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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!

D.Simon
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1 Answers1

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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 $u\in H^2$.

Del
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