1. ## Partial Derivatives....wave equation

Verify that u = t /(a^2 t^2 - x^2) satisfies the wave equation utt = a^2 uxx (Subscript tt and xx)

I've tried many different ways in deriving this...can't seem to get the answer! I tried the quotient rule, but it didn't work...I just tried the product rule by bringing up the denominator as (a^2 t^2 - x^2)^-1 but I get an x that I cannot get rid of!

Help would be appreciated!

2. Hello,
Originally Posted by Qt3e_M3
Verify that u = t /(a^2 t^2 - x^2) satisfies the wave equation utt = a^2 uxx (Subscript tt and xx)
Using partial fractions decomposition, $\displaystyle u(x,t)=\frac{1}{2a}\left( \frac{1}{at-x}+\frac{1}{at+x}\right)$. Now differentiate $\displaystyle u$ twice with respect to $\displaystyle t$, differentiate $\displaystyle u$ twice with respect to $\displaystyle x$ and you're done. If you keep the derivatives under the form $\displaystyle \text{constant}\times \left( \text{a fraction}+\text{another fraction}\right)$ it should be easier than the methods you've tried ; you simply need to know that the derivative of $\displaystyle \frac{1}{f}$ is $\displaystyle -\frac{f'}{f^2}$.

3. Originally Posted by flyingsquirrel
Hello,

Using partial fractions decomposition, $\displaystyle u(x,t)=\frac{1}{2a}\left( \frac{1}{at-x}+\frac{1}{at+x}\right)$. Now differentiate $\displaystyle u$ twice with respect to $\displaystyle t$, differentiate $\displaystyle u$ twice with respect to $\displaystyle x$ and you're done. If you keep the derivatives under the form $\displaystyle \text{constant}\times \left( \text{a fraction}+\text{another fraction}\right)$ it should be easier than the methods you've tried ; you simply need to know that the derivative of $\displaystyle \frac{1}{f}$ is $\displaystyle -\frac{f'}{f^2}$.
Where did the 1/2a come from?

I realise that your u(x,t) is equivalent to my u...I'm just not sure how you transformed it into that

Because $\displaystyle \frac{1}{at - x} + \frac{1}{at+x} = \frac{2at}{a^2t^2 - x^2}$.
There is a factor of $\displaystyle 2a$, you need to cancel it.