= Solution
Use the <singular system of a compact operator> convention $Ax_j=\sigma_jy_j$ and $A^*y_j=\sigma_jx_j$. Starting from zero gives no component in $\ker A$. Write $x(t)=\sum_jc_j(t)x_j$. The coefficient equations are
$$
c_j'=-\sigma_j^2c_j+\sigma_j\langle f,y_j\rangle,\qquad c_j(0)=0.
$$
Solving these scalar <linear ordinary differential equations> gives
$$
c_j(t)=\frac{1-e^{-\sigma_j^2t}}{\sigma_j}\langle f,y_j\rangle.
$$
Consequently <asymptotic regularization> has the <spectral regularization method> representation
$$
\boxed{u_\alpha=\sum_j\frac{1-e^{-\sigma_j^2/\alpha}}{\sigma_j}\langle f,y_j\rangle x_j.}
$$
The component of $f$ orthogonal to the closure of the range of $A$ is annihilated by $A^*$ and contributes nothing. The scalar filter satisfies $(1-e^{-\sigma^2/\alpha})/\sigma\leq\min(\sigma/\alpha,1/\sigma)\leq\alpha^{-1/2}$, so every fixed-$\alpha$ reconstruction is a bounded <linear operator>. Its long-time limit is the <Moore–Penrose inverse of an operator> on admissible data.
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