# Discrepancy in the solution of a nonlinear dynamic system

1. Sep 18, 2009

### dekarman

Hi,

I am solving the following nonlinear dynamical system using Energy Balance Method (EBM*). My intention is to arrive at an approximate analytical expression for the frequency of oscillation and the excitation force.

u''+u=A(1+2*u) with u(0)=u'(0)=0, where A is a constant (Physically it is like a Heaviside step loading).

I first write the Hamiltonian,

0.5*(u')^2+0.5*u^2=Au(1+2*u)

which implies the residue function R= 0.5*(u')^2+0.5*u^2-Au(1+2*u).

Then I assume the initial guess, which satisfies the initial conditions u=A*(1-cos(w*t)).

Then calculate R using the initial guess and collocate it at w*t=(pi/2) to obtain the relation between A and w.

However, the relation which I obtain is w=sqrt(1+4*A), which means that with increasing value of A, the frequency increases and hence indicates a hardening nature of the system stiffness.

However, by looking at the differential equation, this is clearly a softening system since the coefficient of u is (1-2*A).

I am not able to figure out this discrepancy.

Can somebody please point out where I am going wrong exactly.
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*Reference of EBM, APPLICATION OF THE ENERGY BALANCE METHOD
FOR STRONGLY NONLINEAR OSCILLATORS, H. Pashaei et al., Progress In Electromagnetics Research M, Vol. 2, 47–56, 2008, which is available on internet.

Thank you very much in advance.
Manish

2. Sep 19, 2009

### Staff: Mentor

I don't follow this at all. First, the Hamiltonian is not an equation so what you have written cannot possibly be the Hamiltonian. Second, what is your generalized momentum in this problem? And how did you go from your equations of motion to the Hamiltonian.

3. Sep 19, 2009

### dekarman

Hi,

The Hamitonian H which I am writing is the restatement of the energybalance of the system. Since there is no damping, the inputted energy should be equal to the sum of the potential energy and the kinetic energy at any state of oscillation of the system. Subsequently,
H=KE+PE-inputted energy=0.
(You may please refer to the article to which I have referred to)

Now, H will be equal to zero if we use an exact solution of u. For any approximate solution, H is not equal to zero and there will be a residue R. I have collocated this at pi/2 since there is balance of PE and KE at that point.

I hope I have been able to throw a bit more light on the procedure.

Also, can you please tell the expression of the Hamiltonian which you are having in your mind?
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To get additional insight, consider the linear problem of spring mass system excited by a step force
u''+u=A

H function is 0.5*u'^2+0.5*u^2-Au=0, i.e. KE+PE-inputted energy=0
Take a guess of u as u=A*(1-cos(p*t)).

Substitute in H function, and collocate at any point between 0 to 2*pi, you will find that P=1.
Hence the exact solution is u=A*(1-cos(t)), which actually is the exact solution of the system.

4. Sep 19, 2009

### Staff: Mentor

The Hamiltonian is not, in general, equal to 0. Its time derivative is 0, which means that the Hamiltonian is constant.

$$\mathcal{H} = \sum_i p_i {\dot q_i} - \mathcal{L}$$
$$p_i = \frac{\partial \mathcal{L}}{\partial {\dot q_i}}$$