2nd order nonlinear differential equation

In summary: I really appreciate your helps.In summary, this conversation discusses a nonlinear differential equation and the steps taken to solve it analytically. The equation is transformed using a parametric substitution and simplified, but the experts are unable to find a closed analytic solution. A new parameter is introduced to transform the equation into an autonomous system of first order ODEs, but it is still unable to be solved.
  • #1
fatimajan
16
0
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:smile:
thank you
 
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  • #2
If I understand your text correctly, the equation is:

[tex]
\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)
[/tex]

If this is so, then, notice that the equation does not contain the independent variable x explicitly. Therefore, a parametric substitution of the form:

[tex]
p \equiv \frac{du}{dx}
[/tex]

which transforms the derivatives to:
[tex]
\frac{d}{dx} = \frac{du}{dx} \, \frac{d}{du} = p \, \frac{d}{du}
[/tex]
should decrease the order of the ODE by one order. Notice that now [itex]p = p(u)[/itex] or [itex]u = u(p)[/itex], whichever is more convenient.

Please show what is the form of the new ODE after this substitution is made.
 
  • #3
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
 
  • #4
I think you have forgotten the coefficient [itex]a[/itex] 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 [itex]p'(u)[/itex] is equal to 1. Post your simplified equation!
 
  • #5
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
 
Last edited:
  • #6
Yes, this is what I got too. Just rewriting it for further reference:

[tex]
\frac{dp}{du} - \frac{p}{2 u} - \frac{u^{\frac{1}{2}}}{a} - \frac{n u^{2} \, (u - c)}{a \, b \, p} = 0
[/tex]
 
  • #7
What do you mean?
don't you help me anymore? :(
 
  • #8
Dickfore said:
Yes, this is what I got too. Just rewriting it for further reference:

[tex]
\frac{dp}{du} - \frac{p}{2 u} - \frac{u^{\frac{1}{2}}}{a} - \frac{n u^{2} \, (u - c)}{a \, b \, p} = 0
[/tex]

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:

[tex]\frac{dp}{du}-\frac{p}{2u}-\sqrt{u}-\frac{u^3}{p}=0[/tex]

and even that, Mathematica can't solve not to mention you'd have to integrate the solution again to get the final answer.
 
  • #9
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 [itex]u^{-1/2}[/itex], you will get:

[tex]
\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
[/tex]

and introducing a new [itex]p[/itex]:

[tex]
u^{1/2} \, p \rightarrow p \Rightarrow p \rightarrow u^{1/2} \, p
[/tex]

then the equation has one term less:

[tex]
\frac{dp}{du} = \frac{1}{a} + \frac{n \, u \, (u - c)}{a \, b \, p} = \frac{b \, p + n \, u \, (u - c)}{a \, b \, p}
[/tex]

Notice that now the derivative of [itex]x[/itex] with respect to [itex]u[/itex] becomes (because of the redefinition of [itex]p[/itex]):

[tex]
\frac{d x}{d u} = \frac{1}{p} \rightarrow \frac{d x}{d u} = u^{-1/2} \, p^{-1}
[/tex]

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 [itex]d p/d u[/itex] we have a term [itex]u (u - c)[/itex] which is non-monotonic and, thus, has no unique inverse. Also, the equation for [itex]d x/d u[/itex] contains an irrational function which has ill behavior. Therefore, we can rewrite everything in the form:

[tex]
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}
[/tex]

[tex]
\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}
[/tex]

[tex]
\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}
[/tex]

[tex]
\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.
[/tex]

In the process of rewriting the last equation, we see that [itex]x[/itex] is a linear function of [itex]t[/itex]. Therefore, this introduction of a new parameter, if anything, has led us to reinterpret [itex]x[/itex] as a good parameter. As a final transformation, we have the following autonomous system of 2 first order ODEs:

[tex]
\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.
[/tex]

where we also made the transformation [itex]p/2 \rightarrow p[/itex] again.
The boundary conditions that you gave translate to the following:

[tex]
\begin{array}{l}
x \rightarrow -\infty \Rightarrow v = \sqrt{c} \\

x = 0 \Rightarrow v = \sqrt{b + c}
\end{array}
[/tex]
 
  • #10
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)
 
  • #11
fatimajan said:
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).
 
  • #12
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!
 
Last edited:
  • #13
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
 
  • #14
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:

Dickfore said:
[tex]
\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.
[/tex]

where we also made the transformation [itex]p/2 \rightarrow p[/itex] again.
The boundary conditions that you gave translate to the following:

[tex]
\begin{array}{l}
x \rightarrow -\infty \Rightarrow v = \sqrt{c} \\

x = 0 \Rightarrow v = \sqrt{b + c}
\end{array}
[/tex]

The connection with the old variables is given by:

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

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

[tex]
\begin{array}{l}
x = x_{0} \, \bar{x} \\

v = v_{0} \, \bar{v} \\

p = p_{0} \, \bar{p}
\end{array}
[/tex]

Then the equations become:

[tex]
\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}
[/tex]

I think it is most convenient to make this choice:

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

[tex]
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
[/tex]

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

[tex]
\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}
[/tex]

With the boundary conditions being:

[tex]
\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.
[/tex]

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

[tex]
\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}
[/tex]

with the boundary conditions:

[tex]
\left\{\begin{array}{l}
v = \sqrt{1 + \frac{b}{c}} = M, \; x = 0 \\

v \rightarrow 1, \; x \rightarrow \infty \\
\end{array}\right.
[/tex]

Can you find the stationary points for this autonomous non-linear system. What is their type?
 
Last edited:
  • #15
Dear Dickfore
I'm working on your guidance, I think it's so helpful.
thank you very much
 
  • #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
 
  • #18
Hello
Dear Dikfore,
If you remember my problem, you led me till you asked about finding the stationary points and their type for my new nonlinear system. In fact, I couldn't find your answer, so I tried other numerical ways, although I didn't succeed. I want to continue your suggested method. I was wondering if you'd mind helping me again.
 
  • #19
I will, but not today :)
 
  • #20
That's Ok, Thanks before
I'm waiting...
 
  • #21
Dear Fatima Jan,

I completely forgot what we were doing and I don't have time to go back to it right now.

It is obvious we cannot find an analytic general solution for this problem. So, you should clearly state what kind of problem you are trying to solve? Is it an initial value problem? If so, what are the initial conditions? I suspect it is some kind of boundary value problem (perhaps for the current in a tube diode with cylindrical electrodes). If so, what are the boundary conditions?

A stationary point for an autonomous system:

[tex]
\dot{x}_{i} = f_{i}(x_{1}, \ldots, x_{n}), \; (i = 1, \ldots, n)
[/tex]

is a point in [itex]\mathbb{R}^{n}[/itex] that is a constant solution of the above simultaneous system. If it is constant, it means that [itex]\dot{x}_{i} = 0[/itex], so, we can find the stationary points by solving the system of algebraic equations:

[tex]
f_{i}(x_{1}, \ldots, x_{n}) = 0, \; (i = 1, \ldots, n)
[/tex]

After you find the stationary points, the next question you should resolve is to determine their type. You do this by assuming that we are starting from a point very close to a stationary point:

[tex]
x_{i}(t_{0}) = x^{(0)}_{i} + \delta x_{i}(t), \; (i = 1, \ldots, n)
[/tex]

When you substitute this solution and expand the right hand sides in linear powers with respect to the small perturbations, you will get a homogeneous system of linear differential equations with constant coefficients:

[tex]
\delta \dot{x}_{i}(t) = \sum_{k = 1}^{n}{A_{i k} \, \delta x_{k}(t)}, \; (i = 1, \ldots, n)
[/tex]

where the coefficients are:

[tex]
A_{i k} \equiv \left.\frac{\partial f_{i}}{\partial x_{k}}\right|_{x^{(0)}}
[/tex]

You assume that [itex]\delta x_{i}(t) = u_{i} e^{\lambda t}[/itex] and your system transforms into a homogeneous system of linear equations:

[tex]
\sum_{k}{(A_{i k} - \lambda \, \delta_{i k}) \, u_{k}} = 0
[/tex]

which has a nonzero solution if and only if:

[tex]
\mathrm{det}\left[A_{i k} - \lambda \, \delta_{i k}\right] = 0
[/tex]

This is a polynomial equation of order n and it has n complex roots for [itex]\lambda[/itex]. By looking at:

[tex]
|\exp{\lambda \, t}| = \exp(t \, \Re{\lambda})
[/tex]

If [itex]\Re{\lambda} \le 0[/itex] for all roots, then the perturbation decays exponentially with time. Therefore, this stationary point is called stable, since moving away from it returns us to it as t increases. If, on the other hand, at least one [itex]\Re{\lambda} > 0[/itex], it means a small deviation will drift away from the stationary point. Thus, this stationary point is unstable.

If you still think this can help you, find the stationary points for the system I proposed before this recess was taken.
 

What is a 2nd order nonlinear differential equation?

A 2nd order nonlinear differential equation is a mathematical equation that involves a function, its first derivative, and its second derivative, where the function itself is nonlinear. This means that the equation cannot be written in the form of y'' = f(x) or y'' + p(x)y' + q(x)y = g(x), which are common forms for linear differential equations.

How is a 2nd order nonlinear differential equation different from a 2nd order linear differential equation?

A 2nd order linear differential equation can be written in the form of y'' = f(x) or y'' + p(x)y' + q(x)y = g(x), where the function is linear. This means that the equation can be solved using techniques such as separation of variables or the method of undetermined coefficients. In contrast, a 2nd order nonlinear differential equation cannot be solved using these techniques and often requires numerical or approximate methods.

What are some real-world applications of 2nd order nonlinear differential equations?

2nd order nonlinear differential equations are commonly used in physics, engineering, and other scientific fields to model complex systems. Some examples include the motion of a pendulum, population growth and decay, and the behavior of electric circuits.

How can I solve a 2nd order nonlinear differential equation?

There is no one method that can be used to solve all 2nd order nonlinear differential equations. However, some common techniques include using power series, numerical methods such as Euler's method, or finding an equivalent first order system of differential equations.

What are the challenges in solving a 2nd order nonlinear differential equation?

One of the main challenges in solving a 2nd order nonlinear differential equation is that there is no general method that can be applied to all equations. This means that different techniques may need to be used for different equations, and there is often no guarantee of finding an exact solution. Additionally, nonlinear equations can exhibit complex behavior and may have multiple solutions or no solutions at all.

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