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From the Relativistic Mixture of Gases to the Relativistic Cucker–Smale Flocking

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Abstract

We present a relativistic model for a mixture of Euler gases with multiple temperatures. For the proposed relativistic model, we explicitly determine production terms resulting from the interchange of energy–momentum between the constituents via the entropy principle. We use the analogy with the homogeneous solutions of a mixture of gases and the thermomechanical Cucker–Smale (in short TCS) flocking model in a classical setting (Ha and Ruggeri in Arch Ration Mech Anal 223:1397–1425, 2017) to derive a relativistic counterpart of the TCS model. Moreover, we employ the theory of a principal subsystem to derive the relativistic Cucker–Smale (in short CS) model. For the derived relativistic CS model, we provide a sufficient framework leading to the exponential flocking in terms of communication weights and also show that the relativistic CS model reduces to the classical CS model, as the speed of light tends to infinity in any finite-time interval.

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Acknowledgements

The work of S.-Y. Ha is partially supported by National Research Foundation of Korea Grant (NRF-2017R1A2B2001864) funded by the Korea Government, and the work of T. Ruggeri is supported by the National Group of Mathematical Physics GNFM-INdAM.

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Correspondence to Seung-Yeal Ha.

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Appendix A: Proof of Lemma 6.6

Appendix A: Proof of Lemma 6.6

In this appendix, we provide a detailed proof of Lemma 6.6. Since we have two vectors \({\mathbf{x}}\) and \({\mathbf{y}}\), we only need to consider when \({\mathbf{x}},{\mathbf{y}}\in {\mathbb {R}}^2\) by changing the basis. Moreover, it is possible to choose \({\mathbf{x}}\) as a first direction. Therefore, we set

$$\begin{aligned} {\mathbf{x}}:=(x_1,0) \quad \text{ and } \quad {\mathbf{y}}:=(y_1,y_2). \end{aligned}$$

Then, we have

$$\begin{aligned} \begin{aligned}&({\mathbf{x}}-{\mathbf{y}})\cdot \left( \frac{{\mathbf{x}}}{F(x^2)}-\frac{{\mathbf{y}}}{F(y^2)}\right) \\&\quad =(x_1-y_1)\left( \frac{x_1}{F(x^2)}-\frac{y_1}{F(y^2)}\right) +\frac{y_2^2}{F(y^2)}\\&\quad =(x_1-y_1)\left( \frac{x_1F(y^2)-y_1F(x^2)}{F(x^2)F(y^2)}\right) +\frac{y_2^2}{F(y^2)}\\&\quad =(x_1-y_1)\left( \frac{(x_1-y_1)F(y^2)+y_1(F(y^2)-F(x^2))}{F(x^2)F(y^2)}\right) +\frac{y_2^2}{F(y^2)}\\&\quad =\frac{(x_1-y_1)^2}{F(x^2)F(y^2)}\left( F(y^2)+y_1\frac{F(y^2)-F(x^2)}{x_1-y_1}\right) +\frac{y_2^2}{F(y^2)}\\&\quad =\frac{(x_1-y_1)^2}{F(x^2)F(y^2)}\left( F(y^2)-y_1\frac{x^2-y^2}{x_1-y_1}\frac{F(y^2)-F(x^2)}{y^2-x^2}\right) +\frac{y_2^2}{F(y^2)}. \end{aligned} \end{aligned}$$

On the other hand, note that

$$\begin{aligned} \begin{aligned} x^2-y^2&=({\mathbf{x}}-{\mathbf{y}})\cdot ({\mathbf{x}}+{\mathbf{y}})=(x_1-y_1)(x_1+y_1)-y^2_2, \\ F(r)-F(s)&=\sqrt{\frac{r}{c^2}+\frac{r}{g(r)}}-\sqrt{\frac{s}{c^2}+\frac{s}{g(s)}}=\frac{\frac{r}{c^2}-\frac{s}{c^2}+\frac{r}{g(r)}-\frac{s}{g(s)}}{F(r)+F(s)}\\&=:\frac{r-s}{F(r)+F(s)}\left( \frac{1}{c^2}+G(r,s)\right) , \end{aligned} \end{aligned}$$

since g(r) is an increasing function, and it follows from (6.7)\(_2\) that \(\frac{r}{g(r)}\) is also an increasing function and G(rs) is nonnegative. Thus, we have

$$\begin{aligned}&({\mathbf{x}}-{\mathbf{y}})\cdot \left( \frac{{\mathbf{x}}}{F(x^2)}-\frac{{\mathbf{y}}}{F(y^2)}\right) \\&\quad =\frac{(x_1-y_1)^2}{F(x^2)F(y^2)}\left( F(y^2)-y_1\frac{x^2-y^2}{x_1-y_1}\frac{F(y^2)-F(x^2)}{y^2-x^2}\right) +\frac{y_2^2}{F(y^2)} \\&\quad =\frac{(x_1-y_1)^2}{F(x^2)F(y^2)}\left( F(y^2)-y_1(x_1+y_1)\frac{\frac{1}{c^2}+G(x^2,y^2)}{F(x^2)+F(y^2)}\right) \\&\qquad +\frac{y_1(x_1-y_1)y_2^2}{F(x^2)F(y^2)}\frac{\frac{1}{c^2}+G(x^2,y^2)}{F(x^2)+F(y^2)}+\frac{y_2^2}{F(y^2)}\\&\quad =\frac{(x_1-y_1)^2}{F(x^2)F(y^2)(F(x^2)+F(y^2))}\\&\qquad \times \left( F(x^2)F(y^2)+F(y^2)^2-y_1(x_1+y_1)\left( \frac{1}{c^2}+G(x^2,y^2)\right) \right) \\&\qquad +\frac{y_2^2}{F(x^2)F(y^2)(F(x^2)+F(y^2))}\\&\qquad \times \left( F(x^2)^2+F(x^2)F(y^2)+y_1(x_1-y_1)\left( \frac{1}{c^2}+G(x^2,y^2)\right) \right) . \end{aligned}$$

However, one has

$$\begin{aligned} \begin{aligned}&F(y^2)^2 -y_1^2\left( \frac{1}{c^2}+G(x^2,y^2)\right) =y^2\left( \frac{1}{c^2}+\frac{1}{g(y^2)}\right) -y_1^2\left( \frac{1}{c^2}+G(x^2,y^2)\right) \\&\quad =y^2\left( \frac{1}{g(y^2)}-G(x^2,y^2)\right) +y_2^2\left( \frac{1}{c^2}+G(x^2,y^2)\right) , \\&y^2\left( \frac{1}{g(y^2)}-G(x^2,y^2)\right) =y^2\left( \frac{1}{g(y^2)}-\frac{\frac{y^2}{g(y^2)}-\frac{x^2}{g(x^2)}}{y^2-x^2}\right) \\&\quad =\frac{x^2y^2(g(y^2)-g(x^2))}{g(x^2)g(y^2)(y^2-x^2)}. \end{aligned} \end{aligned}$$

We now use the estimate of \(\frac{r}{g(r)}\) in Lemma 6.5 to obtain

$$\begin{aligned} \begin{aligned}&\frac{r}{g(r)}\ge 1+\frac{1}{c^4}, \quad \frac{g(r)-g(s)}{r-s}=g'(a),\quad \min \{r,s\}\leqq a\leqq \max \{r,s\},\\&\quad \text{ and } \text{ thus, } \quad y^2\left( \frac{1}{g(y^2)}-G(x^2,y^2)\right) \ge \left( 1+\frac{1}{c^4}\right) ^2\min _{0\leqq a\leqq R} g'(a)=:C(R) \end{aligned} \end{aligned}$$

to see that

$$\begin{aligned}&({\mathbf{x}}-{\mathbf{y}})\cdot \left( \frac{{\mathbf{x}}}{F(x^2)}-\frac{{\mathbf{y}}}{F(y^2)}\right) \\&\quad \ge \frac{(x_1-y_1)^2}{F(x^2)F(y^2)(F(x^2)+F(y^2))}\left( F(x^2)F(y^2) \right. \\&\qquad \left. -(x_1y_1)\left( \frac{1}{c^2}+G(x^2,y^2)\right) +C(R)\right) \\&\qquad +\frac{y_2^2}{F(x^2)F(y^2)(F(x^2)+F(y^2))}\\&\qquad \times \left( ((x_1-y_1)^2+y_1(x_1-y_1))\left( \frac{1}{c^2}+G(x^2,y^2)\right) \right. \\&\qquad \left. +F(x^2)^2+F(x^2)F(y^2)\right) \\&\quad =\frac{(x_1-y_1)^2}{F(x^2)F(y^2)(F(x^2)+F(y^2))}\left( F(x^2)F(y^2) \right. \\&\qquad \left. -(x_1y_1)\left( \frac{1}{c^2}+G(x^2,y^2)\right) +C(R)\right) \\&\qquad +\frac{y_2^2}{F(x^2)F(y^2)(F(x^2)+F(y^2))}\\&\qquad \times \left( (x_1-y_1)x_1\left( \frac{1}{c^2}+G(x^2,y^2)\right) +F(x^2)^2+F(x^2)F(y^2)\right) \\&\quad =\frac{(x_1-y_1)^2}{F(x^2)F(y^2)(F(x^2)+F(y^2))}\left( F(x^2)F(y^2) \right. \\&\qquad \left. -(x_1y_1)\left( \frac{1}{c}+G(x^2,y^2)\right) +C(R)\right) \\&\qquad +\frac{y_2^2}{F(x^2)F(y^2)(F(x^2)+F(y^2))}\\&\qquad \times \left( x_1^2\left( \frac{1}{c^2}+G(x^2,y^2)\right) +F(x^2)^2+F(x^2)F(y^2) \right. \\&\qquad \left. -x_1y_1\left( \frac{1}{c^2}+G(x^2,y^2)\right) \right) . \end{aligned}$$

On the other hand, since g(r) is an increasing function of r, we also have the following estimate:

$$\begin{aligned} \frac{1}{c^2}+G(r,s)=\frac{1}{c^2}+\frac{\frac{r}{g(r)}-\frac{s}{g(s)}}{r-s}=\frac{1}{c^2}+\frac{1}{g(r)}+s\frac{\frac{1}{g(r)}-\frac{1}{g(s)}}{r-s}\leqq \frac{1}{c^2}+\frac{1}{g(r)}. \end{aligned}$$

By the symmetry between r and s, we have

$$\begin{aligned} \frac{1}{c^2}+G(r,s)\leqq \min \left\{ \frac{1}{c^2}+\frac{1}{g(r)},\frac{1}{c^2}+\frac{1}{g(s)}\right\} \leqq \sqrt{\frac{1}{c^2}+\frac{1}{g(r)}}\sqrt{\frac{1}{c^2}+\frac{1}{g(s)}}. \end{aligned}$$

Thus, we use \(x_1=|{\mathbf{x}}|=x\) and \(y_1\leqq |{\mathbf{y}}|=y\) to obtain

$$\begin{aligned} x_1y_1\left( \frac{1}{c^2}+G(x^2,y^2)\right)&\leqq xy\left( \frac{1}{c^2}+G(x^2,y^2)\right) \\&\leqq \sqrt{\frac{x^2}{c^2}+\frac{x^2}{g(x^2)}}\sqrt{\frac{y^2}{c^2}+\frac{y^2}{g(x^2)}}=F(x^2)F(y^2). \end{aligned}$$

This implies that

$$\begin{aligned}&({\mathbf{x}}-{\mathbf{y}})\cdot \left( \frac{{\mathbf{x}}}{F(x^2)}-\frac{{\mathbf{y}}}{F(y^2)}\right) \\&\quad \ge \frac{(x_1-y_1)^2}{F(x^2)F(y^2)(F(x^2)+F(y^2))}C(R)\\&\qquad +\frac{y_2^2}{F(x^2)F(y^2)(F(x^2)+F(y^2))}\left( x_1^2\left( \frac{1}{c^2}+G(x^2,y^2)\right) +F(x^2)^2\right) \\&\quad \ge (x_1-y_1)^2 \frac{C(R)}{2F(R)^3}+y_2^2 \frac{F(x^2)}{F(y^2)(F(x^2)+F(y^2))}\quad (\because G(x^2,y^2)\ge 0)\\&\quad \ge (x_1-y_1)^2 \frac{C(R)}{2F(R)^3}+y_2^2 \frac{1}{2F(R)^2}\ge |{\mathbf{x}}-\mathbf{y}|^2\min \left\{ \frac{C(R)}{2F(R)^3},\frac{1}{2F(R)^2}\right\} , \end{aligned}$$

where we used \(F\ge 1\) in the last inequality. Finally, we choose

$$\begin{aligned} {\varLambda }(R):=\min \left\{ \frac{C(R)}{2F(R)^3},\frac{1}{2F(R)^2}\right\} \end{aligned}$$

to obtain the desired estimate. This completes the proof.

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Ha, SY., Kim, J. & Ruggeri, T. From the Relativistic Mixture of Gases to the Relativistic Cucker–Smale Flocking. Arch Rational Mech Anal 235, 1661–1706 (2020). https://doi.org/10.1007/s00205-019-01452-y

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