We can solve the equation for \( w(\mathbf{y},t) \) by making a Fourier transform to momentum space. The PDF \( w(\mathbf{x},t) \) is related to its Fourier transform \( \tilde{w}(\mathbf{k},t) \) through
$$ w(\mathbf{x},t) = \int_{-\infty}^{\infty}d\mathbf{k} \exp{(i\mathbf{kx})}\tilde{w}(\mathbf{k},t), $$and using the definition of the \( \delta \)-function
$$ \delta(\mathbf{x}) = \frac{1}{2\pi} \int_{-\infty}^{\infty}d\mathbf{k} \exp{(i\mathbf{kx})}, $$we see that
$$ \tilde{w}(\mathbf{k},0)=1/2\pi. $$