How to solve the Lienard equation $v'' = av -bvv'$?












1












$begingroup$


For a Physics project that I'm doing in school, I have the following differential equation:



$$v'' = av -bvv'$$



where $v$ is the velocity, $v'$ is the rate of change of velocity with respect to time, and $v''$ is the second derivative of the velocity with respect to time.
$a$ and $b$ are both positive constants.
I also know the initial velocity and the initial rate of change of velocity when $t=0$.



It's really important that I solve it, and I'm not able to find the solution anywhere.



This is really urgent, and I hope one of you guys can solve this equation soon.
Thanks






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 4:50












  • $begingroup$
    The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:12










  • $begingroup$
    I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 5:14










  • $begingroup$
    No, I have not taken any classes on it. But I have studied basic first order differential equations
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:19










  • $begingroup$
    Because of the $v v_{}^prime$ term, this is not a linear differential equation.
    $endgroup$
    – John Wayland Bales
    Jan 19 at 5:21


















1












$begingroup$


For a Physics project that I'm doing in school, I have the following differential equation:



$$v'' = av -bvv'$$



where $v$ is the velocity, $v'$ is the rate of change of velocity with respect to time, and $v''$ is the second derivative of the velocity with respect to time.
$a$ and $b$ are both positive constants.
I also know the initial velocity and the initial rate of change of velocity when $t=0$.



It's really important that I solve it, and I'm not able to find the solution anywhere.



This is really urgent, and I hope one of you guys can solve this equation soon.
Thanks






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 4:50












  • $begingroup$
    The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:12










  • $begingroup$
    I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 5:14










  • $begingroup$
    No, I have not taken any classes on it. But I have studied basic first order differential equations
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:19










  • $begingroup$
    Because of the $v v_{}^prime$ term, this is not a linear differential equation.
    $endgroup$
    – John Wayland Bales
    Jan 19 at 5:21
















1












1








1





$begingroup$


For a Physics project that I'm doing in school, I have the following differential equation:



$$v'' = av -bvv'$$



where $v$ is the velocity, $v'$ is the rate of change of velocity with respect to time, and $v''$ is the second derivative of the velocity with respect to time.
$a$ and $b$ are both positive constants.
I also know the initial velocity and the initial rate of change of velocity when $t=0$.



It's really important that I solve it, and I'm not able to find the solution anywhere.



This is really urgent, and I hope one of you guys can solve this equation soon.
Thanks






ordinary-differential-equations







share|cite|improve this question











$endgroup$




For a Physics project that I'm doing in school, I have the following differential equation:



$$v'' = av -bvv'$$



where $v$ is the velocity, $v'$ is the rate of change of velocity with respect to time, and $v''$ is the second derivative of the velocity with respect to time.
$a$ and $b$ are both positive constants.
I also know the initial velocity and the initial rate of change of velocity when $t=0$.



It's really important that I solve it, and I'm not able to find the solution anywhere.



This is really urgent, and I hope one of you guys can solve this equation soon.
Thanks






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 9:04









Rodrigo de Azevedo

13k41958




13k41958










asked Jan 19 at 4:43









Rama VelidandaRama Velidanda

111




111








  • 2




    $begingroup$
    What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 4:50












  • $begingroup$
    The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:12










  • $begingroup$
    I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 5:14










  • $begingroup$
    No, I have not taken any classes on it. But I have studied basic first order differential equations
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:19










  • $begingroup$
    Because of the $v v_{}^prime$ term, this is not a linear differential equation.
    $endgroup$
    – John Wayland Bales
    Jan 19 at 5:21
















  • 2




    $begingroup$
    What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 4:50












  • $begingroup$
    The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:12










  • $begingroup$
    I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 5:14










  • $begingroup$
    No, I have not taken any classes on it. But I have studied basic first order differential equations
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:19










  • $begingroup$
    Because of the $v v_{}^prime$ term, this is not a linear differential equation.
    $endgroup$
    – John Wayland Bales
    Jan 19 at 5:21










2




2




$begingroup$
What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:50






$begingroup$
What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:50














$begingroup$
The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
$endgroup$
– Rama Velidanda
Jan 19 at 5:12




$begingroup$
The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
$endgroup$
– Rama Velidanda
Jan 19 at 5:12












$begingroup$
I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 5:14




$begingroup$
I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 5:14












$begingroup$
No, I have not taken any classes on it. But I have studied basic first order differential equations
$endgroup$
– Rama Velidanda
Jan 19 at 5:19




$begingroup$
No, I have not taken any classes on it. But I have studied basic first order differential equations
$endgroup$
– Rama Velidanda
Jan 19 at 5:19












$begingroup$
Because of the $v v_{}^prime$ term, this is not a linear differential equation.
$endgroup$
– John Wayland Bales
Jan 19 at 5:21






$begingroup$
Because of the $v v_{}^prime$ term, this is not a linear differential equation.
$endgroup$
– John Wayland Bales
Jan 19 at 5:21












2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:10










  • $begingroup$
    No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
    $endgroup$
    – Christoph
    Jan 19 at 5:28










  • $begingroup$
    The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
    $endgroup$
    – Christoph
    Jan 19 at 5:31












  • $begingroup$
    thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:39






  • 1




    $begingroup$
    Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
    $endgroup$
    – Christoph
    Jan 19 at 5:52





















0












$begingroup$

First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.



Writing $v' = w(v)$ with a new function $w$, $v'' = w frac{dw}{dv}$, we obtain the first-order separable ODE
begin{equation}
w frac{dw}{dv} = a v - b v w = v (a - b w).
end{equation}
Assuming that $w not equiv frac{a}{b}$ and using separation of variables we obtain
begin{equation}
frac{w}{a-bw} , mathrm{d}w = v , mathrm{d}v quad Leftrightarrow quad lnleft|frac{b}{a}w-1right| + frac{b}{a} w - 1 = - frac{b^2}{2 a} v^2 + C_1, quad C_1 in mathbb{R}.
end{equation}
Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.



Finally, the solution of the separable ODE $frac{dv}{dt} = w(v)$ yields an implicit representation
begin{equation}
int frac{1}{w(v)} , mathrm{d}v = t
end{equation}
for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.



Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = frac{a}{b}$, is given by the linear function $v(t) = frac{a}{b} t + v_0$. This is the special case of a constant $w equiv frac{a}{b}$, for which we don't need to do the separation of variables.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3079029%2fhow-to-solve-the-lienard-equation-v-av-bvv%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:10










    • $begingroup$
      No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
      $endgroup$
      – Christoph
      Jan 19 at 5:28










    • $begingroup$
      The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
      $endgroup$
      – Christoph
      Jan 19 at 5:31












    • $begingroup$
      thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:39






    • 1




      $begingroup$
      Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
      $endgroup$
      – Christoph
      Jan 19 at 5:52


















    1












    $begingroup$

    Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:10










    • $begingroup$
      No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
      $endgroup$
      – Christoph
      Jan 19 at 5:28










    • $begingroup$
      The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
      $endgroup$
      – Christoph
      Jan 19 at 5:31












    • $begingroup$
      thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:39






    • 1




      $begingroup$
      Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
      $endgroup$
      – Christoph
      Jan 19 at 5:52
















    1












    1








    1





    $begingroup$

    Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.






    share|cite|improve this answer









    $endgroup$



    Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 19 at 4:54









    AlexanderJ93AlexanderJ93

    6,173823




    6,173823












    • $begingroup$
      Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:10










    • $begingroup$
      No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
      $endgroup$
      – Christoph
      Jan 19 at 5:28










    • $begingroup$
      The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
      $endgroup$
      – Christoph
      Jan 19 at 5:31












    • $begingroup$
      thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:39






    • 1




      $begingroup$
      Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
      $endgroup$
      – Christoph
      Jan 19 at 5:52




















    • $begingroup$
      Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:10










    • $begingroup$
      No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
      $endgroup$
      – Christoph
      Jan 19 at 5:28










    • $begingroup$
      The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
      $endgroup$
      – Christoph
      Jan 19 at 5:31












    • $begingroup$
      thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:39






    • 1




      $begingroup$
      Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
      $endgroup$
      – Christoph
      Jan 19 at 5:52


















    $begingroup$
    Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:10




    $begingroup$
    Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:10












    $begingroup$
    No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
    $endgroup$
    – Christoph
    Jan 19 at 5:28




    $begingroup$
    No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
    $endgroup$
    – Christoph
    Jan 19 at 5:28












    $begingroup$
    The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
    $endgroup$
    – Christoph
    Jan 19 at 5:31






    $begingroup$
    The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
    $endgroup$
    – Christoph
    Jan 19 at 5:31














    $begingroup$
    thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:39




    $begingroup$
    thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:39




    1




    1




    $begingroup$
    Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
    $endgroup$
    – Christoph
    Jan 19 at 5:52






    $begingroup$
    Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
    $endgroup$
    – Christoph
    Jan 19 at 5:52













    0












    $begingroup$

    First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.



    Writing $v' = w(v)$ with a new function $w$, $v'' = w frac{dw}{dv}$, we obtain the first-order separable ODE
    begin{equation}
    w frac{dw}{dv} = a v - b v w = v (a - b w).
    end{equation}
    Assuming that $w not equiv frac{a}{b}$ and using separation of variables we obtain
    begin{equation}
    frac{w}{a-bw} , mathrm{d}w = v , mathrm{d}v quad Leftrightarrow quad lnleft|frac{b}{a}w-1right| + frac{b}{a} w - 1 = - frac{b^2}{2 a} v^2 + C_1, quad C_1 in mathbb{R}.
    end{equation}
    Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.



    Finally, the solution of the separable ODE $frac{dv}{dt} = w(v)$ yields an implicit representation
    begin{equation}
    int frac{1}{w(v)} , mathrm{d}v = t
    end{equation}
    for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.



    Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = frac{a}{b}$, is given by the linear function $v(t) = frac{a}{b} t + v_0$. This is the special case of a constant $w equiv frac{a}{b}$, for which we don't need to do the separation of variables.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.



      Writing $v' = w(v)$ with a new function $w$, $v'' = w frac{dw}{dv}$, we obtain the first-order separable ODE
      begin{equation}
      w frac{dw}{dv} = a v - b v w = v (a - b w).
      end{equation}
      Assuming that $w not equiv frac{a}{b}$ and using separation of variables we obtain
      begin{equation}
      frac{w}{a-bw} , mathrm{d}w = v , mathrm{d}v quad Leftrightarrow quad lnleft|frac{b}{a}w-1right| + frac{b}{a} w - 1 = - frac{b^2}{2 a} v^2 + C_1, quad C_1 in mathbb{R}.
      end{equation}
      Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.



      Finally, the solution of the separable ODE $frac{dv}{dt} = w(v)$ yields an implicit representation
      begin{equation}
      int frac{1}{w(v)} , mathrm{d}v = t
      end{equation}
      for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.



      Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = frac{a}{b}$, is given by the linear function $v(t) = frac{a}{b} t + v_0$. This is the special case of a constant $w equiv frac{a}{b}$, for which we don't need to do the separation of variables.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.



        Writing $v' = w(v)$ with a new function $w$, $v'' = w frac{dw}{dv}$, we obtain the first-order separable ODE
        begin{equation}
        w frac{dw}{dv} = a v - b v w = v (a - b w).
        end{equation}
        Assuming that $w not equiv frac{a}{b}$ and using separation of variables we obtain
        begin{equation}
        frac{w}{a-bw} , mathrm{d}w = v , mathrm{d}v quad Leftrightarrow quad lnleft|frac{b}{a}w-1right| + frac{b}{a} w - 1 = - frac{b^2}{2 a} v^2 + C_1, quad C_1 in mathbb{R}.
        end{equation}
        Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.



        Finally, the solution of the separable ODE $frac{dv}{dt} = w(v)$ yields an implicit representation
        begin{equation}
        int frac{1}{w(v)} , mathrm{d}v = t
        end{equation}
        for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.



        Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = frac{a}{b}$, is given by the linear function $v(t) = frac{a}{b} t + v_0$. This is the special case of a constant $w equiv frac{a}{b}$, for which we don't need to do the separation of variables.






        share|cite|improve this answer











        $endgroup$



        First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.



        Writing $v' = w(v)$ with a new function $w$, $v'' = w frac{dw}{dv}$, we obtain the first-order separable ODE
        begin{equation}
        w frac{dw}{dv} = a v - b v w = v (a - b w).
        end{equation}
        Assuming that $w not equiv frac{a}{b}$ and using separation of variables we obtain
        begin{equation}
        frac{w}{a-bw} , mathrm{d}w = v , mathrm{d}v quad Leftrightarrow quad lnleft|frac{b}{a}w-1right| + frac{b}{a} w - 1 = - frac{b^2}{2 a} v^2 + C_1, quad C_1 in mathbb{R}.
        end{equation}
        Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.



        Finally, the solution of the separable ODE $frac{dv}{dt} = w(v)$ yields an implicit representation
        begin{equation}
        int frac{1}{w(v)} , mathrm{d}v = t
        end{equation}
        for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.



        Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = frac{a}{b}$, is given by the linear function $v(t) = frac{a}{b} t + v_0$. This is the special case of a constant $w equiv frac{a}{b}$, for which we don't need to do the separation of variables.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 19 at 9:47

























        answered Jan 19 at 8:35









        ChristophChristoph

        58116




        58116






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3079029%2fhow-to-solve-the-lienard-equation-v-av-bvv%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
            Read more