### Boundary Conditions for the Wave Equation

#### Solving PDEs thru symmetries

There are some simple tricks to find solutions to some partial differential equations (PDEs) which I hope to generalize to use in two slit calculations. I recall details from school here. I limit these notes to linear PDEs and boundary conditions (BCs) where for a particular combination of PDE and BC any linear combination of two solutions also solves the PDE and BC.

### First the rope:

2y/∂t2 = ∂2y/∂x2

x is the horizontal position of some small piece of the rope; t is time. Tension on the rope is such that the velocity of a wave in the rope is 1. We consider here vertical displacements: y which is a function of x and t.

For unbounded x and t the general solutions are y = f(t+x) + g(t−x) where f and g are any functions with 2nd derivatives. If we limit x to positive values (x>0) then we must give a boundary condition (BC) for x = 0.

We restrict our attention to y = f(t+x) + b f(t−x) and consider some special cases with a BC at x=0 and the domain of the PDE is x>0.

#### The rope is tied down at 0.

Let b = −1 in the above. For x=0 we have y = 0 and ∂y/∂x = 2f'(t). The rope is fixed at x=0 and a wave proceeding from positive x’s towards 0 is reflected at 0 and inverted as it returns towards positive x. We thus have the general solution for the domain x>0 and the BC (x=0 → y=0).

#### The rope at x=0 is free to move vertically.

Another case is b = 1. A wave proceeding from positive x’s towards 0 is reflected at 0 and preserves its magnitude but is not inverted. At x=0 ∂y/∂x = 0 and this is a BC which says there is no vertcal force on the rope from x<0. This is the general solution for the domain x>0 with BC (x=0 → ∂y/∂x = 0). The rope’s end is free to move up and down with no friction.

#### The rope at x=0 has a critical friction against vertical motion.

Another example, the BCs are (x=0 → ∂y/∂x = −∂y/∂t). The general solution in this case is b=0 and corresponds to the rope end being free to move up and down but with a specific friction which absorbs all of the energy of the wave incoming from positive x. This is also the solution where the domain is the unbounded range of x’s but stipulating all waves come from positive x. If y represents voltage in a transmission line then this is the termination resistor at x=0 that absorbs all signals and energy.

Other values of b show the solution with partial reflections from 0.

### What’s in a boundary condition?

Often one’s intuition is enough to know what must be specified at the boundary of a region inside of which some differential equations rule. Sometimes it is not. (General relativity is especially opaque in this regard.) Things can be made somewhat clearer by the following transformation when 2nd partial derivatives appear. The trick is to introduce first derivatives as independent dynamic variables upon which the 2nd derivatives disappear. We do this first for the rope equation.

2y/∂t2 = ∂2y/∂x2
We introduce v = ∂y/∂t (vertical velocity). We now have:
∂y/∂t = v
∂v/∂t = ∂2y/∂x2.
With this transformation we focus on the vector z = (y, v) which is a function of x and rewrite as:
∂z/∂t = ∂(y, v)/∂t = (v, ∂2y/∂x2).
Now “∂t2” has disappeared and we have an equation in the form ∂z/∂t = f(z) where f involves spatial derivatives only.

BCs for this amount to providing conditions on z and its spatial first derivative at extremal values of x.

The nature of BCs typically becomes clear when we write difference equations.

Fourier’s Take