The fundamental solution or heat kernel in $n$ dimensions is: $G(x, t) = \frac{1}{(4\pi kt)^{n/2}}e^{-|x|^2/(4kt)}$

This satisfies $G_t = k\nabla^2 G$ for $t > 0$ and $G(x,0^+) = \delta(x)$ (Dirac delta). It represents the temperature distribution from a unit point source at $x = 0$, $t = 0$.

In one dimension: $G(x,t) = \frac{1}{\sqrt{4\pi kt}}e^{-x^2/(4kt)}$

This is a Gaussian with variance $\sigma^2 = 2kt$, showing that diffusion spreads as $\sqrt{t}$.

The solution to the Cauchy problem on $\mathbb{R}^n$ with initial data $u(x,0) = f(x)$ is: $u(x,t) = \int_{\mathbb{R}^n} G(x-y, t)f(y)\,dy = \frac{1}{(4\pi kt)^{n/2}}\int_{\mathbb{R}^n} e^{-|x-y|^2/(4kt)}f(y)\,dy$

This convolution formula shows how initial temperature distribution $f$ is "smeared out" by the Gaussian kernel.

For the heat equation on $[0, L]$ with boundary conditions $u(0,t) = u(L,t) = 0$, separation of variables $u(x,t) = X(x)T(t)$ yields: $u_n(x,t) = \sin\left(\frac{n\pi x}{L}\right)e^{-k(n\pi/L)^2 t}$

The general solution is: $u(x,t) = \sum_{n=1}^\infty b_n \sin\left(\frac{n\pi x}{L}\right)e^{-k(n\pi/L)^2 t}$

where $b_n$ are Fourier coefficients of the initial data. Note the exponential decay of each mode, with higher modes decaying faster.

The Barenblatt solution for the porous medium equation (a nonlinear variant) and certain similarity solutions to the heat equation have the form: $u(x,t) = t^{-\alpha}f(x/t^\beta)$ for appropriate $\alpha, \beta$. These describe spreading of localized initial disturbances.