I am trying to solve a problem using method of undetermined coefficients to derive a second order scheme for ux using three points, c1, c2, c3 in the following way:

ux = c1*u(x) + c2*u(x - h) + c3*u(x - 2h)

Now second order scheme just means to solve the equation for the second order derivative, am I right?

I understand how this problem works for actual numerical functions, but I am unsure how to go about it when everything is theoretical and just variables.

Thanks for some help


Having a second order scheme means that it's accurate for polynomials up to and including second degree. The scheme should calculate the first order derivative $u_x$, as the formula says.

It suffices to make sure that the scheme is accurate for $1$, $x$, and $x^2$; then it will work for all second-degree polynomials by linearity.

To make it work for the function $u(x)=1$, we need $$ 0= c_1+c_2+c_3 \tag1$$ To make it work for the linear function $u(x)=x$, we need $$ 1 = c_1 x +c_2(x-h) +c_3(x-2h) \tag2$$ which in view of (1) simplifies to $$ 1 = c_2(-h) +c_3(-2h) \tag{2'}$$ And to make it work for the quadratic function $u(x)=x^2$, we need $$ 2x = c_1 x^2 +c_2(x-h)^2 +c_3(x-2h)^2 \tag3$$ which in view of (1) and (2') simplifies to $$ 0 = c_2h^2 +c_3(4h^2) \tag{3'}$$ Now you can solve the linear system (1), (2') and (3') for the unknowns $c_1,c_2,c_3$.

This may not be the quickest solution, but it's the most concrete one that I could think of.

  • $\begingroup$ did you miss the u(x), u(x-h), and u(x-2h)? You never showed those in your equation $\endgroup$ – Matt Hintzke Feb 28 '13 at 1:06
  • $\begingroup$ @MattHintzke I worked with three concrete functions: $u(x)=1$, $u(x)=x$, and $u(x)=x^2$. For example, when I work with $u(x)=x^2$, I have $u(x-h)=(x-h)^2$. So I write $c_2(x-h)^2$ in the equation. It's plugging in a formula for the function $u$. $\endgroup$ – user53153 Feb 28 '13 at 1:09
  • $\begingroup$ OHH wow I looked at it wrong haha, thank you I understand now :) $\endgroup$ – Matt Hintzke Feb 28 '13 at 1:11
  • $\begingroup$ in the end should c1, c2, and c3 all be functions of only h? $\endgroup$ – Matt Hintzke Feb 28 '13 at 1:44
  • 1
    $\begingroup$ @MattHintzke Yes, definitely. In any scheme. $\endgroup$ – user53153 Feb 28 '13 at 1:48

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