I have to verify by substitution that u(x,t)=F(x+ct) + G(x-ct) satisfies the one-dimensional wave equation.

I feel really stupid. This is supposed to be easy, but how do you simplify $\displaystyle \frac{\partial u}{\partial x}$ ?

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- Oct 15th 2009, 03:38 AMharbottleWave equation
I have to verify by substitution that u(x,t)=F(x+ct) + G(x-ct) satisfies the one-dimensional wave equation.

I feel really stupid. This is supposed to be easy, but how do you simplify $\displaystyle \frac{\partial u}{\partial x}$ ? - Oct 15th 2009, 05:28 AMDeadstar
The solution is a bit long and I'm trying to find my old notes so I write it down right but it basically involves using D'Alemberts solution.

Look at this site for details and I'll post what I did soon...

D'Alembert's solution of the Wave Equation - Oct 15th 2009, 05:35 AMDeadstar
Actually the site has it the same way as my old notes so yeah not much point in me typing it all up! Make sure you understand all the steps especially the... "This equation is much simpler and can be solved by direct integration. First of all integrate with respect to $\displaystyle \xi$ to give..." part

- Oct 15th 2009, 05:35 AMharbottle
Thanks for the quick response, but I don't actually need help finding the worked solution (I have it here in front of me). Mine is a much simpler question, and more embarrassing.

It is to substitute the solution into the DE and prove that it works... - Oct 15th 2009, 05:43 AMDeadstar
Is your original formula right? Should it not be F(x + ct)?

If it is F(x + ct) then $\displaystyle \frac{\partial u}{\partial x} = F'(x + ct) + G'(x - ct)$. Its that simple. You don't have to worry about what the function actually is, just differentiate F to get F' and multiply by the derivative of whats in the brackets, in this case the derivative of x which is just 1.

So $\displaystyle \frac{\partial u}{\partial t} = cF'(x + ct) - cG'(x - ct)$ since the derivative of ct with respect to t is c.

Ya following me? - Oct 15th 2009, 05:47 AMharbottle
That's what I tried to do first; I don't know why I stopped. I was looking for something deeper, I guess.

Thanks! - Oct 15th 2009, 05:52 AMHallsofIvy
Well, first, I suspect that you have written the problem wrong. While what you give

**does**satisfy the problem, It could be more easily written as u(x,t)= H(x-ct) wher H(x)= F(x)+ G(x). There is no reason to have two functions or x- ct.

I suspect you meant that u(x,y)= F(x- ct)+ G(x+ ct). That is the general solution.

$\displaystyle \frac{\partial u}{\partial x}= \frac{\partial F}{\partial x}+ \frac{\partial G}{\partial x}$, of course.

So you can focus on F and G separately, then add. Let y= x- ct. Then F(x-ct)= F(y) and $\displaystyle \frac{\partial F}{\partial x}= \frac{dF}{dy}\frac{\partial y}{\partial x}$. I hope you know that $\displaystyle \frac{\partial y}{\partial x}= \frac{\partial (x-ct)}{\partial x}= 1$. F is now a function of the single variable y. Since F can be**any**differentiable function of a single variable, the best we can do is write [tex]\frac{dF}{dy}= F'(y)[/itex]. $\displaystyle \frac{\partial F}{\partial x}= F'(y)(1)= F'(x- ct)$

You can do the same thing with G and get a similar result.

More interesting is the derivtive with respect to t. Again, write y= x- ct. Then $\displaystyle \frac{\partial F}{\partial t}= \frac{dF}{dy}\frac{\partial y}{\partial t}$. We still have, of course, $\displaystyle \frac{dF}{dy}= F'(y)$ and now $\displaystyle \frac{\partial y}{\partial t}= \frac{\partial (x- ct)}{\partial t}= -c$. [tex]\frac{\partial F}{\partial t}= F'(y)(-c)= -cF'(x-ct)

Do the same thing with G (be careful with the sign).