Solution to the Initial Value Problem for the Wave Equation with Dirichlet Boundary Conditions
Explanation
$$ \begin{cases} u_{tt} = c^2 u_{xx} \\ u(0,x) = f(x) \\ u_{t}(0,x) = g(x) \\ \end{cases} $$
The equation above is in a $1$-dimensional space with a length of $l$, under the Dirichlet boundary condition from the wave equation.
$$ \begin{cases} u(t,0) = \alpha (t) \\ u(t,l) = \beta (t) \end{cases} $$
It’s given when $\alpha = \beta = 0$ and there’s an initial condition regarding the waveform. Among these types of problems, it’s the simplest and easiest form. Here, $t$ represents time, $x$ represents position, and $u(t,x)$ represents the waveform at position $x$ when the time is $t$. $f$ and $g$ are initial conditions, especially $f$ represents the waveform when it’s $t=0$.
When boundary conditions are given, d’Alembert’s formula becomes unusable, similar ideas as when solving the heat equation are applied.
Solution
Step 1.
Suppose the solution is represented as $u(t,x) = w(t) v(X)$, to solve the wave equation, organizing $w’’(t) v(x) = c^2 w(t) v ''(x)$ neatly results in:
$$ {{w’’(t)} \over {w(t) } } v(x) = c^2 {{v ''(x)} \over {v(x)}} = \lambda $$
Where
$$ {{\partial } \over { \partial x }} \lambda = {{\partial } \over { \partial x }} \left( {{w’’(t)} \over {w(t) } } \right) = 0 $$
And
$$ {{\partial } \over { \partial t }} \lambda = {{\partial } \over { \partial x }} \left( c^2 {{v ''(x)} \over {v(x) } } \right) = 0 $$
Therefore, $\lambda$ is a constant.
Step 2.
Since $\lambda$ being a constant is assured, solving the second-order differential equations $w’’ - \lambda w = 0$ and $\displaystyle v '' - {{\lambda} \over {c^2}} v = 0$ individually should do. The solutions will be non-trivial when $\lambda <0$ because the sign in front of $\lambda$ is different compared to when solving the heat equation.
The solution for $\displaystyle \omega := {{ n \pi c} \over {l}}$ is represented as above. Especially, the fundamental solutions for solving the equations are $\displaystyle u_{n}(t,x) = \cos {{n \pi c t} \over {l}} \sin {{ n \pi x} \over {l}}$ and $\displaystyle \tilde{u} _{n}(t,x) = \sin {{n \pi c t} \over {l}} \sin {{ n \pi x} \over {l}}$. Therefore, the solution for a certain $b_{n}, d_{n}$
$$ u(t,x) = \sum_{n = 1}^{\infty} \left[ b_{n } \cos {{n \pi c t} \over {l}} \sin {{ n \pi x} \over {l}} + d_{n} \sin {{n \pi c t} \over {l}} \sin {{ n \pi x} \over {l}} \right] $$
is represented as above.
Step 3. Fourier Expansion for Initial Conditions
$$ u(0,x) = \sum_{n=1}^{\infty} \left[ b_{n } \sin {{ n \pi x} \over {l}} \right] = f(x) $$
Thus,
$$ b_{n} = \left< f(x) , \sin {{n \pi x } \over {l}} \right> = {{2} \over {l}} \int_{0}^{l} f(x) \sin {{n \pi x} \over {l}} dx $$
And,
$$ u_{t}(0,x) = \sum_{n=1}^{\infty} \left[ d_{n } {{n \pi c} \over {l}} \sin {{ n \pi x} \over {l}} \right] = g(x) $$
Therefore,
$$ d_{n} = {{l} \over {n \pi c}} \left< g(x) , \sin {{n \pi x } \over {l}} \right> = {{2} \over { n \pi c }} \int_{0}^{l} g(x) \sin {{n \pi x} \over {l}} dx $$
It turns out like this.
$$ u(t,x) = \sum_{n=1}^{\infty} \left[ {{2} \over {l}} \int_{0}^{l} f(x) \sin {{n \pi x} \over {l}} dx \cos {{n \pi c t} \over {l}} \sin {{ n \pi x} \over {l}} + {{2} \over { n \pi c }} \int_{0}^{l} g(x) \sin {{n \pi x} \over {l}} dx \sin {{n \pi c t} \over {l}} \sin {{ n \pi x} \over {l}} \right] $$
■