Klein-Gordon Equation with boundary conditions

In summary, the Klein-Gordon equation is a relativistic wave equation that describes scalar fields, such as quantum fields in particle physics. When analyzing this equation, applying boundary conditions is essential to ensure well-posedness and to determine unique solutions in specific physical contexts. These conditions can include fixed values at the boundaries or specifying the behavior of the field at infinity. The study of the Klein-Gordon equation with boundary conditions is crucial for understanding various phenomena in theoretical and applied physics, including quantum field theory and cosmology.
  • #1
dsaun777
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I am trying to find solutions for the Klien-Gordon equations in 1-d particle in a box. The difference here is the box itself oscillating and has boundary conditions that are time dependent, something like this L(t)=L0+ΔLsin(ωt). My initial approach is to use a homogeneous solution and use Eigenfunction expansion to get a solution. But I can't make progress. Are there other methods that are easier? maybe numerical methods...
 
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  • #2
Do you mean that you have homogenous BCs at [itex]x = 0[/itex] and [itex]x = L(t)[/itex]?

Set [itex]\epsilon = \Delta L/L_0[/itex] so that the boundary condition is applied at [itex]x = L_0(1 + \epsilon \sin \omega t)[/itex]. Then for small [itex]\epsilon[/itex], you can seek an asymptotic expansion [itex]\psi = \psi_0 + \epsilon \psi_1 + \dots[/itex] where the boundary conditions on the [itex]\psi_n[/itex] are shifted to [itex]x = L_0[/itex] by expanding [itex]\psi(L_0(1 + \epsilon \sin \omega t),t)[/itex] in Taylor series in [itex]x[/itex] about [itex]x = L_0[/itex] and comparing coefficients of powers of [itex]\epsilon[/itex]. This gives you a series of problems which can be solved by eigenfunction expansion.

EDIT: It might be better to solve for each [itex]\psi_n[/itex] using a Laplace transform in time. The change of variable [itex]\tilde x = x/L_0[/itex] may help to simplify the algebra. Also note that if [itex]\omega[/itex] is a natural frequency of any of these problems then the resulting resonance will cause the asymptotic assumption to break down, since eventually the amplitude will exceed [itex]\epsilon^{-1}[/itex].
 
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