I'm studying the first chapter of Fourier Analysis -- An Introduction and unfortunately I don't understand some transitions.
There is an equation given (wave equation): $$\frac{\partial^2u}{\partial t^2} = \frac{\partial^2u}{\partial x^2}. \tag{1}$$ Moreover we know that: $$u(x,t) = F(x+t) + G(x-t)$$ is the solution of the equation $(1)$, where $F, G$ are twice differentiable functions.
We are to show that every solution takes this form.
Now we define new variables:
- $\xi = x+t$,
- $\eta = x-t$.
We define a new function:
$$v(\xi, \eta) = u(x,t).$$
Everything is quite obvious for the time being.
I don't understand the next step however. My book says, that: The change of variables formula shows that v satisfies:
$$\frac{\partial^2v}{\partial \xi \partial \eta} =0. \tag{2}$$
I don't know why. I would appreciate any explanation.
My book also says that integrating $(2)$ twice will give us:
$$v(\xi, \eta) = F(\xi) + G(\eta).$$
Here again I don't know what the integration should look like.
Thanks for any help!
That's because of this $$\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial \eta}\dfrac{\partial \eta}{\partial x}+\dfrac{\partial u}{\partial \xi}\dfrac{\partial \xi}{\partial x}=\dfrac{\partial u}{\partial \eta}+\dfrac{\partial u}{\partial \xi}$$therefore$$\dfrac{\partial}{\partial x}\dfrac{\partial u}{\partial x}=\dfrac{\partial}{\partial \eta}\dfrac{\partial u}{\partial x}+\dfrac{\partial}{\partial \xi}\dfrac{\partial u}{\partial x}=\dfrac{\partial^2 u}{\partial\eta^2}+2\dfrac{\partial^2 u}{\partial\xi\partial\eta}+\dfrac{\partial^2 u}{\partial\xi^2}$$similarly $$\dfrac{\partial}{\partial t}\dfrac{\partial u}{\partial t}=\dfrac{\partial^2 u}{\partial\eta^2}-2\dfrac{\partial^2 u}{\partial\xi\partial\eta}+\dfrac{\partial^2 u}{\partial\xi^2}$$therefore $$\dfrac{\partial^2 u}{\partial t^2}=\dfrac{\partial^2 u}{\partial x^2}$$ yields to $$\dfrac{\partial^2 u}{\partial\xi\partial\eta}=0$$