Prove a satisfaction with the wave equation

[MKNA]

Homework Statement

i want to prove that the functions u(r,t)=(1/r)f(r-v*t) and u(r,t)=(1/r)f(r+v*t) satisfy the wave equation in spherical coordinates, i have tried a lot to solve it but in each time i would face a problem.

Homework Equations

wave equation : grad^2(u)=(1/v)*(partial ^2 u/partial t ^2)

The Attempt at a Solution

i have tried to solve it in different ways but it does not work with me.[/B]

 

[MKNA]

it is v^2 not v

 

[blue_leaf77]

What did you get after evaluating $\nabla^2u$?
Since $u$ is not a function of $\theta$ and $\phi$, the Laplacian operator will look like
$$
\frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2 \frac{\partial u}{\partial r} \right)
$$

 

[MKNA]

What did you get after evaluating $\nabla^2u$?

i am rely sorry ,i am a new member here and i don't know how to write the symbols correctly ...((i found grad ^2 by saying that it equals the second partial derivative for u with respect to r)) .and i found the second partial derivative for u with respect to time. finally i could not match between them , usually when i evaluate the last i would find the first in it and so we can substitute here and reach to our goal,but it didn't work this time ,the question wants a general solution .a have solved such a question when u=sin(x-v*t) it was easy .
thank you so much ^-^

 

[blue_leaf77]

((i found grad ^2 by saying that it equals the second partial derivative for u with respect to r)) .and i found the second partial derivative for u with respect to time. finally i could not match between them

Your wave equation is correct, taking into account the correction you gave in post #2. If you can't match the left and right hand side then you must be doing something wrong. Since you said that you have calculated $\partial^2 u/\partial t^2$, can you show what you got here?

 

[MKNA]

Your wave equation is correct, taking into account the correction you gave in post #2. If you can't match the left and right hand side then you must be doing something wrong. Since you said that you have calculated $\partial^2 u/\partial t^2$, can you show what you got here?

it goes like :
∂u/∂t=(1/r)(∂f/∂t)+(0)*f
=(1/r)(∂f/∂t)(-v)
∂²u/∂t²=(v²/r)*(∂²f/∂t²)

 

[blue_leaf77]

∂u/∂t=(1/r)(∂f/∂t)+(0)*f
=(1/r)(∂f/∂t)(-v)
∂²u/∂t²=(v²/r)*(∂²f/∂t²)

That doesn't seem to be quite correct, especially the way you arrived at the second line. From the first line you have $\frac{1}{r}\frac{\partial f}{\partial t}$. To do the partial derivative w.r.t $t$, you would have to use the chain rule, upon which it will be
$$
\frac{1}{r}\frac{\partial f}{\partial t} = \frac{1}{r}\frac{\partial f}{\partial (r-vt)} \frac{\partial (r-vt)}{\partial t} = \frac{-v}{r}f'(r-vt)
$$
where $f'(r-vt) = \frac{\partial f}{\partial (r-vt)}$. Calculating $\partial^2 u/\partial t^2$, what did you get?

 

[MKNA]

That doesn't seem to be quite correct, especially the way you arrived at the second line. From the first line you have $\frac{1}{r}\frac{\partial f}{\partial t}$. To do the partial derivative w.r.t $t$, you would have to use the chain rule, upon which it will be
$$
\frac{1}{r}\frac{\partial f}{\partial t} = \frac{1}{r}\frac{\partial f}{\partial (r-vt)} \frac{\partial (r-vt)}{\partial t} = \frac{-v}{r}f'(r-vt)
$$
where $f'(r-vt) = \frac{\partial f}{\partial (r-vt)}$. Calculating $\partial^2 u/\partial t^2$, what did you get?

i went through the first solution that you gave me,Since u is not a function of θ and ϕ as you have said ,and it is going real good ,i guess this is the best way ,i got two equation and i just have to substitute one in the other..thank you so much bro.