How to non-dimensionalise a system of ODEs?
So I have these two equations
begin{align}
frac{dx}{dt} &= k_1A-k_2Bx+k_3x^2y-k_4x tag{1} \
frac{dy}{dt} &= k_2Bx-k_3x^2y tag{2} end{align}
and I am suppose to non-dimensionalise them to get this result
begin{align}
frac{du}{dT} &= 1-(b+1)u+au^2v tag{3} \
frac{dv}{dT} &= bu-au^2v tag{4}
end{align}
and determine constants $a$ and $b$
Where $x$ and $y$ correspond to $u$ and $v$, respectively.
If I use $x=x^* u$, $y=y^* v$ and $T=k_4t$
I can simplfy $(1)$ to
$$ frac{du}{dT}=frac{k_1A}{k_4x^*} - left(frac{k_2B}{k_4}+1right)u+frac{k_3x^* y^* u^2v}{k_4} $$
and if I let $A=frac{k_4x^*}{k_1}$, $b=frac{k_2B}{k_4}$ and $a=frac{k_3x^*y^*}{k_4}$ I get the desired result of $(3)$
However
when I simplify $(2)$ in the same way I get
$$ frac{dv}{dT} = frac{x^*}{y^*} left(frac{k_2B}{k_4}-frac{k_3x^* y^* u^2v}{k_4}right) $$
and if I sub in my values for $b$ and $a$ from above I also kind of get my desired result of $(4)$ but I have $x^*/y^*$ as a factor stuck out in front. I get
$$ frac{dv}{dT}= frac{x^*}{y^*}(bu-au^2v) $$
Where have I gone wrong?
Unless I am allowed to have different $a$ and $b$ for $(3)$ and $(4)$
differential-equations
edited Nov 27 at 8:50
Dylan
12.2k31026
asked Nov 26 at 18:29
countduckula
1
add a comment |
So I have these two equations
begin{align}
frac{dx}{dt} &= k_1A-k_2Bx+k_3x^2y-k_4x tag{1} \
frac{dy}{dt} &= k_2Bx-k_3x^2y tag{2} end{align}
and I am suppose to non-dimensionalise them to get this result
begin{align}
frac{du}{dT} &= 1-(b+1)u+au^2v tag{3} \
frac{dv}{dT} &= bu-au^2v tag{4}
end{align}
and determine constants $a$ and $b$
Where $x$ and $y$ correspond to $u$ and $v$, respectively.
If I use $x=x^* u$, $y=y^* v$ and $T=k_4t$
I can simplfy $(1)$ to
$$ frac{du}{dT}=frac{k_1A}{k_4x^*} - left(frac{k_2B}{k_4}+1right)u+frac{k_3x^* y^* u^2v}{k_4} $$
and if I let $A=frac{k_4x^*}{k_1}$, $b=frac{k_2B}{k_4}$ and $a=frac{k_3x^*y^*}{k_4}$ I get the desired result of $(3)$
However
when I simplify $(2)$ in the same way I get
$$ frac{dv}{dT} = frac{x^*}{y^*} left(frac{k_2B}{k_4}-frac{k_3x^* y^* u^2v}{k_4}right) $$
and if I sub in my values for $b$ and $a$ from above I also kind of get my desired result of $(4)$ but I have $x^*/y^*$ as a factor stuck out in front. I get
$$ frac{dv}{dT}= frac{x^*}{y^*}(bu-au^2v) $$
Where have I gone wrong?
Unless I am allowed to have different $a$ and $b$ for $(3)$ and $(4)$
differential-equations
edited Nov 27 at 8:50
Dylan
12.2k31026
asked Nov 26 at 18:29
countduckula
1
Here's a tutorial on the site's formatting. Your question would be better received with it. I did it for you this time.
– Dylan
Nov 27 at 8:48
add a comment |
So I have these two equations
begin{align}
frac{dx}{dt} &= k_1A-k_2Bx+k_3x^2y-k_4x tag{1} \
frac{dy}{dt} &= k_2Bx-k_3x^2y tag{2} end{align}
and I am suppose to non-dimensionalise them to get this result
begin{align}
frac{du}{dT} &= 1-(b+1)u+au^2v tag{3} \
frac{dv}{dT} &= bu-au^2v tag{4}
end{align}
and determine constants $a$ and $b$
Where $x$ and $y$ correspond to $u$ and $v$, respectively.
If I use $x=x^* u$, $y=y^* v$ and $T=k_4t$
I can simplfy $(1)$ to
$$ frac{du}{dT}=frac{k_1A}{k_4x^*} - left(frac{k_2B}{k_4}+1right)u+frac{k_3x^* y^* u^2v}{k_4} $$
and if I let $A=frac{k_4x^*}{k_1}$, $b=frac{k_2B}{k_4}$ and $a=frac{k_3x^*y^*}{k_4}$ I get the desired result of $(3)$
However
when I simplify $(2)$ in the same way I get
$$ frac{dv}{dT} = frac{x^*}{y^*} left(frac{k_2B}{k_4}-frac{k_3x^* y^* u^2v}{k_4}right) $$
and if I sub in my values for $b$ and $a$ from above I also kind of get my desired result of $(4)$ but I have $x^*/y^*$ as a factor stuck out in front. I get
$$ frac{dv}{dT}= frac{x^*}{y^*}(bu-au^2v) $$
Where have I gone wrong?
Unless I am allowed to have different $a$ and $b$ for $(3)$ and $(4)$
differential-equations
edited Nov 27 at 8:50
Dylan
12.2k31026
asked Nov 26 at 18:29
countduckula
1
So I have these two equations
begin{align}
frac{dx}{dt} &= k_1A-k_2Bx+k_3x^2y-k_4x tag{1} \
frac{dy}{dt} &= k_2Bx-k_3x^2y tag{2} end{align}
and I am suppose to non-dimensionalise them to get this result
begin{align}
frac{du}{dT} &= 1-(b+1)u+au^2v tag{3} \
frac{dv}{dT} &= bu-au^2v tag{4}
end{align}
and determine constants $a$ and $b$
Where $x$ and $y$ correspond to $u$ and $v$, respectively.
If I use $x=x^* u$, $y=y^* v$ and $T=k_4t$
I can simplfy $(1)$ to
$$ frac{du}{dT}=frac{k_1A}{k_4x^*} - left(frac{k_2B}{k_4}+1right)u+frac{k_3x^* y^* u^2v}{k_4} $$
and if I let $A=frac{k_4x^*}{k_1}$, $b=frac{k_2B}{k_4}$ and $a=frac{k_3x^*y^*}{k_4}$ I get the desired result of $(3)$
However
when I simplify $(2)$ in the same way I get
$$ frac{dv}{dT} = frac{x^*}{y^*} left(frac{k_2B}{k_4}-frac{k_3x^* y^* u^2v}{k_4}right) $$
and if I sub in my values for $b$ and $a$ from above I also kind of get my desired result of $(4)$ but I have $x^*/y^*$ as a factor stuck out in front. I get
$$ frac{dv}{dT}= frac{x^*}{y^*}(bu-au^2v) $$
Where have I gone wrong?
Unless I am allowed to have different $a$ and $b$ for $(3)$ and $(4)$
differential-equations
differential-equations
edited Nov 27 at 8:50
Dylan
12.2k31026
asked Nov 26 at 18:29
countduckula
1
edited Nov 27 at 8:50
Dylan
12.2k31026
asked Nov 26 at 18:29
countduckula
1
edited Nov 27 at 8:50
Dylan
12.2k31026
edited Nov 27 at 8:50
Dylan
12.2k31026
edited Nov 27 at 8:50
Dylan
12.2k31026
12.2k31026
asked Nov 26 at 18:29
countduckula
1
asked Nov 26 at 18:29
countduckula
1
asked Nov 26 at 18:29
countduckula
1
1
Here's a tutorial on the site's formatting. Your question would be better received with it. I did it for you this time.
– Dylan
Nov 27 at 8:48
add a comment |
Here's a tutorial on the site's formatting. Your question would be better received with it. I did it for you this time.
– Dylan
Nov 27 at 8:48
Here's a tutorial on the site's formatting. Your question would be better received with it. I did it for you this time.
– Dylan
Nov 27 at 8:48
Here's a tutorial on the site's formatting. Your question would be better received with it. I did it for you this time.
– Dylan
Nov 27 at 8:48
add a comment |
1 Answer
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You didn't make a mistake, you just neglected to specify what $x^*$ and $y^*$ are.
To make the second equation consistent with the first, clearly you need $x^* =y^*$. Also, you need $dfrac{k_1A}{k_4x^*}=1$, therefore $x^* = dfrac{k_1A}{k_4}$
The last step is express $a$ and $b$ in terms of the known constants, hence
$$ a = frac{k_3}{k_4}x^*y^* = frac{k_3}{k_4}left(frac{k_1A}{k_4}right)^2 = frac{k_1^2k_3A^2}{k_4^3} $$
and $b = dfrac{k_2B}{k_4}$ as before.
Hope this helps.
answered Nov 27 at 9:03
Dylan
12.2k31026
add a comment |
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1 Answer
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You didn't make a mistake, you just neglected to specify what $x^*$ and $y^*$ are.
To make the second equation consistent with the first, clearly you need $x^* =y^*$. Also, you need $dfrac{k_1A}{k_4x^*}=1$, therefore $x^* = dfrac{k_1A}{k_4}$
The last step is express $a$ and $b$ in terms of the known constants, hence
$$ a = frac{k_3}{k_4}x^*y^* = frac{k_3}{k_4}left(frac{k_1A}{k_4}right)^2 = frac{k_1^2k_3A^2}{k_4^3} $$
and $b = dfrac{k_2B}{k_4}$ as before.
Hope this helps.
answered Nov 27 at 9:03
Dylan
12.2k31026
add a comment |
You didn't make a mistake, you just neglected to specify what $x^*$ and $y^*$ are.
To make the second equation consistent with the first, clearly you need $x^* =y^*$. Also, you need $dfrac{k_1A}{k_4x^*}=1$, therefore $x^* = dfrac{k_1A}{k_4}$
The last step is express $a$ and $b$ in terms of the known constants, hence
$$ a = frac{k_3}{k_4}x^*y^* = frac{k_3}{k_4}left(frac{k_1A}{k_4}right)^2 = frac{k_1^2k_3A^2}{k_4^3} $$
and $b = dfrac{k_2B}{k_4}$ as before.
Hope this helps.
answered Nov 27 at 9:03
Dylan
12.2k31026
add a comment |
You didn't make a mistake, you just neglected to specify what $x^*$ and $y^*$ are.
To make the second equation consistent with the first, clearly you need $x^* =y^*$. Also, you need $dfrac{k_1A}{k_4x^*}=1$, therefore $x^* = dfrac{k_1A}{k_4}$
The last step is express $a$ and $b$ in terms of the known constants, hence
$$ a = frac{k_3}{k_4}x^*y^* = frac{k_3}{k_4}left(frac{k_1A}{k_4}right)^2 = frac{k_1^2k_3A^2}{k_4^3} $$
and $b = dfrac{k_2B}{k_4}$ as before.
Hope this helps.
answered Nov 27 at 9:03
Dylan
12.2k31026
You didn't make a mistake, you just neglected to specify what $x^*$ and $y^*$ are.
To make the second equation consistent with the first, clearly you need $x^* =y^*$. Also, you need $dfrac{k_1A}{k_4x^*}=1$, therefore $x^* = dfrac{k_1A}{k_4}$
The last step is express $a$ and $b$ in terms of the known constants, hence
$$ a = frac{k_3}{k_4}x^*y^* = frac{k_3}{k_4}left(frac{k_1A}{k_4}right)^2 = frac{k_1^2k_3A^2}{k_4^3} $$
and $b = dfrac{k_2B}{k_4}$ as before.
Hope this helps.
answered Nov 27 at 9:03
Dylan
12.2k31026
answered Nov 27 at 9:03
Dylan
12.2k31026
answered Nov 27 at 9:03
Dylan
12.2k31026
answered Nov 27 at 9:03
Dylan
12.2k31026
12.2k31026
add a comment |
add a comment |
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