Any ideas about the solution to this nonlinear ODE? $-u''-frac{1}{r}u'+e^u-e^{-u} = 0$












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The ODE is $-u''-frac{1}{r}u'+e^u-e^{-u} = 0$, where u is a function of $r$ (i.e. $u(r)$) and $uin C^2{(-infty,0)cup(0,infty)}$. No initial condition is provided, so just try to find the general solution.
This question appears when I tried to find the fundamental solution of $-Delta u + e^u - e^{-u} = delta (vec{x})$ in 2-D, where $Delta$ stands for the Laplace operator and $delta(vec{x})$ stands for the Dirac's function. Since the pde equation is radially symmetric, it's reasonable to think $u$ as a univariate function of $r$ in polar coordinates.






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    The ODE is $-u''-frac{1}{r}u'+e^u-e^{-u} = 0$, where u is a function of $r$ (i.e. $u(r)$) and $uin C^2{(-infty,0)cup(0,infty)}$. No initial condition is provided, so just try to find the general solution.
    This question appears when I tried to find the fundamental solution of $-Delta u + e^u - e^{-u} = delta (vec{x})$ in 2-D, where $Delta$ stands for the Laplace operator and $delta(vec{x})$ stands for the Dirac's function. Since the pde equation is radially symmetric, it's reasonable to think $u$ as a univariate function of $r$ in polar coordinates.






    differential-equations







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      2





      The ODE is $-u''-frac{1}{r}u'+e^u-e^{-u} = 0$, where u is a function of $r$ (i.e. $u(r)$) and $uin C^2{(-infty,0)cup(0,infty)}$. No initial condition is provided, so just try to find the general solution.
      This question appears when I tried to find the fundamental solution of $-Delta u + e^u - e^{-u} = delta (vec{x})$ in 2-D, where $Delta$ stands for the Laplace operator and $delta(vec{x})$ stands for the Dirac's function. Since the pde equation is radially symmetric, it's reasonable to think $u$ as a univariate function of $r$ in polar coordinates.






      differential-equations







      share|cite|improve this question















      The ODE is $-u''-frac{1}{r}u'+e^u-e^{-u} = 0$, where u is a function of $r$ (i.e. $u(r)$) and $uin C^2{(-infty,0)cup(0,infty)}$. No initial condition is provided, so just try to find the general solution.
      This question appears when I tried to find the fundamental solution of $-Delta u + e^u - e^{-u} = delta (vec{x})$ in 2-D, where $Delta$ stands for the Laplace operator and $delta(vec{x})$ stands for the Dirac's function. Since the pde equation is radially symmetric, it's reasonable to think $u$ as a univariate function of $r$ in polar coordinates.






      differential-equations




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      edited Nov 28 '18 at 0:35

























      asked Nov 27 '18 at 8:56









      U2647

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          1 Answer
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          Hint:



          Let $r=e^t$ ,



          Then $t=ln r$



          $dfrac{du}{dr}=dfrac{du}{dt}dfrac{dt}{dr}=dfrac{1}{r}dfrac{du}{dt}=e^{-t}dfrac{du}{dt}$



          $dfrac{d^2u}{dr^2}=dfrac{d}{dr}left(e^{-t}dfrac{du}{dt}right)=dfrac{d}{dt}left(e^{-t}dfrac{du}{dt}right)dfrac{dt}{dr}=left(e^{-t}dfrac{d^2u}{dt^2}-e^{-t}dfrac{du}{dt}right)e^{-t}=e^{-2t}dfrac{d^2u}{dt^2}-e^{-2t}dfrac{du}{dt}$



          $therefore-e^{-2t}dfrac{d^2u}{dt^2}+e^{-2t}dfrac{du}{dt}-e^{-2t}dfrac{du}{dt}+e^u-e^{-u}=0$



          $e^{-2t}dfrac{d^2u}{dt^2}=e^u-e^{-u}$



          $dfrac{d^2u}{dt^2}=e^{2t+u}-e^{2t-u}$






          share|cite|improve this answer





















          • Yes, this transformation is right and beautiful, but I don't know how to carry on. Would you explain the following steps please?
            – U2647
            Nov 28 '18 at 0:28










          • @U2647 You can rewrite the equation as $u'' = 2exp(2t)sinh(u)$, but there will be no closed form solution. If you seek solutions for $u$ near $0$, you can say $sinh(u)approx u$ and find solutions in terms of the modified Bessel functions.
            – AlexanderJ93
            Nov 28 '18 at 1:11












          • @AlexanderJ93 So the approximate solution includes an imaginary part?
            – U2647
            Nov 28 '18 at 1:27










          • @U2647 They are real-valued functions for real and positive arguments.
            – AlexanderJ93
            Nov 28 '18 at 1:33











          Your Answer





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          1 Answer
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          1 Answer
          1






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          active

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          0














          Hint:



          Let $r=e^t$ ,



          Then $t=ln r$



          $dfrac{du}{dr}=dfrac{du}{dt}dfrac{dt}{dr}=dfrac{1}{r}dfrac{du}{dt}=e^{-t}dfrac{du}{dt}$



          $dfrac{d^2u}{dr^2}=dfrac{d}{dr}left(e^{-t}dfrac{du}{dt}right)=dfrac{d}{dt}left(e^{-t}dfrac{du}{dt}right)dfrac{dt}{dr}=left(e^{-t}dfrac{d^2u}{dt^2}-e^{-t}dfrac{du}{dt}right)e^{-t}=e^{-2t}dfrac{d^2u}{dt^2}-e^{-2t}dfrac{du}{dt}$



          $therefore-e^{-2t}dfrac{d^2u}{dt^2}+e^{-2t}dfrac{du}{dt}-e^{-2t}dfrac{du}{dt}+e^u-e^{-u}=0$



          $e^{-2t}dfrac{d^2u}{dt^2}=e^u-e^{-u}$



          $dfrac{d^2u}{dt^2}=e^{2t+u}-e^{2t-u}$






          share|cite|improve this answer





















          • Yes, this transformation is right and beautiful, but I don't know how to carry on. Would you explain the following steps please?
            – U2647
            Nov 28 '18 at 0:28










          • @U2647 You can rewrite the equation as $u'' = 2exp(2t)sinh(u)$, but there will be no closed form solution. If you seek solutions for $u$ near $0$, you can say $sinh(u)approx u$ and find solutions in terms of the modified Bessel functions.
            – AlexanderJ93
            Nov 28 '18 at 1:11












          • @AlexanderJ93 So the approximate solution includes an imaginary part?
            – U2647
            Nov 28 '18 at 1:27










          • @U2647 They are real-valued functions for real and positive arguments.
            – AlexanderJ93
            Nov 28 '18 at 1:33
















          0














          Hint:



          Let $r=e^t$ ,



          Then $t=ln r$



          $dfrac{du}{dr}=dfrac{du}{dt}dfrac{dt}{dr}=dfrac{1}{r}dfrac{du}{dt}=e^{-t}dfrac{du}{dt}$



          $dfrac{d^2u}{dr^2}=dfrac{d}{dr}left(e^{-t}dfrac{du}{dt}right)=dfrac{d}{dt}left(e^{-t}dfrac{du}{dt}right)dfrac{dt}{dr}=left(e^{-t}dfrac{d^2u}{dt^2}-e^{-t}dfrac{du}{dt}right)e^{-t}=e^{-2t}dfrac{d^2u}{dt^2}-e^{-2t}dfrac{du}{dt}$



          $therefore-e^{-2t}dfrac{d^2u}{dt^2}+e^{-2t}dfrac{du}{dt}-e^{-2t}dfrac{du}{dt}+e^u-e^{-u}=0$



          $e^{-2t}dfrac{d^2u}{dt^2}=e^u-e^{-u}$



          $dfrac{d^2u}{dt^2}=e^{2t+u}-e^{2t-u}$






          share|cite|improve this answer





















          • Yes, this transformation is right and beautiful, but I don't know how to carry on. Would you explain the following steps please?
            – U2647
            Nov 28 '18 at 0:28










          • @U2647 You can rewrite the equation as $u'' = 2exp(2t)sinh(u)$, but there will be no closed form solution. If you seek solutions for $u$ near $0$, you can say $sinh(u)approx u$ and find solutions in terms of the modified Bessel functions.
            – AlexanderJ93
            Nov 28 '18 at 1:11












          • @AlexanderJ93 So the approximate solution includes an imaginary part?
            – U2647
            Nov 28 '18 at 1:27










          • @U2647 They are real-valued functions for real and positive arguments.
            – AlexanderJ93
            Nov 28 '18 at 1:33














          0












          0








          0






          Hint:



          Let $r=e^t$ ,



          Then $t=ln r$



          $dfrac{du}{dr}=dfrac{du}{dt}dfrac{dt}{dr}=dfrac{1}{r}dfrac{du}{dt}=e^{-t}dfrac{du}{dt}$



          $dfrac{d^2u}{dr^2}=dfrac{d}{dr}left(e^{-t}dfrac{du}{dt}right)=dfrac{d}{dt}left(e^{-t}dfrac{du}{dt}right)dfrac{dt}{dr}=left(e^{-t}dfrac{d^2u}{dt^2}-e^{-t}dfrac{du}{dt}right)e^{-t}=e^{-2t}dfrac{d^2u}{dt^2}-e^{-2t}dfrac{du}{dt}$



          $therefore-e^{-2t}dfrac{d^2u}{dt^2}+e^{-2t}dfrac{du}{dt}-e^{-2t}dfrac{du}{dt}+e^u-e^{-u}=0$



          $e^{-2t}dfrac{d^2u}{dt^2}=e^u-e^{-u}$



          $dfrac{d^2u}{dt^2}=e^{2t+u}-e^{2t-u}$






          share|cite|improve this answer












          Hint:



          Let $r=e^t$ ,



          Then $t=ln r$



          $dfrac{du}{dr}=dfrac{du}{dt}dfrac{dt}{dr}=dfrac{1}{r}dfrac{du}{dt}=e^{-t}dfrac{du}{dt}$



          $dfrac{d^2u}{dr^2}=dfrac{d}{dr}left(e^{-t}dfrac{du}{dt}right)=dfrac{d}{dt}left(e^{-t}dfrac{du}{dt}right)dfrac{dt}{dr}=left(e^{-t}dfrac{d^2u}{dt^2}-e^{-t}dfrac{du}{dt}right)e^{-t}=e^{-2t}dfrac{d^2u}{dt^2}-e^{-2t}dfrac{du}{dt}$



          $therefore-e^{-2t}dfrac{d^2u}{dt^2}+e^{-2t}dfrac{du}{dt}-e^{-2t}dfrac{du}{dt}+e^u-e^{-u}=0$



          $e^{-2t}dfrac{d^2u}{dt^2}=e^u-e^{-u}$



          $dfrac{d^2u}{dt^2}=e^{2t+u}-e^{2t-u}$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 12:49









          doraemonpaul

          12.5k31660




          12.5k31660












          • Yes, this transformation is right and beautiful, but I don't know how to carry on. Would you explain the following steps please?
            – U2647
            Nov 28 '18 at 0:28










          • @U2647 You can rewrite the equation as $u'' = 2exp(2t)sinh(u)$, but there will be no closed form solution. If you seek solutions for $u$ near $0$, you can say $sinh(u)approx u$ and find solutions in terms of the modified Bessel functions.
            – AlexanderJ93
            Nov 28 '18 at 1:11












          • @AlexanderJ93 So the approximate solution includes an imaginary part?
            – U2647
            Nov 28 '18 at 1:27










          • @U2647 They are real-valued functions for real and positive arguments.
            – AlexanderJ93
            Nov 28 '18 at 1:33


















          • Yes, this transformation is right and beautiful, but I don't know how to carry on. Would you explain the following steps please?
            – U2647
            Nov 28 '18 at 0:28










          • @U2647 You can rewrite the equation as $u'' = 2exp(2t)sinh(u)$, but there will be no closed form solution. If you seek solutions for $u$ near $0$, you can say $sinh(u)approx u$ and find solutions in terms of the modified Bessel functions.
            – AlexanderJ93
            Nov 28 '18 at 1:11












          • @AlexanderJ93 So the approximate solution includes an imaginary part?
            – U2647
            Nov 28 '18 at 1:27










          • @U2647 They are real-valued functions for real and positive arguments.
            – AlexanderJ93
            Nov 28 '18 at 1:33
















          Yes, this transformation is right and beautiful, but I don't know how to carry on. Would you explain the following steps please?
          – U2647
          Nov 28 '18 at 0:28




          Yes, this transformation is right and beautiful, but I don't know how to carry on. Would you explain the following steps please?
          – U2647
          Nov 28 '18 at 0:28












          @U2647 You can rewrite the equation as $u'' = 2exp(2t)sinh(u)$, but there will be no closed form solution. If you seek solutions for $u$ near $0$, you can say $sinh(u)approx u$ and find solutions in terms of the modified Bessel functions.
          – AlexanderJ93
          Nov 28 '18 at 1:11






          @U2647 You can rewrite the equation as $u'' = 2exp(2t)sinh(u)$, but there will be no closed form solution. If you seek solutions for $u$ near $0$, you can say $sinh(u)approx u$ and find solutions in terms of the modified Bessel functions.
          – AlexanderJ93
          Nov 28 '18 at 1:11














          @AlexanderJ93 So the approximate solution includes an imaginary part?
          – U2647
          Nov 28 '18 at 1:27




          @AlexanderJ93 So the approximate solution includes an imaginary part?
          – U2647
          Nov 28 '18 at 1:27












          @U2647 They are real-valued functions for real and positive arguments.
          – AlexanderJ93
          Nov 28 '18 at 1:33




          @U2647 They are real-valued functions for real and positive arguments.
          – AlexanderJ93
          Nov 28 '18 at 1:33


















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