# Weak Form versus Boundary Integral Form

I am considering a fairly basic Boundary Value Problem to expose my question.

Boundary Value Problem (1)

• local equation: $$u_{xx}(x)=0$$, $$\forall x \in [0,1]$$
• boundary conditions: $$u(0)=0$$ and $$u(1)=1$$

Usual weak form (2)

Find $$u\in H^1(0,1)$$ satisfying the BC such that $$\forall v\in H^1_0(0,1)$$, $$\int_0^1 u_x(x)v_x(x) \mathrm{d}x=0$$

Showing that (2) implies (1) is explained in many books.

Boundary Integral Form (3)

1. Find the fundamental solution $$w(x,\xi)$$ solving $$w_{xx}(x,\xi)=\delta_\xi$$, $$\forall (x,\xi)\in[0,1]^2$$

2. Premultiply the local equation by $$w$$ and integrate over the domain $$(0,1)$$

$$\int_0^1 u_{xx}(x)w(x,\xi)\mathrm{d}x=0,\quad \forall \xi\in[0,1]$$

1. Integrate by parts twice along $$x$$ to get

$$\int_0^1 u(x)w_{xx}(x,\xi)\mathrm{d}x= u(x)w_x(x,\xi)\Big|_{x=0}^{x=1}-u_x(x)w(x,\xi)\Big|_{x=0}^{x=1}$$

1. Use the property $$w_{xx}(x,\xi)=\delta_\xi$$ to obtain

$$u(\xi) = u(x)w_x(x,\xi)\Big|_{x=0}^{x=1}-u_x(x)w(x,\xi)\Big|_{x=0}^{x=1}$$

Question How do we know that $$\int_0^1 u_{xx}(x)w(x,\xi)\mathrm{d}x=0,\quad \forall \xi\in[0,1]\qquad \Rightarrow\qquad u_{xx}(x)=0,\quad\forall x \in [0,1]$$ where, by definition, $$w_{xx}(x,\xi)=\delta_\xi$$, $$\forall (x,\xi)\in[0,1]^2$$.

## 1 Answer

You end up with:

$$\forall \xi \in (0,1),\quad u(\xi)=u(1)w_x(1,\xi)-u(0)w_x(0,\xi)-u_x(1)w(1,\xi)+u_x(0)w(0,\xi).$$ So $$u''(\xi)=u(1)w_{xxx}(1,\xi)-u(0) w_{xxx}(0,\xi) -u_x(1)w_{xx}(1,\xi)+u_x(0)w_{xx}(0,\xi)$$

That implies that $\forall x\in(0,1)$, $u''(x)=0$, because for any fixed $\xi$, $\dfrac{\mathrm{d}\delta_\xi}{\mathrm{d}x}$ is zero on an open set of the form $(0,\alpha<\xi)$ or $(\xi<\alpha,1)$, hence $w_{xxx}(0,\xi)=w_{xxx}(1,\xi)=0$.