Multiscale Conservation Laws Driven by Lévy Stable and Linnik Diffusions: Asymptotics, Shock Creation, Preservation and Dissolution

Abstract

Recent work has shown that the solutions of the fractal conservation laws driven by Lévy $\alpha$-stable diffusions exhibit shocks for bounded, odd, and convex on the positive half-line, initial data when the parameter $\alpha < 1.$ We study the analogous situation for the Lévy $\alpha$-Linnik diffusions in which case the local behavior is strikingly different, although we are able to establish analytically that the large time behavior of the two types of conservation laws are similar. But the main new insights obtained via large-scale numerical experiments is that, for any $0<\alpha \le 2$, the conservation laws driven by $\alpha$-Linnik diffusions display shocks that do not dissipate over time, while those for $\alpha$-stable diffusion ($0<\alpha \le 1$) do. We formulate rigorous conjectures based on these numerical experiments.

Appendix: A Numerical Method for Stable and Linnik Conservation Laws

In this section, we present the details of the numerical method used to produce results of Sect. 6, adapting to the Linnik case the methodology developed in [12].

Let $\delta _{t}>0 \$, and $\delta x >0$ be the time and space steps. The scheme consists in computing approximate values $u_{i}^{n} \$ of the solution to (4.1) on the lattice $\ [n\delta t, (n+1)\delta t) \times [i \delta x, (i+1)\delta x) \$, $n \in \mathbb {N}$, and $i \in \mathbb {Z},$

$$\begin{aligned} u_{i}^{0}=\frac{1}{\delta x}\int _{i \delta x}^{(i+1)\delta x}u_{o}(x)dx, \end{aligned}$$

(6.1)

$$\begin{aligned} \frac{\delta t}{\delta x}\left( u_{i}^{n+1}-u_{i}^{n} \right) + F\left( u_{i}^{n}, u_{i+1}^{n}\right) - F\left( u_{i-1}^{n}, u_{i}^{n}\right) + \delta x \mathcal {L}^{\delta x}[u^{n+1}]_{i} =0, \end{aligned}$$

(6.2)

where $F$ is a numerical Burgers flux corresponding to the continuous flux $f$, and $\mathcal {L}^{\delta x}$ is a discretization of the non-local term $\mathcal {L},$ where $\mathcal {L}^{\delta x}$ is the discretization of $\mathcal {L}, \$ and the numerical flux is defined as follows:

$$\begin{aligned} F(a,b)=\left\{ \begin{array}{ll} \displaystyle \min \limits _{ a \le u \le b}\frac{u^2}{2}, &{}\quad \text {if } a \le b ; \\ \displaystyle \max \limits _{b \le u \le a}\frac{u^2}{2}, &{}\quad \text {if } a > b. \end{array} \right. \end{aligned}$$

The assumptions on $\mathcal {L}^{\delta x}$ are as follows:

• (i) $l^{\infty }(\mathbb {Z})\ni \nu \mapsto \mathcal {L}^{\delta x}[\nu ]\in l^{\infty }(\mathbb {Z}) \$ is linear;

• (ii) $\forall \nu \in l^{\infty }(\mathbb {Z}), \$ if $\ \left( i_{k} \right) _{k \in \mathbb {N}} \$ is a sequence in $\mathbb {Z}$ such that $\ \lim _{k\rightarrow \infty }\nu _{i_{k}}= \qquad \sup _{j \in \mathbb {Z}}\nu _{j}, \$ then $\ \lim \inf _{k\rightarrow \infty }\mathcal {L}^{\delta x}[\nu ]_{i_{k}} \ge 0$;

• (iii) If $\tau : l^{\infty }(\mathbb {Z})\mapsto l^{\infty }(\mathbb {Z}) \$ is the left translation $\ \tau (\nu )_{i}=\nu _{i+1},$ then $\ \tau \mathcal {L}^{\delta x}=\mathcal {L}^{\delta x}\tau$;

• (iv) $\exists A^{\delta x} >0 \$ such that, for all $\nu \in l^{\infty }(\mathbb {Z}), \$ $\ \mathcal {L}^{\delta x}[\nu ]_{0} \$ only depends on $\ \left( \nu _{j} \right) _{|j| \le A^{\delta x}}. \$

A detailed description of the implementation of the numerical algorithm is provided below.

For a chosen space step $\delta x >0 , \$ formula (4.2) makes it easy to write a discretization of $\ \mathcal {L}:$ we approximate each integral using the basic quadratic rule on the mesh $\ \left( [j\delta x, (j+1)\delta x) \right) _{j \in \mathbb {Z}} \$, and we use the finite difference approximation of the derivative. However, such an approximation would use all of $\ (\nu _{j})_{j \in \mathbb {Z}} \$ in order to compute $\ \mathcal {L}^{\delta x}[\nu ]_{i}; \$ in practical applications, the considered functions are usually constant near $\ -\infty \$ and $\ +\infty . \$ We take this into account when discretizating $\ \mathcal {L} \$ and use the mesh $\ \left( [j\delta x, (j+1)\delta x) \right) _{j \in \mathbb {Z}} \$ only up to $\ |z|=J_{\delta x}\delta x \$ (for some integer $\ J_{\delta x} \$ such that $\ J_{\delta x}\delta x \rightarrow \infty \$ as $\ \delta x \rightarrow 0 \$), approximating the remaining parts with two unbounded space steps $\ (-\infty , -J_{\delta x}\delta x] \$ and $\ [J_{\delta x}\delta x, +\infty ). \$ This leads to the scheme,

$$\begin{aligned} \mathcal {L}^{\delta x}[\nu ]_{i}= & {} -c(\alpha )\sum _{0 < |j| < r/\delta x}\delta x \left( \nu _{i+j}-\nu _{i} - \frac{\nu _{i+1}-\nu _{i-1}}{2 \delta x} \right) \Lambda ^{'}(j \delta x)\nonumber \\&-c(\alpha )\sum _{r/\delta x < |j| \le J_{\delta x}} \delta x \left( \nu _{i+j}-\nu _{i} \right) \Lambda ^{'}(j \delta x)\nonumber \\&-c(\alpha )\sum _{j <-J_{\delta x}} \delta x \left( \nu _{i-J_{\delta x}-1}-\nu _{i} \right) \Lambda ^{'}(j \delta x)\nonumber \\&-c(\alpha )\sum _{j>J_{\delta x}} \delta x \left( \nu _{i+J_{\delta x}+1}-\nu _{i} \right) \Lambda ^{'}(j \delta x) \end{aligned}$$

(6.3)

where

$$\begin{aligned} \Lambda ^{'}(j \delta x)=\frac{\alpha }{2|j \delta x|}\mathbb E\exp \left( -\left| \frac{j \delta x}{X}\right| ^{\alpha } \right) , \end{aligned}$$

and $X$ has a $\alpha$-stable distribution.

Additionally, we can estimate the approximate value of $\ \sum _{j>J_{\delta x}}\Lambda ^{'}(j \delta x)$ using the formula (2.22) for the Lévy measure of $\alpha$-Linnik distribution,

$$\begin{aligned}&\int _{J_{\delta x}}^{\infty }\int _{0}^{\infty }f_{\alpha }\left( \frac{j \delta x}{w^{1/\alpha }} \right) .\frac{e^{-w}}{w^{1 + 1/ \alpha }}dw dj \\&\quad = \int _{0}^{\infty }\int _{J_{\delta x}}^{\infty }f_{\alpha }\left( \frac{j \delta x}{w^{1/\alpha }} \right) dj.\frac{e^{-w}}{w^{1 + 1/ \lambda }}dw \\&\quad \sim \int _{0}^{\infty }\int _{J_{\delta x}}^{\infty }\left( \frac{w^{1/\alpha }}{\delta x} \right) f_{\alpha }\left( \frac{j \delta x}{w^{1/\alpha }} \right) d\left( \frac{j \delta x}{w^{1/\alpha }} \right) \frac{e^{-w}}{w^{1 + 1/ \alpha }}dw \\&\quad =\int _{0}^{\infty }\left( \frac{w^{1/\lambda }}{\delta x} \right) . \left( \frac{k_{\alpha }}{\left( \frac{j\delta x}{w^{1/\alpha }}\right) ^\alpha } \right) .\frac{e^{-w}}{w^{1 + 1/ \alpha }}dw \\&\quad =\int _{0}^{\infty }\frac{k_{\alpha }}{\delta x^ {\alpha +1} J_{\delta x}^\alpha }e^{-w}dx \\&\quad =\frac{k_{\alpha }}{\delta x^{\alpha +1}J_{\delta x}^{\alpha }}, \end{aligned}$$

where $\ k_{\alpha }= { \sin (\frac{\pi \alpha }{2})\Gamma (\alpha )}/{\pi }. \$ Therefore, the approximate value of $\ \mathcal {L}^{\delta x}[\nu ]_{i} \$ is given by the following formula,

$$\begin{aligned} \mathcal {L}^{\delta x}[\nu ]_{i} =&-c(\alpha )\sum _{0 < |j| < r/\delta x}\delta x \left( \nu _{i+j}-\nu _{i}\right) \frac{\alpha }{2|j \delta x|}\mathbb E\exp \left( -\left| \frac{j \delta x}{X}\right| ^{\alpha } \right) \\&-c(\alpha )\left( \nu _{i-J_{\delta x}-1}-\nu _{i} \right) \frac{k_{\alpha }}{(\delta x J_{\delta x})^\alpha } -c(\alpha )\left( \nu _{i+J_{\delta x}+1}-\nu _{i} \right) \frac{k_{\alpha }}{(\delta x J_{\delta x})^\alpha } \end{aligned}$$

Finally, we have

$$\begin{aligned} \mathcal {L}^{\delta x}[\nu ]_{i} =&-c(\alpha )\sum _{0 < |j| < r/\delta x}\delta x \left( \nu _{i+j}-\nu _{i}\right) \frac{\alpha }{2|j \delta x|}\mathbb E\exp \left( -\left| \frac{j \delta x}{X}\right| ^{\alpha } \right) .\nonumber \\&-c(\alpha )\left( \nu _{i-J_{\delta x}-1}-\nu _{i} \right) \frac{\sin (\frac{\pi \alpha }{2})\Gamma (\alpha )}{\pi (\delta x J_{\delta x})^\alpha } -c(\alpha )\left( \nu _{i+J_{\delta x}+1}-\nu _{i} \right) \frac{\sin (\frac{\pi \alpha }{2})\Gamma (\alpha )}{\pi (\delta x J_{\delta x})^\alpha }.\nonumber \\ \end{aligned}$$

(6.4)

The step-by-step numerical algorithm for computation of the solution of the Linnik conservation law is as follows:

Step 1: Define $\nu ,$ $\ W,$ $\ \mathcal {L}^{\delta x} \$ and $G^{\delta x},$ by the following formulae:

• (i)
  $$\begin{aligned} \nu =\left( h_{i}^{n}1_{\mathbb {Z}\setminus [-m,m]}(i)\right) _{|i|\in \mathbb {Z}}. \end{aligned}$$

  For example, for $\ m=2, \$

  $$\begin{aligned} \nu ^T = \left[ \begin{array}{@{}cc|c@{}} \ldots h_{-4}^{n} , h_{-3}^{n} , 0 ,0 , 0 , 0 ,0 ,h_{3}^{n} , h_{4}^{n} , \ldots \end{array}\right] \end{aligned}$$

  so that $\ \nu _{2}=0, \$ $\ \nu _{-4}= h_{-4}^{n} \$, etc.

• (ii)
  $$\begin{aligned} W&= \left( U_{i}^{n+1}\right) _{|i|\le m} = \left( \begin{array}{ccc} U_{-m}^{n+1} \\ \vdots \\ U_{0}^{n+1} \\ \vdots \\ U_{m}^{n+1} \end{array} \right) _{(2m+1)\times 1} \end{aligned}$$
• (iii)
  $$\begin{aligned} \mathcal {L}^{\delta x}[\nu ]_{i}= & {} -c(\alpha )\sum _{0< |j|\le J_{\delta x}}\delta x (\nu _{i+j}-\nu _{i}) \Lambda ^{'}(j \delta x) \\&-c(\alpha )(\nu _{i-J_{\delta x}-1}-\nu _{i}) \frac{\sin (\frac{\pi \alpha }{2})\Gamma (\alpha )}{\pi (\delta x J_{\delta x})^\alpha }\\&-c(\alpha ) (\nu _{i+J_{\delta x}+1}-\nu _{i}) \frac{\sin (\frac{\pi \alpha }{2})\Gamma (\alpha )}{\pi (\delta x J_{\delta x})^\alpha }\\= & {} -c(\alpha )\sum _{0< |j|\le J_{\delta x}}\delta x (h_{i+j}^{n}\!-\!h_{i}^{n})\Lambda ^{'}(j \delta x)\! -\!c(\alpha ) (h_{i-J_{\delta x}-1}^{n}-h_{i}^{n})\frac{\sin (\frac{\pi \alpha }{2})\Gamma (\alpha )}{\pi (\delta x J_{\delta x})^\alpha }\\&-c(\alpha )(h_{i+J_{\delta x}+1}^{n}-h_{i}^{n})\frac{\sin (\frac{\pi \alpha }{2})\Gamma (\alpha )}{\pi (\delta x J_{\delta x})^\alpha }\\= & {} c_1\sum _{0< |j|\le J_{\delta x}}(h_{i+j}^{n}-h_{i}^{n})\Lambda ^{'}(j \delta x)+c_2 \left( h_{i-J_{\delta x}-1}^{n}-h_{i}^{n} \right) \\&+\, c2 \left( h_{i+J_{\delta x}+1}^{n}-h_{i}^{n} \right) , \end{aligned}$$

  where $c_1=-c(\alpha ) \$, and $\ c_2= {-c(\alpha ) \sin (\frac{\pi \alpha }{2})\Gamma (\alpha )}/({\pi (\delta x J_{\delta x})^\alpha } ).$

• (iv) The formula for the 1st column of the symmetric Toeplitz matrix $G^{\delta x}$ is given below. Recall that an $(n\times n)$ matrix $A$ is said to be Toeplitz if it has the form

  $$\begin{aligned} A=[a_{j-k}]_{j,k=1}^{n} \end{aligned}$$

  (6.5)

  The entries along each diagonal of a Toeplitz matrix are constant. The 1st column of a symmetric Toeplitz matrix $G^{\delta x}$ is as follows:

  $$\begin{aligned} G(1)&= \left( \begin{array}{ccc} -2c1 \displaystyle \sum \limits _{j=1}^{1500}\Lambda ^{'}(j \delta x)-2c2 \\ \Lambda ^{'}(1. \delta x)\\ \Lambda ^{'}(2. \delta x)\\ \Lambda ^{'}(3. \delta x)\\ \vdots \\ \Lambda ^{'}(1500. \delta x) \end{array} \right) _{1501\times 1} \end{aligned}$$

  and the 1st column of the symmetric, positive definite Toeplitz matrix $\left( I+ \delta t G^{\delta x}\right)$

  $$\begin{aligned} \left( I+ \delta t G^{\delta x}\right) (1)&= \left( \begin{array}{ccc} 1-2c1. \delta t \displaystyle \sum \limits _{j=1}^{1500}\Lambda ^{'}(j \delta x)-2c2.\delta t \\ \delta t.\Lambda ^{'}(1. \delta x)\\ \delta t.\Lambda ^{'}(2. \delta x)\\ \delta t.\Lambda ^{'}(3. \delta x)\\ \vdots \\ \delta t.\Lambda ^{'}(1500. \delta x) \end{array} \right) _{1501\times 1} \end{aligned}$$

  Now, the complete symmetric Toeplitz matrix now can be found because it is determined by the first column.

• (v) With the numerical Burgers flux defined at the beginning of this Appendix, we have:

  $$\begin{aligned} F(a,b)=\left\{ \begin{array}{ll} \displaystyle a^2/2, &{}\quad \text {if} a,b>0; \\ \displaystyle b^2/2, &{}\quad \text {if} a,b<0; \\ \displaystyle 0, &{}\quad \text {if} a<0,b>0; \\ \displaystyle b^2/2, &{}\quad \text {if } a>0,b<0, |a|<|b|; \\ \displaystyle a^2/2, &{}\quad \text {otherwise}.\\ \end{array} \right. \end{aligned}$$

Step 2: Assume

$$\begin{aligned} u_{i}^{0}=\left\{ \begin{array}{ll} \ -1, &{}\quad \text {if } i \ge 0 ; \\ \ 1, &{}\quad \text {if } i < 0. \end{array} \right. \end{aligned}$$

and $\ \delta t= 0.00167,$ $\ m=750.$

Step 3: For each $i$, find $h_{i}^{0} \$, and use the equation

$$\begin{aligned} h_{i}^{n}= \left\{ \begin{array}{ll} \ u_{i}^{n} + \frac{\delta t}{\delta x}F(u_{i-1}^{n}, u_{i}^{n})- \frac{\delta t}{\delta x}F(u_{i}^{n}, u_{i+1}^{n}),&{}\quad \text {if} \ \ 3001 \le i \le 4501; \\ \ u_{i+1}^{n}, &{}\quad \text {otherwise.} \end{array} \right. \end{aligned}$$

Since $u_{i}^{0}$ are known for all $i$, we can calculate $h_{i}^{0} \$ for all $i.$

If $\nu \$ and $W \$ are defined as in Step 1, the above equation reduces to a square system of size $2m+1 \$ on $\ W$:

$$\begin{aligned} W+\delta tG^{\delta x}W=\left( h_{i}^{0} - \delta t g^{\delta x}[\nu ]_{i}\right) _{|i|\le 750} \end{aligned}$$

Step 4: Find $u_{i}^{1}, \$ for all i. A Toeplitz matrix $A$ (see, Step 1) is said to be circulant if it has the form

$$\begin{aligned} a_{-k}=a_{n-k}, \quad 1 \le k \le n-1. \end{aligned}$$

(6.6)

Obviously, any circulant matrix is uniquely determined by its first column,

$$\begin{aligned} a=[a_{0},a_{1} , a_{2}, \dots , a_{n-2} , a_{n-1}]^T. \end{aligned}$$

Next, we employ the equation (6.5) to calculate $u_{i}^{1}$, for all $i$, using the standard preconditioned conjugate gradient method, where $A \$ is the symmetric positive definite matrix $\ I + \delta t G^{\delta x} \$, of dimension (1501 $\times$ 1501) (defined in Step 1), and $\ \mathbf x= W \$, is as defined in the Step 1. Taking $n=0,$ we have

$$\begin{aligned} \mathbf x = \left( U_{i}^{1}\right) _{|i|\le 750} = \left( \begin{array}{ccc} U_{-750}^{1} \\ \vdots \\ U_{0}^{1} \\ \vdots \\ U_{750}^{1} \end{array} \right) _{1501 \times 1} \end{aligned}$$

The vector

$$\begin{aligned} \ \mathbf {b}= \left( h_{i}^{0} - \delta t \mathcal {L}^{\delta x}[\nu ]_{i}\right) _{|i|\le 750},\end{aligned}$$

defined by equation (6.5) has the dimension $\ 1501 \times 1. \$ Here, we know the values of $\ h_{i}^{0} \$ for all $i$, and the formula for $\ \mathcal {L}^{\delta x}[\nu ]_{i} \$ is defined in Step 1. For example,

$$\begin{aligned} \mathcal {L}^{\delta x}[\nu ]_{-750}&=\,c1\sum _{0< |j|\le 1500}\left( h_{-750+j}^{0}-h_{-750}^{0}\right) \Lambda ^{'}(j \delta x) +c2 \left( h_{-750-1500-1}^{0}-h_{-750}^{0} \right) \\&\quad +\, c2 \left( h_{-750+1500+1}^{0}-h_{-750}^{0} \right) \end{aligned}$$

Step 5: Find $h_{i}^{1}$, for all i. Since we already know $u_{i}^{1}$, the values for $h_{i}^{1}$, for all $i$, can be calculated by going back to Step 3. Then we move forward to Step 4 to calculate $u_{i}^{2}$ for all $i$. Finally we need repeat the procedure to calculate the values of $u_{i} , i=1,\dots , n$, for $n=300.$