# Why should a solution to the wave equation be finite?

A function which represents a wave must satisfy the following differential equation:

$$\frac{\partial^2 y}{\partial t^2} = k\frac{\partial^2 y}{\partial x^2}$$

Any function that satisfies the wave differential equation represents a wave provided that it is finite everywhere at all times.

What does "it is finite everywhere at all times" mean?

Question:Which of the following functions represent a wave?

a) $(x - vt)^2$

b) $\ln(x + vt)$

c) $e^{-(x - vt)^2}$

d) $(x + vt)^{-1}$

Only option (c) is given as the answer though all 4 satisfy the differential equation.

I believe I did not understand the significance "function should be finite everywhere at all times" which is why I am unable to answer the aforementioned question.

• Basics of Mathjax and i like this for fast editing – Gowtham Jan 1 '15 at 7:08
• Experimental evidence: when was the last time you saw a wave on the sea with a singularity as n (b) or (d) or with unbounded growth in both directions? Less flippantly, wave equations model quantities - such as internal stresses/strains in materials - that must be finite, otherwise we'd see the results. Or, often the square of the disturbance is proportional to the wave's energy, so that all proposals other than (c) would need an infinite amount of energy to bring into being. Conservation of energy rules this out. – WetSavannaAnimal Jan 1 '15 at 7:33
• Note also that b) and c) are not solutions of the PDE you wrote. They are instead weak solutions. – Ruslan Jan 1 '15 at 15:02
• Oh, I meant d), not c) in the comment above. – Ruslan Jan 2 '15 at 12:39
• @Yashas Have you been reading DC Pandey? – Arishta Apr 21 '17 at 3:55

## 3 Answers

It's semantics. Whoever wrote the problem prefers to refer to a wave as "A function which satisfies the wave equation and which is bounded" instead of "a function which satisfies the wave equation".

Unfortunately there are bound to be conventions which you disagree with, but in academics (undergrad and lower) the only way to deal with it is to figure out which conventions the professor (or problem writer) is working with before you read the problems. It's too easy for conversations on convention to turn into, "technically, it is a wave even though it's not physical" countered with "technically, it's not a wave because it's not bounded." The best you can do is recognize an issue in terminology ASAP and deal with it in a constructive way.

A better statement, which is more objectively true, would be: "Functions like $(x-vt)^2$ solve the wave equation, but generally don't come up and are not useful in physical solutions."

• I'd hazard to guess that this is probably a big difference between an undergraduate math and physics course. The math course may say anything that solves the PDE is a wave since it solves the wave equation while physicists will tend toward bounded solutions only. – jkeuhlen Jan 2 '15 at 4:17

What nobody has mentioned so far is that the individual terms in the wave-function usually have a physical interpretation. For example, $\frac{\partial^2 y}{\partial t^2}$ represents an acceleration while $\frac{\partial^2 y}{\partial x^2}$ represents a force. Also, in many cases the amplitude of the wave is related to the energy density (or, in quantum mechanics, the probability density). The statements about finiteness, continuity of first derivatives, etc., all have analogues in terms of finiteness of energy, force, or the ability to localize the wave.

Also, those solutions may be valid on some finite domain that does not include singularities that are too severe.

Any function that satisfies the wave differential equation represents a wave provided that it is finite everywhere and at all times.

Has to be finite. For instance, take $f(x, t) = \frac{1}{x} + \frac{1}{t-1}$. It is not a wave, true, but it is just an example. It is not finite at all points and at all times, because at point $x=0$ we have $f(0, t) = \infty$ and at time $t = 1$ we have $f(x, 1) = \infty$. Since it is infinity, it is not finite.

$(a)$ fails when $x+vt\to\infty$.

$(b)$ fails at $x+vt \le 0$.

$(d)$ fails at $x+vt=0$

• As an addendum: there are a large number of "shortcut" equations you can use once you know a function represents a wave. However, functions like the one Physicist137 provided as a non-wave do not work well with such shortcuts. Sometimes they are kind and simply blow up with a divide by zero. Other times they can be very sneaky and lead you to false but valid looking results. – Cort Ammon Jan 1 '15 at 23:50
• Your function does not satisfy the wave equation. Furthermore, the question was, why does a solution need to be finite to be considered a wave? – pwf Jan 2 '15 at 0:04
• @pwf well, I know it doesn't satisfy the wave equation. I just said that in the answer. Besides, the OP is asking about "the significance "function should be finite everywhere at all times"". This can be interpreted as why does solution need to be finite (as you said). Or can be interpreted as what does mathematically means a solution to be finite. First answer of question, by NeuroFuzzy, explored the why part. Then I answered mine to explore the what part, since OP was not that clear in asking what he wanted. – Physicist137 Jan 2 '15 at 13:16