Redshifted cosine medium

Extend the displayed initial profile to all ${\mathbb{R}}^3$ by the same cosine formula. It is $C^{\mathrm{\infty }}$, even and $2L_i$-periodic in every coordinate. Its values lie between $\bar{\vartheta }-\mathrm{\mid }\delta \mathrm{\mid }$ and $\bar{\vartheta }+\mathrm{\mid }\delta \mathrm{\mid }$, so the strict inequality on $\delta$ gives positive lower and finite upper bounds. Its average on $\mathrm{\Omega }$ is $\bar{\vartheta }$, because the integral of $\cos (\pi x^1\mathrm{/}{L}_1)$ from $0$ to $L_1$ is $0$.

Put $r=A_0{\pi }^2\mathrm{/}(C_V{L}_1^2)>0$. Differentiating the proposed solution gives ${\mathrm{\partial }}_{t_K}\vartheta =-r\delta e^{-r{t}_K}\cos (\pi x^1\mathrm{/}{L}_1)$ and ${\mathrm{\Delta }}_x\vartheta =-({\pi }^2\mathrm{/}{L}_1^2)\delta e^{-r{t}_K}\cos (\pi x^1\mathrm{/}{L}_1)$, hence $C_V{\mathrm{\partial }}_{t_K}\vartheta =A_0{\mathrm{\Delta }}_x\vartheta$. Its initial trace is ${\vartheta }_0$. Also ${\mathrm{\partial }}_{x^1}\vartheta =-(\pi \delta \mathrm{/}{L}_1)e^{-r{t}_K}\sin (\pi x^1\mathrm{/}{L}_1)$ vanishes at $x^1=0,L_1$, and ${\mathrm{\partial }}_{x^2}\vartheta ={\mathrm{\partial }}_{x^3}\vartheta =0$, so every face has zero normal derivative. Uniqueness in [F1] identifies this function with the model solution.

The exponential factor tends to zero, so $\vartheta (t_K,x)\to \bar{\vartheta }$. Because $N$ is time independent and positive, $T=\vartheta \mathrm{/}N\to \bar{\vartheta }\mathrm{/}N$ pointwise. The latter is nonconstant by the stipulated nonconstancy of $N$. Direct integration using ${\int }_0^{L_1}{\cos }^2(\pi x^1\mathrm{/}{L}_1)d{x}^1=L_1\mathrm{/}2$ gives the exact norm ${\parallel \vartheta (t_K,\cdot )-\bar{\vartheta }\parallel }_{L^2(\mathrm{\Omega })}=\mathrm{\mid }\delta \mathrm{\mid }\sqrt{\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }\mathrm{/}2}{e}^{-r{t}_K}$. Thus this example realizes a decaying redshifted mode with a spatially varying proper-temperature equilibrium in the selected model.