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I would like to find the exact solution of the following nonlinear ODE.

$$ A_1\frac{d^2 \rho}{dx^2}+(1-2 \rho + A_2)\frac{d \rho}{dx}-A_3\rho+A4=0, \hspace{10mm} (eq.1)$$
with $$\rho(0)=b_1 , \hspace{10mm} \rho(1)=b_2.$$

Here $\rho$ is a differentiable function and $A_1, A_2, A_3, A_4, b_1$ and $b_2$ are real constants.

My attempt: I tried to convert this equation into first-order equation by substituting

$$w(\rho)=\frac{d\rho}{dx}$$ which leads to $$A_1 w \frac{dw}{dp}=(2\rho-A2-1)w+A3\rho-A4.$$

Letting $2\rho+c=z$ where $c=-A_2-1$ gives

$$2A_1w\frac{dw}{dz}=z(w+\frac{A3}{2})-\frac{A3}{2}c-A4.$$

which can be written as $$c_1ww^{'}=z(w+c_2)+c_3 \hspace{10mm} (eq.2)$$
where $c_1=2A_1$, $c_2=\frac{A_3}{2}$, $c_3=-\frac{A_3}{2}-A_4$.

I don't know how to proceed further and solve (eq.2).

Remark: I have solved this equation numerically. So I want an analytical solution for this equation.

Could anyone among you help me in solving eq.(1) or eq.(2) analytically? Any help is appreciated.

Thank you.

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3 Answers 3

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$$c_1w(z)\frac{dw}{dz}=(w(z)+c_2)z+c_3 \hspace{10mm} (eq.2)$$ Let $\quad w(z)=\frac{1}{u(z)}$

$-c_1\frac{1}{u^3}\frac{du}{dz}=(\frac{1}{u(z)}+c_2)z+c_3 $

$$\frac{du}{dz}= -\frac{1}{c_1}(c_2z+c_3)u^3 -\frac{z}{c_1}u^2$$

This is an Abel's Differential Equation of first kind.

These kind of equations are not solvable in general (except particular cases) with elementary and standard special functions.

https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf

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  • $\begingroup$ Thanks for your help and suggestion. Though I have understood that this type of ODE is not directly solvable. However, I would like to know whether it is possible to convert eq.2 in a variable separable form using some suitable transformation? $\endgroup$
    – Atul Verma
    Jan 13, 2018 at 13:16
  • $\begingroup$ I don't think that it is possible. If there was such a transformation, some particular forms of Abel's equations would have been solved a long time ago with more direct method than the arduous implicite ones appearing in the paper referenced at the end of my answer. $\endgroup$
    – JJacquelin
    Jan 13, 2018 at 14:05
  • $\begingroup$ Thanks for your reply. Another query is can we solve this equation using Lambert W function, series solution or any transform like Laplace or Fourier? Moreover, can we find integrating function for this equation so that we can make it an exact equation? $\endgroup$
    – Atul Verma
    Jan 13, 2018 at 15:06
  • $\begingroup$ The function with Lambert W function is not a solution of (eq.2). It is the solution of a very different ODE, introduced by "Fog_convergence" in comments. There is no true relationship with (eq.2). The Lambert W function is of no help to solve (eq.2). Laplace or Fourier transforms cannot apply to (eq.2) which is not linear. I think that the practical way is numerical calculus, or series approximations, better directly on (eq.1) than on (eq.2). $\endgroup$
    – JJacquelin
    Jan 13, 2018 at 18:23
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Hint:

your last equation in the homogeneous case has the form $$yy'=a(y+b) \Rightarrow \frac{yy'}{y+b}=a \Rightarrow y'(\frac{y+b}{y+b}-\frac{b}{y+b})=a \Rightarrow y'-\frac{by'}{y+b}=a $$ $$ \Rightarrow \int y'-\int \frac{by'}{y+b}= \int a $$

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  • $\begingroup$ The solution of the homogeneous case is of no use in case of non-linear ODE. Moreover, because the $z$ in (eq.2) in the Atul Verma's question, the form that you consider don't corresponds to (eq.2). $\endgroup$
    – JJacquelin
    Jan 13, 2018 at 11:18
  • $\begingroup$ Thank you, if we can get an explicit formula of $y$ from $ y-\log(y+b)-a x =0 $ does it still useless? $\endgroup$
    – ahdahmani
    Jan 13, 2018 at 11:22
  • $\begingroup$ What is the relationship between $y-\log(y+b)-ax=0$ and (eq.2) ? If there is no relationship, it should be useless. $\endgroup$
    – JJacquelin
    Jan 13, 2018 at 11:41
  • $\begingroup$ $y-\log(y+b)-a x =0$ this is the solution of the equation I mentioned in my solution so if we can get an explicit formula of $y$ from it, we can use the technique of variation of the constant then we pick up the solution $\endgroup$
    – ahdahmani
    Jan 13, 2018 at 11:43
  • $\begingroup$ Indeed, we can get an explicit formula of $y(x)$, thanks to the Lambert W function. This will give the solution of the homogenous equation. But it will be of no use and moreover will not help to solve the OP's equation which is very different. Try and see. $\endgroup$
    – JJacquelin
    Jan 13, 2018 at 11:52
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Let $w=\dfrac{d\rho}{dx}$ ,

Then $\dfrac{d^2 \rho}{dx^2}=\dfrac{dw}{dx}=\dfrac{dw}{d\rho}\dfrac{d\rho}{dx}=w\dfrac{dw}{d\rho}$

$\therefore A_1w\dfrac{dw}{d\rho}+(1-2\rho+A_2)w-A_3\rho+A_4=0$

$A_1w\dfrac{dw}{d\rho}=(2\rho-A_2-1)w+A_3\rho-A_4$

Let $s=\rho-\dfrac{A_2+1}{2}$ ,

Then $A_1w\dfrac{dw}{ds}=2sw+A_3s+\dfrac{(A_2+1)A_3}{2}-A_4$

Let $t=\dfrac{s^2}{A_1}$ ,

Then $\dfrac{dw}{ds}=\dfrac{dw}{dt}\dfrac{dt}{ds}=\dfrac{2s}{A_1}\dfrac{dw}{dt}$

$\therefore2sw\dfrac{dw}{dt}=2sw+A_3s+\dfrac{(A_2+1)A_3}{2}-A_4$

$w\dfrac{dw}{dt}=w+\dfrac{A_3}{2}+\dfrac{(A_2+1)A_3-2A_4}{4s}$

$w\dfrac{dw}{dt}=w+\dfrac{A_3}{2}\pm\dfrac{(A_2+1)A_3-2A_4}{4\sqrt{A_1t}}$

This belongs to an Abel equation of the second kind in the canonical form.

Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf

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