I try to understand the Fourier stability analysis for Leap-Frog scheme to solve linear advection equation,
$$\dfrac{\partial u}{\partial t}+a\dfrac{\partial u}{\partial x}=0.$$
Above screenshot is from Morton and Mayers second edition book (Numerical solution of partial differential equation, pp. 123-124).
The scheme of leap frog is
\begin{equation}
\boxed{
u_j^{n+1}=u_j^{n-1}-\nu \left(u_{j+1}^{n}-u_{j-1}^{n}\right),}
\end{equation}
with $\nu=\frac{a\Delta t}{\Delta x}$.
With Fourier analysis
$$u_j^n=\lambda^n e^{ikj\Delta x},$$
we have
\begin{align*}
\lambda^{n+1} e^{ikj\Delta x} = \lambda^{n-1} e^{ikj\Delta x} -\nu \left(\lambda^{n} e^{ik(j+1)\Delta x} – \lambda^{n} e^{ik(j-1)\Delta x}\right).
\end{align*}
By simplify equation, we have
\begin{align*}
\lambda = \lambda^{-1} – \nu(2i\sin(k\Delta x)).
\end{align*}
Multiplying two side by $\lambda$ we have
\begin{align}\label{PKKK}
\lambda^2+i(2\nu\sin(k\Delta x))\lambda-1=0.
\end{align}
By solving last equation, we have
$$\lambda_{1,2}=-i(\nu \sin(k\Delta x))\pm \sqrt{-\nu^2\sin^2(k\Delta x)+1}.$$
Now, I find the modulus of $\lambda_{1,2}$, i.e.
\begin{align*}
\vert\lambda_{1,2}\vert^2 &= \left( \pm \sqrt{-\nu^2\sin^2(k\Delta x)+1} \right)^2 + \left( -\nu \sin(k\Delta x) \right)^2\\
&= -\nu^2\sin^2(k\Delta x)+1+\nu^2 \sin^2(k\Delta x)\\
&= 1.
\end{align*}
So we have $\vert\lambda_1\vert=\vert\lambda_2\vert=1$.
The condition to stable is $\vert\lambda\vert\leq 1$. But I found $\vert\lambda_1\vert=\vert\lambda_2\vert=1$, so I can conclude the scheme is unconditionally stable.
But, if I read from Morton and Mayers book, the leap frog scheme is stable if $-1\leq \nu\leq 1$.
Now, I try to understand the stability analysis from Morton book.
Since the product of these roots is $−1$, we must require both roots
to have modulus 1 for the scheme to be stable. It is easy to verify
that the roots are complex and equal in modulus for all k if and only
if $|\nu| \leq 1$: so for this scheme the Fourier analysis gives the
same result as the CFL condition; and when the stability condition is
satisfied there is no damping.
I don't understand
"since the product of these roots is $−1$, we must require both roots
to have modulus 1"
, because if I try to find modulus of $\lambda$, then the value is 1.
So, I confused to prove this statement:
It is easy to verify that the roots are complex and equal in modulus for all k if and only
if $|\nu| \leq 1$
It's not easy for me because I don't understand it.
Anyone can explain that statement?
Anyone can give me hint to prove the stability of leap frog scheme is $|\nu| \leq 1$ as Morton book?
Best Answer
To get modulus $1$, you seemingly unknowingly assumed that the term under the root is positive, so that the root gives the real part of the complex expression. However, if that term is negative, $|ν\sin(kΔx)|>1$ for some $k$, then the now both imaginary terms just add. If they do not add to $\pm i$, then one eigenvalue is outside the unit circle.