Difference between revisions of "Team:KU Leuven/Modeling/Top"

Line 56: Line 56:
 
   <p>
 
   <p>
 
     The Keller segel model used is <sub> <a href="#ref1">[1] </a></sub>:
 
     The Keller segel model used is <sub> <a href="#ref1">[1] </a></sub>:
    $$\frac{\partial A}{\partial t} = \bigtriangledown^2 A + k_A A(1 - \frac{A}{k_p}),$$ 
 
 
     $$\frac{\partial A}{\partial t} = D_a \bigtriangledown^2 A + k_A A(1 - \frac{A}{k_{p}}),$$
 
     $$\frac{\partial A}{\partial t} = D_a \bigtriangledown^2 A + k_A A(1 - \frac{A}{k_{p}}),$$
 
     $$\frac{\partial B}{\partial t} = D_b \bigtriangledown^2 B + \bigtriangledown (X(B,H,R) \bigtriangledown R)+  k_B B(1 - \frac{B}{k_{p}}), $$
 
     $$\frac{\partial B}{\partial t} = D_b \bigtriangledown^2 B + \bigtriangledown (X(B,H,R) \bigtriangledown R)+  k_B B(1 - \frac{B}{k_{p}}), $$
 
     $$ \frac{\partial R}{\partial t} = D_r \bigtriangledown^2 B +  k_r A - k_lossH R $$
 
     $$ \frac{\partial R}{\partial t} = D_r \bigtriangledown^2 B +  k_r A - k_lossH R $$
 
     $$\frac{\partial H}{\partial t} = D_h \bigtriangledown^2 B +  k_h A - k_lossR H . $$
 
     $$\frac{\partial H}{\partial t} = D_h \bigtriangledown^2 B +  k_h A - k_lossR H . $$
 +
    With:  </br>
 +
    $$ X(B,H,R) = -B K_{c1} (K_{c2} H(t,:)/R(t,:)). $$
  
    </br>
 
 
     When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
 
     When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
 
     $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
 
     $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

Revision as of 13:44, 23 July 2015

1-D continuous model

The Keller segel model used is [1] : $$\frac{\partial A}{\partial t} = D_a \bigtriangledown^2 A + k_A A(1 - \frac{A}{k_{p}}),$$ $$\frac{\partial B}{\partial t} = D_b \bigtriangledown^2 B + \bigtriangledown (X(B,H,R) \bigtriangledown R)+ k_B B(1 - \frac{B}{k_{p}}), $$ $$ \frac{\partial R}{\partial t} = D_r \bigtriangledown^2 B + k_r A - k_lossH R $$ $$\frac{\partial H}{\partial t} = D_h \bigtriangledown^2 B + k_h A - k_lossR H . $$ With:
$$ X(B,H,R) = -B K_{c1} (K_{c2} H(t,:)/R(t,:)). $$ When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

References

Reference 1