The weak Dirichlet Poisson problem on an interval

Differentiation under the integral: fixing $x$ and letting $h\to 0$, the kernel identity $(G(x+h,y)-G(x,y))\mathrm{/}h\to {\mathrm{\partial }}_{x}G(x,y)$ holds for every $y\mathrm{\ne }x$, hence for almost every $y$, and the quotients are bounded by $1$ because $G$ is $1$-Lipschitz in its first variable on $[0,1]$; since $\mathrm{\mid }f\mathrm{\mid }\in L^1$, dominated convergence gives $u^{\mathrm{\prime }}(x)={\int }_0^1{\mathrm{\partial }}_{x}G(x,y)f(y)dy={\int }_x^{1}f(y)dy-{\int }_0^{1}yf(y)dy\mathrm{.}$

Boundary cutoff: prove the criterion of [F3]. Let $w$ be absolutely continuous on $[0,1]$ with $w(0)=w(1)=0$ and $w^{\mathrm{\prime }}\in L^2$. For $m\ge 4$ set ${\chi }_m(x):=\sigma (mx-1)\sigma (m(1-x)-1)$ and $w_m:={\chi }_{m}w$. Then ${\chi }_m$ vanishes on $[0,1\mathrm{/}m]\cup [1-1\mathrm{/}m,1]$, equals $1$ on $[2\mathrm{/}m,1-2\mathrm{/}m]$, takes values in $[0,1]$, and $\mathrm{\mid }{\chi }_m^{\mathrm{\prime }}\mathrm{\mid }\le 2m{C}_{\sigma }$ by the chain and product rules. The product $w_m$ is absolutely continuous with derivative ${\chi }_m^{\mathrm{\prime }}w+{\chi }_m{w}^{\mathrm{\prime }}\in L^2$ and compact support in $I$, so $w_m\in W^{1,2}(I)$.

Regularity and boundary values: the formula of step 1.1 exhibits $u^{\mathrm{\prime }}$ as the sum of the continuous function $x\mapsto {\int }_x^{1}f$ and a constant, so $u\in C^1([0,1])$; $u(0)=\int G(0,y)f=0$ and $u(1)=0$ by [F1], and the fundamental theorem gives $u(x)={\int }_0^x{u}^{\mathrm{\prime }}(t)dt$. Moreover $u^{\mathrm{\prime }}$ is the difference of an absolutely continuous function and a constant, so $u^{\mathrm{\prime }\mathrm{\prime }}=-f$ almost everywhere.

Convergence in $H^1$: put $E_m:=(0,2\mathrm{/}m)\cup (1-2\mathrm{/}m,1)$. Since $w_m=w$ off $E_m$ and $w\in L^2$, $\mathrm{\parallel }{w}_m-w{\mathrm{\parallel }}_{L^2}\to 0$; in $L^2$, $(w_m-w{)}^{\mathrm{\prime }}=({\chi }_m-1)w^{\mathrm{\prime }}+{\chi }_m^{\mathrm{\prime }}w$, and the first term tends to zero because $w^{\mathrm{\prime }}\in L^2$ and the strips shrink to the endpoints. Near $0$, $\mathrm{\mid }w(x){\mathrm{\mid }}^2=\mathrm{\mid }{\int }_0^x{w}^{\mathrm{\prime }}(t)dt{\mathrm{\mid }}^2\le x{\int }_0^x\mathrm{\mid }{w}^{\mathrm{\prime }}(t){\mathrm{\mid }}^{2}dt$, while near $1$, $\mathrm{\mid }w(x){\mathrm{\mid }}^2=\mathrm{\mid }{\int }_x^1{w}^{\mathrm{\prime }}(t)dt{\mathrm{\mid }}^2\le (1-x){\int }_x^1\mathrm{\mid }{w}^{\mathrm{\prime }}(t){\mathrm{\mid }}^{2}dt$. Therefore ${\int }_{1\mathrm{/}m}^{2\mathrm{/}m}\mathrm{\mid }{\chi }_m^{\mathrm{\prime }}w{\mathrm{\mid }}^{2}dx\le 6C_{\sigma }^2{\int }_0^{2\mathrm{/}m}\mathrm{\mid }{w}^{\mathrm{\prime }}{\mathrm{\mid }}^{2}dt\text{and}{\int }_{1-2\mathrm{/}m}^{1-1\mathrm{/}m}\mathrm{\mid }{\chi }_m^{\mathrm{\prime }}w{\mathrm{\mid }}^{2}dx\le 6C_{\sigma }^2{\int }_{1-2\mathrm{/}m}^1\mathrm{\mid }{w}^{\mathrm{\prime }}{\mathrm{\mid }}^{2}dt,$ and both bounds tend to zero. Thus $\mathrm{\parallel }{w}_m-w{\mathrm{\parallel }}_{H^1(0,1)}\to 0$.

Mollification and closure: each $w_m$ has compact support inside $I$, so its zero extension $W_m$ belongs to $W^{1,2}(\mathbb{R})$ by [F3]. Mollifying at sufficiently small scales gives $W_m\ast {\rho }_{\epsilon }\in C_c^{\mathrm{\infty }}(I)$. The weak-derivative identity with test $y\mapsto {\rho }_{\epsilon }(x-y)$ gives $(W_m\ast {\rho }_{\epsilon }{)}^{\mathrm{\prime }}=W_m^{\mathrm{\prime }}\ast {\rho }_{\epsilon }$; applying the $L^2$ approximate-identity result separately to $W_m$ and $W_m^{\mathrm{\prime }}$ proves convergence in $W^{1,2}$. Hence $w_m\in H_0^1(I)$. Since this space is closed and $w_m\to w$ in $H^1$, $w\in H_0^1(I)$. Applying the criterion to step 2.1 gives $u\in H_0^1(I)$.

Weak identity on test functions: for $\phi \in C_c^{\mathrm{\infty }}(I)$, integration by parts on $[0,1]$ and step 2.1 give ${\int }_0^1{u}^{\mathrm{\prime }}\overline{{\phi }^{\mathrm{\prime }}}dx=[u^{\mathrm{\prime }}\bar{\phi }{]}_0^1-{\int }_0^1{u}^{\mathrm{\prime }\mathrm{\prime }}\bar{\phi }dx={\int }_0^{1}f\bar{\phi }dx,$ the boundary term vanishing because $\phi$ has compact support in $I$ and $u^{\mathrm{\prime }\mathrm{\prime }}=-f$ a.e.

Identification and uniqueness: the right-hand side $v\mapsto (f,v{)}_{L^2}$ is an element of $H^{-1}(I)$ by [F4], and both sides of the identity are bounded in $v$ on $H_0^1(I)$ by H"older; since $C_c^{\mathrm{\infty }}(I)$ is dense in $H_0^1(I)$, the identity of step 3.2 extends to every $v\in H_0^1(I)$. Hence $u$ is the unique weak solution of $-u^{\mathrm{\prime }\mathrm{\prime }}=f$ with zero boundary values. For $f=\mathbf{1}$ the formula gives $u^{\mathrm{\prime }}(x)=(1-x)-\frac{1}{2}=\frac{1}{2}-x$ and $u(x)=\frac{1}{2}x(1-x)$.

General $H^{-1}$ datum: let $F(v)=(f_0,v{)}_{L^2}+(f_1,v^{\mathrm{\prime }}{)}_{L^2}$ with $f_0,f_1\in L^2(I)$ and define $w(x):={\int }_0^1{\mathrm{\partial }}_{y}G(x,y)f_1(y)dy={\int }_0^x{f}_1(y)dy-x{\int }_0^1{f}_1(y)dy$. Then $w$ is absolutely continuous with $w(0)=w(1)=0$ and $w^{\mathrm{\prime }}=f_1-{\int }_0^1{f}_1\in L^2$, so $w\in H_0^1(I)$ by the criterion of step 3.1; for $v\in H_0^1(I)$ one computes ${\int }_0^1{w}^{\mathrm{\prime }}\overline{v^{\mathrm{\prime }}}={\int }_0^1{f}_1\overline{v^{\mathrm{\prime }}}-\left({\int }_0^1{f}_1\right)\overline{{\int }_0^1{v}^{\mathrm{\prime }}}$, and ${\int }_0^1{v}^{\mathrm{\prime }}=0$ for $v\in H_0^1$, by approximating $v$ in $H^1$ by compactly supported smooth $v_j$, for which $\int v_j^{\mathrm{\prime }}=0$, and using $\mathrm{\mid }\int (v^{\mathrm{\prime }}-v_j^{\mathrm{\prime }})\mathrm{\mid }\le \mathrm{\parallel }{v}^{\mathrm{\prime }}-v_j^{\mathrm{\prime }}{\mathrm{\parallel }}_2$, so the identity ${\int }_0^1{w}^{\mathrm{\prime }}\overline{v^{\mathrm{\prime }}}={\int }_0^1{f}_1\overline{v^{\mathrm{\prime }}}$ holds. Combined with step 4.1 applied to $f_0$, the function $u=u_0+w$ with $u_0(x)=\int G(x,y)f_0(y)dy$ satisfies the weak equation for the datum $F$, and by uniqueness it is the weak solution; this is the displayed Green representation with ${\mathrm{\partial }}_{y}G$ acting on $f_1$.

Conclusion: the Green function representation produces the unique weak solution on the interval, with the weak identity and the explicit case $f=\mathbf{1}$ giving $u(x)=\frac{1}{2}x(1-x)$, and the general divergence-form datum is handled by the same kernel with ${\mathrm{\partial }}_{y}G$; no general Green-function theory is claimed.