How can I solve second order wave equation by Fourier transformation?

Answer 1

Looking at the idea behind a transform is to substitute (in a well defined manner) variable for another. Now I say substitute but rather it is dependent on the properties of the transform which allow for changing differential equations into algebraic equations...

For the wave equation we have (∇² + ∂²/∂²t) f(x,t) = 0.

Here the x is a vector in 3-space (x,y,z) and t is time. The nabla is the representation of ∂²/∂²x + ∂²/∂²y + ∂²/∂²z. f(x,t) is an arbitrary function which fits this partial differentiation equation. Note the fact these are partial derivatives...

Now using the definition of a Fourier transform...

If we allow for (x,t) to be represented as a 4-vector (space-time vector) we can apply a 4-dimensional Fourier transform to convert the wave equation...

3D case: spatial coordinates only

F(q,t) = L{f(x,t)} where L{} is the Fourier transform operation.

F(q,t) = ∫_[lb=-∞]_[ub=+∞] e^-iq·x * f(x,t) dx

We can then use integration by parts to "undo" the differential equation...

∫_[lb=-∞]_[ub=+∞] e^-iq·x * (∇² + ∂²/∂²t) f(x,t) * dx = 0

∫_[lb=-∞]_[ub=+∞] e^-iq·x * (∇²(f(x,t))) * dx + ∫_[lb=-∞]_[ub=+∞] e^-iq·x * (∂²/∂²t (f(x,t))) * dx = 0

I₁ + I₂ = 0

The second term (I₂) (that representing the time component) does not change and we can use the definition of F(q) --> I₁ + ∂²/∂²t (F(q,t) = 0

Now onto I₁...

∫_[lb=-∞]_[ub=+∞] e^-iq·x * (∇²(f(x,t))) * dx can be solved using integration by parts...

First round --> u = e^-iq·x , dv = ∇²(f(x,t)) * dx , du = -i e^-iq·x * q·dx , v = ∇f(x,t)

∫_[lb=-∞]_[ub=+∞] e^-iq·x * (∇²(f(x,t))) * dx = ∇f(x,t) * e^-iq·x [lb=-∞]_[ub=+∞] + iq ∫_[lb=-∞]_[ub=+∞] e^-iq·x * ∇f(x,t)·dx --> 0 + iq ∫_[lb=-∞]_[ub=+∞] e^-iq·x * ∇f(x,t)·dx

Therefore...

∫_[lb=-∞]_[ub=+∞] e^-iq·x * (∇²(f(x,t))) * dx = iq ∫_[lb=-∞]_[ub=+∞] e^-iq·x * ∇f(x,t)·dx

Second round --> u = e^-iq·x , dv = ∇(f(x,t))·dx , du = -i e^-iq·x * q·dx , v = f(x,t)

∫_[lb=-∞]_[ub=+∞] e^-iq·x * ∇f(x,t)·dx = f(x,t) * e^-iq·x [lb=-∞]_[ub=+∞] + iq· ∫_[lb=-∞]_[ub=+∞] e^-iq·x * f(x,t)dx --> 0 + iq· ∫_[lb=-∞]_[ub=+∞] e^-iq·x * f(x,t)dx

Remembering the factor of iq from the first round of integration and using the definition of the Fourier transform we arrive at...

∫_[lb=-∞]_[ub=+∞] e^-iq·x * (∇²(f(x,t))) * dx = -q² * ∫_[lb=-∞]_[ub=+∞] e^-iq·x * f(x,t) dx = -q²F(q)

Putting this all together with the temporal derivative yields...

-q²F(q,t) + ∂²/∂²t( F(q,t) ) = 0 --> (-q² + ∂²/∂²t) F(q,t) = 0

In the 4-dimensional case you would have the 4-vector of q = (Q,q) where the Q is the temporal component.

This would follow the same procedure except we would do the same with temporal derivatives rather than just the spatial derivatives...

In the 4-D case --> (-q² - Q²) F(q,Q) = 0

Since f(q,t) is arbitrary --> this implies F(q,Q) is also arbitrary and therefore q² + Q² = 0

This can solve for q,Q and then one can transform back. Note the ansatz was using e^-iq·x which are just plane waves.