Heat evolution of a triangular profile ($u(t,x)=\frac{1}{\pi }{\int }_{\mathbb{R}}{e}^{-t{\omega }^2+ix\omega }\frac{{\sin }^2\omega }{{\omega }^2}d\omega$)

On the real line, let $T(x)=\max (1-|x|/2,0)$. The Fourier transform of a triangular function gives $\hat{T}(\omega )=2{\sin }^2\omega /{\omega }^2$, with value two at zero. The Gaussian heat kernel has transform $e^{-t{\omega }^2}$. Hence $u=q_t\ast T$ solves the heat equation with initial profile $T$, and the inverse Fourier transform gives the displayed integral. Differentiating for positive time is justified by Gaussian decay. The uniform continuity of $T$ and the vanishing Gaussian mass outside any fixed neighborhood of zero prove uniform recovery of $T$ as $t\downarrow 0$. Convolution preserves total integral two, keeps the solution nonnegative, and smooths the corners for every positive time. Unlike the finite-interval problem in heat flow from a triangular initial temperature, this evolution has no endpoints through which heat can escape.