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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

$\|u^{n+1}\|\leq K e^{\beta t}\|u^{0}\|$; for $0\leq t=(n+1)\Delta t$, $0\leq \Delta x \leq \Delta x_{0}$ and $0\leq \Delta t \leq \Delta t_{0}$. i.e. this definition allows for exponential growth. But in some literature I found the definition that ''A finite difference approximation is stable if the errors (truncation, round-off etc) decay as the computation proceeds from one marching step to the next .'' The Book which I have mentioned, at the same page in remark 3, author has written that the above definition with exponential growth is general definition and later one can be derived from this.But I am not getting how to find this definition. So my question is which definition is true ? And if other one is not true, then what is reason behind ? and how can we compare these two definition? Thanking You in Advance.

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  • $\begingroup$ I don't have an extensive background on this, but from what I know, some finite difference method has the error part grows exponentially, while others do not. It depends specifically on what method of finite difference you are using. $\endgroup$ – Paichu May 2 '17 at 15:16
  • $\begingroup$ But if error will grow exponentially then how can we say it stable? $\endgroup$ – VIVEK KUMAR May 4 '17 at 13:33
  • $\begingroup$ That really depends on how you define stability. $\endgroup$ – Paichu May 4 '17 at 13:43
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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.

Consider the advection diffusion equation: $$ \partial_t u + a \partial_x u - b\partial^2_x u = 0, \text{ with } u(x,t=0) = f(x) $$ for $x \in [-1,1]$ with $u(x+2,t) = u(x,t)$ for all $x$ and $t$.

Applying the explicit difference method then, $$ U(x,t+\Delta t) = U(x,t) - \frac{a \Delta t}{2 \Delta x} \left(T - T^{-1}\right) U(x,t) + \frac{b \Delta t}{\Delta x^2}\left(T - 2 + T^{-1} \right)U(x,t) $$


Recall the following Discrete Fourier Transform rules:

$$U(x + 2 L) = U(x), \text{ and } L = N \Delta x = 1$$

$$ \hat{U}(\xi + 2L) = \hat{U}(\xi), \text{ and } \hat{L} = N \Delta \xi \text{ with } N\Delta x \Delta \xi = \pi$$

$$\widehat{TU} (\xi) = \hat{U} (\xi) e^{i \xi \Delta x}$$


Applying the previous rules, then our difference method equation becomes:

$$ \hat{U}(\xi, t+\Delta t) = \hat{U}(\xi,t) - \frac{a \Delta t}{2 \Delta x}\underbrace{\left(e^{i \xi \Delta x} - e^{-i \xi \Delta x} \right)}_{2i sin(\xi \Delta x /2)}\hat{U} + \frac{b\Delta t}{\Delta x^2}\underbrace{\left(e^{i \xi \Delta x} - 2 + e^{-i \xi \Delta x} \right)}_{\left(e^{i \xi \Delta x} - e^{-i \xi \Delta x} \right)^2 = \left(2i sin(\xi \Delta x /2)\right)^2}\hat{U} $$

With

$$\xi \in [-\frac{\pi}{\Delta x},\frac{\pi}{\Delta x}], \text{ or } \frac{\Delta x\xi}{2} \in [-\frac{\pi}{2},\frac{\pi}{2}]$$

And after some algebra,

$$ \hat{U}(\xi, t+\Delta t) = \left[1 - \frac{a \Delta t}{\Delta x} i \sin(\xi \Delta /2) - \frac{b \Delta t}{\Delta x^2} 4 \sin^2(\xi \Delta /2)\right]\hat{U}(\xi,t)$$

Here is the Stability part

Require $\|\hat{U}(t+\Delta t)\|_2 \le \|\hat{U}(t)\|_2$ to avoid blowing up (Stability) and recall that the Fourier Transform process is norm-preserving, then $$ \left\|1 - \frac{a \Delta t}{\Delta x} i \sin(\xi \Delta /2) - \frac{b \Delta t}{\Delta x^2} 4 \sin^2(\xi \Delta /2) \right\|_2 \le 1 $$ Which is equivalent to $$ \left( 1 - \frac{b \Delta t}{\Delta x^2} 4 \sin^2(\xi \Delta /2)\right)^2 + \left( \frac{a \Delta t}{\Delta x} \sin(\xi \Delta /2)\right)^2 \le 1 $$

To obtain the constrain on $\Delta t$ and $\Delta x$, which are the time step and the space step for the numerical method, set \begin{align*} \sin(\xi \Delta /2) = 1 &\Rightarrow \frac{\Delta t}{\Delta x^2} (b^2 + a^2) \le 2b\\ &\Rightarrow \Delta t \le \frac{2b \Delta x^2}{a^2 + b^2} \end{align*}

Note there is another condition for stability for this system, but this condition encompass the other one so I don't mention it.

I hope this helps.

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  • $\begingroup$ @VIVEKKUMAR You're welcome. $\endgroup$ – Paichu May 6 '17 at 22:09
  • $\begingroup$ @ Paichu Thank you very much. $\endgroup$ – VIVEK KUMAR Aug 2 '17 at 10:55

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