Is the following equation a wave?

[Aritra Kundu]

We know that a wave is represented by f(x - vt) and it follows the differential wave equation. e^(x - vt) satisfies both the condition. But is it really a wave? Because to sustain the wave we need infinite energy which is not possible. So what's happening here?

 

[NFuller]

We know that a wave is represented by f(x - vt)

Not necessarily, if f=ln this is not a wave.

e^(x - vt) satisfies both the condition.

I assume you are referring to the complex exponential
$$e^{i(kx-\omega t)}$$.

But is it really a wave?

Yes

Because to sustain the wave we need infinite energy which is not possible

Why do you think this wave requires infinite energy?

 

[muscaria]

Because to sustain the wave we need infinite energy which is not possible.

For the wave you have written down, I would agree with you - so long as the energy density of the wave increases with the wave amplitude ;)

 

[Aritra Kundu]

Not necessarily, if f=ln this is not a wave.

I assume you are referring to the complex exponential
$$e^{i(kx-\omega t)}$$.

Yes

Why do you think this wave requires infinite energy?

No its not complex exponential.

And it requires infinite energy because at x --> infinity, the function tends to infinity. It's energy density increases with x. So it need to be supplied with energy for infinity.

 

[NFuller]

No its not complex exponential.

Then it is not a wave. This is not a solution to the wave equation.

 

[Aritra Kundu]

Then it is not a wave. This is not a solution to the wave equation.

It satisfies the wave equation. You can check, by double differentiating it with x and t.

 

[sophiecentaur]

It satisfies the wave equation.

Looking at how the function varies with x or with t it seems to describe a function that increases without limit as x increases and tends to zero as t increases. Is that what you would describe as a wave? (Just giving it a reality check)
Not all Maths has meaning in the Physical World.

 

[NFuller]

It satisfies the wave equation.

Yes, sorry about that. But it is not really a wave because what does $\omega$ signify here if there is no oscillation?
It is common practice to throw out solutions which do not satisfy boundary conditions. In this case it doesn't make sense for the solution to blow up as x goes to infinity.

 

[sophiecentaur]

what does ωω\omega signify here if there is no oscillation?

There needn't be any oscillation. A single pulse can constitute a wave when there is only one single excursion from the equilibrium condition.
But you are right to say that you always need to apply boundary conditions.

 

[NFuller]

A single pulse can constitute a wave when there is only one single excursion from the equilibrium condition.

This would consist of many values of $\omega$ via a Fourier transform of a delta function. The issue here is what does e^($\omega t$) signify being purely real?

 

[sophiecentaur]

This would consist of many values of $\omega$ via a Fourier transform of a delta function. The issue here is what does e^($\omega t$) signify being purely real?

The frequency domain is a valid way to express a wave but a single pulse involves an infinite number of frequency components (the harmonics of zero fundamental) and the pulse is not 'periodic'. The use of a particular symbol (ω) doesn't actually imply periodicity and the OP is referring to a Real Function (afaics).
Btw, the delta function need not be involved here.