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General solution of a second order nonlinear DE

  • Thread starter ephedyn
  • Start date
  • #1
170
1
1. Homework Statement + relevant equations

I have to solve
http://home.vs.moe.edu.sg/linl/eqn1.gif [Broken]

sigma, pinfinity, rho, N are constants. To make things easier for us, we are allowed to treat T as a constant.

2. The attempt at a solution

Treat T as constant:

http://home.vs.moe.edu.sg/linl/eqn2.gif [Broken]

At this point, this might help... Using a Maple-based solver, I get:

dsolve('Rho*(y*D2y+(3*Dy^2)/2)=-(pinf+(2*sig)/R-(N*T)/(R^3))')
Warning: Explicit solution could not be found; implicit solution returned.
> In dsolve at 312

ans =

Int(Rho*_a^2*R^2*3^(1/2)/(Rho*_a*R*(C1-2*_a^3*pinf*R^3-4*_a^3*sig*R^2+2*_a^3*N*T))^(1/2),_a = .. y)-t-C2 = 0
Int(-Rho*_a^2*R^2*3^(1/2)/(Rho*_a*R*(C1-2*_a^3*pinf*R^3-4*_a^3*sig*R^2+2*_a^3*N*T))^(1/2),_a = .. y)-t-C2 = 0
But I'm not sure how to read '_a = .. y)-t-C2 = 0' in symbolic.

Anyway, moving along... using substitution:

http://home.vs.moe.edu.sg/linl/eqn3.gif [Broken]

I don't know how to proceed from here, but using the same solver,

y denotes U:
dsolve('Rho*(R*y*Dy+3*y^2/2)=-(pinf+2*sig/y-N*T/y^3)','R')
Warning: Explicit solution could not be found; implicit solution returned.
> In dsolve at 312

ans =

log(R)-Int(2/(2*N*T-3*Rho*_a^5-2*pinf*_a^3-4*sig*_a^2)*Rho*_a^4,_a = .. y)+C1 = 0
Does anyone know how I can get to either solution or at least simplify the problem further from here? I don't know how to introduce an implicit solution into my working, too.

Thanks.
 
Last edited by a moderator:

Answers and Replies

  • #2
263
0
When you have a DE like this, with a quadratic form of the derivative and the second derivative, you should always try the following. This is also a de-facto method of solving equations of motion in mechanics. So write the second derivative as:

[tex]\ddot R = \frac{d\dot R}{dR}\frac{dR}{dt}=\frac{d\dot R}{dR}\dot R = \frac12 \frac{d\dot R^2}{dR}[/tex]

Now put the new variable as [tex]u=\dot R^2[/tex]

So our original DE is:

[tex]R\frac12 \frac{du}{dR}+\frac32 u = f(R,T)[/tex]

Dividing out:

[tex]\frac{du}{dR}+\frac{3}{R}u=\frac{2f(R,T)}{R}[/tex]

Now what is this? this is a Differential equation for the [tex]u(R)[/tex] function. As we see it is first order, and linear, nonhomogenous. But we are lucky since the non-homogenous term consists of powers of R only. So this can be solved easily using the Lagrange method, of variating the constants..
After we get solve this we get the u(R) function. And then we just need to take the square root and integrate to get the R(t) function, which looks like it can be given in this case.
 
  • #3
170
1
Ahhh! Thanks! I wasn't too confident so I didn't want to jump in for a reply before I got it, but I solved it at last.
 
  • #4
170
1
Oh wait, I was wondering what you meant by "which looks like it can be given in this case". Because I verified that

http://home.vs.moe.edu.sg/linl/eqnsol.gif [Broken]

But its square root cannot be integrated to a form containing standard functions, so I'm certain I have to make assumptions at this point (fluid mechanics problem). Nonetheless, thanks a lot for your help!
 
Last edited by a moderator:
  • #5
263
0
Then I was probably wrong, I didnt actually calculate anything just guessed.. :D Looks like it was a wrong guess :D
 
  • #6
170
1
Then I was probably wrong, I didnt actually calculate anything just guessed.. :D Looks like it was a wrong guess :D
:D No problem, I'll just have to cheat a little.
 

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