# Wave on a string, and the chain rule argh

1. Jul 24, 2008

### physmurf

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. Lets 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.

2. Jul 24, 2008

### Gianni2k

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.

3. Jul 24, 2008

### Dick

When you are going for the second derivative, apply the operator:

$$\frac{\partial}{\partial x} = \frac{\partial \alpha}{\partial x} \frac{\partial}{\partial \alpha} + \frac{\partial \beta}{\partial x} \frac {\partial }{\partial \beta}$$

to $$\frac{\partial u}{\partial x}$$.

Last edited: Jul 24, 2008
4. Jul 24, 2008

### physmurf

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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