% compile with pdflatex demo; mpost portret; pdflatex demo \documentclass{article} \usepackage{multido} \usepackage{amsmath} %\usepackage{color} \usepackage%[clip] {mfpic} \usemetapost \opengraphsfile{portret} \tlabelsep{3pt} \usemplabels \mfpicunit=1cm \usepackage{mfpic4ode} \begin{document} \title{Test file for mfpic4ode package} \author{Robert Ma\v r\'\i k} \date\today \maketitle See the source file \texttt{demo.tex} for comments in the \TeX{} code. \clipmfpic \section{Logistic equation} Here we draw a simple picture which describes stability of stationary points of teh equation and then draw phase portrait of the equation. \begin{center} \begin{equation} x'= {{r}\cdot\left(1-\frac x{{K}}\right)x} \end{equation} \end{center} % We set parameters for logistic equation \mfsrc{r:=1;K:=0.98;} % We set parameters for drawing and for the ODE solver \mfsrc{ODEarrowlength:=0.07; ODEstep:=0.02; ODEstepcount:=500;} % We define the equation \ODEdefineequation{r*y*(1-(y/K))} Stability and sign of the right--hand side. \begin{mfpic}[5][3]{-0.1}{1.5}{-0.1}{0.5} \axes \xmarks{K} \tlabel[tc](K,0){$K$} \tlabel[bc](0,ypos){$\smash{f(x)}$} \tlabel[cl](xpos,0){$x$} % This code draws arrows on x axis, the arrow points to the right if % $f(x)$ is positive and to the left if f(x) is negative. If the % starting point of the arrow is x_0 and the final point x_1 and if % the function changes sign between x_0 and x_1, the arrow from x_0 % is not drawn \multido{\r=0.01+0.2}{8}{\ODEharrow{\r}} \pen{1pt} \draw[rgb(0,0.5,0)]\parafcn{0,1.5,0.1}{(t,ODErhs(1,t))} \draw[red]\parafcn{0,1.5,0.1}{(t,ODErhs(1,t))} \draw[blue]\parafcn{0,K,0.1}{(t,ODErhs(1,t))} \end{mfpic} \mfsrc{ODEarrowlength:=0.3;} Phase portrait \begin{mfpic}[1][4]{-0.1}{10}{-0.1}{1.5} \axes \ymarks{K} \tlabel[cr](0,K){$K$} \penwd{1pt} \tlabel[tc](xpos,0){$t$} \tlabel[bc](0,ypos){$x$} \pen{0.7pt} % This code draws arrows in 19x15 points and draws three integral % curves \multido{\r=0.0+0.1}{15} {\multido{\R=0.0+0.5}{19} {\ODEarrow{\R}{\r}}} \trajectories{0,0.3;0,0.01;0,1.4} \end{mfpic} \section{Logistic equation with harvesting} Similar to the previous picture, but both pictures are drawn together to see the relations between them. \begin{center} \begin{equation} x'= {{r}\cdot\left(1-\frac x{{K}}\right)x}-{p} \end{equation} \end{center} % We set parameters for logistic equation with harvesting and define % this equation. \mfsrc{r:=1;K:=0.98;lov:=0.15;} \ODEdefineequation{r*y*(1-(y/K))-lov} % If the equation possesses stationary points, we store them into % variables meza and mezb. If not, we set these variables to negative % values (and the are not drawn in view of mfpicclip option. \mfsrc{if (r**2*(K**2)-4*r*lov*K)<0: meza:=-1;mezb:=-1.1 else: meza:=(r*K-sqrt(r**2*(K**2)-4*r*lov*K))/(2*r); mezb:=(r*K+sqrt(r**2*(K**2)-4*r*lov*K))/(2*r) fi;} \hbox to \hsize{\hss\begin{mfpic}[1][4]{-0.1}{10}{-0.1}{1.3} \axes \ymarks{K} \tlabel[cr](0,K){$K$} \penwd{1pt} \tlabel[tc](xpos,0){$t$} \tlabel[bc](0,ypos){$x$} % here we draw lines - stationary solutions stored in meza and mezb % variables \draw[gray(0.5)]\dashed\lines{(0,meza),(xpos,meza)} \draw[gray(0.5)]\dashed\lines{(0,mezb),(xpos,mezb)} % We draw direction field using metapost cycle. Another option is to % use multido command as in the previous example. \mfsrc{for j=0 step 0.07 until 1.2: for i:=0 step 0.5 until 10:} \ODEarrow{i}{j} \mfsrc{endfor;endfor;} % We draw trajectories using black color. \drawcolor{black} \trajectories{0,0.1;0,0.2;0,0.4;0,0.6;0,0.8;0,1.1} \end{mfpic}\qquad % On the right we draw the right hand side of the equation \begin{mfpic}[3][4]{-0.15}{0.6}{-0.1}{1.3} \axes \ymarks{K} \tlabel[cr](0,K){$K$} \tlabel[br](xpos,0){$f(x)$} \tlabel[bc](0,ypos){$x$} \pen{1pt} % This code draws the graph of right-hand side of logistic equation % without harvesting. \drawcolor{gray(0.7)}\draw\parafcn{0,1.2,0.1}{(ODErhs(1,t)+lov,t)} % This code draws the graph of right-hand side. We use the blue % color for positivce and red color for negative parts. We draw also % arrows which are determined by the sigh of the right-hand side of % the equation. \drawcolor{red}\draw\parafcn{0,1.2,0.1}{(ODErhs(1,t),t)} \drawcolor{blue}\parafcn{meza,mezb,0.05}{(ODErhs(1,t),t)} \multido{\r=0.1+0.15}{7}{\ODEvarrow{\r}} \end{mfpic}\hss} \section{Three numerical methods for ODEs} Here we draw solution of ODE using all three available numerical methods. We use big step to see the difference between Euler, Runge--Kutta and fourth order Runge--Kutta method. \begin{center} \begin{equation*} \boxed{\begin{aligned}y'&=x+y^3\\y(0)&=1 \end{aligned}}\qquad \boxed{\begin{aligned}x_{n+1}&=x_n+h\\ y_{n+1}&=y_n+kh\\h&=0.2 \end{aligned}} \end{equation*} \leavevmode \begin{mfpic}[20][5]{0}{0.5}{0.9}{2.4} % We set-up parameters \mfsrc{ODEarrowlength:=0.5;} \ODEdefineequation{x+(y**3)} \pen{1pt}\tlabelsep{1pt} % We set up parameters for small step \nomplabels \drawcolor{green} \mfsrc{ODEstep:=0.02; ODEstepcount:=30;} \dashed\trajectoryRKF{0}{1} \tlabel[tr](0.39,2.4){\bf{Exact solution}} % We use bigger step to see the difference between various % methods. \mfsrc{ODEstep:=0.2; ODEstepcount:=2;} % We draw trajectory by Euler method \drawcolor{black} \trajectory{0}{1} \tlabel[bl](0.4,1.6){\bf{Euler}} \tlabel[tl](0,1){\bf{$k_1$}} \tlabel[tl](0.2,1.2){\bf{$k_1$ for second step}} % We draw trajectory by Runge-Kutta method \drawcolor{rgb(0.5,0.5,0.5)} \trajectoryRK{0}{1} \tlabel[cl](0.4,2.05){\bf{RK}} \tlabel[tl](0.1,1.1){\bf{$k_2$}} % We draw trajectory by fourth order Runge-Kutta method \drawcolor{rgb(1,0,0)} \trajectoryRKF{0}{1} \tlabel[tl](0.4,2.15){\bf{RK4}} \tlabelsep{3pt} % We draw direction field using blue arrows and metapost cycle \penwd{1pt} \drawcolor{blue}\headcolor{blue} \mfsrc{for j=0.9 step 0.1 until 2.3: for i:=0 step 0.05 until 0.5:} \ODEarrow{i}{j} \mfsrc{endfor;endfor;} \drawcolor{black} \doaxes{lbrt} \bmarks{0,0.2,0.4} \tmarks{0,0.2,0.4} \lmarks{1,1.1,1.2,1.3,1.4,1.5,1.6,1.7,1.8,2,2.2} \pointcolor{red} \pointfilltrue \point[4pt]{(0,1)} % We draw some other and bigger arrows in the direction field. The % slopes of hese arrows are important for the first step by Euler % and Ruge-Kutta method and the second step by Euler method. \pen{2pt} \mfsrc{ODEarrowlength:=1;} \colorODEarrowfalse \drawcolor{red}\headcolor{red} \ODEarrows{0,1;0.1,1.1} \ODEarrow{0.2}{1.2} \axislabels{b}[tc]{{$0$}0,{$0.2$}0.2,{$0.4$}0.4} \axislabels{l}[cr]{{$0.8$}0.8,{$1$}1,{$1.2$}1.2,{$1.4$}1.4,{$1.6$}1.6,{$1.8$}1.8,{$2$}2,{$2.2$}2.2,{$2.4$}2.4} \end{mfpic} \end{center} \section{Autonomous systems} % The color arrows have no sense in the phase portrait of autonomous % system. \colorODEarrowfalse We draw the phase portrait of autonomous system, nulclines, invariant set between nulclines, trajectories. We draw arrows in regular grid and add few more arrows on nulclines and outside the regular grid. \mfsrc{TIMEstep:=0.05; TIMEend:=30;} \begin{mfpic} [0.5]{-2}{15}{-2}{15} \nomplabels \tlabel[cc](8,15.5){\bf Competing species} \usemplabels % We set up parameters, define equations, define functions which % describe nulclines and store stationary points into variables z10, % z11, z12, z1. \mfsrc{a:=11;b:=1;c:=0.8;k:=10;l:=1.1;m:=1.2;} \mfsrc{ODEarrowlength:=0.3;} \ASdefineequations{x*(a-b*x-c*y)}{y*(k-l*x-m*y)} \fdef{xnulklina}{x}{(a-b*x)/c} \fdef{ynulklina}{x}{(k-l*x)/m} \mfsrc{z10=(0,a/c);z11=(0,k/m);z12=(a/b,0);z13=(k/l,0);} % Here we draw a gray polygon - invariant set fot the system. \pen{0.3pt} \gfill[gray(0.7)]\lclosed\lines{z10,z11,z13,z12} \axes \tlabel[bc](0,ypos){$y$} \tlabel[cl](xpos,0){$x$} \pointsize=3pt \pointfilltrue\pointcolor{red}\point{(a/b,0)} \draw[red]\function{0,a/b,1}{xnulklina(x)} \draw[red]\lines{(0,0),(0,ypos)} \tlabel[cr](x10,y10){$a\over c$} \tlabel[tc](x12,y12){$a\over b$} \pointfilltrue\pointcolor{green}\point{(0,k/m)} \draw[green]\function{0,k/l,1}{ynulklina(x)} \draw[green]\lines{(0,0),(xpos,0)} \tlabel[cr](x11,y11){$\alpha\over \beta$} \tlabel[tc](x13,y13){$\alpha\over \gamma$} \penwd{1.5pt} \drawcolor{gray(0.25)}\headcolor{gray(0.25)} \ASarrows{0,6;0,11;7,0;9,0;4.5,0;13,0;2,0;0,14;0,2} \ASarrows{3,3;8,7;13,2;3,14} \ASarrows{4,ynulklina(4);5,ynulklina(5);6,ynulklina(6); 1.7,ynulklina(1.7)} \ASarrows{0.5,xnulklina(0.5);1.8,xnulklina(1.8);6,xnulklina(6);4,xnulklina(4); 7.5,xnulklina(7.5)} \drawcolor{black} \AStrajectories{12,12;8,12;4,12;0.2,3;1,3;6,12;1,1;1,0.12;0.3,14} \mfsrc{TIMEstep:=-0.05; TIMEend:=5;} \AStrajectories{12,12;8,12;4,12;0.2,3;1,3;6,12;1,1;1,0.12;0.3,14} \penwd{0.5pt} \drawcolor{gray(0.5)} \headcolor{gray(0.5)} \multido{\r=0.5+1}{15}{\multido{\R=0.5+1}{15}{\ASarrow{\R}{\r}}} \end{mfpic} \begin{center} \textbf{Pedator prey system with HollingII response function} \end{center} \begin{mfpic}[2]{-0.1}{4}{-0.1}{3} % we define functions and parameters, right hand sides of the system % and a function which defines nulcline \mfsrc{r:=1;K:=3;a:=1;k:=0.8;P:=1;Alfa:=0.42;} \fdef{funkceV}{x}{a*x/(x+P)} \ASdefineequations{r*x*(1-(x/K))-funkceV(x)*y}{(-Alfa+k*funkceV(x))*y} \fdef{xnulklina}{x}{r*(1-(x/K))*(x+P)/a} \mfsrc{ODEarrowlength:=0.2;} % Here we draw axes and nulclines \axes \tlabel[bc](0,ypos){$y$} \tlabel[cl](xpos,0){$x$} \draw[red]\function{0,K,0.1}{xnulklina(x)} \draw[green]\lines{(P/((k*a/Alfa)-1),0),(P/((k*a/Alfa)-1),ypos)} % Here we draw some arrows on nulclines and then arrows in the plane \penwd{0.5pt} \drawcolor{gray(0.25)}\headcolor{gray(0.25)} \ASarrows{P/((k*a/Alfa)-1),1;P/((k*a/Alfa)-1),2;P/((k*a/Alfa)-1),0.5;P/((k*a/Alfa)-1),1.5} \ASarrows{0,xnulklina(0);0.5,xnulklina(0.5);1,xnulklina(1);1.5,xnulklina(1.5);2,xnulklina(2);2.25,xnulklina(2.25)} \multido{\r=0.1+0.25}{20}{\multido{\R=0.1+0.25}{20}{\ASarrow{\R}{\r}}} %% We draw trajectory with IC x=2, y=2 \drawcolor{black} \mfsrc{TIMEstep:=0.05; TIMEend:=10;} \AStrajectoryRKF{2}{2} %% We continue the trajectory (spiral) from the last point \AStrajectoryRKF{x1}{y1} \AStrajectoryRKF{x1}{y1} \AStrajectoryRKF{x1}{y1} \AStrajectoryRKF{x1}{y1} \AStrajectoryRKF{x1}{y1} \AStrajectoryRKF{x1}{y1} \AStrajectoryRKF{x1}{y1} %% We continue backwards \mfsrc{TIMEstep:=-0.05;} \AStrajectoryRKF{2}{2} \end{mfpic} \end{document}