1. If the function $f(x)=\frac{x}{x^{2}+a}$ has a maximum value of $\frac{\sqrt{2}}{2}$ on $[1,+\infty)$, then $a=$
1. $\sqrt{2}-1$.

First, $f(x)=\frac{x}{x^{2}+a}=\frac{1}{x+\frac{a}{x}}(x \geqslant 1)$. When $a \geqslant 1$, the maximum value of the function $f(x)$ is $\frac{1}{2 \sqrt{a}}=\frac{\sqrt{2}}{2}$, solving for $a$ gives $a=\frac{1}{2}<1$, which is a contradiction. When $0<a<1$, $f(x)$ is monotonically decreasing on $[1,+\infty)$, and its maximum value is $\frac{1}{1+a}=\frac{\sqrt{2}}{2}$, yielding $a=\sqrt{2}-1$. Therefore, $a=\sqrt{2}-1$.