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







share|cite|improve this question











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    2












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      2



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 2 '18 at 4:03







      ht1204

















      asked Dec 2 '18 at 3:35









      ht1204ht1204

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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.






          share|cite|improve this answer











          $endgroup$









          • 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











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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.






          share|cite|improve this answer











          $endgroup$









          • 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
















          2












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$









          • 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














          2












          2








          2





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 2 '18 at 4:09

























          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














          • 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


















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