Physics 374
Fall 2005
Prof. E. F. Redish

MP20 Solution

a.   Given the wave equation   \(TraditionalForm\`∂\^2 y\/\^\(∂x\^2\)\) = (1/((Subscript[v, o])^2))(∂^2) (y/∂)(t^2), we want to switch variables from x and t to 
					ξ = x - (Subscript[v, o]) t
			  	and	η = x + (Subscript[v, o])t
That's going to involve a heavy use of the Chain Rule...I'll write out the steps very explicitly, which is almost always a good approach when things get complicated.  Let's first take a look at how the partial derivative with respect to x goes:
	∂((y(ξ, η))/∂)x = ∂(y/∂)ξ∂(ξ/∂)x+ ∂(y/∂)η∂(η/∂)x  
		= ∂(y/∂)ξ1+ ∂(y/∂)η1 
		= ∂(y/∂)ξ+ ∂(y/∂)η = Ω
Thus:
	(∂^2) (y/∂)(x^2) = ∂(Ω/∂)x =  ∂(Ω/∂)ξ∂(ξ/∂)x+ ∂(Ω/∂)η∂(η/∂)x 
		= ∂(Ω/∂)ξ1+ ∂(Ω/∂)η1 =  ∂(Ω/∂)ξ+ ∂(Ω/∂)η
	         
	         = (∂/∂)ξ(∂(y/∂)ξ+ ∂(y/∂)η ) + (∂/∂)η(∂(y/∂)ξ+ ∂(y/∂)η ) 
	         
	         = (∂^2) 
            (y/∂)(ξ^2) + 2 \(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\) + (∂^2) (y/∂)(η^2)

Now let's take a look at the partial with respect to time:
	∂((y(ξ, η))/∂)t = ∂(y/∂)ξ∂(ξ/∂)t+ ∂(y/∂)η∂(η/∂)t =  ∂(y/∂)ξ(-(Subscript[v, o])) + ∂(y/∂)η((Subscript[v, o])) = Ω
Thus:
	(∂^2) (y/∂)(t^2) = ∂(Ω/∂)t=  ∂(Ω/∂)ξ∂(ξ/∂)t+ ∂(Ω/∂)η∂(η/∂)t = ∂(Ω/∂)ξ(-(Subscript[v, o])) + ∂(Ω/∂)η((Subscript[v, o]))
	
	   = (Subscript[v, o])(∂(Ω/∂)η - ∂(Ω/∂)ξ) 
	                   
	   = (Subscript[v, o]){(∂/∂)η[∂(y/∂)ξ(-(Subscript[v, o])) + ∂(y/∂)η((Subscript[v, o])) ] - (∂/∂)ξ[∂(y/∂)ξ(-(Subscript[v, o])) + ∂(y/∂)η((Subscript[v, o])) ] }
	                   
	   = ((Subscript[v, o])^2) ( (∂^2) (y/∂)(ξ^2)- 2 \(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\) +  (∂^2) (y/∂)(η^2))

Now we can make these substitutions into the wave equation we started with:
			 \(TraditionalForm\`∂\^2 y\/\^\(∂x\^2\)\) = (1/((Subscript[v, o])^2))(∂^2) (y/∂)(t^2)
					 
	           (∂^2) 
            (y/∂)(ξ^2) + 2 \(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\) + (∂^2) (y/∂)(η^2) =  ((Subscript[v, o])^2) ( (∂^2) (y/∂)(ξ^2)- 2 \(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\) +  (∂^2) (y/∂)(η^2))
Which implies that 	
			2 \(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\) = -2 \(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\)
Which can only be true if 
			\(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\) = 0

b.   We showed above that the original wave equation (with derivatives with respect to x and t) is equivalent to the statement \(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\) = 0.  Thinking a little deeper as to what this equation implies, we realize that it means a solution to the wave equation cannot be a function of both ξ and η.  You could have an extremely complicated function of ξ, but that would still satisfy \(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\) = 0 as long as this extremely complicated function has no η-dependence (so that the (∂/∂)ηsends it to zero, as required.  Since any function of ξ alone thus satisfies the wave equation, y(ξ,η) = f (ξ) is a solution.  Similarly, y(ξ,η) = g (η) is also a solution.  The most general solution to the wave equation would be a linear combination of these two linearly independent solutions (so long as both f and g have well-behaved first and second derivatives that exist, otherwise there's not much sense in talking about \(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\) = 0 at all).  Thus:
		y(ξ,η) = f (ξ) + g (η)	
By all means, plug this solution into our wave equation\(TraditionalForm\`∂\^2 y\/\(∂ξ ∂η\)\) = 0 and verify that it is satisfied.

c.   Putting back in the explicit definitions of ξ and η from part "a" makes our general solution look like:
		y(x,t) = f (x - (Subscript[v, o])t) + g (x + (Subscript[v, o])t)
This means that our general wave equation has the general solution of a pulse traveling to the right at speed (Subscript[v, o])and a pulse traveling to the left at speed (Subscript[v, o]).  The "f" function corresponds to the right traveling wave since we'd need successively larger values of x to return the same value of the function's argument (and hence the same value for the function) as time goes on.  Note that all this argument-shifting does nothing to the actual shape of the pulse...it just shifts it farther and farther to the right, again fitting our experience with non-dispersive waves.  Similarly, the "g" function corresponds to a left-moving pulse since you'd need successively smaller values of x to return the same argument of the function as time goes on.

(Solution by T .  Bing)