# 2nd order nonlinear differential equation

1. Aug 18, 2010

### fatimajan

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

2. Aug 18, 2010

### Dickfore

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.

3. Aug 18, 2010

### fatimajan

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?
Thank you again

4. Aug 18, 2010

### Dickfore

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!

5. Aug 18, 2010

### fatimajan

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: Aug 18, 2010
6. Aug 18, 2010

### Dickfore

Yes, this is what I got too. Just rewriting it for further reference:

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

7. Aug 18, 2010

### fatimajan

What do you mean?
don't you help me anymore? :(

8. Aug 19, 2010

### jackmell

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.

9. Aug 19, 2010

### Dickfore

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:

$$\begin{array}{l} x \rightarrow -\infty \Rightarrow v = \sqrt{c} \\ x = 0 \Rightarrow v = \sqrt{b + c} \end{array}$$

10. Aug 19, 2010

### 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)

11. Aug 19, 2010

### Dickfore

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. Aug 20, 2010

### fatimajan

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:

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: Aug 20, 2010
13. Aug 22, 2010

### fatimajan

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. Aug 22, 2010

### Dickfore

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:

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?

Last edited: Aug 22, 2010
15. Aug 22, 2010

### fatimajan

Dear Dickfore
thank you very much

16. Aug 26, 2010

### fatimajan

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

17. Aug 26, 2010

### Dickfore

18. Nov 7, 2010

### fatimajan

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. Nov 7, 2010

### Dickfore

I will, but not today :)

20. Nov 8, 2010

### fatimajan

That's Ok, Thanks before
I'm waiting...