Basic doubt on chain rule in DAlemberts soln to wave equation

[bksree]

TL;DR

Basic doubt on applying chain rule in DAlemberts soln to wave equation

In D Alembert's soln to wave equation two new variables are defined
$\xi$ = x - vt
$\eta$ = x + vt
x is therefore a function of $\xi$ , $\eta$ , v and t.

For fixed phase speed, v and given instant of time, x is a function of $\xi$ and $\eta$.
Therefore partial derivative of x w.r.t y is (using the chain rule)
$\frac {\partial x} {\partial y} = \frac {\partial x} {\partial \xi} \frac {\partial \xi} {\partial y} + \frac {\partial x} {\partial \eta} \frac {\partial \eta} {\partial y}$

But how to find $\frac {\partial y} {\partial x}$ ?
Here x (the independent variable) is function of $\xi$ and $\eta$

(The chain rule says that if f(r,s) - dependent variable - then
$\frac {\partial f} {\partial t} = \frac {\partial f} {\partial r}\frac {\partial r} {\partial t} + \frac {\partial f} {\partial s}\frac {\partial s} {\partial t}$

TIA

 

[vanhees71]

You have to keep in mind, which variables are held fixed when you do a partial derivative. I've no clue what $y$ may be, because you didn't define it. Usually d'Alembert's transformation is used to solve the (1+1)-dimensional wave equation anyway.

So in the independent coordinates $(t,x)$ the equation you want to solve reads
$$\Box u(t,x)=\frac{1}{v^2} \partial_t^2 u(t,x) - \partial_x^2 u(t,x)=0.$$
Now you introduce new indpendent variables $\xi=\xi(t,x)$ and $\eta(t,x)$ and write
$$u(t,x)=\tilde{u}(\xi,eta)=\tilde{u}[\xi(t,x),\eta(t,x)].$$
Then
$$\partial_t u(t,x)=(\partial_t \xi) \partial_{\xi} \tilde{u} + (\partial_t \eta) \partial_{\eta} \partial_{\eta} \tilde{u} = v (\partial_{\eta} \tilde{u} - \partial_{\xi} \tilde{u}).$$
Using the this rule once more
$$\partial_t^2 u(t,x) = v^2 (\partial_{\eta}^2 \tilde{u}-\partial_{\eta} \partial_{\xi} \tilde{u} - \partial_{\xi} \partial_{\eta} \tilde{u} + \partial_{\xi}^2 \tilde{u}) = v^2 (\partial_{\eta}^2 \tilde{u} -2 \partial_{\xi} \partial_{\eta} \tilde{u} + \partial_{\xi}^2 \tilde{u}).$$
Further
$$\partial_x u(t,x)=(\partial_{x} \xi) \partial_{\xi} \tilde{u} + (\partial_{y} \eta) \partial_{\eta} \tilde{u} = \partial_{\xi} \tilde{u} + \partial_{\eta} \tilde{u}$$
and
$$\partial_x^2 u(t,x)=\partial_{\xi}^2 \tilde{u} + 2 \partial_{\xi} \partial_{\eta} \tilde{u} + \partial_{\eta}^2 \tilde{u}.$$
Then you get
$$\Box u(t,x)=-4 \partial_{\xi} \partial_{\eta} \tilde{u} = 0.$$
Now the solution is very simple: Integration of the equation by $\xi$:
$$\partial_{\eta} \tilde{u}(\xi,\eta)=g'(\eta),$$
where I wrote the "integration constant", which here is an arbitrary function of $\eta$, as a derivative for obious reasons of convenience since now integrating wrt. $\eta$ we get
$$\tilde{u}(\xi,\eta)=f(\xi)+g(\eta),$$
where $f$ and $g$ are arbitrary functions to be determined by the inital and boundary conditions.

Expressed in terms of the old coordinates you simply find
$$u(t,x)=\tilde{u}[\xi(t,x),\eta(t,x)] = f(x-v t) + g(x+v t).$$

 

[bksree]

Thank you

 

[bksree]

Continuing the solution :
Suppose it is given that
$V(x) = \frac {\partial y} {\partial t}_{t=0}$

Then differentiating $y(x,t) = f(\eta) + g(\xi)$ = f(x+vt) + g(x-vt) w.r.t time 't'
$\frac {\partial y(x,t)} {\partial x}_{t=0} = \frac {\partial f} {\partial \eta} \frac {\partial \eta} {\partial t}_{t=0} + \frac {\partial g} {\partial \xi} \frac {\partial \xi} {\partial t}_{t=0}$

But $\frac {\partial \eta} {\partial t} = v$ and $\frac {\partial \xi} {\partial t} = -v$.

The next step is confusing.
Setting time to zero
$\frac {\partial f} {\partial \eta}$ = $\frac {\partial f} {\partial x}$ and $\frac {\partial g} {\partial \xi}$ = $\frac {\partial g} {\partial x}$

What I can't understand is $\frac {\partial f} {\partial \eta}$ is differentiation of f w.r.t the variable $\eta$ where $\eta$ = x + vt
t=0 condition will be applied after differentiation which is already completed w.r.t $\eta$. Then how can we conclude $\frac {\partial f} {\partial \eta}$ = $\frac {\partial f} {\partial x}$ (which us done by putting t= 0 in $\eta$ = x + vt = x)
Similarly for the $\xi$ variable

TIA

 

[vanhees71]

Why are you still working with $\eta$ and $\xi$ (note you exchanged $f$ and $g$ in comparison to my notation and set $u=y$; henceforth I use your notation)? Now it's easier to go on with $t$ and $x$ as independent variables again (because the boundary/initial conditions) are defined interms of $t$ and $x$ anyway. Just use the chain rule to get
$$\partial_t y(t,x)=v f'(x+v t)-v g'(x-v t).$$