I know this should be super easy, but, I don't know how to show that the $x$ and $t$ derivatives of $f$ are equal.
I have to show that, $\displaystyle y(x,t) = \frac{f(x+kt)-f(x-kt)}{2}$ satisfies the PDE $\displaystyle \frac{\partial ^2y}{\partial t^2} = k^2 \frac{\partial ^2y}{\partial x^2}$.
I mean, $\displaystyle \frac{\partial ^2y}{\partial t^2} =\frac{k^2(f_{tt}(x+kt)-f_{tt}(x-kt))}{2}$, and $\displaystyle \frac{\partial ^2y}{\partial x^2} = \frac{f_{xx}(x+kt)+f_{xx}(x-kt)}{2}$.
The factor of $k^2$ is accounted for in the PDE but how can we conclude that $f_{xx}$ and $f_{tt}$ are the same? And there's also the negative sign in the middle of the second $y$ derivative. Am I missing the point entirely?
Thank you.
Notice that, although $y$ has two variables, $f$ is a single-variable function. It only has the one derivative, not two partial derivatives. Think of it this way: you have a function $f(v)$ where $v = x + kt$ or $ v = x - kt$ in the two instances. The function $f$ only has the second derivative $\frac{d^2 f}{dv^2}$. You could even use $f^{\prime\prime}$ if you were so inclined.