I'm really anxious about getting this exercise right:
Given the function $\phi = f(x-t) + g(x+t)$, with $f,g$ one variable functions in $C^2$, prove that $\phi$ satisfies the wave equation $\frac{\partial^2\phi}{\partial t^2} = \frac{\partial^2\phi}{\partial x^2}$
I started by defining $a, b: \mathbb{R^2} \to \mathbb{R}$, $a(x,t)=x-t, b(x,t)=x+t$.
Then we have
$$\frac{\partial^2\phi}{\partial t^2} = \frac{d}{dt} \bigg(\frac{\partial \phi}{\partial t}\bigg) = \frac{d}{dt} \bigg(\frac{df}{da} \cdot \frac{\partial a}{\partial t} + \frac{dg}{db} \cdot \frac{\partial b}{\partial t}\bigg)$$
$$ = \frac{d}{dt} \bigg(\frac{df}{da}\bigg) \cdot \frac{\partial a}{\partial t} + \frac{df}{dt} \cdot \frac{\partial^2 a}{\partial t^2} + \frac{d}{dt} \bigg( \frac{dg}{db} \bigg) \cdot \frac{\partial b}{\partial t} + \frac{dg}{db} \cdot \frac{\partial^2 b}{\partial t^2}$$
But we know that $\frac{\partial a}{\partial t} = -1, \frac{\partial b}{\partial t}=1$ and $\frac{\partial^2 a}{\partial t^2} = \frac{\partial^2 b}{\partial t^2} = 0$. Therefore
$$= \frac{d}{dt} \bigg( \frac{dg}{db} \bigg) - \frac{d}{dt} \bigg(\frac{df}{da}\bigg) = \frac{d^2g}{db^2} + \frac{d^2f}{da^2}$$
The only difference when we differentiate w.r.t. $x$ is that $\frac{\partial a}{\partial x} = 1$. And this doesn't make a difference because the negative sign of $\frac{\partial a}{\partial t}$ was cancelled out.
Hence $\frac{\partial^2\phi}{\partial t^2} = \frac{\partial^2\phi}{\partial x^2}$.
I don’t sea any problem there ? I guess what you did is right, the function $f \circ a$ And $g \circ b$ are well defined and a theorem states that if you have $f : R\rightarrow R$, $C^1$ And $u : R^2 \rightarrow R$ a $C^1$ Function, then you can define the function : $$F : R^2 \rightarrow R$$ Such that : $F(x,y)=f(u(x,y))$ And : $$\frac{\partial F}{\partial x}(x,y)=\frac{\partial u}{\partial x}(x,y)\frac{\partial f }{\partial u}(x,y)$$ Which is equivalent to : $$\nabla F(x,y)=\frac{df}{du}(u(x,y)).\nabla u(x,y)=f’(u(x,y))\nabla u(x,y)$$