Finite difference for non-uniform unstructured mesh/stencilHow to obtain (prove) 5-stencil formula for 2nd...

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Finite difference for non-uniform unstructured mesh/stencil


How to obtain (prove) 5-stencil formula for 2nd derivative?Finite difference implicit schema for wave equation 1d not unconditionally stable?Calculating gradient from finite difference resultsDiscretization of the Anisotropic Diffusion Operator for Finite Difference MethodNumerically Solving a Poisson Equation with Neumann Boundary Conditionsfinite difference method with non-uniform meshFinite-Difference Approximation of Mixed Derivative in Spherical Coordinates?Stiffness Matrix for Galerkin Method (Finite Element Approx)Finite difference method for non-uniform gridApproximating finite differences by higher order derivatives of continuous functions













0












$begingroup$


Below I have shown my non-uniform unstructured mesh (as in there is no pattern between the relative size of $h_i$ and $h_{i+1}$ etc.



NU-US-M



I've made the following equations using nodes $i-1$, $i$, and $i+1$.




  1. $phi_{i-1}=phi_i+h_{i-1}frac{dphi}{dx}|_i+frac{h_{i-1}^2}{2!}frac{d^2phi}{dx^2}|_i+frac{h_{i-1}^3}{3!}frac{d^3phi}{dx^3}|_i+...$

  2. $phi_{i+1}=phi_i+h_{i}frac{dphi}{dx}|_i+frac{h_{i}^2}{2!}frac{d^2phi}{dx^2}|_i+frac{h_{i}^3}{3!}frac{d^3phi}{dx^3}|_i+...$


Combining these in a linear combination, then solving for the first derivative
$$frac{dphi}{dx}|_i=frac{Aphi_{i-1}+Bphi_{i+1}-(A+B)phi_i}{Ah_{i-1}+Bh_i}-frac{1}{2}left(frac{Ah_{i-1}^2+Bh_i^2}{Ah_{i-1}+Bh_i}right)frac{d^2phi}{dx^2}|_i-frac{1}{6}left(frac{Ah_{i-1}^3+Bh_i^3}{Ah_{i-1}+Bh_i}right)frac{d^3phi}{dx^3}|_i+/-...$$



According to my notes from class, I should be getting a final answer of
$$frac{dphi}{dx}|_i=frac{phi_{i+1}-phi_{i-1}}{x_{i+1}-x_{i-1}}-frac{left(h_iright)^2-left(h_{i-1}right)^2}{2left(x_{i+1}-x_{i-1}right)}frac{d^2phi}{dx^2}|_i-frac{left(h_iright)^3-left(h_{i-1}right)^3}{6left(x_{i+1}-x_{i-1}right)}frac{d^3phi}{dx^3}|_i+O(h^3).$$



Based on what I am expecting to get, and the fact that I want a second order accurate final equation I can say the following.




  1. $Ah_{i-1}^2+Bh_i^2=0$

  2. $A+B=0$


However, I can't get the expected answer. Can I get some help with the error I'm making?










share|cite|improve this question









$endgroup$












  • $begingroup$
    The Taylor expansion for $phi_{i-1}$ is incorrect. The distance between between the points $x_{i-1}$ and $x_i$ is $x_{i-1} - x_i = -h_{i-1}$, so you need to replace $h_{i-1}$ by $-h_{i-1}$ everywhere in this expansion. Does this get you to the final answer?
    $endgroup$
    – ekkilop
    Apr 3 at 13:52
















0












$begingroup$


Below I have shown my non-uniform unstructured mesh (as in there is no pattern between the relative size of $h_i$ and $h_{i+1}$ etc.



NU-US-M



I've made the following equations using nodes $i-1$, $i$, and $i+1$.




  1. $phi_{i-1}=phi_i+h_{i-1}frac{dphi}{dx}|_i+frac{h_{i-1}^2}{2!}frac{d^2phi}{dx^2}|_i+frac{h_{i-1}^3}{3!}frac{d^3phi}{dx^3}|_i+...$

  2. $phi_{i+1}=phi_i+h_{i}frac{dphi}{dx}|_i+frac{h_{i}^2}{2!}frac{d^2phi}{dx^2}|_i+frac{h_{i}^3}{3!}frac{d^3phi}{dx^3}|_i+...$


Combining these in a linear combination, then solving for the first derivative
$$frac{dphi}{dx}|_i=frac{Aphi_{i-1}+Bphi_{i+1}-(A+B)phi_i}{Ah_{i-1}+Bh_i}-frac{1}{2}left(frac{Ah_{i-1}^2+Bh_i^2}{Ah_{i-1}+Bh_i}right)frac{d^2phi}{dx^2}|_i-frac{1}{6}left(frac{Ah_{i-1}^3+Bh_i^3}{Ah_{i-1}+Bh_i}right)frac{d^3phi}{dx^3}|_i+/-...$$



According to my notes from class, I should be getting a final answer of
$$frac{dphi}{dx}|_i=frac{phi_{i+1}-phi_{i-1}}{x_{i+1}-x_{i-1}}-frac{left(h_iright)^2-left(h_{i-1}right)^2}{2left(x_{i+1}-x_{i-1}right)}frac{d^2phi}{dx^2}|_i-frac{left(h_iright)^3-left(h_{i-1}right)^3}{6left(x_{i+1}-x_{i-1}right)}frac{d^3phi}{dx^3}|_i+O(h^3).$$



Based on what I am expecting to get, and the fact that I want a second order accurate final equation I can say the following.




  1. $Ah_{i-1}^2+Bh_i^2=0$

  2. $A+B=0$


However, I can't get the expected answer. Can I get some help with the error I'm making?










share|cite|improve this question









$endgroup$












  • $begingroup$
    The Taylor expansion for $phi_{i-1}$ is incorrect. The distance between between the points $x_{i-1}$ and $x_i$ is $x_{i-1} - x_i = -h_{i-1}$, so you need to replace $h_{i-1}$ by $-h_{i-1}$ everywhere in this expansion. Does this get you to the final answer?
    $endgroup$
    – ekkilop
    Apr 3 at 13:52














0












0








0





$begingroup$


Below I have shown my non-uniform unstructured mesh (as in there is no pattern between the relative size of $h_i$ and $h_{i+1}$ etc.



NU-US-M



I've made the following equations using nodes $i-1$, $i$, and $i+1$.




  1. $phi_{i-1}=phi_i+h_{i-1}frac{dphi}{dx}|_i+frac{h_{i-1}^2}{2!}frac{d^2phi}{dx^2}|_i+frac{h_{i-1}^3}{3!}frac{d^3phi}{dx^3}|_i+...$

  2. $phi_{i+1}=phi_i+h_{i}frac{dphi}{dx}|_i+frac{h_{i}^2}{2!}frac{d^2phi}{dx^2}|_i+frac{h_{i}^3}{3!}frac{d^3phi}{dx^3}|_i+...$


Combining these in a linear combination, then solving for the first derivative
$$frac{dphi}{dx}|_i=frac{Aphi_{i-1}+Bphi_{i+1}-(A+B)phi_i}{Ah_{i-1}+Bh_i}-frac{1}{2}left(frac{Ah_{i-1}^2+Bh_i^2}{Ah_{i-1}+Bh_i}right)frac{d^2phi}{dx^2}|_i-frac{1}{6}left(frac{Ah_{i-1}^3+Bh_i^3}{Ah_{i-1}+Bh_i}right)frac{d^3phi}{dx^3}|_i+/-...$$



According to my notes from class, I should be getting a final answer of
$$frac{dphi}{dx}|_i=frac{phi_{i+1}-phi_{i-1}}{x_{i+1}-x_{i-1}}-frac{left(h_iright)^2-left(h_{i-1}right)^2}{2left(x_{i+1}-x_{i-1}right)}frac{d^2phi}{dx^2}|_i-frac{left(h_iright)^3-left(h_{i-1}right)^3}{6left(x_{i+1}-x_{i-1}right)}frac{d^3phi}{dx^3}|_i+O(h^3).$$



Based on what I am expecting to get, and the fact that I want a second order accurate final equation I can say the following.




  1. $Ah_{i-1}^2+Bh_i^2=0$

  2. $A+B=0$


However, I can't get the expected answer. Can I get some help with the error I'm making?










share|cite|improve this question









$endgroup$




Below I have shown my non-uniform unstructured mesh (as in there is no pattern between the relative size of $h_i$ and $h_{i+1}$ etc.



NU-US-M



I've made the following equations using nodes $i-1$, $i$, and $i+1$.




  1. $phi_{i-1}=phi_i+h_{i-1}frac{dphi}{dx}|_i+frac{h_{i-1}^2}{2!}frac{d^2phi}{dx^2}|_i+frac{h_{i-1}^3}{3!}frac{d^3phi}{dx^3}|_i+...$

  2. $phi_{i+1}=phi_i+h_{i}frac{dphi}{dx}|_i+frac{h_{i}^2}{2!}frac{d^2phi}{dx^2}|_i+frac{h_{i}^3}{3!}frac{d^3phi}{dx^3}|_i+...$


Combining these in a linear combination, then solving for the first derivative
$$frac{dphi}{dx}|_i=frac{Aphi_{i-1}+Bphi_{i+1}-(A+B)phi_i}{Ah_{i-1}+Bh_i}-frac{1}{2}left(frac{Ah_{i-1}^2+Bh_i^2}{Ah_{i-1}+Bh_i}right)frac{d^2phi}{dx^2}|_i-frac{1}{6}left(frac{Ah_{i-1}^3+Bh_i^3}{Ah_{i-1}+Bh_i}right)frac{d^3phi}{dx^3}|_i+/-...$$



According to my notes from class, I should be getting a final answer of
$$frac{dphi}{dx}|_i=frac{phi_{i+1}-phi_{i-1}}{x_{i+1}-x_{i-1}}-frac{left(h_iright)^2-left(h_{i-1}right)^2}{2left(x_{i+1}-x_{i-1}right)}frac{d^2phi}{dx^2}|_i-frac{left(h_iright)^3-left(h_{i-1}right)^3}{6left(x_{i+1}-x_{i-1}right)}frac{d^3phi}{dx^3}|_i+O(h^3).$$



Based on what I am expecting to get, and the fact that I want a second order accurate final equation I can say the following.




  1. $Ah_{i-1}^2+Bh_i^2=0$

  2. $A+B=0$


However, I can't get the expected answer. Can I get some help with the error I'm making?







numerical-methods finite-differences finite-difference-methods






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 20 at 18:37









WnGatRC456WnGatRC456

10811




10811












  • $begingroup$
    The Taylor expansion for $phi_{i-1}$ is incorrect. The distance between between the points $x_{i-1}$ and $x_i$ is $x_{i-1} - x_i = -h_{i-1}$, so you need to replace $h_{i-1}$ by $-h_{i-1}$ everywhere in this expansion. Does this get you to the final answer?
    $endgroup$
    – ekkilop
    Apr 3 at 13:52


















  • $begingroup$
    The Taylor expansion for $phi_{i-1}$ is incorrect. The distance between between the points $x_{i-1}$ and $x_i$ is $x_{i-1} - x_i = -h_{i-1}$, so you need to replace $h_{i-1}$ by $-h_{i-1}$ everywhere in this expansion. Does this get you to the final answer?
    $endgroup$
    – ekkilop
    Apr 3 at 13:52
















$begingroup$
The Taylor expansion for $phi_{i-1}$ is incorrect. The distance between between the points $x_{i-1}$ and $x_i$ is $x_{i-1} - x_i = -h_{i-1}$, so you need to replace $h_{i-1}$ by $-h_{i-1}$ everywhere in this expansion. Does this get you to the final answer?
$endgroup$
– ekkilop
Apr 3 at 13:52




$begingroup$
The Taylor expansion for $phi_{i-1}$ is incorrect. The distance between between the points $x_{i-1}$ and $x_i$ is $x_{i-1} - x_i = -h_{i-1}$, so you need to replace $h_{i-1}$ by $-h_{i-1}$ everywhere in this expansion. Does this get you to the final answer?
$endgroup$
– ekkilop
Apr 3 at 13:52










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