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Inconsistent finite difference scheme

0

$begingroup$

I have the following:

The Equation
$$bfrac{partial u}{partial t} + frac{partial u}{partial t} - c(x,t) = 0$$
b is a constant. The Equation is approximated at point $(x_i, t^n)$ in the $x-t$ plane by the difference scheme
$$frac{b}{Delta t}left{u^{n+1}_i - frac{1}{2}(u^n_{i+1} + u^n_{i-1})right} + frac{1}{2Delta x}(u^n_{i+1} - u^n_{i-1}) - c^n_i =0$$
If we set $Delta t = lambda(Delta x)^2$ then I have that the truncation error is given by
$$T^n_i = frac{b}{2}left(lambda(Delta x)^2u_{tt} - frac{1}{lambda}u_{xx}right) + frac{(Delta x)^2}{6}u_{xxx} + cdots$$
So as $Delta x rightarrow 0$ then $T^n_i rightarrow -dfrac{b}{2lambda}u_{xx}$.

---

I am asked that if the scheme is inconsistent like this then to obtain the equation it does approximate. How would I do this?

numerical-methods finite-differences finite-difference-methods

share|cite|improve this question

asked Mar 23 at 11:24

MRTMRT

348216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The way the differential equation is written there are only derivatives with respect to $t$ involved. Is this intentional or a typo? If it is intentional, what is the difference between the index $i$ and the index $n$?
  $endgroup$
  – ekkilop
  Apr 3 at 13:41
  

  
  
  
  

add a comment | 

0

$begingroup$

I have the following:

The Equation
$$bfrac{partial u}{partial t} + frac{partial u}{partial t} - c(x,t) = 0$$
b is a constant. The Equation is approximated at point $(x_i, t^n)$ in the $x-t$ plane by the difference scheme
$$frac{b}{Delta t}left{u^{n+1}_i - frac{1}{2}(u^n_{i+1} + u^n_{i-1})right} + frac{1}{2Delta x}(u^n_{i+1} - u^n_{i-1}) - c^n_i =0$$
If we set $Delta t = lambda(Delta x)^2$ then I have that the truncation error is given by
$$T^n_i = frac{b}{2}left(lambda(Delta x)^2u_{tt} - frac{1}{lambda}u_{xx}right) + frac{(Delta x)^2}{6}u_{xxx} + cdots$$
So as $Delta x rightarrow 0$ then $T^n_i rightarrow -dfrac{b}{2lambda}u_{xx}$.

---

I am asked that if the scheme is inconsistent like this then to obtain the equation it does approximate. How would I do this?

numerical-methods finite-differences finite-difference-methods

share|cite|improve this question

asked Mar 23 at 11:24

MRTMRT

348216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The way the differential equation is written there are only derivatives with respect to $t$ involved. Is this intentional or a typo? If it is intentional, what is the difference between the index $i$ and the index $n$?
  $endgroup$
  – ekkilop
  Apr 3 at 13:41
  

  
  
  
  

add a comment | 

$begingroup$

I have the following:

The Equation
$$bfrac{partial u}{partial t} + frac{partial u}{partial t} - c(x,t) = 0$$
b is a constant. The Equation is approximated at point $(x_i, t^n)$ in the $x-t$ plane by the difference scheme
$$frac{b}{Delta t}left{u^{n+1}_i - frac{1}{2}(u^n_{i+1} + u^n_{i-1})right} + frac{1}{2Delta x}(u^n_{i+1} - u^n_{i-1}) - c^n_i =0$$
If we set $Delta t = lambda(Delta x)^2$ then I have that the truncation error is given by
$$T^n_i = frac{b}{2}left(lambda(Delta x)^2u_{tt} - frac{1}{lambda}u_{xx}right) + frac{(Delta x)^2}{6}u_{xxx} + cdots$$
So as $Delta x rightarrow 0$ then $T^n_i rightarrow -dfrac{b}{2lambda}u_{xx}$.

---

I am asked that if the scheme is inconsistent like this then to obtain the equation it does approximate. How would I do this?

numerical-methods finite-differences finite-difference-methods

share|cite|improve this question

asked Mar 23 at 11:24

MRTMRT

348216

$endgroup$

share|cite|improve this question

asked Mar 23 at 11:24

MRTMRT

348216

share|cite|improve this question

asked Mar 23 at 11:24

MRTMRT

348216

asked Mar 23 at 11:24

MRTMRT

348216

asked Mar 23 at 11:24

MRTMRT

348216