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

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

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!

 

[topsquark]

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

 

[ChiralSuperfields]

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!

 

[haruspex]

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 $x$.

 

[PeroK]

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

 

[Redbelly98]

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]

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!

 

[ChiralSuperfields]

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

Thank you for your reply @PeroK !

 

[ChiralSuperfields]

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!

 

[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 $x$.

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

 

[ChiralSuperfields]

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

Thank you for your reply @Redbelly98 !

 

[ChiralSuperfields]

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!

 

[PeroK]

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]

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

Thank you for your reply @PeroK!

 

[vela]

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.

 

[ChiralSuperfields]

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!