# Wave equation problem

1. May 1, 2010

### fluidistic

1. The problem statement, all variables and given/known data
Show that the function $$u(x,y,z,t)=f(\alpha x + \beta y + \gamma z \mp vt)$$ where $$\alpha ^2 + \beta ^2 + \gamma ^2 =1$$ satisfies the tridimensional wave equation if one assume that f is differentiable twice.

2. Relevant equations
$$\frac{\partial ^2 u}{\partial t ^2}-c^2 \triangle u=0$$.

3. The attempt at a solution
I'm not sure how to use the chain rule.
$$\frac{\partial u}{\partial t}=\frac{\partial f}{\partial t}=\mp v \frac{\partial f}{\partial \sigma}$$ where $$\sigma =vt$$. Thus $$\frac{\partial ^2 u}{\partial t ^2}=v^2 \left ( \frac{\partial f}{\partial \sigma} \right ) ^2$$.
I'm 98% sure it's not right.
Am I approaching well the problem?

2. May 1, 2010

You're not too far off. The chain rule tells you that $\frac{\partial}{\partial t}f(q)=f'(q)\frac{\partial q}{\partial t}[/tex], so.... 3. May 2, 2010 ### fluidistic Thank you for your help. My confusion resides in "$$f'(q)$$". Is it equal to $$\frac{\partial f}{\partial \sigma}$$ where $$\sigma = \mp vt$$ in the exercise? 4. May 2, 2010 ### fluidistic Second attempt: $$\frac{\partial u}{\partial t}=\mp v u '$$; $$\frac{\partial ^2 u}{\partial t^2}=v^2 u'$$. $$\nabla u = \frac{\partial u}{\partial x} \hat i + \frac{\partial u}{\partial y} \hat j + \frac{\partial u}{\partial z} \hat k = u'(\alpha \hat i + \beta \hat j + \gamma \hat k)$$ thus $$\triangle u = u''(\alpha ^2 + \beta ^2 + \gamma ^2)=u''$$. Using the wave equation, it follows that the wave equation holds if and only if $$v^2=c^2$$ so $$v=\pm c$$. Does this mean that the speed of the wave given is the one of the light? What kind of wave is that? Ellipsoidal or so? Also, I didn't understand what I've done to reach this result. Each time I wrote u'. To me the u' appearing in $$\frac{\partial u}{\partial t}$$ is different from the one appearing in $$\nabla u$$. There is more than 1 variable, so writing u' is confusing to me. 5. May 2, 2010 ### gabbagabbahey No, the argument of [itex]u$ is $\alpha x+\beta y+\gamma z \mp vt$, so that's what you would set $q$ to (or $\sigma$ if you prefer).

The more general wave equation has $c$ as the speed of the wave, not necessarily the speed of light. So, I would say that $u$ satisfies the wave equation, with speed $v$.

The shape of the wave at any given time will depend on the exact functional form of $u$, as well as the constants $\alpha$, $\beta$, and $\gamma$.

$u$ has the same argument in both derivatives. Just name the variable $q=\alpha x+\beta y+\gamma z \mp vt$

6. May 2, 2010

### fluidistic

Thank you once again, it's clearer to me now. At least the argument q is much clearer and so is u'. So $$u'=\frac{\partial u}{\partial q}$$ where $$q=\alpha x+ \beta y + \gamma z \mp vt$$. Is this right?
Ok but I found out that $$v=\pm c$$. So if v is the speed of the wave, then it's equal to c, which is the speed of light or I'm missing your point?

Ah I see. Say if u is the identity and $$\alpha = \beta = \gamma$$ then it's a spherical wave?
And in the hypothetical case of u is worth the identity and $$\alpha ^2 + \beta ^2 + \gamma ^2 = 1$$ such that $$\alpha \neq \beta \neq \gamma$$, would the wave be ellipsoidal?

7. May 2, 2010

### gabbagabbahey

Yup

You're missing the point; the $c$ in your wave equation isn't necessarily the speed of light, just the speed of the wave. For electromagnetic waves, in vacuum, $c$ will be the speed of light.

Not really. For a spherical wave, you would expect a solution of the form $u=\frac{1}{r}g(r-vt)$.

8. May 2, 2010

### fluidistic

Ah I get it now! Ok.

Ah you're right, it must decay as 1/r to be a spherical wave. Not sure what happens if it decays as 1/r^2 but I get the point. Thanks for all.
Problem solved.