Aliyah Thompson

2022-11-04

Stability of Finite Difference schemes.
I was reading a book ''Numerical partial differential equation: Finite difference methods'' by J. W. Thomas. Here at page no. 74, definition of stability is given as
$‖{u}^{n+1}‖\le K{e}^{\beta t}‖{u}^{0}‖$ for $0\le \mathrm{\Delta }x\le \mathrm{\Delta }{x}_{0}$ and $0\le \mathrm{\Delta }t\le \mathrm{\Delta }{t}_{0}$. i.e. this definition allows for exponential growth. however in a few literature i found the definition that ''A finite distinction approximation is solid if the errors (truncation, spherical-off and so forth) decay as the computation proceeds from one marching step to the next .'' The ebook which i've noted, at the equal page in remark 3, creator has written that the above definition with exponential increase is trendy definition and later one may be derived from this.but i am not getting the way to locate this definition. So my question is which definition is true ? And if different one is not proper, then what is motive at the back of ? and the way can we examine those two definition?

lavarcar2d2

To address the concern of your stability question, one often has to enforce some sort of constrain on the time step. Here is an example.
, with
for $x\in \left[-1,1\right]$ with $u\left(x+2,t\right)=u\left(x,t\right)$ for all x and t.
Applying the explicit difference method then,
$U\left(x,t+\mathrm{\Delta }t\right)=U\left(x,t\right)-\frac{a\mathrm{\Delta }t}{2\mathrm{\Delta }x}\left(T-{T}^{-1}\right)U\left(x,t\right)+\frac{b\mathrm{\Delta }t}{\mathrm{\Delta }{x}^{2}}\left(T-2+{T}^{-1}\right)U\left(x,t\right)$
Recall the following Discrete Fourier Transform rules:

$\stackrel{^}{TU}\left(\xi \right)=\stackrel{^}{U}\left(\xi \right){e}^{i\xi \mathrm{\Delta }x}$
Applying the previous rules, then our difference method equation becomes:
$\stackrel{^}{U}\left(\xi ,t+\mathrm{\Delta }t\right)=\stackrel{^}{U}\left(\xi ,t\right)-\frac{a\mathrm{\Delta }t}{2\mathrm{\Delta }x}\underset{2isin\left(\xi \mathrm{\Delta }x/2\right)}{\underset{⏟}{\left({e}^{i\xi \mathrm{\Delta }x}-{e}^{-i\xi \mathrm{\Delta }x}\right)}}\stackrel{^}{U}+\frac{b\mathrm{\Delta }t}{\mathrm{\Delta }{x}^{2}}\underset{{\left({e}^{i\xi \mathrm{\Delta }x}-{e}^{-i\xi \mathrm{\Delta }x}\right)}^{2}={\left(2isin\left(\xi \mathrm{\Delta }x/2\right)\right)}^{2}}{\underset{⏟}{\left({e}^{i\xi \mathrm{\Delta }x}-2+{e}^{-i\xi \mathrm{\Delta }x}\right)}}\stackrel{^}{U}$
With

And after some algebra,
$\stackrel{^}{U}\left(\xi ,t+\mathrm{\Delta }t\right)=\left[1-\frac{a\mathrm{\Delta }t}{\mathrm{\Delta }x}i\mathrm{sin}\left(\xi \mathrm{\Delta }/2\right)-\frac{b\mathrm{\Delta }t}{\mathrm{\Delta }{x}^{2}}4{\mathrm{sin}}^{2}\left(\xi \mathrm{\Delta }/2\right)\right]\stackrel{^}{U}\left(\xi ,t\right)$
Here is the Stability part
Require $‖\stackrel{^}{U}\left(t+\mathrm{\Delta }t\right){‖}_{2}\le ‖\stackrel{^}{U}\left(t\right){‖}_{2}$ to avoid blowing up (Stability) and recall that the Fourier Transform process is norm-preserving,
then
${‖1-\frac{a\mathrm{\Delta }t}{\mathrm{\Delta }x}i\mathrm{sin}\left(\xi \mathrm{\Delta }/2\right)-\frac{b\mathrm{\Delta }t}{\mathrm{\Delta }{x}^{2}}4{\mathrm{sin}}^{2}\left(\xi \mathrm{\Delta }/2\right)‖}_{2}\le 1$
Which is equivalent to
${\left(1-\frac{b\mathrm{\Delta }t}{\mathrm{\Delta }{x}^{2}}4{\mathrm{sin}}^{2}\left(\xi \mathrm{\Delta }/2\right)\right)}^{2}+{\left(\frac{a\mathrm{\Delta }t}{\mathrm{\Delta }x}\mathrm{sin}\left(\xi \mathrm{\Delta }/2\right)\right)}^{2}\le 1$
To obtain the constrain on $\mathrm{\Delta }t$ and $\mathrm{\Delta }x$, which are the time step and the space step for the numerical method, set
$\begin{array}{rl}\mathrm{sin}\left(\xi \mathrm{\Delta }/2\right)=1& ⇒\frac{\mathrm{\Delta }t}{\mathrm{\Delta }{x}^{2}}\left({b}^{2}+{a}^{2}\right)\le 2b\\ & ⇒\mathrm{\Delta }t\le \frac{2b\mathrm{\Delta }{x}^{2}}{{a}^{2}+{b}^{2}}\end{array}$
Note there is another condition for stability for this system, but this condition encompass the other one so I don't mention it.

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