An Asymptotic-Preserving IMEX Method for Nonlinear Radiative Transfer Equation

Abstract

We present an asymptotic preserving method for the radiative transfer equations in the framework of $P_N$ method. An implicit and explicit numerical scheme is proposed to solve the $P_N$ system based on the order analysis of the expansion coefficients of the specific intensity, where the order of each expansion coefficient is derived by the Chapman-Enskog method. The coefficients at higher-order are treated explicitly while those at lower-order are treated implicitly in each equation of the $P_N$ system. Energy inequality is proved for this numerical scheme. Several numerical examples validate the efficiency of this scheme in both optically thick and thin regions.

Appendix

1.1 The Gray Approximation of the Radiative Transfer Equations for 1D Angle Problem and Related $P_N$ Equations

The time-dependent gray approximation of the radiative transfer equations [31] in a one-dimensional planar geometry medium has the form as

$$\begin{aligned}&\frac{\epsilon ^2}{c} \dfrac{\partial {I}}{\partial {t}} +\epsilon \mu \dfrac{\partial {I}}{\partial {x}} = \sigma \left( \frac{1}{2}a c T^4 - I\right) , \qquad x \in [0, L], \end{aligned}$$

(A.1a)

$$\begin{aligned}&\epsilon ^2 C_{v} \dfrac{\partial {T}}{\partial {t}} = \sigma \left( \int _{-1}^1 I \,\mathrm {d}\mu - ac T^4\right) , \end{aligned}$$

(A.1b)

where $I(t, x, \mu )$ is the specific intensity of radiation, $\mu = \cos \theta \in [-1, 1]$ is the internal coordinate associated with the angle $\theta \in [0, \pi ]$. T(t, x) is the material temperature, and $\sigma$ is the absorption opacity. Moreover, the one-dimensional form of (2.10) is reduced into

$$\begin{aligned} \frac{\epsilon ^2}{c} \dfrac{\partial {I}}{\partial {t}} +\epsilon \mu \dfrac{\partial {I}}{\partial {x}} = \sigma \left( \frac{1}{2} \int I \,\mathrm {d}\mu - I\right) , \qquad x \in [0, L]. \end{aligned}$$

(A.2)

For (A.1), the basis function for the $P_N$ method is the Legendre polynomials. The moments are defined as

$$\begin{aligned} I_l = \int _{-1}^1 P_l(\mu ) I(t, x, \mu ) \,\mathrm {d}\mu , \qquad l = 0, \cdots M, \end{aligned}$$

(A.3)

where $P_l$ is the Legendre polynomial. Then, the $P_N$ equations for (A.1) are

$$\begin{aligned} \begin{aligned}&\frac{\epsilon ^2}{c}\dfrac{\partial {\bar{\mathbf{I}}}}{\partial {t}} + \epsilon \mathbf{B}^{\mathrm{low }} \dfrac{\partial {\bar{\mathbf{I}}}}{\partial {x}} + \epsilon {\mathbf{B}}^{\mathrm{up}} \dfrac{\partial {{\bar{\mathbf{I}}}}}{\partial {x}} = -\sigma {\bar{\mathbf{I}}} + \sigma a c T^4 e_1, \\&\epsilon ^2 C_{v} \dfrac{\partial {T}}{\partial {t}} = \sigma \left( I_0 - ac T^4\right) , \end{aligned} \end{aligned}$$

(A.4)

where $\bar{\mathbf{I}}= (I_0, I_1, \cdots , I_M)$ and $e_1 = (1, 0, \cdots , 0)^T$. $\mathbf{B}^{\mathrm{low}}$ and $\mathbf{B}^{\mathrm{up}}$ are triangular matrix with the non-zero entries as

$$\begin{aligned} \mathbf{B}^{\mathrm{low}}(i+1, i) = \frac{i}{2 i+1}, \qquad \mathbf{B}^{\mathrm{up}}(i, i+1) = \frac{i}{2(i-1)+1}, \qquad i = 1, \cdots M. \end{aligned}$$

(A.5)

1.2 Inflow Boundary Condition for PN Equations

We implement the inflow boundary condition for the $P_N$ equations by specifying the values of coefficients of the $P_N$ system in ghost cells. Choosing the left boundary as an example, the incoming specific intensity incident on the boundary interface is

$$\begin{aligned} I(\mu ) = I^b(\mu ), \quad \text {for}~\mu > 0. \end{aligned}$$

(A.6)

For the $P_N$ method, the numerical boundary can be rewritten as

$$\begin{aligned} I^{\mathrm{ghost}}(\mu ) = \left\{ \begin{array}{cc} I^{b}(\mu ), &{}\mu > 0, \\ I^{i}(\mu ), &{} \mu < 0, \end{array} \right. \end{aligned}$$

(A.7)

where $I^{i}(\mu )$ is the specific intensity at the left boundary of the area. Then, the expansion coefficient at the ghost cell is

$$\begin{aligned} I^{\mathrm{ghost}}_{l} = \int _{-1}^1 I^{\mathrm{ghost}}(\mu ) P_l(\mu )\,\mathrm {d}\mu . \end{aligned}$$

(A.8)

The implementation of the inflow boundary condition in 2D is similar in spirit to that of 1D. Supposing ${\varvec{n}}$ is the outward normal of the boundary interface, the incident specific intensity on the boundary is

$$\begin{aligned} I({\varvec{\Omega }}) = I^b({\varvec{\Omega }}), \quad \text {for}~{\varvec{\Omega }}\cdot {\varvec{n}}< 0. \end{aligned}$$

(A.9)

For the $P_N$ method, the numerical boundary can be rewritten as

$$\begin{aligned} I^{\mathrm{ghost}}({\varvec{\Omega }}) = \left\{ \begin{array}{cc} I^{b}({\varvec{\Omega }}), &{} {\varvec{\Omega }}\cdot {\varvec{n}}< 0, \\ I^{i}({\varvec{\Omega }}), &{} {\varvec{\Omega }}\cdot {\varvec{n}}> 0, \end{array} \right. \end{aligned}$$

(A.10)

where $I^{i}({\varvec{\Omega }})$ is the specific intensity on the interior side of the boundary interface. Thus, the expansion coefficient at the ghost cell is

$$\begin{aligned} \begin{aligned} I^{m, \mathrm ghost}_{l}&= \int _{{\mathbb {S}}^2} I^{\mathrm{ghost}}({\varvec{\Omega }}) \overline{Y^m_l}({\varvec{\Omega }}) \,\mathrm {d}{\varvec{\Omega }}\\&= \int _{{\varvec{\Omega }}\cdot {\varvec{n}}< 0} I^{b}({\varvec{\Omega }}) \overline{Y^m_l}({\varvec{\Omega }}) \,\mathrm {d}{\varvec{\Omega }}+ \int _{{\varvec{\Omega }}\cdot {\varvec{n}}> 0} I^{i}({\varvec{\Omega }}) \overline{Y^m_l}({\varvec{\Omega }}) \,\mathrm {d}{\varvec{\Omega }}\\&= \int _{{\varvec{\Omega }}\cdot {\varvec{n}}< 0} I^{b}({\varvec{\Omega }}) \overline{Y^m_l}({\varvec{\Omega }}) \,\mathrm {d}{\varvec{\Omega }}+ \sum \limits _{j=0}^M\sum \limits _{k=-j}^j I^{k, i}_j \int _{{\varvec{\Omega }}\cdot {\varvec{n}}> 0} Y^k_j({\varvec{\Omega }}) \overline{Y^m_l}({\varvec{\Omega }}) \,\mathrm {d}{\varvec{\Omega }}. \end{aligned} \end{aligned}$$

(A.11)

The integration $\int _{{\varvec{\Omega }}\cdot {\varvec{n}}> 0} Y^k_j({\varvec{\Omega }}) \overline{Y^m_l}({\varvec{\Omega }}) \,\mathrm {d}{\varvec{\Omega }}$ does not depend on the specific numerical solutions, and is pre-computed.

Fig. 13

The stability region for the numerical scheme (3.18) of $P_1$ system under the condition (A.12). The x-axis is $\beta _2 = \log _{10}(\epsilon / \Delta x)$, and the y-axis is the CFL number C. The blue region is the area where the numerical scheme is stable and the yellow region is the area where the numerical scheme is unstable

1.3 Proof of Proposition 2

In this section, the proof of the Proposition 2 is proposed here.

Proof

Following the method in [39], we will begin the Fourier analysis of (3.18) for the $P_1$ system of the linear equation system (A.2). The result can be extended to the generalized $P_N$ system naturally. We will first discuss two special cases where $\xi = 0, \pi$. Therein, $\mathbf{C}$ is reduced into a real diagonal matrix with the maximum eigenvalues equaling 1. According to the principle, the numerical scheme is stable.

Then, we study the general case by considering two scenarios according to the time step length:

• 1. $\epsilon < \Delta x$, where

  $$\begin{aligned} \Delta t = C \Delta x^2. \end{aligned}$$

  (A.12)

  Substituting the time step length (A.12) into (3.36), we can find that $\lambda _i, i=1,2$ are functions of C, $\frac{\epsilon }{\Delta x}$, $\alpha$ and $\xi$. Introducing two variables as $\beta _1 = \log _{10}(C)$ and $\beta _2 = \log _{10} (\epsilon / \Delta x)$, the stability regions are plotted in Fig. 13 with fixed $\alpha$. Here the discrete wave number $\xi$ is uniformly taken from $[0, 2\pi ]$ with 200 samples. In this case, due to the definition of C which is the CFL number and $\epsilon < \Delta x$, the range for $\beta _1$ and $\beta _2$ is changed into

  $$\begin{aligned} \beta _1< 0, \qquad \beta _2 < 0. \end{aligned}$$

  (A.13)

  Moreover, it is natural to demand that $\epsilon< \Delta x < 0.4$. Thus, $\alpha$ is taken uniformly from $[0, \exp (-1/0.16)]$ with 100 samples and six cases are shown in Fig. 13 due to their similar behavior. From Fig. 13, we can find that when $\alpha = 0$, the numerical scheme is always stable. However, with the increase of $\alpha$, the stability region is becoming smaller, especially when the CFL number C is large and the radio $\epsilon / \Delta x$ is small. We find that when $\alpha = \exp (-1/0.16)$, the numerical scheme is stable when $\log _{10}(\epsilon / \Delta x) > -2.5$. Noting that when $\alpha = \exp (-1/0.16)$, which means $\epsilon = 0.4$, $\log _{10}(\epsilon / \Delta x)$ is always larger than $-2.5$ for $\Delta x < 1$. This indicates that in the simulation of benchmark problems, the stability condition is always satisfied.

• 2. $\epsilon > \Delta x$, where

  $$\begin{aligned} \Delta t = C \epsilon \Delta x. \end{aligned}$$

  (A.14)

  Substituting the time step length (A.14) into (3.36), we can easily find that $\lambda _i, i = 1, 2$ are also the function of C, $\frac{\epsilon }{\Delta x}$, $\alpha$ and $\xi$. Introducing the same two variables $\beta _{i}, i = 1,2$, we plot the stability regions in Fig. 14 with fixed $\alpha$. Here the discrete wave number $\xi$ is uniformly taken from $[0, 2\pi ]$ with 200 samples. Since

  $$\begin{aligned} \alpha = \exp \left( -\frac{1}{\epsilon ^2}\right) , \end{aligned}$$

  (A.15)

  and assuming $\epsilon < 1$ in the numerical test, $\alpha$ is taken uniformly from [0, 0.5] with 100 samples. As their behavior is similar, the six cases $\alpha = 0, 0.05, 0.1, 0. 2, 0.3$ and 0.5 are plotted here to illustrate the result. Moreover, due to the definition of C which is the CFL number, and the condition that $\epsilon > \Delta x$, it holds that

  $$\begin{aligned} \beta _1 < 0, \qquad \beta _2 > 0. \end{aligned}$$

  (A.16)

  From Fig. 14, we can find that the numerical scheme is stable under the time step length (A.14). In the numerical tests, the upper bound of $\epsilon$ is $\epsilon = 10^5 \Delta x$, which is large enough for the computational parameter.

$\square$

Fig. 14

The stability region for the numerical scheme (3.18) of $P_1$ system under the condition (A.14). The x-axis is $\beta _2 = \log _{10}(\epsilon / \Delta x)$, and the y-axis is the CFL number C. The blue region is the area where the numerical scheme is stable

1.4 Proof of Theorem 2

In this section, the proof of Theorem 2 is proposed here.

Proof

We will take $M=2$ as an example, and it could be extended to the general case naturally. Moreover, without loss of generality, we set $a = c = C_v = \sigma = 1$ in the proof. When $M=2$, (4.10) is reduced into

$$\begin{aligned}&\epsilon ^2 \frac{I_{0,j}^{n+1} - I_{0,j}^n}{\Delta t} + \epsilon \frac{I_{1, j+1}^{n} - I_{1, j-1}^{n}}{2 \Delta x} - \frac{\alpha \epsilon }{2} \frac{I_{0, j+1}^n - 2 I_{0,j}^n + I_{0, j-1}^n}{\Delta x} = \left( (T_j^4)^{n+1} - I_{0,j}^{n+1}\right) , \end{aligned}$$

(A.17a)

$$\begin{aligned}&\epsilon ^2 \frac{I_{1,j}^{n+1} - I_{1,j}^n}{\Delta t} + \frac{\epsilon }{3} \frac{I_{0, j+1}^{n+1} - I_{0, j-1}^{n+1}}{2 \Delta x} +\frac{2\epsilon }{3} \frac{I_{2, j+1}^{n} - I_{2, j-1}^{n}}{2 \Delta x} - \frac{\alpha \epsilon }{2} \frac{I_{1, j+1}^n - 2 I_{1,j}^n + I_{1, j-1}^n}{\Delta x} \nonumber \\&\quad = - I_{1,j}^{n+1}, \end{aligned}$$

(A.17b)

$$\begin{aligned}&\epsilon ^2 \frac{I_{2,j}^{n+1} - I_{2,j}^n}{\Delta t} + \frac{2\epsilon }{5} \frac{I_{1, j+1}^{n+1} - I_{1, j-1}^{n+1}}{2 \Delta x} - \frac{\alpha \epsilon }{2} \frac{I_{2, j+1}^n - 2 I_{2,j}^n + I_{2, j-1}^n}{\Delta x} = - I_{2,j}^{n+1}, \end{aligned}$$

(A.17c)

$$\begin{aligned}&\epsilon ^2 \frac{T_j^{n+1} - T_j^n}{\Delta t} +\epsilon \frac{I_{0,j}^{n+1} - I_{0,j}^n}{\Delta t} + \epsilon \frac{I_{1, j+1}^{n} - I_{1, j-1}^{n}}{2 \Delta x} - \frac{\alpha \epsilon }{2} \frac{I_{0, j+1}^n - 2 I_{0,j}^n + I_{0, j-1}^n}{\Delta x} = 0. \end{aligned}$$

(A.17d)

For (A.17a), multiplying it by $I_{0,j}^{n+1}$, we can get that

$$\begin{aligned}&\epsilon ^2 I_{0,j}^{n+1} \frac{I_{0,j}^{n+1} - I_{0,j}^n}{\Delta t} + \epsilon I_{0,j}^{n+1} \frac{I_{1, j+1}^{n} - I_{1, j-1}^{n}}{2 \Delta x} - \frac{\alpha \epsilon }{2} I_{0,j}^{n+1} \frac{I_{0, j+1}^n - 2 I_{0,j}^n + I_{0, j-1}^n}{\Delta x} \nonumber \\&\quad = I_{0,j}^{n+1} \left( (T_j^4)^{n+1} - I_{0,j}^{n+1}\right) . \end{aligned}$$

(A.18)

For (A.17b) and (A.17c), shifting it backward one time step and multiplying $3I_{1,j}^{n}$ and $5I_{2, j}^{n}$ respectively, we can derive that

$$\begin{aligned} \begin{aligned}&3\epsilon ^2 I_{1,j}^n \frac{I_{1,j}^{n} - I_{1,j}^{n-1}}{\Delta t} + \epsilon I_{1,j}^n \frac{I_{0, j+1}^{n} - I_{0, j-1}^{n}}{2 \Delta x} + \epsilon I_{1,j}^n \frac{I_{2, j+1}^{n-1} - I_{2, j-1}^{n-1}}{ \Delta x}\\&\quad - \frac{3\alpha \epsilon }{2} I_{1,j}^n\frac{I_{1, j+1}^{n-1} - 2 I_{1,j}^{n-1} + I_{1, j-1}^{n-1}}{\Delta x} = - 3(I_{1,j}^{n})^2, \\&5 \epsilon ^2 I_{2,j}^n\frac{I_{2,j}^{n} - I_{2,j}^{n-1}}{\Delta t} + \epsilon I_{2,j}^n \frac{I_{1, j+1}^{n} - I_{1, j-1}^{n}}{ \Delta x} \\&\quad -\frac{5 \alpha \epsilon }{2} I_{2,j}^n \frac{I_{2, j+1}^{n-1} - 2 I_{2,j}^{n-1} + I_{2, j-1}^{n-1}}{\Delta x} = - 5 (I_{2,j}^{n})^2. \end{aligned} \end{aligned}$$

(A.19)

Summing (A.18) and (A.19) over j, then it holds that

$$\begin{aligned}&\frac{\epsilon ^2}{2\Delta t} \sum _j\Big [ (I_{0,j}^{n+1})^2 - (I_{0,j}^{n})^2 + 3\left( (I_{1,j}^{n})^2 - (I_{1,j}^{n-1})^2 \right) + 5 \left( (I_{2,j}^{n})^2 - (I_{2,j}^{n-1})^2\right) \Big ]\nonumber \\&\quad + A_0 = A_1 + A_2 + A_3, \end{aligned}$$

(A.20)

where

$$\begin{aligned} A_0&= \frac{\epsilon ^2}{2\Delta t} \sum _j\Big [ (I_{0,j}^{n+1}- I_{0,j}^{n})^2 + 3(I_{1,j}^{n} - I_{1,j}^{n-1})^2 + 5(I_{2,j}^{n} - I_{2,j}^{n-1})^2 \Big ], \end{aligned}$$

(A.21a)

$$\begin{aligned} A_1&= -\frac{\epsilon }{2\Delta x}\sum _j \Big [ I_{0,j}^{n+1} (I_{1, j+1}^{n} - I_{1, j-1}^{n}) + I_{1,j}^n (I_{0, j+1}^{n} - I_{0, j-1}^{n}) + 2I_{1,j}^n (I_{2, j+1}^{n-1} - I_{2, j-1}^{n-1}) \nonumber \\&\quad + 2I_{2,j}^n (I_{1, j+1}^{n} - I_{1, j-1}^{n})\Big ], \end{aligned}$$

(A.21b)

$$\begin{aligned} A_2&= \frac{\alpha \epsilon }{2\Delta x}\sum _j \Big [ I_{0,j}^{n+1} (I_{0, j+1}^{n} - 2I_{0, j}^{n} +I_{0, j-1}^{n}) + 3I_{1,j}^n (I_{1, j+1}^{n-1} - 2 I_{1,j}^{n-1} + I_{1, j-1}^{n-1})\end{aligned}$$

(A.21c)

$$\begin{aligned}&\quad + 5I_{2,j}^n (I_{2, j+1}^{n-1} - 2I_{2, j}^{n-1} + I_{2, j-1}^{n-1})\Big ],\nonumber \\ A_3&= -\sum _j \Big [-I_{0,j}^{n+1}(T_{j}^4)^{n+1} + (I_{0,j}^{n+1})^2 + 3(I_{1, j}^n)^2 + 5(I_{2,j}^n)^2\Big ]. \end{aligned}$$

(A.21d)

Then we will begin from the approximation of $A_1$ and $A_2$. With some arrangement and the periodic boundary condition, $A_1$ is changed into

$$\begin{aligned} \begin{aligned} A_1 =&-\frac{\epsilon }{2\Delta x}\sum _j \Big [ (I_{0,j}^{n+1} - I_{0,j}^{n}) (I_{1, j+1}^{n} - I_{1, j-1}^{n}) + 2(I_{1,j+1}^n - I_{1,j-1}^{n}) (I_{2, j}^{n} - I_{2, j}^{n-1}) \Big ]\\ \le&\frac{\epsilon }{2\Delta x}\sum _j \Big [ \frac{1}{2}\beta _1^2 (I_{0,j}^{n+1} - I_{0,j}^{n})^2 + \frac{1}{2 \beta _1^2} (I_{1, j+1}^{n} - I_{1, j-1}^{n})^2 + \beta _2^2 (I_{2,j}^n - I_{2,j}^{n-1})^2 \\&+ \frac{1}{\beta _2^2} (I_{1, j+1}^{n} - I_{1, j-1}^{n})^2 \Big ] \\ \le&\frac{\epsilon }{2\Delta x}\sum _j \Big [ \frac{1}{2}\beta _1^2 (I_{0,j}^{n+1} - I_{0,j}^{n})^2 + \beta _2^2 (I_{2,j}^n - I_{2,j}^{n-1})^2 + \left( \frac{2}{\beta _1^2} + \frac{4}{\beta _2^2}\right) (I_{1, j}^{n})^2 \Big ]. \end{aligned} \end{aligned}$$

(A.22)

With the estimation that

$$\begin{aligned} \begin{aligned}&\sum _j I_{0,j}^{n+1}(I_{0,j+1}^n - 2 I_{0,j}^n + I_{0,j-1}^n)\\&\qquad = \sum _j \Big [(I_{0,j}^{n+1} - I_{0,j}^n)(I_{0,j+1}^n - 2 I_{0,j}^n + I_{0,j-1}^n) + I_{0,j}^n(I_{0,j+1}^n - 2 I_{0,j}^n + I_{0,j-1}^n) \Big ] \\&\qquad \le \sum _j\Big [\frac{1}{2}\beta _3^2 (I_{0,j}^{n+1} - I_{0,j}^n)^2 + \frac{2}{\beta _3^2}(I_{0,j}^n - I_{0,j-1}^n)^2\Big ] - \sum _j (I_{0,j}^n - I_{0,j-1}^n)^2 \\&\qquad = \sum _j\Big [\frac{1}{2}\beta _3^2 (I_{0,j}^{n+1} - I_{0,j}^n)^2 + (\frac{2}{\beta _3^2} - 1)\left( I_{0,j}^n - I_{0,j-1}^n\right) ^2\Big ]. \end{aligned} \end{aligned}$$

(A.23)

Similarly, it also holds that

$$\begin{aligned} \begin{aligned}&\sum _j I_{1,j}^{n}(I_{1,j+1}^{n-1} - 2 I_{1,j}^{n-1} + I_{1,j-1}^{n-1}) = \sum _j I_{1,j}^{n-1}(I_{1,j+1}^{n} - 2 I_{1,j}^{n} + I_{1,j-1}^{n}) \\&\le \sum _j\Big [\frac{1}{2}\beta _4^2 (I_{1,j}^{n} - I_{1,j}^{n-1})^2 + 4\left( \frac{2}{\beta _4^2} - 1\right) (I_{1,j}^n)^2\Big ], \\&\sum _j I_{2,j}^{n}(I_{2,j+1}^{n-1} - 2 I_{2,j}^{n-1} + I_{2,j-1}^{n-1}) = \sum _j I_{2,j}^{n-1}(I_{2,j+1}^{n} - 2 I_{2,j}^{n} + I_{2,j-1}^{n}) \\&\le \sum _j\Big [\frac{1}{2}\beta _5^2 (I_{2,j}^{n} - I_{2,j}^{n-1})^2 + \left( \frac{2}{\beta _5^2}-1\right) (I_{2,j}^n - I_{2,j-1}^n)^2\Big ]. \end{aligned} \end{aligned}$$

(A.24)

Let

$$\begin{aligned} \beta _1^2 = \frac{2 \epsilon \Delta x}{\Delta t}- 2 \alpha , \qquad \beta _2^2 = \frac{5\epsilon \Delta x}{\Delta t}- 5 \alpha , \qquad \beta _3^2 = 2, \qquad \beta _4^2 = \frac{2 \Delta x \epsilon }{\alpha \Delta t}, \qquad \beta _5^2 = 2,\nonumber \\ \end{aligned}$$

(A.25)

then together with (A.21), (A.22), (A.23), (A.24) and (A.25), (A.20) is reduced into

$$\begin{aligned} \begin{aligned} \frac{\epsilon ^2}{2} \sum _j\Big [ (I_{0,j}^{n+1})^2 - (I_{0,j}^{n})^2 + 3\left( (I_{1,j}^{n})^2 - (I_{1,j}^{n-1})^2 \right) + 5 \left( (I_{2,j}^{n})^2 - (I_{2,j}^{n-1})^2\right) \Big ]\\ \le \beta _6\sum _j (I_{1,j}^{n})^2 + \Delta t \sum _j \Big [I_{0,j}^{n+1}(T_{j}^4)^{n+1} - (I_{0,j}^{n+1})^2 - 5(I_{2,j}^{n})^2\Big ], \end{aligned} \end{aligned}$$

(A.26)

with

$$\begin{aligned} \beta _6 = \frac{9}{10}\dfrac{\epsilon \Delta t}{\Delta x} \left( \frac{1}{\frac{\epsilon \Delta x}{\Delta t} - \alpha }\right) + \dfrac{6 \alpha ^2 (\Delta t)^2}{(\Delta x)^2}-\dfrac{6\alpha \epsilon \Delta t}{\Delta x} - 3 \Delta t. \end{aligned}$$

(A.27)

If it holds for $\beta _6$ that

$$\begin{aligned} \beta _6 \le 0, \end{aligned}$$

(A.28)

with the time step length (3.32), then we can derive the stability result (4.16). Precisely, with (A.17a) and (A.17d), we can derive that

$$\begin{aligned} \epsilon ^2 \frac{T_j^{n+1} - T_j^n}{\Delta t} = -(T_j^4)^{n+1} + I_{0,j}^{n+1}. \end{aligned}$$

(A.29)

Multiplying (A.29) with $(T_{j}^4)^{n+1}$ and summing over j, it holds with (A.26)

$$\begin{aligned} \begin{aligned}&\sum _j \left[ \frac{\epsilon ^2}{2\Delta t}\Big ( (I_{0,j}^{n+1})^2 - (I_{0,j}^{n})^2 + 3\left[ (I_{1,j}^{n})^2 - (I_{1,j}^{n-1})^2 \right] + 5 \left[ (I_{2,j}^{n})^2 - (I_{2,j}^{n-1})^2\right] \Big ) \right. \\&\qquad \left. +\frac{\epsilon ^2}{5 \Delta t} \Big [(T_j^5)^{n+1} - (T_j^5)^{n}\Big ] \right] \le - \sum _j \left[ I_{0,j}^{n+1} - (T_{j}^4)^{n+1}\right] ^2 \le 0. \end{aligned} \end{aligned}$$

(A.30)

We derive the energy stability (4.16). The only point left is to prove (A.28), which we will be done in two cases:

• 1. $\epsilon > \Delta x$, in which case,

  $$\begin{aligned} \Delta t= C \epsilon \Delta x. \end{aligned}$$

  (A.31)

  Substituting (A.31) into (A.27), we can deduce that

  $$\begin{aligned} \beta _6 = \frac{9C^2\epsilon ^2}{10}\left( \frac{1}{1 - \alpha C }\right) + 6\alpha \epsilon ^2C(\alpha C - 1) - 3C \Delta x \epsilon . \end{aligned}$$

  (A.32)

  Thus if

  $$\begin{aligned} 0< C < \min \left( \frac{\epsilon }{\alpha \Delta x},\frac{10 \Delta x}{3 \epsilon + 10 \Delta x \alpha } \right) , \end{aligned}$$

  (A.33)

  it holds that $\beta _6 \le 0$.

  For the coefficients $\beta _i^2, i = 1,\cdots 5$, it requires that $\beta _i^2 > 0$. Thus, from (A.25), it demands that

  $$\begin{aligned} \frac{\epsilon \Delta x}{\Delta t} - \alpha > 0. \end{aligned}$$

  (A.34)

  Substituting (A.31) into (A.34), we can obtain that

  $$\begin{aligned} C < \frac{1}{\alpha }. \end{aligned}$$

  (A.35)

  Thus, the constrain on C is changed into

  $$\begin{aligned} 0< C < \min \left( \frac{1}{\alpha },\frac{10 \Delta x}{3 \epsilon + 10 \Delta x \alpha } \right) . \end{aligned}$$

  (A.36)

• 2. $\epsilon < \Delta x$, in which case

  $$\begin{aligned} \Delta t= C \Delta x^2. \end{aligned}$$

  (A.37)

  Substituting (A.37) into (A.27), we can deduce that

  $$\begin{aligned} \beta _6 = \frac{9}{10}C^2\epsilon \Delta x^2\left( \frac{1}{\epsilon - \alpha C \Delta x}\right) + 6\alpha \Delta x C(\alpha C \Delta x - \epsilon ) - 3C \Delta x^2. \end{aligned}$$

  (A.38)

  Thus if

  $$\begin{aligned} 0< C < \min \left( \frac{\epsilon }{\alpha \Delta x},\frac{10 \epsilon }{3 \epsilon + 10 \Delta x \alpha } \right) , \end{aligned}$$

  (A.39)

  it holds that $\beta _6 \le 0$. Similarly, we can verify that the constrain $\beta _i^2 > 0, i = 1, \cdots 5$ will not affect the condition (A.39), then the proof is finished.

  For $\epsilon < \Delta x$, it is always true that $\alpha = \exp (-1/\epsilon ^2)$ is quite small, and (A.39) could be reduced into

  $$\begin{aligned} 0< C < \frac{10}{3}. \end{aligned}$$

  (A.40)

$\square$

1.5 Analysis of the Higher-Order Scheme

From the test of the AP property for the numerical scheme, we found that even for the IMEX3 scheme with WENO reconstruction, the convergence order is only two. Analysis of the numerical scheme shows that when solving $T^{n+1}$, the fourth-order polynomial equation of $T^{n+1}$ is solved, where $(T^{n+1})^{4}$ is approximated as

$$\begin{aligned} (T^{4})_{i} \approx (T_{i})^{4} \end{aligned}$$

(A.41)

instead of

$$\begin{aligned} (T)^{4}_{i} \approx \frac{\int _{x_{i-\frac{1}{2}}}^{x_{i + \frac{1}{2}}}T^{4}dx}{\Delta x}, \end{aligned}$$

(A.42)

where $T_{i}$ is the cell average of cell i. Noting that

$$\begin{aligned} \frac{\int _{ x_{i - \frac{1}{2} }}^{ x_{i + \frac{1}{2} }} T(x) dx }{\Delta x} = T(x_{i}) + \frac{1}{24}(T(\xi _{i}))^{''}\Delta x^2,\quad \xi \in [x_{i - \frac{1}{2}}, x_{i + \frac{1}{2}}], \end{aligned}$$

(A.43)

and

$$\begin{aligned} \frac{ \int _{x_{i - \frac{1}{2}}}^{x_{i + \frac{1}{2}}} T^{4}(x) dx }{\Delta x} = T^{4}(x_{i}) + \frac{1}{24}(T^{4}(\eta _{i}))^{''}\Delta x^2,\eta _{i}\in [x_{i - \frac{1}{2}}, x_{i + \frac{1}{2}}], \end{aligned}$$

(A.44)

thus, it holds

$$\begin{aligned} \frac{\int _{x_{i - \frac{1}{2}}}^{x_{i + \frac{1}{2}}} T^{4} dx}{\Delta x} - (T_{i})^{4} = {\mathcal {O}}(\Delta x^{2}). \end{aligned}$$

(A.45)

Therefore, the convergence order of the whole numerical scheme is at most two.