I am dealing with the following problem: I am interested in the regularity of a weak solution $u\in H^1_0(I)$ of the following elliptic differential equation:
$$-u_{xx}=f(x) \text{ on } I$$
where $I=(a,b)$ is a bounded open interval and $f\in L^2(I)$. I want to conclude that in fact $u\in H^2(I)$. To do so I want to use theorem 4/5 chapter 6.3 of Lawrence C. Evans Partial Differential Equations which states:
$\textbf{Theorem:}$ Let $L$ be an elliptic linear differential operator of second order, that is:
$$Lu=-\sum_{i,j=1}^n(a_{ij}(x)u_{x_i})_{x_j}+\sum_{i=1}^nb_i(x)u_{x_i}+c(x)u $$ with symmetric coefficients $a_{ij}=a_{ji}$ and the property $\sum_{ij}a_{ij}\xi_i\xi_j \geq \theta |\xi|^2$ for some $\theta>0$ and all $\xi \in \mathbb{R}^n$.
Let further $U\subset \mathbb{R}^n$ be an open bounded subset and $u\in H^1_0(U)$ be a weak solution of:
$$ Lu=f \text{ in } U \text{, }u=0 \text{ on } \partial U $$
Assume $a_{ij},b_i,c \in C^{m+1}(\overline{U}) \text{, } i,j=1,\dots,n$ and $f\in H^m(U)$ for some $m\geq 0$ ($H^0(U)\equiv L^2(U)$). Assume finally that \begin{align} \partial U \text{ is } C^{2+m} \end{align} then we have $u\in H^{m+2}(U)$.
My question now is whether the condition $\partial U\in C^2$ holds true for my particular problem $n=1,U=I=(a,b)$ or not. By definition in Evan's book (Appendix C1) there has to be a $C^2$ function $\gamma: \mathbb{R}^{n-1}\rightarrow \mathbb{R}$ with certain properties and I wonder how (if at all) this definition is applicable to the case $n=1$.
So my precise question is whether or not the quoted theorem may be applied to my 1-dimensional problem.
Thank you all in advance!
You should check if there's some disclaimer in the beginning, for example in the introduction, or at the beginning of the chapter, where it's said for which $n$ the results hold. Anyway in dimension 1 usually results become simpler so there's no need for the full theory. In this case, by definition of weak solution $u$, we have that for any test function $\phi$ $$\int_a^b u_x\phi_x=\int_a^b f\phi$$ but this is exactly the equation that tells you that $-f$ is the weak derivative of $u_x$! Since $f\in L^2$ this means that $u\in H^2$.
For a regular elliptic linear operator you can do the same: the ellipticity is simply $a(x)\geq \theta >0$ and by definition of weak solution $$\int au_x \phi_x=-\int (bu_x+ c u-f)\phi$$ thus $v=bu_x+c u-f\in L^2$ is the weak derivative of $au_x$. To obtain the weak derivative of $u_x$ we observe that $$au_x\phi_x=u_x(a\phi)_x-u_xa_x\phi$$ and substituting above $$\int u_x(a\phi)_x=-\int(-u_xa_x+bu_x+cu-f)\phi=-\int \frac1a (-u_xa_x+bu_x+cu-f)(a\phi).$$ Now $a$ is regular so it attains a maximum, therefore $0<\delta\leq\frac1a\leq \frac{1}{\theta}$ which implies that the multiplication by $a$ is a bijection on the space of test functions. Therefore calling $\psi=a\phi$ we have $$\int u_x\psi_x=-\int \frac1a (-u_xa_x+bu_x+cu-f)\psi$$ for any test function $\psi$, and the weak derivative of $u_x$ is the integrand on the right hand side (without $\psi$) which is in $L^2$, and this means $u\in H^2$.