Let $ u(x,t) $ satisfy the initial boundary value problem
$$
\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}, \quad x \in (0,1), t>0
$$
given that
$$
u(x, 0) = \sin(\pi x), \quad x \in [0,1], \quad u(0,t) = u(1,t) = 0, \quad t>0.
$$
Then $ u \left( x, \frac{1}{\pi^2} \right) $ for $ x \in (0,1) $ is