{x+\sqrt{2 x-1}}}$, squaring both sides yields
$$
2 x+2|x-1|=a \text {. }
$$

When $x>1$, the solution is $x=\frac{2+a}{4}$.
When $x \leqslant 1$, we get $2=a$, i.e., when $a=2$, $\frac{1}{2} \leqslant x \leqslant 1$.
Conclusion (1) When $a>2$, there is one root $x=\frac{2+a}{4}$;
(2) When $a=2$, the equation has infinitely many solutions $\frac{1}{2} \leqslant x \leqslant 1$;
(3) When $a<2$, the equation has no solution.