%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Behavior of Piece-wise Self-similar Solutoins and A Numerical % Scheme Based on it : a scheme without time marching % % a scheme without time marching and space discretization % % Yong-Jung Kim % % Aug 2000 % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % This is the sample paper for the AmSTeX SIAM style file, (amstex)siam.sty % for use with AmSTeX version 2.1 or later and amsppt.sty, version 2.1a. % RCS information: $Revision: 1.1 $, $Date: 93/01/25 15:33:19 $. \input amstex \documentstyle{amstexsi} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Macro definitions for running heads and first page % %\accepted\SINUM % %\firstpageno{4} % %\lastpageno{116} % %\issuevolume{1} % %\issuenumber{2} % %\issuemonth{February} % %\placenumber{002} % place of paper in this issue % %\issueyear{1999} % \shortauthor{Y-J Kim} % \shorttitle{A scheme based on self-similairty} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Macros specific to this paper % \font\small=cmr8 at 8pt \font\smbf=cmbx8 at 8pt \font\smit=cmssi8 at 8pt \font\smallfont=cmr8 \define\loner{{L^1(\Bbb R)}} % \define\linfr{{L^\infty(\Bbb R)}} % \define\bvr{{\roman{BV}(\Bbb R)}} % \define\TV{{\roman {TV}}} % \define\sdot{\,\cdot\,} % \def\grad{\nabla} \def\supp{\text {supp}} \def\sign{\text {sign}} \def\loc{\text {loc}} \def\Z{{\Bbb Z}} \def\R{{\Bbb R}} \def\Rn{{\Bbb R^n}} \def\N{{\Bbb N}} \def\DD{{\Cal D}} \def\xx{{\underline x}} \define\e{\varepsilon} \define\eps{\varepsilon} \define\s{\sigma} \define\g{\gamma} \define\r{\rho} \define\sgn{\text{sgn }} %---------------------------------------------------------- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Macros %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \def\jp12{{j+1/2}} \def\jm12{{j-1/2}} \def\sig{{\sigma}} \def\del{{\partial}} \def\GRAD{{\nabla}} \def\DIV{{\nabla \cdot}} \def\H {{\bold H}} %\def\R {{\bold R}} %%already defined as \Bbb R \def\O {{\bold O}} \def\u {{\bold u}} \def\l {{\bold l}} \def\J {{\bold J}} \def\n {{\bold n}} \def\s {{\bold s}} \def\p {{\bold p}} \def\q {{\bold q}} \def\x {{\bold x}} %\def\RD{{\R^\D}} \def\tt{{\bold t}} \def\tu{\tilde u} \def\tp{\tilde p} \def\tq{\tilde q} \def\tH{\tilde H} \def\bH{\bar H} \def\btH{\bar{\tilde{H}}} \def\xP{{x^+}} \def\xM{{x^-}} \def\xPM{{x^\pm}} \def\LL{{\Cal L}} \def\BB{{\Cal B}} \def\EE{{\Cal E}} \def\HH{{\Cal H}} \def\JJ{{\Cal J}} \def\AA{{\Cal A}} \def\SS{{\Cal S}} \def\OO{{\Cal O}} %\def\dx{{d^{^\D}\!\!\!x}} \def \dx {\Delta x} \def \dy {\Delta y} \def \dt {\Delta t} \def \vectorx{{\text{\bf x}}} \def \vectory{{\text{\bf y}}} \def \vectorz{{\text{\bf z}}} \def \tfrac#1#2{{\textstyle {#1 \over #2}}} \def \frac#1#2{{#1 \over #2}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \def\OneFigPlace#1#2#3#4#5#6#7{{\hbox{} \vbox to #2 true in {} \special{psfile=#1 hscale=#3 vscale=#4 hoffset=#5 voffset=#6} \centerline{#7} }} \def\TwoFigPlace#1#2#3#4#5#6#7#8#9{{\hbox{} \vbox to #3 true in {} \special{psfile=#1 hscale=#4 vscale=#5 hoffset=#6 voffset=#8} \special{psfile=#2 hscale=#4 vscale=#5 hoffset=#7 voffset=#8} \noindent{#9} \medskip }} \pageno 1 \topmatter \title Behavior of Piece-wise Self-similar Solutoins and A Numerical Scheme Based on it : a scheme without time marching \endtitle \author Yong-Jung Kim\footnote{Institute for Mathematics and its Applications, University of Minnesota, Minneapolis, MN 55455-0436 (yjkim\@ima.umn.edu).} \endauthor \abstract A note. \endabstract \endtopmatter \document \vskip 0.5 truein \centerline{\bf Contents} \noindent 1. Introduction and questions \noindent 2. Exact Solutions of Cauchy Problems 2.1 Base functions and Cauchy solver for Burgers 2.2 K-summation of base functions 2.3 K-summation of 2 base functions \noindent 3. Coding the scheme \noindent 4. positive solution 4.1 Data discretization 4.2 Short time behavior \noindent 5. Solutions with sign change \noindent 6. The cubic law \noindent 7. General scalar conservation laws in 1-D \noindent 8. Example : Burgers + the cubic law : $u_t+(u^2+u^3)_x=0$ \noindent 9. Phase change : $u_t+(u^2-u^3)_x=0$ \noindent 11. General systems \noindent 12. Rotational model from magnato-hydro dynamics(?) \noindent 13. Planer model from magnato-hydro dynamics(?) \noindent 20. Multi-dimensional Burgers Equation \noindent 30. Boundary problem(?) \vskip 0.3 in \head 1. Introduction and remaining questions\endhead First we consider the Cauchy problem for Burgers, $$u_t+(u^2/2)_x=0, \tag 1.1 $$ with $L^1$ initial data $$u(x,0)=u_0(x). \tag 1.2 $$ Let $u_1(x,t),u_2(x,t)$ be solutions of (1.1). Because of the non-linearity of the problem, $u_1+u_2$ is not a solution of (1.1). Here we consider a special summation `$\oplus$' (call it `K-summation' for now) between piecewise `self-similar' (it is linear in the case of Burgers) solutions, $u_1,u_2$, which makes $u_1\oplus u_2$ be a solution of (1.1). (Precise arguments are in Section 2 for Burgers and in Section 6 for the cubic law. The cubic law is the easier case.) It is actually based on finding the place of a shock using the characteristics of the problem. It is explained in Section 2. This idea can be directly coded for a numerical scheme which has no time marching step. So we can get the solution at time $t=0.1$ or $t=1,000,000$ after same amount of computational time, which is almost instant. Numerical schemes which use piecewise constant (not piecewise linear) functions as the approximation of the real solution, which is most of the cases, actually solve Riemann problem locally in some sense. So CFL condition should be considered and hence it is necessary to do it with time marching. But the scheme based on Cauchy solver, Lemma 2.1, and K-summation $\oplus$ is free from the CFL condition. We never solve Riemann problem. The benefite of the scheme is more than that. From Figure 4.5 we can clearly see that it is not possible to make an error estimate from a single compuation based on Godunov scheme. But Figure 4.6 shows that the analytic solution of (1.1,2) passes through the artificial shocks of numerical solutions based on Cauchy solver and K-summation. Furthermore this artificial shocks decreases in time with order $O(1/t)$, (4.3). This idea can be applied to general scalar conservation laws, $$u_t+f(u)_x=0, \tag 1.3 $$ where the flux satisfies $$f''(u)\ne 0,\qquad u\ne 0. \tag H $$ To Support this behavior theoretically we need to understand characteristics well. If we consider a system, for example a simple MHD model, $$\eqalign{u_t+(u(u^2+v^2))_x=0,\cr v_t+(v(u^2+v^2))_x=0,\cr} \tag 1.7 $$ there will be interactions between $u$ and $v$. The figure in Section 12 does not represents the real solution correctly. It seems possible to apply this scheme to simple multi-dimensional problems. But it will be hard to consider with complexity. \vfil\eject \head 2. Exact Solutions of Cauchy Problems\endhead In this section we consider Cauchy solver for base functions and consider a algorithm solving Burgers equation without time marching. \subhead 2.1 Base functions and Cauchy solver for Burgers\endsubhead We consider three indexed functions $$B(x)=B_{t,z,s}(x)=\left\{ \eqalign{(x-z)/t&,\quad\text{$x$ is between $z$ and $s$},\cr 0\quad&,\qquad\qquad\text{otherwise},\cr}\right. \tag 2.1 $$ which is used as a base function in this section. So the base function is a linear function over an interval between the zero point $z$ and the shock place $s$. The signed area $m$ enclosed by $x$-axis and the function is given by $$m=|s-z|(s-z)/(2t). \tag 2.2 $$ In following we consider base functions as if it is four indexed function $B(x)=B_{m,t,z,s}(x)$ for our convenience under the assumption that the index $m,t,z,s$ satisfies the relation (2.2). So if any three of them is decided, the other one is given by (2.2). Note that the indexes should satisfy $t>0,m(s-z)>0$. Now we consider a Cauchy problem for Burgers equation, $$\eqalign{u_t+uu_x=0,\cr u(x,0)=B_{m_0,t_0,z_0,s_0}(x).\cr} \tag 2.3 $$ \proclaim{Lemma 2.1} (Cauchy Solver) Let $u(x,t)$ be the entropy solution of the Cauchy problem (2.3). Then $u(x,T)=B_{m=m_0,t=T+t_0,z=z_0}(x)$. \endproclaim \demo{Proof} I've got this idea from the characteristics. We can also see this algebraically. Consider variables $$w=\sqrt{t+t_0}\,u,\xi=(x-z_0)/\sqrt{t+t_0},\tau=\ln(t+t_0), \tag 2.4 $$ then (2.3) is transformed to $$\eqalign{w_\tau+{1\over2}(w^2-\xi w)_\xi=0,\cr w(\xi,\ln(t_0)\,)=B_{m=m_0,t=1,z=0}(\xi).\cr} \tag 2.5 $$ We can easily check that $B_{m=m_0,t=1,z=0}(\xi)$ is an admissible steady state of the equation and hence $w(\xi,\tau)=B_{m_0,t=1,z=0}(\xi)$ is the solution of (2.5). If we transform the variables back to $u,t,x$, then we get $u(x,T)=B_{m=m_0,t=T+t_0,z=z_0}(x)$. \qed \enddemo \remark{Remark 2.1} If we consider a $\delta$-function as the initial data, for example $u_0(x)=m_0\delta(x-z_0)$, then the solution is given by $u(x,t_0)=B_{m=m_0,t=t_0,z=z_0}(x)$. So the reciprocal of the slope of the base function represents the time, and that is why we are using index $t$ for the base function. \endremark \remark{Remark 2.2} We can easily get one of the four variables $m,t,z,s$ from the other three, $$\eqalign{t&=(s-z)^2/(2|m|),\cr s&=z+\sign(m)\sqrt{2|m|t}\,,\cr z&=s-\sign(m)\sqrt{2|m|t}\,.\cr } \tag 2.6 $$ \endremark \remark{Remark 2.3} Lemma 2.1 shows that if the initial data are given by the base function (2.1), the solution of the Burgers equation is given exactly for any given time with a simple algebra. In the following sections we see that this is also true for even more general cases and develop a numerical scheme in the spirit of this Cauchy solver for general initial data. It will be turned to be a scheme without time marching. \endremark \subhead 2.2 K-summation of base functions\endsubhead We can easily guess that if the initial data are given by $$u_0(x)=\sum_{k=1}^n B_{m_k,t_k,z_k}(x),\quad z_1\le z_2\le ...\le z_n, \tag 2.10 $$ then the solution should be, $$u(x,t)=\sum_{k=1}^n B_{m_k,t_k+t,z_k}(x), \tag 2.11 $$ if the support of the base functions of (2.11) are separated. But we can not expect that in general. Even though the right hand side of (2.11) contains enough information to single out the correct solution. What we have to do is to interpret it correctly. We may define a proper way of summation of base functions for the interpretation, $$B_n(x)=\bigoplus_{k=1}^n B_{m_k,t_k,z_k}(x), \quad z_1\le z_2\le ...\le z_n, \tag 2.12 $$ in the following way. Note that this summation is not associative and hence the order of the summation is important. \medskip Case 1. $m_k>0$ for all k : In this case all the characteristics move to right hand side and hence all the shock wave move to the riight hand side. So we do the summation from the direction. Suppose that $B_{n-1}(x)=\bigoplus_{k=2}^n B_{m_k,t_k,z_k}(x)$ be well defined and piecewise linear with zero points or shocks at the end of each of the intervals of linear part. Suppose there exists a point $\xi>z_1$ such that $$B_{n-1}(x)<(x-z_1)/t_1,\quad z_1z_1$, we define $B_n(x)$ as $$B_n(x)=\bigoplus_{k=1}^n B_{m_k,t_k,z_k}(x)= \left\{\eqalign { (x-z_1)/t_1&,\quad z_10$ : In the case the supports of two base functions are separated and hence the K-summation is the usual summation. \midinsert\qquad\quad\special{psfile="scheme21.ps" hscale=90 vscale=80 hoffset=-270 voffset=-600 } \vskip 3.1 in \centerline{\bf Figure 2.1} \medskip \endinsert \medskip Case B, $m_1>0$ and $m_2>0$ : Suppose that there exists $s_1<\xit_2 (\xis_2$, then (2.15) implies that the emerging K-summation is another base function, $$B_2(x)=B_{m_1+m_2,t_1,z_1}(x). \tag 2.35 $$ \midinsert\qquad\quad\special{psfile="scheme22.ps" hscale=90 vscale=80 hoffset=-270 voffset=-600 } \vskip 3.1 in \centerline{\bf Figure 2.2} \medskip \endinsert \medskip Case C, $m_1<0$ and $m_2<0$ : (This case is symmetric to the previous case if we switch index $1,2$, we get the K-summation for this case.) Suppose that there exists $s_1<\xis_1). \tag 2.37 $$ If $t_1\ne t_2$, we get $$\xi={(t_2z_1-t_1z_2)+\sqrt{D} \over t_2-t_1}, \quad\text { if } t_2s_1), \tag 2.38 $$ $$\xi={(t_2z_1-t_1z_2)-\sqrt{D} \over t_2-t_1}, \quad\text { if } t_2>t_1 (\xi>s_1), \tag 2.39 $$ where $$D=(t_2z_1-t_1z_2)^2 -(t_2-t_1)(t_2z_1^2+t_1s_2^2-2t_1s_2z_2)\ge 0. $$ Example of this case are given in Figure 2.1. If there is no such a point $\xi$ in the interval $(s_1,s_2)$, which implies $\xi0$ and $m_2<0$ : Suppose there exists $s_2<\xis_1). \tag 2.42 $$ If $t_1\ne t_2$, we get $$\xi={(t_2z_1-t_1z_2)+\sqrt{D} \over t_2-t_1}, \quad\text { if } t_2t_1 (s_1<\xi0,\cr B_2(x)=B_{m_1+m_2,t_2,z_2}(x),\quad\text{if } m_1+m_2<0.\cr} \tag 2.45 $$ \vfil\eject \head 3. Coding the scheme\endhead In this section we briefly show the algorithm of the scheme. Since the base function is the unit of the computation, we consider an array of base functions $B[j],j=1,2,...,n$. Each of the node consists of 4 data $B[j].m,B[j].t,B[j].z,B[j].s$. \medskip \remark{Step 1, An assumption} We need an assumption for the initial data that K-summation of $B[j]$ is well defined. \endremark \medskip \remark{Step 2, Evolution } For any given time $T>0$, the solution evolves to simply $$\tilde B[j].m=B[j].m,\quad \tilde B[j].t=B[j].t+T,\tilde B[j].z=B[j].z, \tag 3.1 $$ and $B[j].s$ is given by (2.6), i.e., $$\tilde B[j].s=\tilde B[j].z+\sign(\tilde B[j].m) \sqrt{2|\tilde B[j].m|\tilde B[j].t}. \tag 3.2 $$ In this step we don't need time marching. That was possible we are using a Cauchy solver. \endremark \medskip \remark{Step 3, Sorting } Displaying K-summation is not easy to do directly. So we do the sorting first. This step is basically comparing two adjacent base functions and check the point $\xi$ of Section 2.3 and decide if they merged into a single base function like (2.35,40,45). If they do, combine them and reduce the number of base function. The result should be reduced number of base functions such that none of them is covered by any of them. \endremark \medskip \remark{Step 4, Displaying } Now we can easily display the solution with real shocks. \endremark %\vfil\eject %\head 4. Piecewise linear initial data\endhead \vfil\eject \head 4. Positive initial data\endhead In this section we consider initial data $$u_0(x)=\left\{\eqalign{ \sin(\pi x)/\pi,\quad 01$, and we consider the solution a little after the appearance of the shock, say $t=1.2$. We compare the numerical solutions based on Godunov scheme and our scheme. \midinsert\qquad\quad\special{psfile="scheme45.ps" hscale=70 vscale=30 hoffset=-78 voffset=-240 } \vskip 3.3 in \centerline{\bf Figure 4.5 \it Godunov with different meshes at $t=1.2$} \medskip \endinsert In Figure 4.5 numerical solution for the scheme with different mesh points for the interval $(0,1)$ have been displayed. Here we can see that the solution converges as we increase the number of mesh points. As we decrease the mesh size $\Delta x$, it is clear that the solution has a shock placed just after $x=1$. Convergence to the correct weak solution (i.e., satisfying entropy condition) of the numerical solution of the scheme is known, but we don't have an idea how the numerical approximation is close to the correct solution. Now we consider numerical solutions of K-scheme. First we can easily check that the numerical solutions converges to the correct solution as we increase the number of base functions. Let $u_n(x)$ be the data decomposition of the smooth initial $u_0(x)$ given by shock approximation (2.12), see Figure 4.4, such that $||u_n(x)-u_0(x)||_1\to 0$ as $n\to\infty$. Then the solution $u_n(x,t)$ of the Burgers equation with initial data $u_n(x)$ is given by (2.21). From the contraction theory of hyperbolic conservation laws, we get $$||u_n(x,t)-u(x,t)||_1\le ||u_n(x)-u_0(x)||_1\to 0, \tag 4.2 $$ where $u(x,t)$ is the correct solution with initial data $u_0(x)$. In Figure 4.6 numerical solutions of the scheme at $t=1.2$ is displayed using different number of data decomposition. The first thing we can see is that all of them have very close real shock just after the point $x=1$. The other tooth like shocks are created by the scheme, but the last one is in some sense an accumulation of these small shocks and represents the real shock. In Figure 4.7 the portion of the shock has been magnified and we can easily see how close the shocks are. \midinsert\qquad\quad\special{psfile="scheme46.ps" hscale=70 vscale=30 hoffset=-78 voffset=-240 } \vskip 3.3 in \centerline{\bf Figure 4.6 \it K-summation at $t=1.2$} \medskip \endinsert The second thing we can observe is that the sizes of artificial shocks are smaller than the initial one. Suppose that an artificial shock is generated by an initial base function $B_{m_k,t_0,z_k}(x)$. Then we can easily see that the height of the shock is $h_k(t)=(z_{k+1}-z_k)/(t+t_0)$ at time $t>0$. So the height of artificial shocks are estimated by, $$\max h_k(t)\le {(\text{length of domain of data})(1-t_0\min_x(u_0'(x)) \over (\# \text{ of base functions })(t+t_0)}. \tag 4.3 $$ So the height of the artificial shocks decreases with decay rate $1/t$. Since the solution decreases with decay rate $1/\sqrt{t}$, the ratio of the size of the artificial shock over the size of the solution decreases to zero with rate $1/\sqrt{t}$. The last thing we want to mention is that the solutions with finer mesh pass the shocks of coarser ones. It is clearly observed in Figure 4.7, which has 4 cases with 10,40,160 and 10,000 base functions. If it is true for any case and at any given time $t>0$, it implies that the correct solution of the original problem always passes through the small shocks, and hence we can see that how close the correct solution is to he approximation. We can conclude that like following. \midinsert\qquad\quad\special{psfile="scheme47.ps" hscale=70 vscale=30 hoffset=-78 voffset=-240 } \vskip 3.3 in \centerline{\bf Figure 4.7 \it A magnification near the real shock} \medskip \endinsert \proclaim{Lemma 4.1} Let $u_n(x)$ be a piecewise linear approximation of compactly supported initial data $u_0(x)$, see Figure 4.4, such that the area of a triangle like shape between them is exactly same as the one across the shock. Then the solution $u(\cdot,t)$ of (1.1) always passes through artificial shocks of $u_n(\cdot,t)$, which is the solution of (1.1) with initial data $u_n(x)$. \endproclaim \demo{Proof} The proof is basically from the fact that the area change is always along the shock and if the areas across the shock are same they can not be disapeard and it implies the lemma. \qed \enddemo \remark{Remark 4.1} In Figure 4.1 we can see that the initial data $u_0(x)$ passes through the top portion of each of the rectagle. But we can clearly see that it is not true for general time $t>0$ from Figure 4.5. We can consider two reasons. First Godunov does not solve the solutions exactly with modified initial data. Furthermore the small areas enclosed by the initial data $u_0$ and approximation, Figure 4.1, are not same across the shock but they are same across the continuity points. So even if we can solve it exactly with cell average approximation, we can not expect propety of Lemma 4.1 for the case. \endremark Now it is clear that the correct solution is estimated by $$||u(x,t)-u_n(x,t)||_\infty\le \max h_k(t)=O(1/t), \tag 4.5 $$ at the point of the artificial shock. Since $u_n(\cdot,t)-u(\cdot,t)$ is increasing except the shock points of $u_n$ (4.5) is true for all points. Since the height of the artificial shock is of order $O(1/t)$ and the real shock propagate with speed $O(1/\sqrt{t})$, (4.2) can be improved to $$||u_n(x,t)-u(x,t)||_1\le O(1/\sqrt{t})\,||u_n(x)-u_0(x)||_1. \tag 4.6 $$ \vfil\eject \head 5. Solutions with sign change\endhead Here we consider couple of examples of numerical runs with initial data which have positive and negative parts simutaniously. The purpose of that is to convince people that the K-summations and the numerical scheme based on it is also good for these cases. \midinsert\qquad\quad\special{psfile="scheme51.ps" hscale=70 vscale=30 hoffset=-78 voffset=-240 } \vskip 3.3 in \centerline{\bf Figure 5.1 \it Initial Discretization} \medskip \endinsert \midinsert\qquad\quad\special{psfile="scheme52.ps" hscale=70 vscale=30 hoffset=-78 voffset=-240 } \vskip 3.3 in \centerline{\bf Figure 5.2 \it Solutions at $t=1.2$} \medskip \endinsert \midinsert\qquad\quad\special{psfile="scheme53.ps" hscale=70 vscale=30 hoffset=-78 voffset=-240 } \vskip 3.3 in \centerline{\bf Figure 5.3 \it A magnification} \medskip \endinsert From these three figures, 5.1-3, we can obseve the same phemomina which is observed in the previous section. \midinsert\qquad\quad\special{psfile="scheme54.ps" hscale=70 vscale=30 hoffset=-78 voffset=-240 } \vskip 3.3 in \centerline{\bf Figure 5.4 \it Solutions at $t=10$} \medskip \endinsert \midinsert\qquad\quad\special{psfile="scheme55.ps" hscale=70 vscale=30 hoffset=-78 voffset=-240 } \vskip 3.3 in \centerline{\bf Figure 5.5 \it Solutions at $t=30$} \medskip \endinsert \vfil\eject .\vfil\eject .\vfil\eject \head 6. The cubic law\endhead In this section we consider Cauchy solver and base functions for the cubic law, $$u_t+(u^3)_x=0. \tag 6.1 $$ \subhead 6.1 Base functions and Cauchy solver for the cubic law\endsubhead In this case base functions are given with three index, $$B(x)=B_{t,a,b}(x)=\left\{ \eqalign{\sqrt{(x-a)/3t}&,\quad a0,m(b-a)>0$. Now we consider a Cauchy problem for the cubic law (6.1) with initial data, $$ u(x,0)=B_{m_0,t_0,a_0,b_0}(x). \tag 6.4 $$ \proclaim{Lemma 6.1} (Cauchy Solver) Let $u(x,t)$ be the entropy solution of the Cauchy problem (6.1-4). Then $u(x,T)=B_{m=m_0,t=t_0+T,a=a_0}(x)$ for $m_0>0$ and $u(x,T)=B_{m=m_0,t=t_0+T,b=b_0}(x)$ for $m_0<0$. \endproclaim \demo{Proof} Consider the case $m_0>0$ only. Variables, $$w=\root3\of{t+t_0} \,u,\xi=(x-a_0)/\root3\of{t+t_0},\tau=\ln(t+t_0), \tag 6.5 $$ transform (6.1-4) to $$\eqalign{w_\tau+(w^3-{1\over2}\xi w)_\xi=0,\cr w(\xi,\ln(t_0)\,)=B_{m=m_0,t=1,a=0}(\xi).\cr} \tag 6.6 $$ We can easily check that $B_{m=m_0,t=1,a=0}(\xi)$ is an admissible steady state of the equation and hence $w(\xi,\tau)=B_{m_0,t=1,a=0}(\xi)$ is the solution of (6.6). If we transform the variables back to $u,t,x$, then we get $u(x,T)=B_{m=m_0,t=T+t_0,a=a_0}(x)$. \qed \enddemo \remark{Remark 6.1} We can easily get one of the four variables $m,t,a,b$ from the other three, $$\eqalign{t&=4|b-a|^3/27m^2,\cr b&=a+\sign(m)\root3\of{27tm^2/4}\,,\cr a&=b-\sign(m)\root3\of{27tm^2/4}\,.\cr } \tag 6.7 $$ \endremark \subhead 6.2 K-summation of base functions\endsubhead We know that K-summation is not defined in general. But in the numerical schemes it is enough to consider K-summation between base functions with same slope index $t$, say $t_k=t_0$ for all $k$, and in the case K-summation is always well defined. In this section we consider this case only. For the convenience of the notation we take a new notation $z=\min(a,b)$ which actually implies the zero point of the base function. We define K-summation of base functions, $$B_n(x)=\bigoplus_{k=1}^n B_{m_k,t_0,z_k}(x), \quad z_1\le z_2\le ...\le z_n, \tag 6.10 $$ inductively. Note that this summation is not associative. We consider the case $m_1>0$ only. Suppose that $B_{n-1}(x)=\bigoplus_{k=2}^n B_{m_k,t_k,z_k}(x)$ be well defined and piecewise parabolic having structure of (6.2) with $t=t_0$ with zero points or shocks at the end of each of intervals and have total mass, $$\int_{\infty}^\infty B_{n-1}(x)dx=\sum_{k=2}^n m_k. \tag 6.11 $$ It is clear that $$B_{n-1}(x)<\sqrt{(x-z_1)/3t_0},\quad z_1z_1$ such that $F(\xi_1)=\sum_{k=1}^n m_k$. $B_n(x)$ is defined by $$B_n(x)=\bigoplus_{k=1}^n B_{m_k,t_k,z_k}(x)= \left\{\eqalign { \sqrt{(x-z_1)/3t_0}&,\quad z_10$ for all $u>0$ or $f'(u)<0$ for all $u>0$. For the definitness we assume $f'(u)>0$ for all $u>0$. For the sign changeing solutions there are more possible cases. Here we write down the hypothesis for the flux, $$f''(u)>0,\quad f'(u)>0,\quad f'(0)=0. \tag H $$ \subhead 7.1 Base functions and Cauchy solver\endsubhead In this case base functions are given with three index, $$B(x)=B_{t,a,b}(x)=\left\{ \eqalign{g((x-a)/t)&,\quad aa_1, \tag 7.13 $$ and hence $$F(\xi)=\int_{a_1}^\xi g({x-a_1\over t_0}) dx+\int_\xi^\infty B_{n-1}(x) dx, \quad \xi\ge a_1, \tag 7.14 $$ is an increasing function with $F(a_1)=\sum_{k=2}^n m_k$. So there exists a unique $\xi_1>a_1$ such that $$F(\xi_1)=\sum_{k=1}^n m_k. \tag 7.15 $$ Now we define, $$B_n(x)=\bigoplus_{k=1}^n B_{m_k,t_k,a_k}(x)= \left\{\eqalign { g({x-a_1\over t_0})&,\quad a_1b_1$ that solves (7.31). But in many cases there is no such a point $\xi$ in the interval $(b_1,b_2)$, which implies $\xi>b_2$. Then the emerging K-summation is another base function, $$B_2(x)=B_{m_1+m_2,t_0,a_1}(x). \tag 7.32 $$ \vfil\eject \head 8. Burgers + the cubic law : $u_t+(u^3+u^2)_x=0$ \endhead We consider positive solutions. In this case the flux is $f(u)=u^3+u^2$ and $f'(u)=3u^2+2u$. So the inverse function $g$ of $f'$ is given by $$g(x)={-1+\sqrt{3x+1}\over 3}, \tag 8.1 $$ and its antiderivative is $$G(x)=-{x\over3}+{2\over27}(3x+1)^{3/2}. \tag 8.2 $$ So the mass function of (7.5) is given $$m(a,b,t)=t[G({b-a\over t})-G(0)], \tag 8.3 $$ where $G(0)=2/27$. The inverse function $G^{-1}$ is given in a implicit form and hence we solve the equation (7.5) to get (7.11). Now we show some numerical runs. In the picture we can observe the property of the Burgers and the cubic law clearly. \midinsert\qquad\quad\special{psfile="scheme81.ps" hscale=70 vscale=30 hoffset=-78 voffset=-240 } \vskip 3.3 in \centerline{\bf Figure 8.1 \it Solution of $u_t+(u^2+u^3)_x=0$ } \medskip \endinsert \vfil\eject \head 11. General systems\endhead Consider conservatioin laws, $$U_t+f(U)_x=0,\quad U\in\Bbb R^n,x\in\Bbb R. \tag 11.1 $$ Suppose there exists a solution $U(x,t)=g(x/t)$ for $0\le x\le b$. Then (11.1) is written as $$-{x\over t^2}g'+\grad_U f(g) g'{1\over t} =0. \tag 11.2 $$ Fix $t=1$. Then the self-similar solutions turns out to be an eigen-value problem, $$\grad_U f(g) g'=xg', \tag 11.3 $$ where $x$ is the eigen value and $g'$ is the eigen-vector. \vfil\eject \head 12. Rotational model from magnato-hydro dynamics\endhead We consider $$\eqalign{u_t+(u(u^2+v^2))_x=0,\cr v_t+(v(u^2+v^2))_x=0,\cr} \tag 12.1 $$ where $u,v,x\in\Bbb R$ and $t>0$. We can easily check that the gradient of the flux, $$\grad_U f(U)=\left(\eqalign{ 3u^2+v^2\qquad 2uv\quad\cr \quad 2uv\qquad u^2+3v^2\cr}\right), \tag 12.2 $$ has eigen-values, $$\lambda_1=u^2+v^2,\quad\lambda_2=3(u^2+v^2), \tag 12.3 $$ and corresponding right eigen-vectors, $$r_1=\left(\eqalign{v\cr-u\cr}\right),\quad r_2=\left(\eqalign{u\cr v\cr}\right). \tag 12.4 $$ Let $g(x)=(u(x),v(x)), 0\le x\le b$ be a self-similar profile corresponding to the first eigen-family. Then it satisfies $$x=u^2+v^2,\quad {u'\over v'}=-{v\over u}. \tag 12.5 $$ We can easily check that (12.5) does not have a solution. Let $g$ be a self-similar profile corresponding to the second eigen-family. Then it satisfies $$x=3(u^2+v^2),\quad {u'\over v'}={u\over v}. \tag 12.6 $$ (12.6) has solutions $$(u,v)=\left\{\eqalign{(a\sqrt{x},b\sqrt{x})&,\quad 0\le x0$. We can easily check that the gradient of the flux, $$\grad_U f(U)=\left(\eqalign{ 2cu\quad 2v\cr 2v\,\quad 2u\cr}\right), \tag 13.2 $$ has eigen-values, $$\lambda_1=(c+1)u-\sqrt{D},\quad\lambda_2=(c+1)u+\sqrt{D}, \tag 13.3 $$ and corresponding right eigen-vectors, $$r_1=\left(\eqalign{\lambda_1-2u\cr 2v\quad\cr}\right),\quad r_2=\left(\eqalign{\lambda_2-2u\cr 2v\quad\cr}\right), \tag 13.4 $$ where $D$ is given by $$D=(1-c)^2u^2+4v^2. $$ Also note that the gradient matrix has a unique singular point $(u,v)=(0,0)$. Since the singularity appears only at the origin, the self-similar profiles can be considered seperately for positive ones and negatives ones. From now on we consider positive solutions only. Let $g(x)=(u(x),v(x)), 0\le x\le b$ be a self-similar profile satisfying $$v(x)=au(x). $$ Then eigen values are given by $$\lambda_{1,2}=(c+1)u\pm\sqrt{(1-c)^2+4a^2}\,u.$$ Suppose that $(u',v')$ is an eigen vector of the first family, then it implies $$2u'=(c+1)-\sqrt{(1-c)^2+4a^2}, $$ which implies that $u$ is decreasing and hence it should be negative. Since we consider positive profile it has no solutions. Suppose that $(u',v')$ is an eigen vector of the second family, then we get $$2u'=(c+1)+\sqrt{(1-c)^2+4a^2}, $$ which gives a self-similar profile, $$\eqalign{ u(x)={(c+1)+\sqrt{(1-c)^2+4a^2}\over 2}x,\cr v(x)=a{(c+1)+\sqrt{(1-c)^2+4a^2}\over 2}x.\cr} \tag 13.5 $$ \vfil\eject \head 20. Multi-dimensional Burgers Equation\endhead Inviscid Burgers equation is generalized in N-dimensional space by $$w_t+a\cdot\grad_\vectorz(|w|^q)=0,\quad q>1, \tag 20.1 $$ where $w\in\Bbb R, \vectorz=(z_1,...,z_N)\in, t>0$ and $a=(a_1,...,a_N)$ is a constant vector. Consider new variables, $$\vectory(\vectorz)=(y_1(\vectorz),...,y_N(\vectorz)),\quad v(\vectorz,t)=w(\vectory(\vectorz),t), \tag 20.2 $$ where $y_k(\vectorz)=a_k z_k$. Then $v(\vectorz,t)$ satisfies $$v_t+\text{div}(|v|^q)=0. \tag 20.3 $$ So (20.1) can be transformed into a simpler form. We can rotate and normalize the axises to get $$u_t+(|u|^q)_x=0, \tag 20.3 $$ where $\vectorx=(x=x_1,x_2,...,x_N), u=u(\vectorx,t)$. So the problem turned out to be a tube problem of 1-dimensional structure. We can easily apply the method develped in previous sections for one-dimensional scalar conservation laws. \subhead 20.1 2-D case, an example\endsubhead Cosider the solution of 2-dimensional Cauchy problem $$u_t+uu_x=0,\qquad u=u(x,y,t),x,y\in \Bbb R,t>0, \tag 20.4 $$ with initial data, $$u_0(x,y)=\left\{\eqalign{\sin(x)\sin(y)&,\quad 0\le x,y\le \pi,\cr 0\qquad&,\qquad\text{otherwise.}\cr}\right. \tag 20.5 $$ Basically we can solve (20.4) along a tube seperatly. Following is an numerical run. \midinsert \special{psfile="scheme201.ps" hscale=50 vscale= 50 hoffset=-90 voffset=-330 } \special{psfile="scheme202.ps" hscale=50 vscale= 50 hoffset= 160 voffset=-330 } \vskip 3.2 in \centerline{\bf Figure 20.1} \endinsert \subhead 20.2 2-D case with general vector field\endsubhead \enddocument