Showing that this equation is a solution to the linear wave equation

  • #1
ChiralSuperfields
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Homework Statement:
Please see below
Relevant Equations:
Please see below
For this problem,
1675825070687.png

Where equation 16.27 is the wave equation.

The solution is
1675825115838.png

1675825132452.png


I don't understand how they got the second partial derivative of ##y## with respect to
##x## circled in red.

I thought it would be ##1## since ##v## and ##t## are constants

Many thanks!
 

Answers and Replies

  • #2
topsquark
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Homework Statement:: Please see below
Relevant Equations:: Please see below

For this problem,
View attachment 321932
Where equation 16.27 is the wave equation.

The solution is
View attachment 321933
View attachment 321934

I don't understand how they got the second partial derivative of ##y## with respect to
##x## circled in red.

I thought it would be ##1## since ##v## and ##t## are constants

Many thanks!
You are likely over-thinking this. Use the chain rule.

Big hint:
##\dfrac{ \partial }{ \partial x} \dfrac{1}{x - vt} \rightarrow \dfrac{ \partial }{ \partial u } \dfrac{1}{u} \cdot \dfrac{ \partial u}{ \partial x}##

What would you use for u?

-Dan
 
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  • #3
ChiralSuperfields
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You are likely over-thinking this. Use the chain rule.

Big hint:
##\dfrac{ \partial }{ \partial x} \dfrac{1}{x - vt} \rightarrow \dfrac{ \partial }{ \partial u } \dfrac{1}{u} \cdot \dfrac{ \partial u}{ \partial x}##

What would you use for u?

-Dan
Thank you for your reply @topsquark !

##u = x- vt##

Sorry do you please know why we need to use the chain rule?

Thank you!
 
  • #4
haruspex
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why we need to use the chain rule?
Because that is what tells you how to differentiate ##\frac 1{x-vt}## wrt ##x##. First, you differentiate it wrt ##x-vt##, 'cos that's easy, then you multiply by the derivative of ##x-vt## wrt ##x##.
 
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  • #5
PeroK
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Isn't ##y = f(x - vt)## a solution to the wave equation, for any twice differentiable ##f##?
 
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  • #6
Redbelly98
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I thought it would be ##1## since ##v## and ##t## are constants
Perhaps you "misspoke" here? ##t## is the time variable, as you must be aware?
 
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  • #7
ChiralSuperfields
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Because that is what tells you how to differentiate ##\frac 1{x-vt}## wrt ##x##. First, you differentiate it wrt ##x-vt##, 'cos that's easy, then you multiply by the derivative of ##x-vt## wrt ##x##.
Thank you for your reply @haruspex !

Sorry could you please explain a bit more why we need to use the chain rule? I have not done partial derivatives from multivariate calculus yet.

Many thanks!
 
  • #8
ChiralSuperfields
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Isn't ##y = f(x - vt)## a solution to the wave equation, for any twice differentiable ##f##?
Thank you for your reply @PeroK !
 
  • #9
ChiralSuperfields
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Perhaps you "misspoke" here? ##t## is the time variable, as you must be aware?
Thank you for your reply @Redbelly98 !

Yeah I agree that ##t## is a time variable, however, I thought I was meant to treat it has a constant along with ##v## since we are partially differentiating ##y## with respect to ##x##.

Many thanks!
 
  • #10
Redbelly98
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Yeah I agree that ##t## is a time variable, however, I thought I was meant to treat it has a constant along with ##v## since we are partially differentiating ##y## with respect to ##x##.
Ah, okay. Yes, for the purposes of doing partial derivatives w.r.t. other variables, that's right.
 
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  • #11
ChiralSuperfields
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Ah, okay. Yes, for the purposes of doing partial derivatives w.r.t. other variables, that's right.
Thank you for your reply @Redbelly98 !
 
  • #12
ChiralSuperfields
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Because that is what tells you how to differentiate ##\frac 1{x-vt}## wrt ##x##. First, you differentiate it wrt ##x-vt##, 'cos that's easy, then you multiply by the derivative of ##x-vt## wrt ##x##.
I guess one way of thinking about it is thinking that ##x -vt## is in brackets to the power of 1 since you must always use chain rule for brackets. I guess we also have to use chain rule here because we can't bring the x up because it is with the other terms.

Let me know if I am correct!

Many thanks!
 
  • #13
PeroK
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I guess one way of thinking about it is thinking that ##x -vt## is in brackets to the power of 1 since you must always use chain rule for brackets. I guess we also have to use chain rule here because we can't bring the x up because it is with the other terms.

Let me know if I am correct!

Many thanks!
$$\frac d {dx}\bigg (\frac 1 {f(x)}\bigg ) = -\frac {f'(x)}{(f(x))^2}$$
 
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  • #14
ChiralSuperfields
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$$\frac d {dx}\bigg (\frac 1 {f(x)}\bigg ) = -\frac {f'(x)}{(f(x))^2}$$
Thank you for your reply @PeroK!
 
  • #15
vela
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I guess one way of thinking about it is thinking that ##x -vt## is in brackets to the power of 1 since you must always use chain rule for brackets.
You don't need the chain rule for
$$\frac{d}{dx}(x^2+x).$$ Parentheses, brackets, etc. don't automatically mean the chain rule.

Based on what you initially wrote, you seemed to be thinking
$$\frac{\partial}{\partial x} \frac{1}{x-vt} =\frac{1}{\frac{\partial}{\partial x}{(x-vt)}} = \frac 11 = 1.$$ I hope you see that the first equality is wrong. That's not how differentiation works.

Because you're differentiating a quotient, you could use the quotient rule and not use the chain rule. Most people, however, would use the chain rule because the numerator is a constant. It might help to rewrite the problem as
$$\frac{\partial}{\partial x} (x-vt)^{-1}.$$

I guess we also have to use chain rule here because we can't bring the x up because it is with the other terms.
Right.
 
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  • #16
ChiralSuperfields
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You don't need the chain rule for
$$\frac{d}{dx}(x^2+x).$$ Parentheses, brackets, etc. don't automatically mean the chain rule.

Based on what you initially wrote, you seemed to be thinking
$$\frac{\partial}{\partial x} \frac{1}{x-vt} =\frac{1}{\frac{\partial}{\partial x}{(x-vt)}} = \frac 11 = 1.$$ I hope you see that the first equality is wrong. That's not how differentiation works.

Because you're differentiating a quotient, you could use the quotient rule and not use the chain rule. Most people, however, would use the chain rule because the numerator is a constant. It might help to rewrite the problem as
$$\frac{\partial}{\partial x} (x-vt)^{-1}.$$


Right.
Thank you for your reply @vela! That is very helpful!
 

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