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

• ChiralSuperfields
In summary: 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.Many thanks for explaining that!I guess we also have to use chain rule here because we can't bring the x up because
ChiralSuperfields
Homework Statement
Relevant Equations
For this problem,

Where equation 16.27 is the wave equation.

The solution is

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!

Callumnc1 said:

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

ChiralSuperfields
topsquark said:
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

##u = x- vt##

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

Thank you!

Callumnc1 said:
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##.

ChiralSuperfields and topsquark
Isn't ##y = f(x - vt)## a solution to the wave equation, for any twice differentiable ##f##?

ChiralSuperfields, topsquark and haruspex
Callumnc1 said:
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?

ChiralSuperfields and topsquark
haruspex said:
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##.

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!

PeroK said:
Isn't ##y = f(x - vt)## a solution to the wave equation, for any twice differentiable ##f##?

Redbelly98 said:
Perhaps you "misspoke" here? ##t## is the time variable, as you must be aware?

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!

Callumnc1 said:
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.

ChiralSuperfields
Redbelly98 said:
Ah, okay. Yes, for the purposes of doing partial derivatives w.r.t. other variables, that's right.

haruspex said:
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!

Callumnc1 said:
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!
$$\frac d {dx}\bigg (\frac 1 {f(x)}\bigg ) = -\frac {f'(x)}{(f(x))^2}$$

ChiralSuperfields
PeroK said:
$$\frac d {dx}\bigg (\frac 1 {f(x)}\bigg ) = -\frac {f'(x)}{(f(x))^2}$$

Callumnc1 said:
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
$$\frac{\partial}{\partial x} (x-vt)^{-1}.$$

Callumnc1 said:
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.

ChiralSuperfields and PeroK
vela said:
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.

## 1. What is the linear wave equation?

The linear wave equation is a mathematical equation that describes the propagation of a wave through a medium. It is a second-order partial differential equation that relates the second derivative of a function to its first derivative and the independent variables of time and space.

## 2. How do you show that an equation is a solution to the linear wave equation?

To show that an equation is a solution to the linear wave equation, you must substitute the equation into the equation and verify that it satisfies the equation. This involves taking the second derivative of the equation and comparing it to the first derivative and independent variables in the original equation.

## 3. What are the key properties of a solution to the linear wave equation?

A solution to the linear wave equation must satisfy the equation and have a continuous second derivative. It must also have a finite speed of propagation, meaning that changes in the solution at one point cannot affect points that are too far away in space or time.

## 4. How is the linear wave equation used in real-world applications?

The linear wave equation is used in a variety of fields, including physics, engineering, and geology. It can be used to model the behavior of waves in different media, such as sound waves in air or seismic waves in the Earth. It is also used in signal processing and image analysis.

## 5. What are some common techniques for solving the linear wave equation?

Some common techniques for solving the linear wave equation include separation of variables, Fourier analysis, and Green's functions. These methods involve breaking the equation down into simpler parts and using mathematical tools to solve for the solution. Numerical methods, such as finite difference or finite element methods, can also be used to approximate solutions to the equation.

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