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

• I
• bksree
In summary: Now determine ##f'## and ##g'## from the conditions you are given. If you don't know them, you may have to apply the chain rule again for the second derivatives.
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

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
$$u(t,x)=\tilde{u}[\xi(t,x),\eta(t,x)] = f(x-v t) + g(x+v t).$$

Last edited:
Thank you

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

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

## 1. What is the chain rule in D'Alembert's solution to the wave equation?

The chain rule in D'Alembert's solution to the wave equation is a mathematical concept that allows us to find the derivative of a function composed of two or more functions. In this case, it is used to find the solution to the wave equation by breaking it down into simpler functions.

## 2. How does the chain rule work in D'Alembert's solution to the wave equation?

The chain rule in D'Alembert's solution to the wave equation works by breaking down the equation into simpler functions and finding the derivative of each function. These derivatives are then combined using the chain rule to find the overall solution to the wave equation.

## 3. Why is the chain rule important in D'Alembert's solution to the wave equation?

The chain rule is important in D'Alembert's solution to the wave equation because it allows us to find the solution to a complex equation by breaking it down into simpler functions. This makes it easier to solve and understand the underlying principles of the wave equation.

## 4. Can you provide an example of the chain rule in D'Alembert's solution to the wave equation?

Yes, an example of the chain rule in D'Alembert's solution to the wave equation is when we break down the equation into the form of two simpler functions, such as f(x) = x^2 and g(x) = sin(x). The chain rule allows us to find the derivative of each function and combine them to find the solution to the wave equation.

## 5. Are there any limitations to using the chain rule in D'Alembert's solution to the wave equation?

Yes, there are limitations to using the chain rule in D'Alembert's solution to the wave equation. It can only be applied to equations that can be broken down into simpler functions, and it may not always provide an exact solution. Additionally, the chain rule can become more complex when applied to higher-order derivatives.

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