# 2nd order nonlinear differential equation

by fatimajan
Tags: differential, equation, nonlinear, order
 P: 16 Hello everybody, could you please direct me how to solve this nonlinear differential equation analytically, so by mathematica or matlab? I really need to solve it for my research project, so please help me du/dx=d/dx[a*u^(-1/2)*du/dx]-n*u^(3/2)*(u-c)/b boundry conditions are: u(0)=b+c u(-infinity)=c where: a, n,c &b are constants by the way, I'm a student of mechanical engineering in master thank you
 P: 3,014 If I understand your text correctly, the equation is: $$\frac{du}{dx} = \frac{d}{dx}\left[a \, u^{-\frac{1}{2}} \, \frac{du}{dx}\right] - \frac{n}{b} \, u^{\frac{3}{2}}(u - c)$$ If this is so, then, notice that the equation does not contain the independent variable x explicitly. Therefore, a parametric substitution of the form: $$p \equiv \frac{du}{dx}$$ which transforms the derivatives to: $$\frac{d}{dx} = \frac{du}{dx} \, \frac{d}{du} = p \, \frac{d}{du}$$ should decrease the order of the ODE by one order. Notice that now $p = p(u)$ or $u = u(p)$, whichever is more convenient. Please show what is the form of the new ODE after this substitution is made.
 P: 16 Hi, Thanks for your help, Dickfore. my equation after that substitution transforms to: P(u)= P(u)*[p'(u)*u^(-1/2)-P(u)*u^(-3/2)/2]-n/b*u^(3/2)*(u-c) Where: a, n,c &b are constants like you said: P=du/dx, P=P(u) So how can I keep solving that? please help me... Thank you again
 P: 3,014 2nd order nonlinear differential equation I think you have forgotten the coefficient $a$ in front of the first term on the right hand side. Other than that, the equation is correct. Do some algebra to simplify the equation so that the coefficient in fron of $p'(u)$ is equal to 1. Post your simplified equation!
 P: 16 I did what you said, Dickfore. the simplified equation is: P'(u)=u^(1/2)/a+P(u)*u^(-1)/2+n*u^2*(u-c)/(P(u)*a*b) what should I do after that? actually I don't know! Thank you
 P: 3,014 Yes, this is what I got too. Just rewriting it for further reference: $$\frac{dp}{du} - \frac{p}{2 u} - \frac{u^{\frac{1}{2}}}{a} - \frac{n u^{2} \, (u - c)}{a \, b \, p} = 0$$
 P: 16 What do you mean? don't you help me anymore? :(
P: 1,666
 Quote by Dickfore Yes, this is what I got too. Just rewriting it for further reference: $$\frac{dp}{du} - \frac{p}{2 u} - \frac{u^{\frac{1}{2}}}{a} - \frac{n u^{2} \, (u - c)}{a \, b \, p} = 0$$
We can't go any further with this can we? Guess if I had to work with it, I'd first get rid of a, b, c, and n:

$$\frac{dp}{du}-\frac{p}{2u}-\sqrt{u}-\frac{u^3}{p}=0$$

and even that, Mathematica can't solve not to mention you'd have to integrate the solution again to get the final answer.
 P: 3,014 Perhaps it would have been better not to expand the derivative using the product rule in the first place. If you multiply the equation by $u^{-1/2}$, you will get: $$\left(u^{-1/2}\frac{dp}{du} + (-\frac{1}{2}) \, u^{-3/2} \, p\right) - \frac{1}{a} - \frac{n u \, (u - c)}{a \, b} u^{1/2} \, p^{-1} = 0$$ and introducing a new $p$: $$u^{1/2} \, p \rightarrow p \Rightarrow p \rightarrow u^{1/2} \, p$$ then the equation has one term less: $$\frac{dp}{du} = \frac{1}{a} + \frac{n \, u \, (u - c)}{a \, b \, p} = \frac{b \, p + n \, u \, (u - c)}{a \, b \, p}$$ Notice that now the derivative of $x$ with respect to $u$ becomes (because of the redefinition of $p$): $$\frac{d x}{d u} = \frac{1}{p} \rightarrow \frac{d x}{d u} = u^{-1/2} \, p^{-1}$$ I cannot think of any way to find a solution of this equation in a closed analytic form. (*** MY SUGGESTION ***) The only thing I thought of doing is introducing a new parameter on which u, p and x depend. This was because in the equation for $d p/d u$ we have a term $u (u - c)$ which is non-monotonic and, thus, has no unique inverse. Also, the equation for $d x/d u$ contains an irrational function which has ill behavior. Therefore, we can rewrite everything in the form: $$u^{1/2} \frac{d}{d u} \equiv A \frac{d}{d v} \Leftrightarrow d v = A \, u^{-1/2} \, du \Rightarrow v = 2 \, A \, u^{1/2}, \; A = \frac{1}{2}, \; v = u^{1/2} \Rightarrow u = v^{2}$$ $$\frac{d p}{d u} = \frac{1}{2 v} \, \frac{d p}{d v} = \frac{b \, p + n \, v^{2} \, (v^{2} - c)}{a \, b \, p} \Rightarrow \frac{d p}{d v} = \frac{2 \, v \left[b \, p + n \, v^{2} \, (v^{2} - c)\right]}{a \, b \, p}$$ $$\frac{d x}{d u} = \frac{1}{2 \, v} \, \frac{d x}{d v} = \frac{1}{v \, p} \Rightarrow \frac{d x}{d v} = \frac{2}{p}$$ $$\left\{\begin{array}{rcl} \frac{d p}{d t} & = & 2 \, v \, \left[b \, p + n \, v^{2} \, (v^{2} - c)\right] \\ \frac{d v}{d t} & = & a \, b \, p \\ \frac{d x}{d t} & = & \frac{d v}{d t} \, \frac{d x}{d v} = a \, b \, p \, 2 \, p^{-1} = 2 \, a \, b \end{array}\right.$$ In the process of rewriting the last equation, we see that $x$ is a linear function of $t$. Therefore, this introduction of a new parameter, if anything, has led us to reinterpret $x$ as a good parameter. As a final transformation, we have the following autonomous system of 2 first order ODEs: $$\left\{\begin{array}{rcl} \frac{d v}{d x} & = & p \\ \frac{d p}{d x} & = & \frac{v}{a} \, \left[p + \frac{n}{2 \, b} \, v^{2} \, (v^{2} - c) \right] \end{array}\right.$$ where we also made the transformation $p/2 \rightarrow p$ again. The boundary conditions that you gave translate to the following: $$\begin{array}{l} x \rightarrow -\infty \Rightarrow v = \sqrt{c} \\ x = 0 \Rightarrow v = \sqrt{b + c} \end{array}$$
 P: 16 Hi, Dickfore Thank you very much for your help again. you've introduced a new p, If I understand correctly, that is: u^(1/2)*p~p but didn't you mean, u^(-1/2)*p~p? ( Actually I mentioned the power of u)
P: 3,014
 Quote by fatimajan Hi, Dickfore Thank you very much for your help again. you've introduced a new p, If I understand correctly, that is: u^(1/2)*p~p but didn't you mean, u^(-1/2)*p~p? ( Actually I mentioned the power of u)
Yes. You are right. However, everything else is correct (I made p -> u^{1/2} p as it should be in the continuation of that line).
 P: 16 Hi, Dickfore but I don't think everything else is correct: Like I said I think you mean u^(-1/2)*P ->P NOT u^(1/2)*P ->P otherwise the equation doesn't have one term less. so, dx/du=u^(1/2)*P^(-1) then like you said dx/du contains an irrational function. Therefore, we can rewrite in the form: u^(1/2)*d/du=A*d/dv but according to what I said we'll have: dx/dv=2*v^2/P so I think we should write: u^(-1/2)d/du=Ad/dv =>v=2/3*A*u^(3/2) ,A=3/2, =>u=v^(2/3) so dP/dv=2/3*v^(-1/3)*[bP+nv^(2/3)*(v^(2/3)-c)]/abP Then dx/dv=2/(3P) finally like the way you suggested: dx/dt=dv/dt*dx/dv=abP*2/(3P)=(2/3)ab the autonomous system of 2 first order ODEs transforms to: dv/dx=P (3P/2 -> P) and dP/dx=v^(-1/3)/a*[P+(n3/2b)v^(2/3)*(v^(2/3)-c)] what do you think? was I right? by the way, I don't know how I can type better like you, you really write clearly. excuse me!
 P: 16 Hi, Dickfore don't you help me anymore? I was hoping to solve my problem with the help of you! you really helped me, would you mind helping me again, please? thank you
P: 3,014
Well, I can't seem to find an error in my derivation and I can't decipher your writing, so I will assume the derivation I posted is correct. As someone already mentioned in this thread, it seems impossible that the equation can be solved in a closed form. So, I'm afraid you are left with solving it numerically.

Last time we ended at this step:

 Quote by Dickfore $$\left\{\begin{array}{rcl} \frac{d v}{d x} & = & p \\ \frac{d p}{d x} & = & \frac{v}{a} \, \left[p + \frac{n}{2 \, b} \, v^{2} \, (v^{2} - c) \right] \end{array}\right.$$ where we also made the transformation $p/2 \rightarrow p$ again. The boundary conditions that you gave translate to the following: $$\begin{array}{l} x \rightarrow -\infty \Rightarrow v = \sqrt{c} \\ x = 0 \Rightarrow v = \sqrt{b + c} \end{array}$$
The connection with the old variables is given by:

$$u(x) = [v(x)]^{2}$$

When working with numerics, it is best to get rid of as many parameters as possible. Let's scale everything:

$$\begin{array}{l} x = x_{0} \, \bar{x} \\ v = v_{0} \, \bar{v} \\ p = p_{0} \, \bar{p} \end{array}$$

Then the equations become:

$$\begin{array}{rcl} \frac{v_{0}}{x_{0}} \, \frac{d \bar{v}}{d \bar{x}} & = & p_{0} \, \bar{p} \\ \frac{p_{0}}{x_{0}} \, \frac{d \bar{p}}{d \bar{x}} & = & \frac{v_{0}}{a} \, \left[ p_{0} \, \bar{p} + \frac{n}{2 b} \, v^{2}_{0} \, \bar{v}^{2} \left(v^{2}_{0} \, \bar{v}^{2} - c \right) \right] \end{array}$$

I think it is most convenient to make this choice:

$$v^{2}_{0} = c, \; p_{0} = \frac{n}{2 b} \,v^{4}_{0}. \; \frac{p_{0}}{x_{0}} = \frac{v_{0} \, p_{0}}{a}$$

$$x_{0} = \frac{a}{\sqrt{c}}, p_{0} = \frac{n \, c^{2}}{2 b}, \; v_{0} = \sqrt{c} \Rightarrow \frac{x_{0} \, p_{0}}{v_{0}} = \frac{a n c}{2 b} \equiv k$$

Also, let's get rid of the bars above the symbols again:

$$\begin{array}{rcl} \frac{d v}{d x} & = & k \, p \\ \frac{d p}{d x} & = & v \, \left[ p + v^{2} \, (v^{2} - 1) \right] \end{array}$$

With the boundary conditions being:

$$\left\{\begin{array}{l} \sqrt{c} \, v = \sqrt{c}, \; x \rightarrow -\infty \\ \sqrt{c} \, v = \sqrt{b + c}, \; x = 0 \end{array}\right. \Rightarrow \left\{\begin{array}{l} v = 1, \; x \rightarrow -\infty \\ v = \sqrt{1 + \frac{b}{c}} = M, \; x = 0 \end{array}\right.$$

Instead of having this limit as $x \rightarrow -\infty$, let us make the simultaneous substitution $x \rightarrow -x, p \rightarrow -p$. The equations become:

$$\begin{array}{rcl} \frac{d v}{d x} & = & k \, p \\ \frac{d p}{d x} & = & v \, \left[ p + v^{2} \, (v^{2} - 1) \right] \end{array}$$

with the boundary conditions:

$$\left\{\begin{array}{l} v = \sqrt{1 + \frac{b}{c}} = M, \; x = 0 \\ v \rightarrow 1, \; x \rightarrow \infty \\ \end{array}\right.$$

Can you find the stationary points for this autonomous non-linear system. What is their type?
 P: 16 Dear Dickfore I'm working on your guidance, I think it's so helpful. thank you very much
 P: 16 Dear Dickfore Actually I didn't understand what you meant about "stationary points". I tried to find out that. thus, today I found an article "Stationary points iteration method for periodic solution to nonlinear system" . is it the way you mean? I couldn't download it yet, however I'll do that soon for responding what you asked. though if you know an easier also faster way direct me ,please. Thank you