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I have approximated the differential equation using finite difference approximation and have the vector $u$. To find the error norm, it says I need the exact solution and the piecewise linear interpolant from $u$. Can anyone explain how I can get the piecewise linear interpolant? Would it be something like $u_{i+1}-u_{i}$ divided by I am not sure what. Also how do I get the exact solution? (Analytically??)

btw the bvp in question is:

$-u'' + u' = 1$ for $0<x<1$

$u(0) = u(1) = 0$

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The basis solutions of the homogeneous equation are $1, e^x$, thus the particular solution is of the form $ax$ and by inserting $a=1$. The solution with the left boundary condition satisfied is thus $$u(x)=c(e^x-1)+x,$$ $0=u(1)=c(e-1)+1$ then gives $c=-\frac1{e-1}$.

The piecewise linear interpolant is a piecewise linear function that interpolates between the values of the numerical solution $$ u_h((1-s)x_k+sx_{k+1})=(1-s)u_k+su_{k+1}~\text{ for all }~s\in[0,1]. $$

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