Well, tensors aren't really necessary... I'm just using rules of changing variables in
partial differentiation. Let
t = \frac{1}{2}(u + v)
x = \frac{1}{2}(u - v)
Then
\frac{\partial}{\partial t} = \frac{\partial u}{\partial t} \frac{\partial}{\partial u} + \frac{\partial v}{\partial t} \frac{\partial}{\partial v} = \frac{\partial}{\partial u} + \frac{\partial}{\partial v}
\frac{\partial}{\partial x} = \frac{\partial u}{\partial x} \frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v} = \frac{\partial}{\partial u} -\frac{\partial}{\partial v}
Now, tensors transform in the same way... Specifically, upstairs indices transform the same way as dx, while downstairs indices transform the same way as \tfrac{\partial}{\partial x}. Let me demonstrate.
T_x = \frac{\partial u}{\partial x} T_u + \frac{\partial v}{\partial x} T_v
T_t = \frac{\partial u}{\partial t} T_u + \frac{\partial v}{\partial t} T_v
T^x = \frac{\partial x}{\partial u} T^u + \frac{\partial x}{\partial v} T^v
T^t = \frac{\partial t}{\partial u} T^u + \frac{\partial t}{\partial v} T^v
You can keep going with more indices, this doesn't really have anything to do with wave equations anymore =)
Here's an exercise for you... Express T^x,T^t in terms of T^u,t^v, which you can in turn express in terms of T^x,T^t again... When you're done, you'll have something like T^x = \dots = T^x, and the stuff in between will give you an identity between the partial derivatives. It is this identity that makes \frac{\partial}{\partial x^\mu} \frac{\partial}{\partial x_\mu} = \partial_\mu \partial^\mu a Lorentz invariant.
