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\lhead{Technische Universität Berlin\\Numerische Mathematik I - WS 2019/2020}
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%{\LARGE Numerische Mathematik}\\[1.5ex]
%%{\large \CurrentWeekNb. \"Ubungsblatt}\\ 
{\Large Lösungsskizzen für die \CurrentWeekNb. Tutorien}\\ 
{Woche 25.11-29.11.19}\\
%\\{Diese Aufgaben müßen spätestens am 24.10.19 in der Vorlesung abgegeben werden.}
%{Abgabe:} Spätestens am 24.10.19 in der Vorlesung\\[0.5cm]
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\paragraph{Solution \arabic{taskNb}} \stepcounter{taskNb}
Without lost of generality we consider the case of homogeneous boundary conditions. (The Dirichlet boundary value problem can be shifted if the BC are not homogeneous).

Discretized system:
\[
U_{i+1} = U_{i} + \Delta t D^2 U_i
\]
where 
\[
D^2=\frac{1}{\Delta x^2}\begin{bmatrix}
-2 &1 \\
1&-2&1\\
&\ddots&\ddots&\ddots \\
&&1& -2&1\\
&&&1& -2
\end{bmatrix} \in \mathbb{R}^{n_x \times n_x},
\]
$U_i \in\mathbb{R}^{n_x}$ for all $i=0,1,2,\ldots$.

Initial condition:
\[
U_0 = [S(x_1); \ldots; S(x_n{_x})].
\]

The numerical system is stable (solution is bounded for all $s$ and all $t$) if and only if \textit{all} eigenvalues $\lambda$ of $M = I_{n_x} + \Delta t D^2$ lie in the unit circle. Indeed $D^2, I$ and thus also $M$ are symmetric matrices, so they are diagonalizable and have only semi-simple eigenvalues. The numerical system is asymptotically stable if and only if the eigenvalues lie strictly inside the unit circle (not on the boundary). $\rightarrow$ see also 1st homework sheet.

Clearly, the eigenvalues $\lambda$ of $M$ and $\mu$ of $D^2$ are connected through
\[
\lambda-1=\Delta t\mu.
\]  
Therefore we need $|\lambda|=|(\Delta t \mu+1)| \leq 1$. This is satisfied for all eigenvalues if and only if $\Delta t \mu\geq -2$ since all $\mu$ are negative.

We know that $4\leq-\mu\leq 4/ \Delta x^2$. (The upper bound gets always tighter as $\Delta x^2$ tends to 0).
Thus a sufficient condition for stability is that $\Delta t \leq 2/ |\mu_{max}|\leq 2\Delta x^2/4=\Delta x^2/2$.

This means that time step should be adapted to the spatial mesh. Using big time steps on a fine mesh can lead on the long run to an exploding solution. This would be a very bad approximation of the stationary state (= equilibrium point) $\hat{y}(x) = (1-x) (a-b)+b$ ($ \equiv 0$ in the case of homogeneous BC) to which tends the solution $y(t,x)$ of the PDE.
 
Note that the eigenvalues of $D^2$ can be easily computed exactly, but this is the topic of this week's homeworks. So it is on purpose that we did not write them explicitly here. 

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