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Prove:

$ | u(x) - u_h(x) | \leq h \ \underset{0 \leq y \leq 1}{\max} |u''(y)| $

for $ 0 \leq x \leq 1 $

Using the fact that

$|| (u-u_h)' || \leq \underset{0 \leq y \leq 1}{\max} |u''(y)| $ and the boundary condition $ u(0) = u_h(0) = 0 $

the exercise also suggests

use the relation

$$ (u-u_h)(x) = \int_0^x (u-u_h)'(y) dy $$ and the cauchy inequality

where

$$ || w || = (w,w)^{\frac{1}{2}} = ( \int_0^1 w^2 )^{\frac{1}{2}} dx $$

I have tried to start with this:

I raise both sides squarely

$$ || (u-u_h)' || = ( \int_0^1 (u-u_h)'^2 dx )^{\frac{1}{2}} $$

getting

$$ || (u-u_h)' ||^2 = | \int_0^1 (u-u_h)'^2 dx | $$

and then reached a dead end

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  • 3
    $\begingroup$ What is $u_h(x)$? $\endgroup$ – Robert Z Feb 2 at 18:12
  • $\begingroup$ $u$ is the analytical solution to a pde problem, and $ u_h(x) $ is the solution of a finite element problem associated with the same pde $\endgroup$ – Saiten Feb 2 at 18:35
  • $\begingroup$ I imagine that knowing the exact PDE as well as the equation satisfied by the approximation would be helpful when searching for any relation between them. $\endgroup$ – Carl Christian Feb 2 at 19:02
  • $\begingroup$ I guess not, beacause $u$ is a general solution as $ u_h $ and i guess that is not the way to go. For me its just algebraic work, that i cannot do it btw. But i will give a look from that perspective as well $\endgroup$ – Saiten Feb 2 at 19:11
  • $\begingroup$ I guess $u_h$ comes from using linear finite elements, correct? $\endgroup$ – VorKir Feb 3 at 20:34

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