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, $w_{xx}(x,\xi)=\delta_\xi$, $\forall (x,\xi)\in[0,1]^2$.


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 $w_{xxx}(0,\xi)=w_{xxx}(1,\xi)=0$.


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