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

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     $$\frac{\partial A}{\partial t} = D_a \bigtriangledown^2 A + k_A A(1 - \frac{A}{k_{p}}),$$
 
     $$\frac{\partial A}{\partial t} = D_a \bigtriangledown^2 A + k_A A(1 - \frac{A}{k_{p}}),$$
 
     $$\frac{\partial B}{\partial t} = D_b \bigtriangledown^2 B + \bigtriangledown (X(B,H,R) \bigtriangledown R)+  k_B B(1 - \frac{B}{k_{p}}), $$
 
     $$\frac{\partial B}{\partial t} = D_b \bigtriangledown^2 B + \bigtriangledown (X(B,H,R) \bigtriangledown R)+  k_B B(1 - \frac{B}{k_{p}}), $$
−
     $$ \frac{\partial R}{\partial t} = D_r \bigtriangledown^2 B +  k_r A - k_lossH R $$
+
     $$ \frac{\partial R}{\partial t} = D_r \bigtriangledown^2 B +  k_r A - k_{lossH} R $$
−
     $$\frac{\partial H}{\partial t} = D_h \bigtriangledown^2 B +  k_h A - k_lossR H . $$
+
     $$\frac{\partial H}{\partial t} = D_h \bigtriangledown^2 B +  k_h A - k_{lossR} H . $$
 
     With:  </br>
 
     With:  </br>
−
     $$ X(B,H,R) = -B K_{c1} (K_{c2} H(t,:)/R(t,:)). $$
+
     $$ X(B,H,R) = \frac{-B K_{c1} H}{K_{c2} R}. $$
  
 
     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

Revision as of 13:50, 23 July 2015

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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:
$$ X(B,H,R) = \frac{-B K_{c1} H}{K_{c2} R}. $$ 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

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