Is D(x,t) = ln(ax+bt) a solution to the wave function?

[Jamie_Pi]

Homework Statement

Show that the displacement D(x,t) = ln(ax+bt), where a and b are constants, is a solution to the wave function.

Homework Equations

I'm not sure which one to use:
D(x,t) = Asin(kx+ωt+φ)
∂²D/∂t² = v²⋅∂²D/∂x²

The Attempt at a Solution

I'm completely lost on where to start with this one. I'm not sure how it's even possible, considering a wave is supposed to be an oscillating function and this solution would not give an oscillating displacement. Any help would be greatly appreciated! Thank you!

 

[kuruman]

You are thinking of harmonic waves. Here you are asked to show that D(x,t) = ln(ax+bt) is a solution to the wave equation. Just take the derivatives, substitute in the wave equation and forget Asin(kx+ωt+φ). The wave equation has more solutions than just sines and cosines.

 

[eys_physics]

I suppose you should start with wave equation in one dimension, i.e.
$$\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}$$,
and show that the equation is satisfied for $u(x,t)=D(x,t)$.

 

[Jamie_Pi]

Ok, I understand. This is what I have so far:

ln(ax+bt) = D(x,t)

∂²D/∂t² = b²2/(ax+bt)

∂²D/∂x² = a²/(ax+bt)

b²/(ax+bt) = v² a²/(ax+bt)
This is true as long as v = b²/a²

Do you think that this is enough for an answer?

 

[haruspex]

Ok, I understand. This is what I have so far:

ln(ax+bt) = D(x,t)

∂²D/∂t² = b²2/(ax+bt)

∂²D/∂x² = a²/(ax+bt)

b²/(ax+bt) = v² a²/(ax+bt)
This is true as long as v = b²/a²

Do you think that this is enough for an answer?

It is the right method, but your second derivatives are incorrect. Take it one step at a time.

 

[Jamie_Pi]

∂2D/∂t2 = -b²/(ax+bt)²

∂2D/∂x2 = -a²/(ax+bt)²

b²/(ax+bt) = v² a²/(ax+bt)

I think this is correct this time.

 

[haruspex]

∂2D/∂t2 = -b²/(ax+bt)²

∂2D/∂x2 = -a²/(ax+bt)²

b²/(ax+bt) = v² a²/(ax+bt)

I think this is correct this time.

Not quite. The second line is right, but not the third.

 

[Jamie_Pi]

Oh I see, I just wrote it in wrong. If the first two are correct, then the last one should just be:

b²/(ax+bt)² = v² a²/(ax+bt)²

where v² = b²/a²

 

[haruspex]

Oh I see, I just wrote it in wrong. If the first two are correct, then the last one should just be:

b²/(ax+bt)² = v² a²/(ax+bt)²

where v² = b²/a²

Right.

 

[kuruman]

It is generally true that any function of the form $f(ax \pm bt)$ is a solution to the wave equation.
Proof: Define $u=ax \pm bt$. Then $$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial u} \frac{\partial u}{\partial t}=\pm b\frac{\partial f}{\partial u} \\ \frac{\partial^2 f}{\partial t^2}=+b^2\frac{\partial^2 f}{\partial u^2}$$Similarly, $$\frac{\partial^2 f}{\partial x^2}=a^2\frac{\partial^2 f}{\partial u^2}$$Substitute in the wave equation to get $$b^2\frac{\partial^2 f}{\partial u^2}=v^2a^2\frac{\partial^2 f}{\partial u^2}~\Rightarrow~b^2=v^2a^2.$$ Thus, the function is a solution to the wave equation if the speed of propagation is identified as $v=\pm (b/a)$. The top ($+$) sign is used for waves propagating in the decreasing x-direction as time evolves and the bottom ($-$) is used for waves propagating in the increasing x-direction.