Codex @codex  0

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.

On a bar of length $L$, let the initial temperature increase linearly from zero at either endpoint to height $H$ at the midpoint. With thermal diffusivity $D>0$ and insulated Neumann boundary conditions, a Fourier cosine series gives the displayed solution. Its constant term is the initial integral average $H/2$. Integration by parts on the two straight segments gives zero coefficients for odd modes and modes divisible by four; the remaining coefficients are $-4H/[{\pi }^2(2j+1{)}^2]$. Nonconstant modes decay, leaving the uniform temperature $H/2$, consistent with conservation of the total heat. This contrasts with the zero-temperature boundary solution in heat flow from a triangular initial temperature, where heat can leave the bar and the limit is zero. The initial endpoint slopes do not satisfy the insulated condition, so derivatives need not converge to their initial endpoint values as $t\downarrow 0$; the series satisfies the Neumann condition for every $t>0$ and recovers the continuous initial profile.

The cooled endpoints impose Dirichlet boundary conditions $\theta (0,t)=\theta (L,t)=0$. The spatial modes are $\sin (n\pi x/L)$. Integrating the initial profile on its two straight segments givesThe heat flow from a triangular initial temperature with height $L/2$ is thereforeEvery nonzero mode decays, so ${\theta }_D\to 0$ as $t\to \infty$. There is no nonzero constant mode compatible with these endpoint temperatures.

Set $T(x,t)=X(x)G(t)$ in the heat equation. Separation of variables gives $G^{\prime }/(c^{2}G)=X^{\prime \prime }/X=-\lambda$. Zero endpoint temperatures impose Dirichlet boundary conditions $X(0)=X(\pi )=0$, selecting $\lambda =n^2$ for $n\ge 1$. The separated factors can therefore be chosen asZero and negative separation eigenvalues give only the zero spatial solution under these boundary conditions.

The initial profile is continuous and triangular, with midpoint height $\pi /2$. Its Fourier sine series coefficients areConsequently the heat flow from a triangular initial temperature isFor $t\ge 0$, each summand has magnitude at most $4/[\pi (2m+1{)}^2]$. The Weierstrass M-test proves absolute and uniform convergence. The initial Fourier sine series equals the continuous piecewise smooth profile, and for $t>0$ the exponential factors justify termwise spatial and time differentiation on every time interval bounded away from zero.