How Does the Wave Equation Derive and What Solutions Exist?

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The wave equation, represented as $$\frac{\partial^2}{\partial t^2}u(x,t)=c^2\Delta u(\vec{x},t)$$, requires specific initial and boundary conditions for unique solutions, such as fixed endpoints and initial displacement and velocity. The uniqueness of solutions can be demonstrated through methods like separation of variables, d'Alembert's solution, or Fourier series, all of which yield consistent results. The discussion highlights that the one-dimensional case simplifies understanding, as sine and cosine functions naturally satisfy the equation. The boundary conditions play a crucial role in defining the solution, and periodic extension of functions can help meet these conditions. Overall, various mathematical approaches provide insights into solving the wave equation effectively.
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$$\frac{\partial^2}{\partial t^2}u(x,t)=c^2\Delta u(\vec{x},t)\qquad \vec{x}\in \mathbb{R}^n$$
is known as the wave equation. It seems not very trivial, so is there any derivations or inspirations of it?

To solve this equation, we have to know the initial value and boundary conditions:
\begin{equation*}
\begin{cases}
u(0,t)=u(\vec{l},t)=0\\
u(\vec{x},0)=f(\vec{x})\\
u_t(\vec{x},0)=g(\vec{x})\\
\end{cases}
\end{equation*}
This above can be solved uniquely, with separation of variables.
And also see these conditions:
\begin{equation*}
\begin{cases}
u(\vec{x},0)=f(\vec{x})\\
u_t(\vec{x},0)=g(\vec{x})\\
\end{cases}
\end{equation*}
Why this above can be also solved uniquely with d'Alembert or Kirchhoff's method? Why the boundary conditions can be removed easily? So it seems that it has no influence?
 
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Small bugs said:
$$\frac{\partial^2}{\partial t^2}u(x,t)=c^2\Delta u(\vec{x},t)\qquad \vec{x}\in \mathbb{R}^n$$
is known as the wave equation. It seems not very trivial, so is there any derivations or inspirations of it?

Obviously, there are. Have you tried looking for them?

Google "wave equation"
 
If you look at the one dimensional version it become more transparent that sine or cosine of the appropriate arguments satisfy the equation.
 
mathman said:
If you look at the one dimensional version it become more transparent that sine or cosine of the appropriate arguments satisfy the equation.
Im just wondering why the different initial conditions as i posted give both an unique solution?
 
They first of the three is telling you that the ends are fixed, so it is relevant.
 
To make the solution unique you need all three conditions (the first is called a boundary and the 2nd and 3rd initial contitions). The (1+1)D case is indeed most simple. First you should show that the general solution of the wave equation in this case reads
$$u(t,x)=u_1(x-c t)+u_2(x+c t),$$
where ##u_1## and ##u_2## are arbitrary functions that are at least twice differentiable.

Hint: Introduce the new independent variables ##\xi=x-c t## and ##\eta=x+c t## and show that the wave equation is equivalent to
$$\frac{\partial^2}{\partial \xi \partial \eta} u=0.$$

Now think about the boundary conditions and how to work in the initial conditions.

Hint: You should start with the initial conditions, plugging in the above given general solution. What conclusions can you draw from them on the definition of the functions on the interval ##[0,L]##? Then you should think about how to periodically continue the function to the entire real axis to fulfill also the boundary conditions.

That's called the d'Alembertian approach. Another very illuminating way is to use Fourier series, starting with the boundary condition, then using the wave equation to constrain the coefficients and finally use the initial conditions to fully determine them. Of course, both ways lead to the same result!
 
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