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Need help for solving a 2nd order nonlinear differential equation

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fkendoul
#1
Feb1-12, 11:21 PM
P: 4
Hi,

I need some help to find the analytical solution of the following DE:
x" - k x/x' = at + b, with x' = dx/dt and x" = d(dx/dt)/dt

Any kind oh help or advices on where I can find some useful resources are really appreciated.

Thank you
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HallsofIvy
#2
Feb2-12, 08:21 AM
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Thanks
PF Gold
P: 39,568
There are NO general methods for solving non-linear differential equations. All methods that I know are approximation methods.
bigfooted
#3
Feb2-12, 02:04 PM
P: 293
Actually, for nonlinear ODE's of order 2 and more there is a general solution method due to Lie. It is based on reducing the order of the ODE until you can reduce it to quadrature, by consecutively applying point symmetries of the ODE. A lot of 'tricks' to solve specific ODE's are doing nothing more than applying a known symmetry to solve the equation. Finding a symmetry can be quite a tremendous task that you usually don't undertake without something like Maple.

When using Maple on this ODE however, it yields a horrible expression as a solution that is pretty much useless for all practical purposes.

Quote Quote by HallsofIvy View Post
There are NO general methods for solving non-linear differential equations. All methods that I know are approximation methods.

JJacquelin
#4
Feb2-12, 03:54 PM
P: 759
Need help for solving a 2nd order nonlinear differential equation

x" - k (x/x') = at + b
A solution of the ODE :
x(t) = (1/18)(k+3a)(t+(b/a))^3
fkendoul
#5
Feb2-12, 11:31 PM
P: 4
Thank you guys and many thanks for you JJacquelin.

The solution you propose (x(t) = (1/18)(k+3a)(t+(b/a))^3) satisfies the DE but does not satisfy the boundary conditions or initial conditions for x and x'. I don't know if there is a way to find a similar or another solution that satisfies: x(t=0) = x0.

Thanks
JJacquelin
#6
Feb3-12, 01:24 AM
P: 759
Quote Quote by fkendoul View Post
Thank you guys and many thanks for you JJacquelin.
The solution you propose (x(t) = (1/18)(k+3a)(t+(b/a))^3) satisfies the DE but does not satisfy the boundary conditions or initial conditions for x and x'. I don't know if there is a way to find a similar or another solution that satisfies: x(t=0) = x0.
Thanks
But you didn't state any initial condition in your first wording.
I need some help to find the analytical solution of the following DE:
x" - k x/x' = at + b, with x' = dx/dt and x" = d(dx/dt)/dt
Any kind oh help or advices on where I can find some useful resources are really appreciated.
Thank you
fkendoul
#7
Feb3-12, 04:44 AM
P: 4
Sorry for that JJacquelin. Here is a brief description of my problem. In fact,it is a control problem, we want to control the braking (x, x') of an unmanned small helicopter using the Time-To-Contact information "tau = x/x' " So we have:
tau = x/x', with x(0) = x0 < 0 and x'(0) = x0 > 0
x" = u = k*(tau - tau_ref) , u is a control input
tau_ref = a1*t + b1 with 0 < a1 < 0.5 and b1 = tau0 = x0/x'0

We want to prove that there is a time T where Limit(x(t)) = 0 and Limit(x'(t)) = 0 when t converges to zero.

An option to prove that is to find a solution to the following differential equation:
x" = k*x/x' - k*a1*t - k*b1
JJacquelin
#8
Feb4-12, 02:09 AM
P: 759
I cannot understand the wording :
We want to prove that there is a time T where Limit(x(t)) = 0 and Limit(x'(t)) = 0 when t converges to zero.
The symbol T is not defined and is not used in the equations.
fkendoul
#9
Feb4-12, 03:38 AM
P: 4
Correction:
"We want to prove that there is a time T where Limit(x(t)) = 0 and Limit(x'(t)) = 0 when t converges to T(not to zero as it was written before)."

T is the time where the solution x(T) = 0 and x'(T) = 0; We want to prove that such time T exists and this can be done by using nonlinear control tools or by solving the following DE:
x" = k*x/x' - k*a1*t - k*b1 with x(0) = x0 and x'(0) = x'0


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