The stability problems associated with explicit Euler methods are best illustrated by an example in Matlab.
Remember we are solving
$$U_m^{n+1} = U_m^n + \mu\Bigl(U_{m+1}^n – 2U_m^n + U_{m-1}^n \Bigr)$$
With boundary conditions which we can illustrate again as we did in the explicit method derivation
$$
\begin{array}{r|rr|l}
t&⩩⩩⩩⩩&\quad⩩⩩⩩⩩& \\
&⩩⩩⩩⩩&H \quad⩩⩩⩩⩩& \\
&⩩⩩⩩⩩&O\quad⩩⩩⩩⩩& \\
&⩩⩩⩩⩩&T\quad⩩⩩⩩⩩& \\
&⩩⩩⩩⩩&\;⩩⩩⩩⩩& \\
u=0 &⩩⩩⩩⩩&C\quad⩩⩩⩩⩩& u=0 \\
\text{Cold} &⩩⩩⩩⩩&O\quad⩩⩩⩩⩩&\text{Cold}\\
\text{sleeve} &⩩⩩⩩⩩&R\quad⩩⩩⩩⩩&\text{sleeve}\\
&⩩⩩⩩⩩&E\quad⩩⩩⩩⩩&\\
\hline
\textbf{0}&\leftarrow h\rightarrow & | \leftarrow h\rightarrow \mathbf{1} & \quad{x}\\
\end{array}
$$
Initial and Boundary Conditions
We are are solving the partial differential equation
$$\dfrac{\partial{u}}{\partial{t}} = \dfrac{\partial^2{u}}{\partial{x^2}}$$
on a certain domain
\begin{align}
1\leqslant & m \leqslant M-1\\
n &\geqslant 0\\
U_o^n &= U_M^n = 0
\end{align}
Matlab Exercise MM14_2016
Starting with $\mu = 0.7$ we see that $T(a,b)$ has two eigenvalues $>1$ in modulus i.e outside the range $[-1,1]$. It is these eigenvalues that are responsible for the exponential instability.
Plotting the eigenvlues of a large matrix $T$, say $M=100$ we get a \sine curve. Software uses the Power Method to obtain the eignevalues as opposed to computing the roots of a polynomial.
Proving $\lambda_k$, the eigenvalues of $T(a,b)$
The eigenvalues satisfy the following equation. The proof is in the notes (201710031653 pg 53) but not required in an exam. But you must be able to use the result:
$$\lambda_k = 1 – 2\mu + 2\mu \cos \frac{\pi k}{M} $$
where $k = 1,2,\cdots ,M-1$. We can generalise this for any TST matrix
\begin{align}
T (a,b)&=
\left(\begin{array}{ccccc} a & b & & \\ b & a & b \\ & \ddots & \ddots & \ddots & \\ & & b & a & b \\ & & & & a\end{array}\right)
\quad \in \mathbb{R}^{(M-1) \times (M -1)}\\ \\
\end{align}
$$\lambda_k = a + 2b \cos \frac{\pi k}{n + 1}\; , \qquad k = 1,2,\cdots ,n $$
This result would be given in an exam if required. You are expected to be able to use the result.
Eigenvectors of $T(a,b)$
$$\mathbf{V}^{(k)} = V_l^{(k)} = \sin \frac{\pi kl}{M} $$
where $k = 1,2,\cdots ,M-1$. Its generalised form being
$$\mathbf{V}^{(k)} \in \mathbb{R}^{n}, \quad V_l^{(k)} = \sin \frac{\pi kl}{n+1} $$
Notice that the $a$ and $b$ of the TST diagonals appear in the eigenvalues but not in the eigenvectors. That is because all the eigenvalues have the same eigenvectors.
Theorem: All $n \times n$ TST matrices have the same eigenvectors
Observe that
\begin{align}
T (a,b)&=
\left(\begin{array}{ccccc} a & b & & \\ b & a & b \\ & \ddots & \ddots & \ddots & \\ & & b & a & b \\ & & & & a\end{array}\right)
\quad \in \mathbb{R}^{n \times n}
= \left(\begin{array}{ccccc} a & & & \\ & a & \\ & & \ddots & & \\ & & & a & \\ & & & & a\end{array}\right)
+
\left(\begin{array}{ccccc} 0 & b & & \\ b & 0 & b \\ & \ddots & \ddots & \ddots & \\ & & b & 0 & b \\ & & & & 0\end{array}\right) \\ \\
&= a \mathbf{I} + b \mathbf{T}(0,1)
\end{align}
$\lambda$ values that correspond to eigenvalues [-1,1] – EXAM
Notice that when $k$ is close to $1$, $\dfrac{\pi k}{M}$ is small and close to zero. Hence $\cos\dfrac{\pi k}{M}$ is close to $1$. Remember that \cosine oscillates between $-\pi (-1)$, $0(1)$ and $\pi(-1)$ and outside this range are just treated as multiples of $\pi$. As a result $\lambda_k$, with $\cos\dfrac{\pi k}{M}$ close to $1$,
$$\lambda_k = 1 – 2\mu + 2\mu \cos \frac{\pi k}{M} = 1 $$
Clearly this gives the upper range $1$ of $\lambda$.
When $k$ is large e.g $M-1$, $\dfrac{M-1}{M}$ is close to $1$, such that $\cos\dfrac{\pi k}{M}$ is close to $\cos\pi$ which is $-1$.
$$\lambda_k = 1 – 2\mu + 2\mu \cos \dfrac{\pi k}{M} = 1 – 2\mu – 2\mu = 1 – 4\mu $$
Clearly this gives the lower range $ 1 – 4\mu$ of $\lambda$.
Hence,
$$ 1 – 4\mu \leqslant \lambda \leqslant 1 $$
But a requirement of ALL the eigenvalues is that
$$\lambda_k \leqslant | 1 | $$
for stability. Less that $1$ in modulus!
This means that
$$ 1 – 4\mu \geqslant -1$$
for stability of this method. In other words, the interval $ [1 – 4\mu, 1]$ is contained in the interval $[-1,1]$
$$ [1 – 4\mu, 1 ]\subseteq [-1,1]$$
We can obtain a value for $\mu$ as follows
\begin{align}
1 – 4\mu & \geqslant -1\\
-4\mu & \geqslant -2\\
\mu & \leqslant \dfrac{1}{2}
\end{align}
To ensure explicit Euler is spectrally stable but without any indication as to whether it is a good or bad method.