Assumption on traveling wave solutions of Fisher's equation
$begingroup$
I have a question about Fisher's equation in a biology context.
For example, in Fisher's equation $u_{t} = Du_{xx} + u(1-u)$, where $u$ is a density of cell, the logistic term explains that the capacity is 1.
When we look for a traveling wave solution $U(x-ct)$ (or a heteroclinic orbit), $c$ is a speed, connecting $u=0$ and $u=1,$ do we assume that a traveling wave ansatz must satisfy $0leq U(x-ct) leq 1?$ Or, could it be $U(x-ct)>1?$
Thank you so much!
ordinary-differential-equations pde mathematical-modeling biology
asked Aug 18 '18 at 22:42
jp31433jp31433
333
$endgroup$
add a comment |
$begingroup$
I have a question about Fisher's equation in a biology context.
For example, in Fisher's equation $u_{t} = Du_{xx} + u(1-u)$, where $u$ is a density of cell, the logistic term explains that the capacity is 1.
When we look for a traveling wave solution $U(x-ct)$ (or a heteroclinic orbit), $c$ is a speed, connecting $u=0$ and $u=1,$ do we assume that a traveling wave ansatz must satisfy $0leq U(x-ct) leq 1?$ Or, could it be $U(x-ct)>1?$
Thank you so much!
ordinary-differential-equations pde mathematical-modeling biology
asked Aug 18 '18 at 22:42
jp31433jp31433
333
$endgroup$
add a comment |
$begingroup$
I have a question about Fisher's equation in a biology context.
For example, in Fisher's equation $u_{t} = Du_{xx} + u(1-u)$, where $u$ is a density of cell, the logistic term explains that the capacity is 1.
When we look for a traveling wave solution $U(x-ct)$ (or a heteroclinic orbit), $c$ is a speed, connecting $u=0$ and $u=1,$ do we assume that a traveling wave ansatz must satisfy $0leq U(x-ct) leq 1?$ Or, could it be $U(x-ct)>1?$
Thank you so much!
ordinary-differential-equations pde mathematical-modeling biology
asked Aug 18 '18 at 22:42
jp31433jp31433
333
$endgroup$
I have a question about Fisher's equation in a biology context.
For example, in Fisher's equation $u_{t} = Du_{xx} + u(1-u)$, where $u$ is a density of cell, the logistic term explains that the capacity is 1.
When we look for a traveling wave solution $U(x-ct)$ (or a heteroclinic orbit), $c$ is a speed, connecting $u=0$ and $u=1,$ do we assume that a traveling wave ansatz must satisfy $0leq U(x-ct) leq 1?$ Or, could it be $U(x-ct)>1?$
Thank you so much!
ordinary-differential-equations pde mathematical-modeling biology
ordinary-differential-equations pde mathematical-modeling biology
asked Aug 18 '18 at 22:42
jp31433jp31433
333
asked Aug 18 '18 at 22:42
jp31433jp31433
333
asked Aug 18 '18 at 22:42
jp31433jp31433
333
asked Aug 18 '18 at 22:42
jp31433jp31433
333
asked Aug 18 '18 at 22:42
jp31433jp31433
333
333
add a comment |
add a comment |
1 Answer
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Assuming this is a biological problem and the domain is infinite, then the form of the model that you used indicates a re-scale version, where carrying capacity $u$ is normalized to 1. If the initial profile $u(x,0) leq 1$ (biological), then $U(x-ct) leq 1$ for all $tgeq 0$. This is because $u(1-u)leq 1$ and $Du_{xx}$ ($D>0$) contributes to the diffusion process (e.g. diffusing $u$ from high to lower value across space). If the initial profile $u(x,0) > 1$ (not biological), then it is possible that $U(x-ct) > 1$ for small $t$. However, asymptotically, $U(x-ct)$ will still be bounded by $0$ and $1$, e.g. $0 leq liminf U(x-ct) leq limsup U(x-ct) leq 1$.
If the boundary is finite, then the answer depends on the specific boundary conditions and the question needs to be examined case by case.
answered Dec 3 '18 at 8:39
PaichuPaichu
751616
$endgroup$
$begingroup$
Thank you for your answer! It totally makes sense! I haven't connected it with the initial profile. You are right. I've seen some cases where the boundary is finite. That's why I was wondering about the question! Again, thank you for your help!
$endgroup$
– jp31433
Dec 4 '18 at 15:50
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming this is a biological problem and the domain is infinite, then the form of the model that you used indicates a re-scale version, where carrying capacity $u$ is normalized to 1. If the initial profile $u(x,0) leq 1$ (biological), then $U(x-ct) leq 1$ for all $tgeq 0$. This is because $u(1-u)leq 1$ and $Du_{xx}$ ($D>0$) contributes to the diffusion process (e.g. diffusing $u$ from high to lower value across space). If the initial profile $u(x,0) > 1$ (not biological), then it is possible that $U(x-ct) > 1$ for small $t$. However, asymptotically, $U(x-ct)$ will still be bounded by $0$ and $1$, e.g. $0 leq liminf U(x-ct) leq limsup U(x-ct) leq 1$.
If the boundary is finite, then the answer depends on the specific boundary conditions and the question needs to be examined case by case.
answered Dec 3 '18 at 8:39
PaichuPaichu
751616
$endgroup$
$begingroup$
Thank you for your answer! It totally makes sense! I haven't connected it with the initial profile. You are right. I've seen some cases where the boundary is finite. That's why I was wondering about the question! Again, thank you for your help!
$endgroup$
– jp31433
Dec 4 '18 at 15:50
add a comment |
$begingroup$
Assuming this is a biological problem and the domain is infinite, then the form of the model that you used indicates a re-scale version, where carrying capacity $u$ is normalized to 1. If the initial profile $u(x,0) leq 1$ (biological), then $U(x-ct) leq 1$ for all $tgeq 0$. This is because $u(1-u)leq 1$ and $Du_{xx}$ ($D>0$) contributes to the diffusion process (e.g. diffusing $u$ from high to lower value across space). If the initial profile $u(x,0) > 1$ (not biological), then it is possible that $U(x-ct) > 1$ for small $t$. However, asymptotically, $U(x-ct)$ will still be bounded by $0$ and $1$, e.g. $0 leq liminf U(x-ct) leq limsup U(x-ct) leq 1$.
If the boundary is finite, then the answer depends on the specific boundary conditions and the question needs to be examined case by case.
answered Dec 3 '18 at 8:39
PaichuPaichu
751616
$endgroup$
$begingroup$
Thank you for your answer! It totally makes sense! I haven't connected it with the initial profile. You are right. I've seen some cases where the boundary is finite. That's why I was wondering about the question! Again, thank you for your help!
$endgroup$
– jp31433
Dec 4 '18 at 15:50
add a comment |
$begingroup$
Assuming this is a biological problem and the domain is infinite, then the form of the model that you used indicates a re-scale version, where carrying capacity $u$ is normalized to 1. If the initial profile $u(x,0) leq 1$ (biological), then $U(x-ct) leq 1$ for all $tgeq 0$. This is because $u(1-u)leq 1$ and $Du_{xx}$ ($D>0$) contributes to the diffusion process (e.g. diffusing $u$ from high to lower value across space). If the initial profile $u(x,0) > 1$ (not biological), then it is possible that $U(x-ct) > 1$ for small $t$. However, asymptotically, $U(x-ct)$ will still be bounded by $0$ and $1$, e.g. $0 leq liminf U(x-ct) leq limsup U(x-ct) leq 1$.
If the boundary is finite, then the answer depends on the specific boundary conditions and the question needs to be examined case by case.
answered Dec 3 '18 at 8:39
PaichuPaichu
751616
$endgroup$
Assuming this is a biological problem and the domain is infinite, then the form of the model that you used indicates a re-scale version, where carrying capacity $u$ is normalized to 1. If the initial profile $u(x,0) leq 1$ (biological), then $U(x-ct) leq 1$ for all $tgeq 0$. This is because $u(1-u)leq 1$ and $Du_{xx}$ ($D>0$) contributes to the diffusion process (e.g. diffusing $u$ from high to lower value across space). If the initial profile $u(x,0) > 1$ (not biological), then it is possible that $U(x-ct) > 1$ for small $t$. However, asymptotically, $U(x-ct)$ will still be bounded by $0$ and $1$, e.g. $0 leq liminf U(x-ct) leq limsup U(x-ct) leq 1$.
If the boundary is finite, then the answer depends on the specific boundary conditions and the question needs to be examined case by case.
answered Dec 3 '18 at 8:39
PaichuPaichu
751616
answered Dec 3 '18 at 8:39
PaichuPaichu
751616
answered Dec 3 '18 at 8:39
PaichuPaichu
751616
answered Dec 3 '18 at 8:39
PaichuPaichu
751616
751616
$begingroup$
Thank you for your answer! It totally makes sense! I haven't connected it with the initial profile. You are right. I've seen some cases where the boundary is finite. That's why I was wondering about the question! Again, thank you for your help!
$endgroup$
– jp31433
Dec 4 '18 at 15:50
add a comment |
$begingroup$
Thank you for your answer! It totally makes sense! I haven't connected it with the initial profile. You are right. I've seen some cases where the boundary is finite. That's why I was wondering about the question! Again, thank you for your help!
$endgroup$
– jp31433
Dec 4 '18 at 15:50
$begingroup$
Thank you for your answer! It totally makes sense! I haven't connected it with the initial profile. You are right. I've seen some cases where the boundary is finite. That's why I was wondering about the question! Again, thank you for your help!
$endgroup$
– jp31433
Dec 4 '18 at 15:50
$begingroup$
Thank you for your answer! It totally makes sense! I haven't connected it with the initial profile. You are right. I've seen some cases where the boundary is finite. That's why I was wondering about the question! Again, thank you for your help!
$endgroup$
– jp31433
Dec 4 '18 at 15:50
add a comment |
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