Basic doubt on chain rule in DAlemberts soln to wave equation

[bksree]

TL;DR Summary

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) - dependant 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
$$\partial_t y(t,x)=v f'(x+v t)-v g'(x-v t).$$