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Hyperbolicity of the ballistic-conductive model of heat conduction: the reverse side of the coin

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Abstract

The heat equation, based on Fourier’s law, is commonly used for description of heat conduction. However, Fourier’s law is valid under the assumption of local thermodynamic equilibrium, which is violated in very small dimensions and short timescales, and at low temperatures. In the paper Kovács and Ván (Int J Heat Mass Transf 83:613–620, 2015), a ballistic-conductive (BC) model of heat conduction was developed. In this paper, we study the behavior of solutions to an initial value problem in 1D in the framework of the linearized BC model. As a result of the study, an unphysical effect has been found when part of the initial thermal energy does not spread anywhere.

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No datasets were generated or analyzed during the current study.

Notes

  1. Refs. [14, 20] discuss mass transfer. However, the obtained unphysical solutions of the telegraph equation in 2D and 3D, when they may take negative values, even if the initial values are non-negative, apply to the hyperbolic heat equation as well.

  2. In Refs. [49,50,51,52] the dimensionless temperature T is multiplied by a coefficient \(\tau _{\Delta }^{}\). Its presence or absence does not affect the conclusions of this paper.

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Authors and Affiliations

  1. Ioffe Institute, 26 Polytekhnicheskaya, St. Petersburg, Russia, 194021

    S. A. Rukolaine

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  1. S. A. Rukolaine
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Appendices

The Fourier transform of the solution to the initial value problem

1.1 General heat source f

The Fourier transform of the equation (4) and the initial conditions (7) results in the equation

$$\begin{aligned}{} & {} \tau _q^{} \tau _Q^{} {\mathcal {F}}T_{ttt}^{} + \left( \tau _q^{} + \tau _Q^{} + \gamma \tau _q^{} \tau _Q^{} \right) {\mathcal {F}}T_{tt}^{} + \left[ 1 + \gamma \left( \tau _q^{} + \tau _Q^{} \right) + \left( \tau _Q^{} + \kappa ^2 \right) \xi ^2 \right] {\mathcal {F}}T_t^{} \nonumber \\{} & {} \quad + \left[ \gamma + \left( 1 + \gamma \kappa ^2 \right) \xi ^2 \right] {\mathcal {F}}T = \tau _q^{} \tau _Q^{} {\mathcal {F}}f_{tt}^{} + \left( \tau _q^{} + \tau _Q^{} \right) {\mathcal {F}}f_t^{} + {\mathcal {F}}f + \kappa ^2 \xi ^2 {\mathcal {F}}f \end{aligned}$$
(20)

and initial conditions

$$\begin{aligned} {\mathcal {F}}T |_{t=0}^{} = 0, \quad {\mathcal {F}}T_t^{} |_{t=0}^{} = {\mathcal {F}}f |_{t=0}^{}, \quad {\mathcal {F}}T_{tt}^{} |_{t=0}^{} = \left( {\mathcal {F}}f_t^{} - \gamma {\mathcal {F}}f \right) |_{t=0}^{}, \end{aligned}$$
(21)

where the Fourier transform is given by \({\mathcal {F}}T(\xi , \cdot ) = \int _{-\infty }^\infty T(x, \cdot ) \mathop {\textrm{e}}\nolimits ^{\textrm{i}\xi x} \mathop {}\!\textrm{d}x\). The Laplace transform of the equation (20) with the initial conditions (21) results in the equation

$$\begin{aligned}{} & {} \big \{ \tau _q^{} \tau _Q^{} s^3 + \left( \tau _q^{} + \tau _Q^{} + \gamma \tau _q^{} \tau _Q^{} \right) s^2 + \left[ 1 + \gamma \left( \tau _q^{} + \tau _Q^{} \right) + \left( \tau _Q^{} + \kappa ^2 \right) \xi ^2 \right] s \\{} & {} \quad + \left[ \gamma + \left( 1 + \gamma \kappa ^2 \right) \xi ^2 \right] \big \} {\mathcal {L}}{\mathcal {F}}T = \left[ \tau _q^{} \tau _Q^{} s^2 + \left( \tau _q^{} + \tau _Q^{} \right) s + 1 + \kappa ^2 \xi ^2 \right] {\mathcal {L}}{\mathcal {F}}f, \end{aligned}$$

where the Laplace transform is given by \({\mathcal {L}}T(\cdot , s) = \int _0^\infty T(\cdot , t) \mathop {\textrm{e}}\nolimits ^{- s t} \mathop {}\!\textrm{d}t\). Hence, the Laplace–Fourier transform of the solution is given by

$$\begin{aligned}{} & {} {\mathcal {L}}{\mathcal {F}}T(\xi ,s) \nonumber \\{} & {} \quad =\frac{ [\tau _q^{} \tau _Q^{} s^2 + (\tau _q^{} + \tau _Q^{}) s + 1 + \kappa ^2 \xi ^2] {\mathcal {L}}{\mathcal {F}}f(\xi ,s) }{ \tau _q^{} \tau _Q^{} s^3 + (\tau _q^{} + \tau _Q^{} + \gamma \tau _q^{} \tau _Q^{}) s^2 + [1 + \gamma (\tau _q^{} + \tau _Q^{}) + (\tau _Q^{} + \kappa ^2) \xi ^2] s + \gamma + \left( 1 + \gamma \kappa ^2 \right) \xi ^2 }. \end{aligned}$$
(22)

1.2 Instantaneous heat source \(f = \varphi (x) \delta (t)\)

If the heat source term is equal to \(f = \varphi (x) \delta (t)\), then its Laplace–Fourier transform is equal to \({\mathcal {L}}{\mathcal {F}}f(\xi ,s) = {\mathcal {F}}\varphi (\xi )\).

If \(\varphi (x) = \delta (x)\), then \({\mathcal {F}}\varphi (\xi ) = 1\). In this case, the solution is the fundamental one, and its Laplace–Fourier transform is given by (see Eq. (22))

$$\begin{aligned} {\mathcal {L}}{\mathcal {F}}\varPhi (\xi ,s)= & {} \frac{ s^2 + a_0^{} s + (\tau _q^{} \tau _Q^{})^{-1} \left( 1 + \kappa ^2 \xi ^2 \right) }{ s^3 + a s^2 + b s + c } \nonumber \\= & {} \frac{u^2 + C u + D}{(u - 2 A) \left[ \left( u + A \right) ^2 + B^2 \right] } = \frac{E}{u - 2 A} + \frac{F (u + A) + G}{\left( u + A \right) ^2 + B^2}, \end{aligned}$$
(23)

where

$$\begin{aligned} a = a_0^{} + \gamma , \quad b(\xi ) = \frac{1}{\tau _q^{} \tau _Q^{}} + \gamma a_0^{} + v^2 \xi ^2, \quad c(\xi ) = \frac{\gamma + (1 + \gamma \kappa ^2) \xi ^2}{\tau _q^{} \tau _Q^{}}, \quad a_0^{} = \frac{1}{\tau _q^{}} + \frac{1}{\tau _Q^{}}, \end{aligned}$$
(24)

v is given by Eq. (3), \(s^3 + a s^2 + b s + c = u^3 + \chi u + \psi \),

$$\begin{aligned}{} & {} s = u - \frac{a}{3}, \end{aligned}$$
(25)
$$\begin{aligned}{} & {} \chi (\xi ) = - \frac{a^2}{3} + b, \qquad \psi (\xi ) = 2 \left( \frac{a}{3} \right) ^3 - \frac{a}{3} b + c, \end{aligned}$$
(26)
$$\begin{aligned}{} & {} A(\xi ) = \frac{\alpha + \beta }{2}, \quad B(\xi ) = \sqrt{3} \,\frac{\alpha - \beta }{2}, \quad C = \frac{a_0^{} - 2\gamma }{3}, \quad D(\xi ) = \frac{-2 a_0^2 - \gamma a_0^{} + \gamma ^2}{9} + \frac{1 + \kappa ^2 \xi ^2}{\tau _q^{} \tau _Q^{}}, \end{aligned}$$
(27)
$$\begin{aligned}{} & {} \alpha (\xi ) = \root 3 \of {-\frac{\psi }{2} + \sqrt{\varDelta }}, \qquad \beta (\xi ) = \root 3 \of {-\frac{\psi }{2} - \sqrt{\varDelta }}, \qquad \varDelta (\xi ) = \left( \frac{\chi }{3} \right) ^3 + \left( \frac{\psi }{2} \right) ^2, \end{aligned}$$
(28)

the roots \(\alpha \) and \(\beta \) are chosen so that the equality \(\alpha \beta = - \chi /3\) is valid and the value A is real, and

$$\begin{aligned} E(\xi )= & {} \frac{4 A^2 + 2 A C + D}{9 A^2 + B^2}, \quad F(\xi ) = 1 - E(\xi ), \nonumber \\ G(\xi )= & {} \frac{-3 A^3 + A B^2 + (3 A^2 + B^2) C - 3 A D}{9 A^2 + B^2}. \end{aligned}$$
(29)

Taking into account Eq. (25), one can conclude that the inverse Laplace transform is equal to

$$\begin{aligned} {\mathcal {L}}^{-1} \left[ \frac{E}{u - 2 A} + \frac{F (u + A) + G}{\left( u + A \right) ^2 + B^2} \right] = \mathop {\textrm{e}}\nolimits ^{-\mu _1^{} t} E + \mathop {\textrm{e}}\nolimits ^{-\mu _2^{} t} \left[ F \cos B t + G \frac{\sin B t}{B} \right] , \end{aligned}$$

where

$$\begin{aligned} \mu _1^{} \equiv \mu _1^{}(\xi ) = - 2A + \frac{a}{3}, \quad \mu _2^{} \equiv \mu _2^{}(\xi ) = A + \frac{a}{3}. \end{aligned}$$
(30)

As a result, we obtain from Eq. (23) that the Fourier transform of the fundamental solution is given by Eq. (9).

In the case of a general distribution \(\varphi \) the Fourier transform of the solution T is given by

$$\begin{aligned} {\mathcal {F}}T(\xi ,t) = {\mathcal {F}}\varPhi (\xi ,t) {\mathcal {F}}\varphi (\xi ). \end{aligned}$$
(31)

Asymptotic behavior of the coefficients as \(\xi \rightarrow \infty \)

It follows from Eqs. (26) and (24) that the coefficients \(\chi \) and \(\psi \) are given by

$$\begin{aligned} \chi (\xi ) = v^2 \xi ^2 + c' \quad \text {and}\quad \psi (\xi ) = C_\psi ^{} \xi ^2 + c'', \end{aligned}$$

where v is given by Eq. (3),

$$\begin{aligned} C_\psi ^{} = \frac{1 + \gamma \kappa ^2}{\tau _q^{} \tau _Q^{}} - \frac{1}{3} \left( \frac{1}{\tau _q^{}} + \frac{1}{\tau _Q^{}} + \gamma \right) v^2, \end{aligned}$$
(32)

the expressions for the coefficients \(c'\) and \(c''\) are not significant. Hence the asymptotic behavior of \(\varDelta \) (see Eq. (28)) is given by

$$\begin{aligned} \varDelta = \left( \frac{v^2}{3} \right) ^3 \xi ^6 \left[ 1 + \left( \frac{v^4 c'}{9} + \frac{C_\psi ^2}{4} \right) \left( \frac{v^2}{3} \right) ^{-3} \frac{1}{\xi ^2} + O\left( \frac{1}{\xi ^4} \right) \right] , \end{aligned}$$

and, therefore,

$$\begin{aligned} \sqrt{\varDelta } = \left( \frac{v^2}{3} \right) ^{3/2} |\xi |^3 \left[ 1 + \frac{1}{2} \left( \frac{v^4 c'}{9} + \frac{C_\psi ^2}{4} \right) \left( \frac{v^2}{3} \right) ^{-3} \frac{1}{\xi ^2} + O\left( \frac{1}{\xi ^4} \right) \right] . \end{aligned}$$

This results in the asymptotic behavior of the radicals \(\alpha \) and \(\beta \) (Eq. (28))

$$\begin{aligned} \alpha (\xi ) = \left( \frac{v^2}{3} \right) ^{1/2} |\xi |\left[ 1 - \left( \frac{v^2}{3} \right) ^{-3/2} \frac{C_\psi ^{}}{6} \frac{1}{|\xi |} + \frac{K}{\xi ^2} + O \left( \frac{1}{|\xi |^3} \right) \right] \end{aligned}$$

and

$$\begin{aligned} \beta (\xi ) = - \left( \frac{v^2}{3} \right) ^{1/2} |\xi |\left[ 1 + \left( \frac{v^2}{3} \right) ^{-3/2} \frac{C_\psi ^{}}{6} \frac{1}{|\xi |} + \frac{K}{\xi ^2} + O \left( \frac{1}{|\xi |^3} \right) \right] , \end{aligned}$$

where the expression for the coefficient K is not significant.

The asymptotic behavior of the coefficients A, B and D (Eq. (27)) follows from the above asymptotics and is given by

$$\begin{aligned} A= & {} - \frac{C_\psi ^{}}{2 v^2} + O \left( \frac{1}{\xi ^2} \right) \equiv \frac{1}{6} \left( \frac{1}{\tau _q^{}} + \frac{1}{\tau _Q^{}} + \gamma \right) - \frac{1 + \gamma \kappa ^2}{2 (\tau _Q^{} + \kappa ^2)} + O \left( \frac{1}{\xi ^2} \right) , \\ B= & {} v |\xi |\left[ 1 + O\left( \frac{1}{\xi ^2} \right) \right] \quad \text {and}\quad D = \frac{\kappa ^2}{\tau _q^{} \tau _Q^{}} \xi ^2 \left[ 1 + O\left( \frac{1}{\xi ^2} \right) \right] . \end{aligned}$$

Therefore, the asymptotic behavior of the coefficients E, F and G (Eq. (29)) is given by

$$\begin{aligned} E = E_\infty ^{} + O \left( \frac{1}{\xi ^2} \right) , \quad F = F_\infty ^{} + O \left( \frac{1}{\xi ^2} \right) \quad \text {and}\quad G = G_\infty ^{} + O \left( \frac{1}{\xi ^2} \right) , \end{aligned}$$
(33)

where

$$\begin{aligned} E_\infty ^{} = \frac{\kappa ^2}{\tau _Q^{} + \kappa ^2}, \quad F_\infty ^{} = 1 - E_\infty ^{}, \end{aligned}$$
(34)

and the expression for the coefficient \(G_\infty ^{}\) is unimportant.

The asymptotic behavior of the coefficients \(\mu _1^{}\) and \(\mu _2^{}\) (Eq. (30)) is given by

$$\begin{aligned} \mu _1^{}(\xi ) = \mu _{1,\infty }^{} + O \left( \frac{1}{\xi ^2} \right) , \qquad \mu _2^{}(\xi ) = \mu _{2,\infty }^{} + O \left( \frac{1}{\xi ^2} \right) . \end{aligned}$$

where

$$\begin{aligned} \mu _{1,\infty }^{} = \frac{1 + \gamma \kappa ^2}{\tau _Q^{} + \kappa ^2}, \qquad \mu _{2,\infty }^{} = \frac{1}{2} \left( \frac{1}{\tau _q^{}} + \frac{1}{\tau _Q^{}} + \gamma - \frac{1 + \gamma \kappa ^2}{\tau _Q^{} + \kappa ^2} \right) . \end{aligned}$$
(35)

As a result, we obtain the asymptotics

$$\begin{aligned} \begin{aligned} \mathop {\textrm{e}}\nolimits ^{-\mu _1^{} t} E&= \mathop {\textrm{e}}\nolimits ^{-\mu _{1,\infty }^{} t} E_\infty ^{} + O \left( \frac{1}{\xi ^2} \right) \quad \text {as}\quad \xi \rightarrow \infty , \\ \mathop {\textrm{e}}\nolimits ^{-\mu _2^{} t} F \cos B t&= \mathop {\textrm{e}}\nolimits ^{-\mu _{2,\infty }^{} t} F_\infty ^{} \cos v \xi t + O \left( \frac{1}{\xi ^2} \right) \quad \text {as}\quad \xi \rightarrow \infty , \\ \mathop {\textrm{e}}\nolimits ^{-\mu _2^{} t} G \frac{\sin B t}{B}&= \mathop {\textrm{e}}\nolimits ^{-\mu _{2,\infty }^{} t} G_\infty ^{} \frac{\sin v \xi t}{v \xi } + O \left( \frac{1}{\xi ^2} \right) \quad \text {as}\quad \xi \rightarrow \infty . \end{aligned} \end{aligned}$$
(36)

The fundamental solution

Taking into account the asymptotics (36), we obtain the asymptotic behavior of the Fourier transform

$$\begin{aligned} {\mathcal {F}}\varPhi = \mathop {\textrm{e}}\nolimits ^{-\mu _{1,\infty }^{} t} E_\infty ^{} + \mathop {\textrm{e}}\nolimits ^{-\mu _{2,\infty }^{} t} \left[ F_\infty ^{} \cos v \xi t + G_\infty ^{} \frac{\sin v \xi t}{v \xi } \right] + O \left( \frac{1}{\xi ^2} \right) \quad \text {as}\quad \xi \rightarrow \infty , \end{aligned}$$

where the coefficients are given by Eqs. (34) and (35). Note, that

$$\begin{aligned}{} & {} {\mathcal {F}}^{-1} 1 = \delta (x), \qquad {\mathcal {F}}^{-1} \left[ \cos \!\left( v \xi t \right) \right] = \frac{1}{2} \left[ \delta \left( x - v t \right) + \delta \left( x + v t \right) \right] ,\\{} & {} {\mathcal {F}}^{-1} \left[ \frac{1}{\xi } \sin \!\left( v \xi t \right) \right] = \frac{1}{2} \textbf{1}_{(-v t, \,v t)}^{}, \end{aligned}$$

where \({\mathcal {F}}^{-1}\) is the inverse Fourier transform, and

$$\begin{aligned} \textbf{1}_S^{}(x) = {\left\{ \begin{array}{ll} 1, &{} x \in S, \\ 0, &{} x \notin S, \end{array}\right. } \end{aligned}$$
(37)

is the indicator function of a set S. Therefore, one can conclude that the fundamental solution can be represented in the form

$$\begin{aligned} \varPhi (x,t) = \varPhi _{\textrm{sing},1}^{}(x,t) + \varPhi _{\textrm{sing},2}^{}(x,t) + \varPhi _\textrm{disc}^{}(x,t) + \varPhi _\textrm{cont}^{}(x,t), \end{aligned}$$

where \(\varPhi _{\textrm{sing},1}^{}\) and \(\varPhi _{\textrm{sing},2}^{}\) are singular parts given by

$$\begin{aligned} \varPhi _{\textrm{sing},1}^{}(x,t) = \mathop {\textrm{e}}\nolimits ^{-\mu _{1,\infty }^{} t} E_\infty ^{} \delta (x) \end{aligned}$$

and

$$\begin{aligned} \varPhi _{\textrm{sing},2}^{}(x,t) = \mathop {\textrm{e}}\nolimits ^{-\mu _{2,\infty }^{} t} F_\infty ^{} \frac{1}{2} \left[ \delta \left( x - v t \right) + \delta \left( x + v t \right) \right] , \end{aligned}$$

\(\varPhi _\textrm{disc}^{}\) is a discontinuous part given by

$$\begin{aligned} \varPhi _\textrm{disc}^{}(x,t) = \mathop {\textrm{e}}\nolimits ^{-\mu _{2,\infty }^{} t} G_\infty ^{} \frac{1}{2 v} \textbf{1}_{(-v t, \,v t)}^{}, \end{aligned}$$

\(\varPhi _\textrm{cont}^{}(x,t)\) is a continuous function, since its Fourier transform has the asymptotic behavior \(O(1/\xi ^2)\) as \(\xi \rightarrow \infty \) [64].

Note, that the system (1) is hyperbolic and the maximum absolute value of the eigenvalues is equal to v. Therefore, perturbations propagate at the speed v. Since the initial disturbance is at the origin, one can conclude that \(\varPhi (x,t) = 0\) for \(|x |> v t\), and hence \(\varPhi _\textrm{cont}^{}(x,t) = 0\) for \(|x |> v t\) as well. Finally, we define the regular part by \(\varPhi _\textrm{reg}^{} = \varPhi _\textrm{disc}^{} + \varPhi _\textrm{cont}^{}\), which is equal to zero for \(|x |> v t\) and continuous for \(|x |< v t\). As a result, the solution \(\varPhi \) takes the form of Eq. (10).

The values of the dimensionless parameters

In this section, we denote dimensional quantities with an asterisk, as in Refs. [41, 42]. The quantities have been taken from Refs. [49,50,51,52].

The speed of sound (in a transverse phonon gas) is \(v^* = 3186 \,\text {m/s}\). The length of a sample is \(L^* = 7.9 \,\text {mm}\). The heat pulse duration is \(\Delta t^* = 0.24\,\mu \text {s}\). The mass density is \(\rho = 2866 \,\text {kg/m}^3\). The thermal diffusivity is given by

$$\begin{aligned} \alpha = \frac{k}{\rho c}, \end{aligned}$$

where k, \(\rho \) and c are the thermal conductivity, mass density and specific heat, respectively. The dimensional propagation speed is given by [51]

$$\begin{aligned} v^* = \sqrt{\frac{1}{\tau _q^*} \left( \alpha + \frac{{\kappa ^*}^2}{\tau _Q^*} \right) }. \end{aligned}$$

Therefore, the dimensional dissipation parameter \(\kappa ^*\) can be calculated from this formula.

Dimensionless values and parameters are given by [49,50,51,52]

$$\begin{aligned} x = \frac{x^*}{L}, \quad t = \frac{\alpha t^*}{L^2}, \quad \tau _{q,Q}^{} = \frac{\alpha \tau _{q,Q}^*}{L^2}, \quad \kappa = \frac{\kappa ^*}{L}, \quad \gamma = \frac{\Delta t^*}{\rho c} \gamma ^*, \quad v = \frac{L v^*}{\alpha }. \end{aligned}$$

Dimensional parameters for the sample at 11 K are \(k = 8573 \,\text {W/(m K)}\), \(c = 1.118 \,\text {J/(kg K)}\), \(\tau _q^* = 0.471\,\mu \text {s}\), \(\tau _Q^* = 0.18\,\mu \text {s}\), \(\gamma ^* = 3.34 \,\text {W/(mm}^3 \text { K)}\). Then \(\alpha = 2.68 \,\text {m}^2\text {/s}\) and, therefore, \(\tau _q^{} = 0.0202\), \(\tau _Q^{} = 0.00772\), \(\kappa = 0.0780\), \(\gamma = 0.250\), see Fig. 1.

Dimensional parameters for the sample at 13 K are \(k = 10{,}200 \,\text {W/(m K)}\), \(c = 1.8 \,\text {J/(kg K)}\), \(\tau _q^* = 0.586\,\mu \text {s}\), \(\tau _Q^* = 0.22\,\mu \text {s}\), \(\gamma ^* = 2.8 \,\text {W/(mm}^3\text { K)}\). Then \(\alpha = 1.98 \,\text {m}^2\text {/s}\) and, therefore, \(\tau _q^{} = 0.0186\), \(\tau _Q^{} = 0.0070\), \(\kappa = 0.118\), \(\gamma = 0.130\), see Fig. 2.

Dimensional parameters for the sample at 14.5 K are \(k = 10{,}950 \,\text {W/(m K)}\), \(c = 2.543 \,\text {J/(kg K)}\), \(\tau _q^* = 0.65\,\mu \text {s}\), \(\tau _Q^* = 0.24\,\mu \text {s}\), \(\gamma ^* = 2.31 \,\text {W/(mm}^3 \text { K)}\). Then \(\alpha = 1.50 \,\text {m}^2\text {/s}\) and, therefore, \(\tau _q^{} = 0.0156\), \(\tau _Q^{} = 0.0058\), \(\kappa = 0.140\), \(\gamma = 0.076\), see Fig. 3.

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Rukolaine, S.A. Hyperbolicity of the ballistic-conductive model of heat conduction: the reverse side of the coin. Z. Angew. Math. Phys. 75, 50 (2024). https://doi.org/10.1007/s00033-023-02176-6

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