Partial Differential Equation Mathematical Modelling
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Salutations, I have been trying to approach a modelling case about organism propagation which reproducing with velocity $alpha$ spreading randomly according these equations:
$$frac{du(x,t)}{dt}=kfrac{d^2u}{dx^2} +alpha u(x,t)\ \ u(x,0)=delta(x)\ limlimits_{x to pminfty} u(x,t)=0$$
This studying case requires to demonstrate that isoprobability contours, it means, in the points (x,t) which P(x,t)=P=constant is verified that
$$frac{x}{t}=pm [4alpha k-2kfrac{log(t)}{t}-frac{4k}{t}log(sqrt{4pi k} P)]^frac{1}{2}$$
Another aspect to demonstrate is that $t to infty$, the spreading velocity of these contours, it means, the velocity which these organisms are spreading is aproximated to
$$frac{x}{t}=pm(4alpha k)^frac{1}{2}$$
Finally, how to compare this spreading velocity with purely diffusive process $(alpha=0)$, it means , x is aproximated to $sqrt{kt}$
This is just for academical curiosity and I would like to understand better this kind of cases with Partial Differential Equations. So, I require any guidance or starting steps or explanations to find the solutions because it's an interesting problem.
Thanks very much for your attention.
pde mathematical-modeling
asked Dec 2 '18 at 3:35
ht1204ht1204
744
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add a comment |
$begingroup$
Salutations, I have been trying to approach a modelling case about organism propagation which reproducing with velocity $alpha$ spreading randomly according these equations:
$$frac{du(x,t)}{dt}=kfrac{d^2u}{dx^2} +alpha u(x,t)\ \ u(x,0)=delta(x)\ limlimits_{x to pminfty} u(x,t)=0$$
This studying case requires to demonstrate that isoprobability contours, it means, in the points (x,t) which P(x,t)=P=constant is verified that
$$frac{x}{t}=pm [4alpha k-2kfrac{log(t)}{t}-frac{4k}{t}log(sqrt{4pi k} P)]^frac{1}{2}$$
Another aspect to demonstrate is that $t to infty$, the spreading velocity of these contours, it means, the velocity which these organisms are spreading is aproximated to
$$frac{x}{t}=pm(4alpha k)^frac{1}{2}$$
Finally, how to compare this spreading velocity with purely diffusive process $(alpha=0)$, it means , x is aproximated to $sqrt{kt}$
This is just for academical curiosity and I would like to understand better this kind of cases with Partial Differential Equations. So, I require any guidance or starting steps or explanations to find the solutions because it's an interesting problem.
Thanks very much for your attention.
pde mathematical-modeling
asked Dec 2 '18 at 3:35
ht1204ht1204
744
$endgroup$
add a comment |
$begingroup$
Salutations, I have been trying to approach a modelling case about organism propagation which reproducing with velocity $alpha$ spreading randomly according these equations:
$$frac{du(x,t)}{dt}=kfrac{d^2u}{dx^2} +alpha u(x,t)\ \ u(x,0)=delta(x)\ limlimits_{x to pminfty} u(x,t)=0$$
This studying case requires to demonstrate that isoprobability contours, it means, in the points (x,t) which P(x,t)=P=constant is verified that
$$frac{x}{t}=pm [4alpha k-2kfrac{log(t)}{t}-frac{4k}{t}log(sqrt{4pi k} P)]^frac{1}{2}$$
Another aspect to demonstrate is that $t to infty$, the spreading velocity of these contours, it means, the velocity which these organisms are spreading is aproximated to
$$frac{x}{t}=pm(4alpha k)^frac{1}{2}$$
Finally, how to compare this spreading velocity with purely diffusive process $(alpha=0)$, it means , x is aproximated to $sqrt{kt}$
This is just for academical curiosity and I would like to understand better this kind of cases with Partial Differential Equations. So, I require any guidance or starting steps or explanations to find the solutions because it's an interesting problem.
Thanks very much for your attention.
pde mathematical-modeling
asked Dec 2 '18 at 3:35
ht1204ht1204
744
$endgroup$
Salutations, I have been trying to approach a modelling case about organism propagation which reproducing with velocity $alpha$ spreading randomly according these equations:
$$frac{du(x,t)}{dt}=kfrac{d^2u}{dx^2} +alpha u(x,t)\ \ u(x,0)=delta(x)\ limlimits_{x to pminfty} u(x,t)=0$$
This studying case requires to demonstrate that isoprobability contours, it means, in the points (x,t) which P(x,t)=P=constant is verified that
$$frac{x}{t}=pm [4alpha k-2kfrac{log(t)}{t}-frac{4k}{t}log(sqrt{4pi k} P)]^frac{1}{2}$$
Another aspect to demonstrate is that $t to infty$, the spreading velocity of these contours, it means, the velocity which these organisms are spreading is aproximated to
$$frac{x}{t}=pm(4alpha k)^frac{1}{2}$$
Finally, how to compare this spreading velocity with purely diffusive process $(alpha=0)$, it means , x is aproximated to $sqrt{kt}$
This is just for academical curiosity and I would like to understand better this kind of cases with Partial Differential Equations. So, I require any guidance or starting steps or explanations to find the solutions because it's an interesting problem.
Thanks very much for your attention.
pde mathematical-modeling
pde mathematical-modeling
asked Dec 2 '18 at 3:35
ht1204ht1204
744
asked Dec 2 '18 at 3:35
ht1204ht1204
744
edited Dec 2 '18 at 4:03
ht1204
asked Dec 2 '18 at 3:35
ht1204ht1204
744
asked Dec 2 '18 at 3:35
ht1204ht1204
744
asked Dec 2 '18 at 3:35
ht1204ht1204
744
744
add a comment |
add a comment |
1 Answer
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Multiply both sides of your equation by $e^{-alpha t}$. Then you get $$frac{partial u}{partial t}e^{-alpha t}-alpha u, e^{-alpha t}=frac{partial^2 u}{partial x^2}e^{-alpha t}.$$ We recognize the product rule on the left-hand side: $$frac{partial}{partial t}(ue^{-alpha t})=frac{partial^2}{partial x^2}(ue^{-alpha t}).$$ So if $v$ solves the diffusion equation (with the same boundary conditions), then the function you want is $u=ve^{alpha t}.$ Therefore, to solve your problem, you only need to solve the diffusion equation, whose solutions are well-understood.
answered Dec 2 '18 at 4:03
Alex SAlex S
17.9k12160
$endgroup$
1
$begingroup$
Hi, thanks for your commentary, well, I found u(x,t) through laplace transform but I'm not familiarized with the other points of this study case, for example isoprobability contours, specially to find that equation $frac{x}{t}$, I'm starting to approach PDE, that's the reason why I require help, just for mathematical curiosity. thanks again.
$endgroup$
– ht1204
Dec 2 '18 at 4:17
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Multiply both sides of your equation by $e^{-alpha t}$. Then you get $$frac{partial u}{partial t}e^{-alpha t}-alpha u, e^{-alpha t}=frac{partial^2 u}{partial x^2}e^{-alpha t}.$$ We recognize the product rule on the left-hand side: $$frac{partial}{partial t}(ue^{-alpha t})=frac{partial^2}{partial x^2}(ue^{-alpha t}).$$ So if $v$ solves the diffusion equation (with the same boundary conditions), then the function you want is $u=ve^{alpha t}.$ Therefore, to solve your problem, you only need to solve the diffusion equation, whose solutions are well-understood.
answered Dec 2 '18 at 4:03
Alex SAlex S
17.9k12160
$endgroup$
1
$begingroup$
Hi, thanks for your commentary, well, I found u(x,t) through laplace transform but I'm not familiarized with the other points of this study case, for example isoprobability contours, specially to find that equation $frac{x}{t}$, I'm starting to approach PDE, that's the reason why I require help, just for mathematical curiosity. thanks again.
$endgroup$
– ht1204
Dec 2 '18 at 4:17
add a comment |
$begingroup$
Multiply both sides of your equation by $e^{-alpha t}$. Then you get $$frac{partial u}{partial t}e^{-alpha t}-alpha u, e^{-alpha t}=frac{partial^2 u}{partial x^2}e^{-alpha t}.$$ We recognize the product rule on the left-hand side: $$frac{partial}{partial t}(ue^{-alpha t})=frac{partial^2}{partial x^2}(ue^{-alpha t}).$$ So if $v$ solves the diffusion equation (with the same boundary conditions), then the function you want is $u=ve^{alpha t}.$ Therefore, to solve your problem, you only need to solve the diffusion equation, whose solutions are well-understood.
answered Dec 2 '18 at 4:03
Alex SAlex S
17.9k12160
$endgroup$
1
$begingroup$
Hi, thanks for your commentary, well, I found u(x,t) through laplace transform but I'm not familiarized with the other points of this study case, for example isoprobability contours, specially to find that equation $frac{x}{t}$, I'm starting to approach PDE, that's the reason why I require help, just for mathematical curiosity. thanks again.
$endgroup$
– ht1204
Dec 2 '18 at 4:17
add a comment |
$begingroup$
Multiply both sides of your equation by $e^{-alpha t}$. Then you get $$frac{partial u}{partial t}e^{-alpha t}-alpha u, e^{-alpha t}=frac{partial^2 u}{partial x^2}e^{-alpha t}.$$ We recognize the product rule on the left-hand side: $$frac{partial}{partial t}(ue^{-alpha t})=frac{partial^2}{partial x^2}(ue^{-alpha t}).$$ So if $v$ solves the diffusion equation (with the same boundary conditions), then the function you want is $u=ve^{alpha t}.$ Therefore, to solve your problem, you only need to solve the diffusion equation, whose solutions are well-understood.
answered Dec 2 '18 at 4:03
Alex SAlex S
17.9k12160
$endgroup$
Multiply both sides of your equation by $e^{-alpha t}$. Then you get $$frac{partial u}{partial t}e^{-alpha t}-alpha u, e^{-alpha t}=frac{partial^2 u}{partial x^2}e^{-alpha t}.$$ We recognize the product rule on the left-hand side: $$frac{partial}{partial t}(ue^{-alpha t})=frac{partial^2}{partial x^2}(ue^{-alpha t}).$$ So if $v$ solves the diffusion equation (with the same boundary conditions), then the function you want is $u=ve^{alpha t}.$ Therefore, to solve your problem, you only need to solve the diffusion equation, whose solutions are well-understood.
answered Dec 2 '18 at 4:03
Alex SAlex S
17.9k12160
edited Dec 2 '18 at 4:09
answered Dec 2 '18 at 4:03
Alex SAlex S
17.9k12160
answered Dec 2 '18 at 4:03
Alex SAlex S
17.9k12160
answered Dec 2 '18 at 4:03
Alex SAlex S
17.9k12160
17.9k12160
1
$begingroup$
Hi, thanks for your commentary, well, I found u(x,t) through laplace transform but I'm not familiarized with the other points of this study case, for example isoprobability contours, specially to find that equation $frac{x}{t}$, I'm starting to approach PDE, that's the reason why I require help, just for mathematical curiosity. thanks again.
$endgroup$
– ht1204
Dec 2 '18 at 4:17
add a comment |
1
$begingroup$
Hi, thanks for your commentary, well, I found u(x,t) through laplace transform but I'm not familiarized with the other points of this study case, for example isoprobability contours, specially to find that equation $frac{x}{t}$, I'm starting to approach PDE, that's the reason why I require help, just for mathematical curiosity. thanks again.
$endgroup$
– ht1204
Dec 2 '18 at 4:17
1
1
$begingroup$
Hi, thanks for your commentary, well, I found u(x,t) through laplace transform but I'm not familiarized with the other points of this study case, for example isoprobability contours, specially to find that equation $frac{x}{t}$, I'm starting to approach PDE, that's the reason why I require help, just for mathematical curiosity. thanks again.
$endgroup$
– ht1204
Dec 2 '18 at 4:17
$begingroup$
Hi, thanks for your commentary, well, I found u(x,t) through laplace transform but I'm not familiarized with the other points of this study case, for example isoprobability contours, specially to find that equation $frac{x}{t}$, I'm starting to approach PDE, that's the reason why I require help, just for mathematical curiosity. thanks again.
$endgroup$
– ht1204
Dec 2 '18 at 4:17
add a comment |
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