Resonance pde wave equation u(\phi,t) involving lagrange polynomials

[chavo004]

1/sin(phi) * d/d\phi(sin(phi) * du/d\phi) - d^2u/dt^2 = -sin 2t

for 0<\phi < pi, 0<t<\inf

Init. conditions:

u(\phi,0) = 0
du(\phi,0)/dt = 0 for 0<\phi<pi


How do I solve this problem and show if it exhibits resonance?

the natural frequencies are w = w_n = sqrt(/\_n) =2, correct?

 

[dextercioby]

Can u use the LaTex code...?Just type it using [ tex ] ...[ /tex ] (without the spaces,of course) and the compiler will do the rest.


Daniel.

 

[M98Ranger]

To solve this problem, we can use the method of separation of variables. We can assume that the solution can be written as u(\phi,t) = X(\phi)T(t). Substituting this into the given equation, we get:

1/sin(phi) * d/d\phi(sin(phi) * X'(\phi)T(t)) - d^2(X(\phi)T(t))/dt^2 = -sin 2t

Rearranging and dividing by X(\phi)T(t), we get:

1/sin(phi) * d/d\phi(sin(phi) * X'(\phi))/X(\phi) = -d^2T(t)/dt^2 - sin 2t/T(t)

The left side of the equation only depends on \phi, while the right side only depends on t. Therefore, both sides must be equal to a constant, which we will call -\lambda. This gives us two separate equations:

1/sin(phi) * d/d\phi(sin(phi) * X'(\phi))/X(\phi) = -\lambda

d^2T(t)/dt^2 + (sin 2t + \lambda)T(t) = 0

Solving the first equation, we get:

X(\phi) = A_n * sin(sqrt{\lambda_n}\phi) + B_n * cos(sqrt{\lambda_n}\phi)

where \lambda_n = n^2, n = 1,2,3,...

Solving the second equation, we get:

T(t) = C_n * sin(sqrt{\lambda_n}t) + D_n * cos(sqrt{\lambda_n}t)

where C_n and D_n are constants determined by the initial conditions.

Thus, the general solution is given by:

u(\phi,t) = \sum_{n=1}^{\infty} (A_n * sin(sqrt{\lambda_n}\phi) + B_n * cos(sqrt{\lambda_n}\phi)) * (C_n * sin(sqrt{\lambda_n}t) + D_n * cos(sqrt{\lambda_n}t))

To determine if this solution exhibits resonance, we need to look at the natural frequencies, which are given by \omega_n = \sqrt{\lambda_n} = n. In this case, the natural frequencies are not equal to the forcing frequency, which is \omega = \sqrt{2}. Therefore, we can conclude that