Mixing
How to get the chemical form of the Lotka-Volterra ODEs
$begingroup$
I know how to work from a chemical equation to an ODE, as described here:
http://brunel.ac.uk/~cspgoop/uploads/ode_chemical_network.pdf
How do I go the other way? I want to convert the Lotka-Volterra ODE system to chemical equations so I can use it in a cell simulation program.
ordinary-differential-equations mathematical-modeling
edited Jan 11 at 22:18
Did
248k23223460
asked Oct 29 '14 at 12:31
RNs_GhostRNs_Ghost
172119
$endgroup$
add a comment |
$begingroup$
I know how to work from a chemical equation to an ODE, as described here:
http://brunel.ac.uk/~cspgoop/uploads/ode_chemical_network.pdf
How do I go the other way? I want to convert the Lotka-Volterra ODE system to chemical equations so I can use it in a cell simulation program.
ordinary-differential-equations mathematical-modeling
edited Jan 11 at 22:18
Did
248k23223460
asked Oct 29 '14 at 12:31
RNs_GhostRNs_Ghost
172119
$endgroup$
$begingroup$
Do this first for some simple polynomial scalar equations to get a feel, e.g. $dot{x} = ax - bx^2$.
$endgroup$
– Hans Engler
Oct 29 '14 at 12:44
add a comment |
$begingroup$
I know how to work from a chemical equation to an ODE, as described here:
http://brunel.ac.uk/~cspgoop/uploads/ode_chemical_network.pdf
How do I go the other way? I want to convert the Lotka-Volterra ODE system to chemical equations so I can use it in a cell simulation program.
ordinary-differential-equations mathematical-modeling
edited Jan 11 at 22:18
Did
248k23223460
asked Oct 29 '14 at 12:31
RNs_GhostRNs_Ghost
172119
$endgroup$
I know how to work from a chemical equation to an ODE, as described here:
http://brunel.ac.uk/~cspgoop/uploads/ode_chemical_network.pdf
How do I go the other way? I want to convert the Lotka-Volterra ODE system to chemical equations so I can use it in a cell simulation program.
ordinary-differential-equations mathematical-modeling
ordinary-differential-equations mathematical-modeling
edited Jan 11 at 22:18
Did
248k23223460
asked Oct 29 '14 at 12:31
RNs_GhostRNs_Ghost
172119
edited Jan 11 at 22:18
Did
248k23223460
asked Oct 29 '14 at 12:31
RNs_GhostRNs_Ghost
172119
edited Jan 11 at 22:18
Did
248k23223460
edited Jan 11 at 22:18
Did
248k23223460
edited Jan 11 at 22:18
Did
248k23223460
248k23223460
asked Oct 29 '14 at 12:31
RNs_GhostRNs_Ghost
172119
asked Oct 29 '14 at 12:31
RNs_GhostRNs_Ghost
172119
asked Oct 29 '14 at 12:31
RNs_GhostRNs_Ghost
172119
172119
$begingroup$
Do this first for some simple polynomial scalar equations to get a feel, e.g. $dot{x} = ax - bx^2$.
$endgroup$
– Hans Engler
Oct 29 '14 at 12:44
add a comment |
$begingroup$
Do this first for some simple polynomial scalar equations to get a feel, e.g. $dot{x} = ax - bx^2$.
$endgroup$
– Hans Engler
Oct 29 '14 at 12:44
$begingroup$
Do this first for some simple polynomial scalar equations to get a feel, e.g. $dot{x} = ax - bx^2$.
$endgroup$
– Hans Engler
Oct 29 '14 at 12:44
$begingroup$
Do this first for some simple polynomial scalar equations to get a feel, e.g. $dot{x} = ax - bx^2$.
$endgroup$
– Hans Engler
Oct 29 '14 at 12:44
add a comment |
1 Answer
1
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$begingroup$
This is direct if one analyzes the dynamics Lotka-Volterra systems are supposed to model: some population $X$ of preys, some population $Y$ of predators, in isolation preys reproduce and predators die, and when in contact predators kill preys and multiply.
This fits the system of "chemical" reactions $$Xto2X,qquad Yto Z,qquad X+Yto2Y,$$ where $Z$ is a junk species accounting for the dead predators.
Minor caveat: the rates $(a,b,c)$ of the three "reactions" above yield the differential equations $$X'=aX-cXY,qquad Y'=-bY+cXY,tag{$ast$}$$ hence, to solve/simulate the general system $$X'=aX-cXY,qquad Y'=-bY+dXY,$$ depending on four rates $(a,b,c,d)$, one should consider "reduced" populations $$((c/d)X,Y),$$ where $(X,Y)$ solves $(ast)$.
answered Oct 29 '14 at 12:46
DidDid
248k23223460
$endgroup$
$begingroup$
Thanks for your answer, however, why isn't there a few squared terms because of the 2 in the first and third equations?
$endgroup$
– RNs_Ghost
Oct 29 '14 at 13:45
$begingroup$
Scalar factors on the RHS of the chemical reactions translate into scalar factors of the concentrations, not into powers. The reaction $A+3Bto2C+5D$ at rate $r$, say, contributes for $-rAcdot B^3$ in $A'$, for $-3rAcdot B^3$ in $B'$, for $2rAcdot B^3$ in $C'$ and for $5rAcdot B^3$ in $D'$.
$endgroup$
– Did
Oct 29 '14 at 14:42
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
This is direct if one analyzes the dynamics Lotka-Volterra systems are supposed to model: some population $X$ of preys, some population $Y$ of predators, in isolation preys reproduce and predators die, and when in contact predators kill preys and multiply.
This fits the system of "chemical" reactions $$Xto2X,qquad Yto Z,qquad X+Yto2Y,$$ where $Z$ is a junk species accounting for the dead predators.
Minor caveat: the rates $(a,b,c)$ of the three "reactions" above yield the differential equations $$X'=aX-cXY,qquad Y'=-bY+cXY,tag{$ast$}$$ hence, to solve/simulate the general system $$X'=aX-cXY,qquad Y'=-bY+dXY,$$ depending on four rates $(a,b,c,d)$, one should consider "reduced" populations $$((c/d)X,Y),$$ where $(X,Y)$ solves $(ast)$.
answered Oct 29 '14 at 12:46
DidDid
248k23223460
$endgroup$
$begingroup$
Thanks for your answer, however, why isn't there a few squared terms because of the 2 in the first and third equations?
$endgroup$
– RNs_Ghost
Oct 29 '14 at 13:45
$begingroup$
Scalar factors on the RHS of the chemical reactions translate into scalar factors of the concentrations, not into powers. The reaction $A+3Bto2C+5D$ at rate $r$, say, contributes for $-rAcdot B^3$ in $A'$, for $-3rAcdot B^3$ in $B'$, for $2rAcdot B^3$ in $C'$ and for $5rAcdot B^3$ in $D'$.
$endgroup$
– Did
Oct 29 '14 at 14:42
add a comment |
$begingroup$
This is direct if one analyzes the dynamics Lotka-Volterra systems are supposed to model: some population $X$ of preys, some population $Y$ of predators, in isolation preys reproduce and predators die, and when in contact predators kill preys and multiply.
This fits the system of "chemical" reactions $$Xto2X,qquad Yto Z,qquad X+Yto2Y,$$ where $Z$ is a junk species accounting for the dead predators.
Minor caveat: the rates $(a,b,c)$ of the three "reactions" above yield the differential equations $$X'=aX-cXY,qquad Y'=-bY+cXY,tag{$ast$}$$ hence, to solve/simulate the general system $$X'=aX-cXY,qquad Y'=-bY+dXY,$$ depending on four rates $(a,b,c,d)$, one should consider "reduced" populations $$((c/d)X,Y),$$ where $(X,Y)$ solves $(ast)$.
answered Oct 29 '14 at 12:46
DidDid
248k23223460
$endgroup$
$begingroup$
Thanks for your answer, however, why isn't there a few squared terms because of the 2 in the first and third equations?
$endgroup$
– RNs_Ghost
Oct 29 '14 at 13:45
$begingroup$
Scalar factors on the RHS of the chemical reactions translate into scalar factors of the concentrations, not into powers. The reaction $A+3Bto2C+5D$ at rate $r$, say, contributes for $-rAcdot B^3$ in $A'$, for $-3rAcdot B^3$ in $B'$, for $2rAcdot B^3$ in $C'$ and for $5rAcdot B^3$ in $D'$.
$endgroup$
– Did
Oct 29 '14 at 14:42
add a comment |
$begingroup$
This is direct if one analyzes the dynamics Lotka-Volterra systems are supposed to model: some population $X$ of preys, some population $Y$ of predators, in isolation preys reproduce and predators die, and when in contact predators kill preys and multiply.
This fits the system of "chemical" reactions $$Xto2X,qquad Yto Z,qquad X+Yto2Y,$$ where $Z$ is a junk species accounting for the dead predators.
Minor caveat: the rates $(a,b,c)$ of the three "reactions" above yield the differential equations $$X'=aX-cXY,qquad Y'=-bY+cXY,tag{$ast$}$$ hence, to solve/simulate the general system $$X'=aX-cXY,qquad Y'=-bY+dXY,$$ depending on four rates $(a,b,c,d)$, one should consider "reduced" populations $$((c/d)X,Y),$$ where $(X,Y)$ solves $(ast)$.
answered Oct 29 '14 at 12:46
DidDid
248k23223460
$endgroup$
This is direct if one analyzes the dynamics Lotka-Volterra systems are supposed to model: some population $X$ of preys, some population $Y$ of predators, in isolation preys reproduce and predators die, and when in contact predators kill preys and multiply.
This fits the system of "chemical" reactions $$Xto2X,qquad Yto Z,qquad X+Yto2Y,$$ where $Z$ is a junk species accounting for the dead predators.
Minor caveat: the rates $(a,b,c)$ of the three "reactions" above yield the differential equations $$X'=aX-cXY,qquad Y'=-bY+cXY,tag{$ast$}$$ hence, to solve/simulate the general system $$X'=aX-cXY,qquad Y'=-bY+dXY,$$ depending on four rates $(a,b,c,d)$, one should consider "reduced" populations $$((c/d)X,Y),$$ where $(X,Y)$ solves $(ast)$.
answered Oct 29 '14 at 12:46
DidDid
248k23223460
edited Jun 27 '15 at 10:34
answered Oct 29 '14 at 12:46
DidDid
248k23223460
answered Oct 29 '14 at 12:46
DidDid
248k23223460
answered Oct 29 '14 at 12:46
DidDid
248k23223460
248k23223460
$begingroup$
Thanks for your answer, however, why isn't there a few squared terms because of the 2 in the first and third equations?
$endgroup$
– RNs_Ghost
Oct 29 '14 at 13:45
$begingroup$
Scalar factors on the RHS of the chemical reactions translate into scalar factors of the concentrations, not into powers. The reaction $A+3Bto2C+5D$ at rate $r$, say, contributes for $-rAcdot B^3$ in $A'$, for $-3rAcdot B^3$ in $B'$, for $2rAcdot B^3$ in $C'$ and for $5rAcdot B^3$ in $D'$.
$endgroup$
– Did
Oct 29 '14 at 14:42
add a comment |
$begingroup$
Thanks for your answer, however, why isn't there a few squared terms because of the 2 in the first and third equations?
$endgroup$
– RNs_Ghost
Oct 29 '14 at 13:45
$begingroup$
Scalar factors on the RHS of the chemical reactions translate into scalar factors of the concentrations, not into powers. The reaction $A+3Bto2C+5D$ at rate $r$, say, contributes for $-rAcdot B^3$ in $A'$, for $-3rAcdot B^3$ in $B'$, for $2rAcdot B^3$ in $C'$ and for $5rAcdot B^3$ in $D'$.
$endgroup$
– Did
Oct 29 '14 at 14:42
$begingroup$
Thanks for your answer, however, why isn't there a few squared terms because of the 2 in the first and third equations?
$endgroup$
– RNs_Ghost
Oct 29 '14 at 13:45
$begingroup$
Thanks for your answer, however, why isn't there a few squared terms because of the 2 in the first and third equations?
$endgroup$
– RNs_Ghost
Oct 29 '14 at 13:45
$begingroup$
Scalar factors on the RHS of the chemical reactions translate into scalar factors of the concentrations, not into powers. The reaction $A+3Bto2C+5D$ at rate $r$, say, contributes for $-rAcdot B^3$ in $A'$, for $-3rAcdot B^3$ in $B'$, for $2rAcdot B^3$ in $C'$ and for $5rAcdot B^3$ in $D'$.
$endgroup$
– Did
Oct 29 '14 at 14:42
$begingroup$
Scalar factors on the RHS of the chemical reactions translate into scalar factors of the concentrations, not into powers. The reaction $A+3Bto2C+5D$ at rate $r$, say, contributes for $-rAcdot B^3$ in $A'$, for $-3rAcdot B^3$ in $B'$, for $2rAcdot B^3$ in $C'$ and for $5rAcdot B^3$ in $D'$.
$endgroup$
– Did
Oct 29 '14 at 14:42
add a comment |
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$begingroup$
Do this first for some simple polynomial scalar equations to get a feel, e.g. $dot{x} = ax - bx^2$.
$endgroup$
– Hans Engler
Oct 29 '14 at 12:44