Solve Wave Equation Problem: f(\alpha x + \beta y + \gamma z \mp vt)

[fluidistic]

Homework Statement

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.

Homework Equations

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

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?

 

[gabbagabbahey]

You're not too far off. The chain rule tells you that [itex]\frac{\partial}{\partial t}f(q)=f'(q)\frac{\partial q}{\partial t}[/tex], so...

 

[fluidistic]

You're not too far off. The chain rule tells you that [itex]\frac{\partial}{\partial t}f(q)=f'(q)\frac{\partial q}{\partial t}[/tex], so...

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?

 

[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.

 

[gabbagabbahey]

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?

No, the argument of $u$ is $\alpha x+\beta y+\gamma z \mp vt$, so that's what you would set $q$ to (or $\sigma$ if you prefer).

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?

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 $v$.

What kind of wave is that? Ellipsoidal or so?

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 $\gamma$.

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.

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

 

[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?

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 $v$.

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?

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 $\gamma$.

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?

 

[gabbagabbahey]

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?

Yup

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?

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.

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?

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

 

[fluidistic]

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.

Ah I get it now! Ok.

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

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.