Separation.tex

\symbolicsubsection{Separation}\label{subsec:separation}
Conditions, when separation may be successful:

\begin{itemize}
    \item Homogeneous, linear PDE
    \item Homogeneous boundary conditions
    \item Domain must be a cartesian product (i.e. some form of rectangle when in cartesian coordinate system)
\end{itemize}

if these are given, we can try:

\begin{enumerate}
    \item Assume the structure of the solution to be in the form of some ansatz with separable variables, usually a product $u(x,y)=X(x)\cdot Y(y)$
    \item{
        Substitute $u$ in PDE with ansatz by variable:
        \begin{align*}
            \text{all terms of }x = \text{all terms of }y = {\color{red}\lambda}
        \end{align*}
        and solve the ordinary differential equations for $X(x)$ and $Y(y)$.
    }
    \item Use \emph{homogeneous} boundary conditions to determine admissible values ${\color{red}\lambda}_k$.
    \item Solve equations: $X_k(x)$ and $Y_k(y)$ $\Rightarrow u_k(x,y)=X_k(x)\cdot Y_k(y)$ using ansatz.
    \item Combine solutions: $u(x,y)=\sum_{k\in \mathbb{Z}}\left(a_ku_k(x,y)\right)$
    \item Use remaining boundary conditions to determine $a_k$.
\end{enumerate}

\paragraph{Separation Example}

Given the wave equation $\frac{\partial^2u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}$
for a string mounted between $u(0,t)=u(\pi,t)=0$ and is in the rest position at $t=0$: $u(x,0)=0$ but has initial velocity
$\frac{\partial u}{\partial t}(0,x)=\sin^3(x)=\frac{3}{4}\sin x-\frac{1}{4}\sin 3x$.
We choose the Ansatz $u(x,t)=X(x)T(t)$:

\begin{align*}
    \begin{matrix}
        X(x)T''(t)=X''(x)T(x) & |\div X(x)T(t) \\
        T''(t)\div T(t) = X''(x)\div X(x) = \mu
    \end{matrix} \\
\end{align*}
Hence, we receive the ODEs

\begin{align*}
    \mu T(t) = T''(t) \\
    \mu X(x) = X''(x)
\end{align*}
with $\mu < 0$ to receive oscillating solutions, the ODEs have the solutions

\begin{align*}
    X(x) & = C \sin(\sqrt\mu x)+\cancel{ D\cos(\sqrt\mu x) } \\
    T(t) & = A \sin(\sqrt\mu t)+B \cos(\sqrt\mu t)
\end{align*}
by taking into account the boundary conditions $X(0)=X(\pi)=0$, yielding $C=1,D=0$ and $\sqrt{n}\in\mathbb{N}$.
Therefore, we have the general solution
\begin{align*}
    u(x,t) & = \sum_{n=1}^\infty \sin(nx)\cdot\left( A_n\sin(nt)+B_n\cos(nt) \right) \\
    \ & = \sum_{n=1}^\infty A_n\sin(nx)\sin(nt) + \sum_{n=1}^\infty B_n\sin(nx)\cos(nt)
\end{align*}

considering the boundary conditions:
\begin{align*}
    u(x,0) = 0 = \sum_{n=1}^\infty B_n\sin(nx)\underbrace{\cos(0)}_{=1} \Rightarrow B_n=0
\end{align*}
and

\begin{align*}
    \ & \frac{\partial u}{\partial t} = \sum_{n=1}^\infty nA_n\sin(nx)\cos(nt)\\
    \ & \rightarrow \frac{\partial u}{\partial t}(x,0) = \sum_{n=1}^\infty nA_n\sin(nx) = \frac{3}{4}\sin x-\frac{1}{4}\sin 3x \\
    \ & \Rightarrow A_1=3/4, A_3=-1/12, A_{n\setminus \{1,3\}}=0 \\
    \ & \Rightarrow u(x,t)=(3/4)\sin(x)\sin(t)-(1/12)\sin(3x)\sin(3t)
\end{align*}