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I'm trying to solve the following nonlinear second-order ODE \begin{align} 0&=\tilde{v} u' + \frac{d}{d z} \left ( u (1- u)^2 \frac{d}{d z} \Sigma \right ) + u(u_m-u), \\ \Sigma(u) &= \frac{u^2(u-u_e)}{1-u}, \end{align} on $(-\infty,0]$ where $\tilde{v}$ is the constant velocity and $u_m,u_e$ are constants as well. I need the derivative of $u$ for further computations. Since the derivative is extremely steep (seen numerically on a bounded interval) I'd prefer an analytical solution. Any advice on how to start with this equation? One of my thoughts was the Laplace transform due to working on the left half plane, but I don't know how to deal with the nonlinearity.

UPDATE: @Eli Bartlett your first equation differs in the exponent to mine. Therefore I'm posting my computation, maybe you see where I'm being wrong. \begin{align} \frac{d}{dz} \Sigma = u_z'\frac{-2u^3+(u_e+3)u^2 -2u_eu }{(u-1)^2} \end{align} and then \begin{align} \frac{d}{d z} \left ( u (1- u)^2 \frac{d}{d z} \Sigma \right ) = \frac{d}{d z} \left ( u (u_z'(-ux^3+(u_e+3)u^2 -2u_eu ) \right ) = \frac{d}{d z} \left ( u_z'(-2u^4+(u_e+3)u^3 -2u_eu^2 ) \right ) \end{align}

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Expanding the ODE we have that \begin{align} \left[(-2u^5+(u_e+3)u^4-2u_eu^3)u_z'\right]_z'+\tilde vu_z'+u(u_m-u)=0. \end{align} Taking $(-2u^5+(u_e+3)u^4-2u_eu^3)u_z'=p(u)$ we have the first order equation \begin{align} pp'_u+\tilde v p=u^4(u-u_m)(-2u^2+(u_e+3)u-2u_e), \end{align} which is an Abel Equation of the second kind. This equation does not have a general solution, and I nor Wolfram Alpha can find a solution for this particular equation.

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  • $\begingroup$ Thanks for your reply! I'm not entirely sure how to solve it from here: Should I stick with the Laplace transform or could I find an appropriate integrating factor? My boundary conditions are just Dirichlet boundary conditions. $\endgroup$
    – Minusch
    Commented Jan 24, 2023 at 21:23
  • $\begingroup$ @Minusch I've missed a factor of $p$ when taking $u'=p(u)$, give me a moment to correct. $\endgroup$ Commented Jan 25, 2023 at 1:16
  • $\begingroup$ I'm getting one exponent lower in your first equation. I updated my question in regard of the computation. Do you see by any chance where I'm going wrong? And could you tell me what you are exactly doing the second equation? I'm getting $p_z' = u_z' p_u'$ and don't see how you derived your equation. Thanks a lot! $\endgroup$
    – Minusch
    Commented Jan 26, 2023 at 11:06

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