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Please help to solve this question:

Instability of a difference scheme under small perturbations does not exclude the possibility that in special cases the scheme converges towards the correct function, if no errors are permitted in the data or the computation. In particular let $f(x)=e^{\alpha x}$ with a complex constant ${\alpha }$. Show that for fixed $x,t$ and any fixed positive $\lambda = k/h$ whatsoever both the expression $(3.9)$ and $(3.14)$ converge for $n\rightarrow \infty$ towards the correct limit $e^{\alpha (x-ct)}$. (This is consistent with the Courant-Friedrichs-Lewy test, since for an analytic $f$ the values of $f$ in any interval determine those at the point $\xi=x-ct $ uniquely).

The PDE is a scalar linear conservation law $$ v_t + c v_x = 0$$ with initial data $v(x,0) = f(x)$, and the method of finite differences is used. The equations are

$$\begin{aligned} v(x,t)&=v(x,nk) \\ & = \dots \\ &={\sum_{m=0}^{n}}\binom{n}{m}(1+\lambda c)^m(-\lambda c)^{n-m}f (x+(n-m)h) \end{aligned} \tag{3.9}$$ and $$v(x,t)=v(x,nk)={\sum_{m=0}^{n}}\binom{n}{m}(1-\lambda c)^m(\lambda c)^{n-m}f (x-(n-m)h) \, . \tag{3.14}$$

This is problem 3 page 8 from the book by Fritz John [1].

Please help. Thanks


[1] F. John: Partial Differential Equations, 4th Edition, Springer, 1991.

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Combining $t=nk$ and $\lambda = k/h$, we get $h=t/(n\lambda)$. With $f(x) = e^{\alpha x}$, the first expression $(3.9)$ rewrites as \begin{aligned} v(x,nk) &= \sum_{m=0}^n \binom{n}{m} (1+\lambda c)^m (-\lambda c)^{n-m} e^{\alpha (x+(n-m)h)} \\ &= e^{\alpha x} \sum_{m=0}^n \binom{n}{m} (1+\lambda c)^m (-\lambda c e^{\alpha h})^{n-m} \\ &= e^{\alpha x} \left( 1+ \lambda c (1 - e^{\alpha t/(n\lambda)}) \right)^n \, . \end{aligned} The limit as $n\to\infty$ gives the expected result. The proof is similar with $(3.14)$.

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  • $\begingroup$ Thank for your help $\endgroup$ – Vui Tinh Sep 16 '17 at 3:38

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