# I Solution of the wave equation

1. Jan 3, 2017

### Rick16

Text books often give an expression like Asin(kx-ωt) as a solution of the wave equation, but they don’t show how to arrive at this solution. Other textbooks, which go through the complete solution process of the wave equation, determine the coefficients using Fourier series. My goal was to get directly from ∂2y/∂x2 = 1/v22y/∂t2 to y(x,t) = Asin(kx-ωt) without determining the coefficient A, because this is the solution given in many texts and it is very useful to solve problems.

So I went through the separation of variables process and used the boundary conditions and arrived at the solution y(x,t) = Csin(nkx)(Asin(nωt) + Bcos(nωt)), n = 1,2,3... If I am only interested in the fundamental mode I can drop the n-terms and get y(x,t) = Csin(kx)(Asin(ωt)+Bcos(ωt)) = Asin(kx)sin(ωt) + Bsin(kx)cos(ωt). Using the product formulas for sine and cosine I finally get y(x,t) = A(cos(kx-ωt) - cos(kx+ωt)) + B(sin(kx-ωt) + sin(kx+ωt)). (I absorbed the constant C into the constants A and B, and then I replaced A/2 and B/2 by A and B, because anyway they are just arbitrary constants, so it does not matter what I call them).

Now I reason that this last solution is a linear combination of individual solutions. For instance, Asin(kx-ωt) is a wave moving to the right, Asin(kx+ωt) is a wave moving to the left, and the combination of the two is a standing wave. The cosine terms are just shifted versions of the sine terms, so in order to represent a wave moving in one direction the equation y(x,t) = Asin(kx-ωt) is all that is needed.

Is this reasoning correct? I also wonder if the n-terms in the complete solution above should not always be equal to 1 in the case of light waves. Light waves don’t have modes, do they?

2. Jan 4, 2017

### BvU

You mean how to find it, right? Because you are quite capable to show that it's a solution -- am I right there ?
That's good because, as you can see, the equation is satisfied for all possible values of A. That is becasue A is an integration constant and it follows from the initial (or boundary) conditions. Since this is a second order differential equation, there must be yet another such constant. Can you guess where it's hiding ?
Correct, but a simpler expression is generally prefreable.
Not always (because it has only one integration constant). Sometimes you may need something like y(x,t) = Asin(kx-ωt+φ) or Asin(kx-ωt) + Bcos(kx-ωt)
fairly correct, yes.
They certainly do. Once you get to looking at black body radiation and Planck's law to deal with the ultraviolet catastrophe they will come to the rescue.

As an aside: kudos for delving into this subject. A very good investment it is.

3. Jan 4, 2017

### Rick16

Thank you very much. Your answer helped me understand that I stopped too early. I stopped at the solution y(x,t) = Asin(kx-ωt) + Asin(kx+ωt) + Bcos(kx-ωt) - Bcos(kx+ωt). I can combine the first sine and the first cosine term to get y(x,t) = Asin(kx-ωt+φ) -- a wave moving to the right -- and I can combine the second sine and the second cosine to get y(x,t) = Asin(kx+ωt-φ) -- a wave moving to the left. If I combine these two I get the complete solution in very compact form y(x,t) = 2Asin(kx)cos(ωt-φ) -- a standing wave. It was clear to me before that the solution Asin(kx-ωt), that I was looking for, is simply a special case of a wave moving in one direction with zero phase angle, but now I see exactly how it comes about.