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.

  • $\begingroup$ 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

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:

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


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

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.

  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.