Wave on a string and the chain rule Argh

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The discussion focuses on solving the wave equation using a change of variables, specifically defining \(\alpha = x + ct\) and \(\beta = x - ct\). The user seeks clarification on evaluating the second partial derivative of \(u\) with respect to \(x\) after applying the chain rule. The correct approach involves recognizing that \(u\) can be expressed as \(u = f(\alpha) + g(\beta)\), which leads to the proper application of the chain rule for both first and second derivatives. The key takeaway is that the second derivative requires applying the chain rule twice, which the user initially overlooked.

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Wave on a string and the chain rule...Argh

So, I am working through the wave equation for a review before my friend and I go off to grad school. It has been a couple of years since we both graduated with our BS in Physics.

So, here is the question:

Suppose I want to solve the wave equation using a change of variables. Let's use [tex]\alpha = x+ct[/tex], and [tex]\beta = x-ct, and\: u = \alpha + \beta[/tex]

The wave equation is
[tex]\frac{\partial^{2} u}{\partial t^{2}}=c^{2}\frac{\partial^{2} u}{\partial x^{2}}[/tex]


Now, if we take the partial derivative of u with respect to x and applying the chain rule one gets:

[tex]\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \alpha}\frac{\partial \alpha}{\partial x}+\frac{\partial u}{\partial \beta}\frac{\partial \beta}{\partial x}[/tex]

Now if we evaluate [tex]\frac{\partial \alpha}{\partial x}, and \frac{\partial \beta}{\partial x}[/tex] we get
[tex]\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \alpha} + \frac {\partial u}{\partial \beta}[/tex]

So, what and how do I evaluate the second partial differential with respect to x? I get

[tex]\frac{\partial^{2}u}{\partial x^{2}} = \frac{\partial^{2}u}{\partial \alpha^{2}}\ \frac{\partial \alpha}{\partial x}+\frac{\partial^{2}u}{\partial \beta^{2}}\ \frac{\partial \beta}{\partial x}[/tex]

Now, I know this isn't quite right. I am supposed to get:
[tex]\frac{\partial^{2}u}{\partial x^{2}} = \frac{\partial^{2}u}{\partial \alpha^{2}}\ \frac{\partial \alpha}{\partial x}+ \frac{\partial^{2}u}{\partial \alpha \partial \beta}}\ \frac{\partial \beta}{\partial x} + \frac{\partial^{2}u}{\partial \beta \partial \alpha}}\ \frac{\partial \alpha}{\partial x} +\frac{\partial^{2}u}{\partial \beta^{2}}\ \frac{\partial \beta}{\partial x}[/tex]

Can anyone help me? Thanks.
 
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Hmm...I think there is an error at the beginning. It seems to be that you ought to have

[tex]u = f(\alpha) + g(\beta)[/tex]

as this would correspond to the general solution of the one-dimensional wave equation,

[tex]u = f(x - ct) + g(x + ct)[/tex]

Then, taking derivatives, you get

[tex]\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \alpha} \frac{\partial \alpha}{\partial x} + \frac{\partial u}{\partial \beta} \frac{\partial \beta}{\partial x}[/tex]

and then

[tex]\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial \alpha} \left(\frac{\partial u}{\partial x}\right) \frac{\partial \alpha}{\partial x} + \frac{\partial}{\partial \beta} \left(\frac{\partial u}{\partial x}\right) \frac{\partial \beta}{\partial x}[/tex]

I think this is where you made your mistake. You have to apply the chain rule twice in order to take the second derivative!
 

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