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Taking the spatial Fourier transform of the diffusion equation gives ${\tilde{\theta }}_t=-D{k}^2\tilde{\theta }$ and $\tilde{\theta }(k,0)=\tilde{g}(k)$. Hence $\tilde{\theta }=e^{-D{k}^{2}t}\tilde{g}$. Fourier inversion and the Gaussian transform from part (a), or direct evaluation of the inverse Gaussian integral, give the heat kernelSubstituting the integral for $\tilde{g}$ and interchanging absolutely integrable terms produces the convolution solutionFor the stated nonnegative compactly supported data, understood to be nonzero on a set of positive measure, the kernel is strictly positive everywhere. Therefore $\theta (x,t)>0$ for every real $x$ and every $t>0$. Property P holds: diffusion has infinite propagation speed. The distant value can be extremely small, but is not zero. For continuous initial data, any nontrivial nonnegative $g$ automatically has the required positive-measure set.

The thermal diffusion length $\ell (t)=\sqrt{\alpha t}$ is the characteristic length over which thermal conduction changes a temperature distribution during time $t$ in the constant-thermal diffusivity model. It is a scale, not a sharp front or distance travelled at a constant speed: the heat equation has infinite propagation speed mathematically. Different definitions of penetration depth introduce numerical factors. In particular, the normalized one-dimensional heat kernel has spatial variance $2\alpha t$, so its root-mean-square width is $\sqrt{2\alpha t}$.

At $t=10\mathrm{s}$, the model materials have thermal diffusion lengths $4.47$ and $2.24\mathrm{mm}$. For $L=10\mathrm{mm}$ their thermal diffusion times are $50$ and $200\mathrm{s}$. These values describe the same heat equation as the thermal diffusion rate of a Fourier mode; the different numerical factors come from the definition of the scale.

A thermal diffusion wave is a periodically forced solution of the heat equation, with spatially decreasing amplitude and varying wave phase. In a semi-infinite rod $x\ge 0$, assume constant thermal diffusivity, negligible side losses, a nearly uniform temperature across each cross-section, and the boundary conditions $\theta (0,t)={\Theta }_0\cos \omega t$ and $\theta (x,t)\to 0$ as $x\to \infty$. A steady periodic thermal state then satisfiesChoose the square root in the complex numbers with $a>0$ so that the amplitude decays into the rod. Substitution of the phasor into the heat equation proves the formula: $i\omega \hat{\theta }=\alpha {\hat{\theta }}^{\prime \prime }$. Taking the real part gives the displayed cosine. In this convention $a$ is the spatial attenuation slope and $b$ is the spatial lag slope, both in ${\mathrm{m}}^{-1}$; $\gamma$ combines them into one complex number. If a complex wavenumber $K$ is defined by $e^{i\omega t-iKx}$ instead, $K=b-ia$.

With $\Delta x=x_2-x_1>0$, let $A_j=|\hat{\theta }(x_j)|$ and define a positive downstream lag $\Delta \varphi$. Thenand either measurement determines thermal diffusivity:The logarithm uses an amplitude ratio, not a ratio of squared amplitudes. The lag is in radians and must be tracked continuously across cycles: a measured phase difference modulo $2\pi$ leaves an integer-cycle ambiguity. Use closer sensor spacings or vary frequency continuously to resolve that ambiguity. Equal attenuation and lag slopes provide a useful diagnostic of the no-side-loss model.

The apparent phase velocity is $\omega /b=\sqrt{2\alpha \omega }$, which depends on frequency. A thermal diffusion wave therefore does not give a material-independent travel speed for heat; it is a forced pattern in a dissipative diffusion equation. The ideal heat equation still has infinite propagation speed, and the observed lag is not the arrival time of a sharp front.