finding the nullclines of a two connected ODE's











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so I've given the following problem:



$dn_1/dt=N(t)n_1-n_1$



$dn_2/dt=N(t)n_2-4n_2$



$N(t)=25-6n_1-3n_2$



so I've found the fixed points of the 2 equations and got:
[(0,0),(0,7),(4,0)]



now I want to find the nullclines of them:



so for $dn_1/dt=0$ we get $n_1(24-6n_1-3n_2)$ which means:




  1. $n_1=0$ for all $n_2$


  2. $n_1=4-0.5n_2$



so for $dn_2/dt=0$ we get $n_2(21-6n_1-3n_2)$ which means:




  1. $n_2=0$ for all $n_1$


  2. $n_2=7-2n_1$



plotting the following using matlab in order to find the phase plane and basin of attraction gives peculiar results which I can't find to be able to understand. (fig attached below)
I thought the the 3 nullcline need to be intersecting both the fixed points which is more intuitive to me but instead I get the following.



enter image description here



the Matlab code generating this plot is as follows:



clear, close all; clc
%Const
N0=25;
G1=1;G2=1;
a1=6;a2=3;
k1=1;k2=4;
%Params
t0 = 0; % starting time
dt = 0.5; % step size
tEnd = 50; % end time
timeVec=t0:dt:tEnd;
NumSteps=floor(tEnd./dt);
%Function Handlers
N_t=@(n1,n2) N0-a1.*n1-a2.*n2;
n1_dot=@(n1,n2) G1.*N_t(n1,n2).*n1-k1.*n1;
n2_dot=@(n1,n2) G2.*N_t(n1,n2).*n2-k2.*n2;

%%finding the fixed point's
%%%%%%%%%
system=@(n1,n2) [n1_dot(n1,n2);n2_dot(n1,n2)];
s = solve(system);
%%%%%%%%%

%TODO RECHECK!!!!
%%nulclines
%%%%%%%%%
%n1_dot=0
%case 1: n1 = 0
%case 2: n1 != 0

%n2_dot=0
%case 1: n2 = 0
%case 2: n2 != 0

%%%%%%%%%
ptsMesh=-10:0.2:25;
figure
[X,Y]=meshgrid(ptsMesh);
U=n1_dot(X,Y);
V=n2_dot(X,Y);
normalizedFactor=sqrt(U.^2+V.^2);
quiver(X,Y,U./normalizedFactor,V./normalizedFactor,1,'b')
hold on
plot(s.n1,s.n2,'k*','lineWidth',8)
%plot the trivial nullclines on the axis
[x,y]=size(ptsMesh);
nullclineN1_zero=[0,min(ptsMesh);0,max(ptsMesh)];
nullclineN2_zero=[min(ptsMesh),0;max(ptsMesh),0];
plot(nullclineN2_zero(:,1),nullclineN2_zero(:,2),'g','lineWidth',2);
plot(nullclineN1_zero(:,1),nullclineN1_zero(:,2),'r','lineWidth',2);

t1=@(x2) 4-(0.5).*x2;
res1=t1(ptsMesh);
plot(res1,ptsMesh,'c','lineWidth',2);


t2=@(x1) 7-(2).*x1;
res2=t2(ptsMesh);
plot(ptsMesh,res2,'k','lineWidth',2);


will appreciate some help here
thanks










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    up vote
    0
    down vote

    favorite












    so I've given the following problem:



    $dn_1/dt=N(t)n_1-n_1$



    $dn_2/dt=N(t)n_2-4n_2$



    $N(t)=25-6n_1-3n_2$



    so I've found the fixed points of the 2 equations and got:
    [(0,0),(0,7),(4,0)]



    now I want to find the nullclines of them:



    so for $dn_1/dt=0$ we get $n_1(24-6n_1-3n_2)$ which means:




    1. $n_1=0$ for all $n_2$


    2. $n_1=4-0.5n_2$



    so for $dn_2/dt=0$ we get $n_2(21-6n_1-3n_2)$ which means:




    1. $n_2=0$ for all $n_1$


    2. $n_2=7-2n_1$



    plotting the following using matlab in order to find the phase plane and basin of attraction gives peculiar results which I can't find to be able to understand. (fig attached below)
    I thought the the 3 nullcline need to be intersecting both the fixed points which is more intuitive to me but instead I get the following.



    enter image description here



    the Matlab code generating this plot is as follows:



    clear, close all; clc
    %Const
    N0=25;
    G1=1;G2=1;
    a1=6;a2=3;
    k1=1;k2=4;
    %Params
    t0 = 0; % starting time
    dt = 0.5; % step size
    tEnd = 50; % end time
    timeVec=t0:dt:tEnd;
    NumSteps=floor(tEnd./dt);
    %Function Handlers
    N_t=@(n1,n2) N0-a1.*n1-a2.*n2;
    n1_dot=@(n1,n2) G1.*N_t(n1,n2).*n1-k1.*n1;
    n2_dot=@(n1,n2) G2.*N_t(n1,n2).*n2-k2.*n2;

    %%finding the fixed point's
    %%%%%%%%%
    system=@(n1,n2) [n1_dot(n1,n2);n2_dot(n1,n2)];
    s = solve(system);
    %%%%%%%%%

    %TODO RECHECK!!!!
    %%nulclines
    %%%%%%%%%
    %n1_dot=0
    %case 1: n1 = 0
    %case 2: n1 != 0

    %n2_dot=0
    %case 1: n2 = 0
    %case 2: n2 != 0

    %%%%%%%%%
    ptsMesh=-10:0.2:25;
    figure
    [X,Y]=meshgrid(ptsMesh);
    U=n1_dot(X,Y);
    V=n2_dot(X,Y);
    normalizedFactor=sqrt(U.^2+V.^2);
    quiver(X,Y,U./normalizedFactor,V./normalizedFactor,1,'b')
    hold on
    plot(s.n1,s.n2,'k*','lineWidth',8)
    %plot the trivial nullclines on the axis
    [x,y]=size(ptsMesh);
    nullclineN1_zero=[0,min(ptsMesh);0,max(ptsMesh)];
    nullclineN2_zero=[min(ptsMesh),0;max(ptsMesh),0];
    plot(nullclineN2_zero(:,1),nullclineN2_zero(:,2),'g','lineWidth',2);
    plot(nullclineN1_zero(:,1),nullclineN1_zero(:,2),'r','lineWidth',2);

    t1=@(x2) 4-(0.5).*x2;
    res1=t1(ptsMesh);
    plot(res1,ptsMesh,'c','lineWidth',2);


    t2=@(x1) 7-(2).*x1;
    res2=t2(ptsMesh);
    plot(ptsMesh,res2,'k','lineWidth',2);


    will appreciate some help here
    thanks










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      so I've given the following problem:



      $dn_1/dt=N(t)n_1-n_1$



      $dn_2/dt=N(t)n_2-4n_2$



      $N(t)=25-6n_1-3n_2$



      so I've found the fixed points of the 2 equations and got:
      [(0,0),(0,7),(4,0)]



      now I want to find the nullclines of them:



      so for $dn_1/dt=0$ we get $n_1(24-6n_1-3n_2)$ which means:




      1. $n_1=0$ for all $n_2$


      2. $n_1=4-0.5n_2$



      so for $dn_2/dt=0$ we get $n_2(21-6n_1-3n_2)$ which means:




      1. $n_2=0$ for all $n_1$


      2. $n_2=7-2n_1$



      plotting the following using matlab in order to find the phase plane and basin of attraction gives peculiar results which I can't find to be able to understand. (fig attached below)
      I thought the the 3 nullcline need to be intersecting both the fixed points which is more intuitive to me but instead I get the following.



      enter image description here



      the Matlab code generating this plot is as follows:



      clear, close all; clc
      %Const
      N0=25;
      G1=1;G2=1;
      a1=6;a2=3;
      k1=1;k2=4;
      %Params
      t0 = 0; % starting time
      dt = 0.5; % step size
      tEnd = 50; % end time
      timeVec=t0:dt:tEnd;
      NumSteps=floor(tEnd./dt);
      %Function Handlers
      N_t=@(n1,n2) N0-a1.*n1-a2.*n2;
      n1_dot=@(n1,n2) G1.*N_t(n1,n2).*n1-k1.*n1;
      n2_dot=@(n1,n2) G2.*N_t(n1,n2).*n2-k2.*n2;

      %%finding the fixed point's
      %%%%%%%%%
      system=@(n1,n2) [n1_dot(n1,n2);n2_dot(n1,n2)];
      s = solve(system);
      %%%%%%%%%

      %TODO RECHECK!!!!
      %%nulclines
      %%%%%%%%%
      %n1_dot=0
      %case 1: n1 = 0
      %case 2: n1 != 0

      %n2_dot=0
      %case 1: n2 = 0
      %case 2: n2 != 0

      %%%%%%%%%
      ptsMesh=-10:0.2:25;
      figure
      [X,Y]=meshgrid(ptsMesh);
      U=n1_dot(X,Y);
      V=n2_dot(X,Y);
      normalizedFactor=sqrt(U.^2+V.^2);
      quiver(X,Y,U./normalizedFactor,V./normalizedFactor,1,'b')
      hold on
      plot(s.n1,s.n2,'k*','lineWidth',8)
      %plot the trivial nullclines on the axis
      [x,y]=size(ptsMesh);
      nullclineN1_zero=[0,min(ptsMesh);0,max(ptsMesh)];
      nullclineN2_zero=[min(ptsMesh),0;max(ptsMesh),0];
      plot(nullclineN2_zero(:,1),nullclineN2_zero(:,2),'g','lineWidth',2);
      plot(nullclineN1_zero(:,1),nullclineN1_zero(:,2),'r','lineWidth',2);

      t1=@(x2) 4-(0.5).*x2;
      res1=t1(ptsMesh);
      plot(res1,ptsMesh,'c','lineWidth',2);


      t2=@(x1) 7-(2).*x1;
      res2=t2(ptsMesh);
      plot(ptsMesh,res2,'k','lineWidth',2);


      will appreciate some help here
      thanks










      share|cite|improve this question















      so I've given the following problem:



      $dn_1/dt=N(t)n_1-n_1$



      $dn_2/dt=N(t)n_2-4n_2$



      $N(t)=25-6n_1-3n_2$



      so I've found the fixed points of the 2 equations and got:
      [(0,0),(0,7),(4,0)]



      now I want to find the nullclines of them:



      so for $dn_1/dt=0$ we get $n_1(24-6n_1-3n_2)$ which means:




      1. $n_1=0$ for all $n_2$


      2. $n_1=4-0.5n_2$



      so for $dn_2/dt=0$ we get $n_2(21-6n_1-3n_2)$ which means:




      1. $n_2=0$ for all $n_1$


      2. $n_2=7-2n_1$



      plotting the following using matlab in order to find the phase plane and basin of attraction gives peculiar results which I can't find to be able to understand. (fig attached below)
      I thought the the 3 nullcline need to be intersecting both the fixed points which is more intuitive to me but instead I get the following.



      enter image description here



      the Matlab code generating this plot is as follows:



      clear, close all; clc
      %Const
      N0=25;
      G1=1;G2=1;
      a1=6;a2=3;
      k1=1;k2=4;
      %Params
      t0 = 0; % starting time
      dt = 0.5; % step size
      tEnd = 50; % end time
      timeVec=t0:dt:tEnd;
      NumSteps=floor(tEnd./dt);
      %Function Handlers
      N_t=@(n1,n2) N0-a1.*n1-a2.*n2;
      n1_dot=@(n1,n2) G1.*N_t(n1,n2).*n1-k1.*n1;
      n2_dot=@(n1,n2) G2.*N_t(n1,n2).*n2-k2.*n2;

      %%finding the fixed point's
      %%%%%%%%%
      system=@(n1,n2) [n1_dot(n1,n2);n2_dot(n1,n2)];
      s = solve(system);
      %%%%%%%%%

      %TODO RECHECK!!!!
      %%nulclines
      %%%%%%%%%
      %n1_dot=0
      %case 1: n1 = 0
      %case 2: n1 != 0

      %n2_dot=0
      %case 1: n2 = 0
      %case 2: n2 != 0

      %%%%%%%%%
      ptsMesh=-10:0.2:25;
      figure
      [X,Y]=meshgrid(ptsMesh);
      U=n1_dot(X,Y);
      V=n2_dot(X,Y);
      normalizedFactor=sqrt(U.^2+V.^2);
      quiver(X,Y,U./normalizedFactor,V./normalizedFactor,1,'b')
      hold on
      plot(s.n1,s.n2,'k*','lineWidth',8)
      %plot the trivial nullclines on the axis
      [x,y]=size(ptsMesh);
      nullclineN1_zero=[0,min(ptsMesh);0,max(ptsMesh)];
      nullclineN2_zero=[min(ptsMesh),0;max(ptsMesh),0];
      plot(nullclineN2_zero(:,1),nullclineN2_zero(:,2),'g','lineWidth',2);
      plot(nullclineN1_zero(:,1),nullclineN1_zero(:,2),'r','lineWidth',2);

      t1=@(x2) 4-(0.5).*x2;
      res1=t1(ptsMesh);
      plot(res1,ptsMesh,'c','lineWidth',2);


      t2=@(x1) 7-(2).*x1;
      res2=t2(ptsMesh);
      plot(ptsMesh,res2,'k','lineWidth',2);


      will appreciate some help here
      thanks







      mathematical-modeling stability-in-odes






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      edited Nov 14 at 18:57

























      asked Nov 14 at 17:53









      chemist

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      143



























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