Wave on a string, and the chain rule argh

AI Thread Summary
The discussion focuses on solving the wave equation using a change of variables, specifically defining \(\alpha\) and \(\beta\) in terms of \(x\) and \(t\). The user is attempting to compute the second partial derivative of \(u\) with respect to \(x\) but encounters confusion regarding the application of the chain rule. Clarification is provided on how to correctly apply the chain rule as an operator to derive the mixed terms in the second derivative. The user acknowledges the helpful input and realizes the importance of treating the chain rule appropriately in this context. This exchange emphasizes the significance of understanding differential operators in solving wave equations.
physmurf
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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 \alpha = x+ct, and \beta = x-ct, and\: u = \alpha + \beta

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


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

\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}

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

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

\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}

Now, I know this isn't quite right. I am supposed to get:
\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}

Can anyone help me? Thanks.
 
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just think about what d/dx is from your first chain rule equation (factor out the u on which the differential acts on). then use this to find the second derivative and u'll get the mixed terms.
 
When you are going for the second derivative, apply the operator:

<br /> \frac{\partial}{\partial x} = \frac{\partial \alpha}{\partial x} \frac{\partial}{\partial \alpha} + \frac{\partial \beta}{\partial x} \frac {\partial }{\partial \beta} <br />

to \frac{\partial u}{\partial x}.
 
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Thanks for the input guys. That works. I just didn't think to look at it that way and treat the chain rule as an operator.
 
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