Why solutions to the wave equation with constant frequency in time have exponential form? I am reading about the solutions to the wave equation through a separation of variables (time and position): $H(x,t)= A(x)B(t)$.
But since the solutions we are seeking are solutions with constant frequency then the temporal part takes the form $B(t)=e^{-i\omega t}$.
My doubt is why the temporal part must have this exponential form? is there a way to prove this?
 A: Frequency is the rate of change of phase with respect to time. As an operator, it's:
$$ \hat\omega = -i\frac{\partial}{\partial t} $$
so that:
$$ \hat\omega B(t)  =  -i\frac{\partial}{\partial t}e^{i\omega t} = -i^2\omega e^{i\omega t} = \omega B(t) $$
is the definition of "constant frequency".
A: The wave equation is
$$ \frac{\partial^2 H}{\partial t^2} = v^2 \frac{\partial^2 H}{\partial x^2},$$
and you are assuming $H(x,t) = A(x) B(t)$.
If you plug this in,
$$ \frac{\partial^2 A(x) B(t)}{\partial t^2} = v^2 \frac{\partial^2 A(x) B(t)}{\partial x^2}\\
A(x) \frac{\partial^2 B(t)}{\partial t^2} = v^2 B(t) \frac{\partial^2 A(x) }{\partial x^2}\\
 \frac{1}{B(t) }\frac{\partial^2 B(t)}{\partial t^2} = v^2 \frac{1}{A(x)} \frac{\partial^2 A(x) }{\partial x^2}.$$
Since the left side is a function of only $t$ and the right side is a function of only $x$, the two sides can only be equal if they equal the same constant. Using $-n^2$ to denote a constant,
$$
\frac{1}{A(x)} \frac{\partial^2 A(x) }{\partial x^2} = - n^2\\
\frac{1}{B(t)}\frac{\partial^2 B(t)}{\partial t^2} = -v^2 n^2.
$$
The second equation can be re-arranged into:
$$\frac{\partial^2 B(t)}{\partial t^2} = -v^2 n^2 B(t). 
$$
If you know about eigenvalues, you will see that the solution $B(t)$ must be an eigenfunction of the second derivative with eigenvalue $-v^2 n^2$. Alternatively, you may know that the derivative of $e^{at}$ is $ae^{at}$ and the second derivative is therefore $a^2e^{at}$, i.e., $e^{at}$ is an eigenfunction of $\frac{\partial^2 }{\partial t^2}$ with eigenvalue $a^2$. So if $a^2 = -v^2 n^2$, $a = \pm iv n$, and the solution is $e^{\pm iv nt}$.
A: "proof" comes from understanding the existence and uniqueness theorems of differential equations.  The answers so far have just spat out the steps found in most texts.  First one should state that is does not have to have the exponential form you have provided, it could also have $ e^{i\omega t}$.  For that matter one could have $sin(\omega t)$ or $cos(\omega t)$.  The idea is to assume a solution and if it works then it is the solution.  Since the wave equation is a second order equation one should expect two independent solutions, hence the + and - sign for each complex time solution, or $sin$ and $cos$.  In a sense all one needs to do to prove this is show that trying it does not lead to a contradiction and it must be the solution.  I'd review an undergraduate text of ordinary and partial differential equations for more on this.  A better question might be, how did anyone guess to try such a solution, as opposed to say $t^8$ or something.  In fact one can assume an infinite series as a solution and derive a set of relationships that determine the constants in the series.  After some effort one will arrive at your proposed solution, or some equivalent involving trig functions.  At this point in history we are so accustomed to jumping to the exponential form we hardly think of the other paths.  For linear equations the exponential form is very useful as its derivatives are all proportional to the original function.  If you are dealing with a diff eq with constant coefficients you cannot go wrong with an exponential, $e^{\alpha t}$, as a starting point.  This will reduce the diff eq to an algebraic equation that determines the allowed values of $\alpha$.  This is all standard procedure and is related to Laplace transform or Fourier transform.
