# variational formulation of second order differential equation

I am given the following differential equation. Let $$\Omega = (a,b)\subset\mathbb{R},\ f:\Omega \rightarrow\mathbb{R},\ \alpha,\beta \in \mathbb{R}$$ and $$-u'' + u = f \\ u(a)= \alpha, u(b) = \beta$$ Since this is an inhomogeneous problem, I choose a function $$u_\varphi$$ with $$u_\varphi|_\Gamma = u|_\Gamma$$ such that I can treat the problem like a homogeneous one, by finding a function $$u_0$$ with $$u_0(a)=u_0(b)=0$$, such that $$u = u_\varphi + u_0.$$ The DEQ can then be written as

$$-(u_0 + u_\varphi)'' + u_0 + u_\varphi = f \\$$ To find the variational formulation for $$u_0$$, I test with an arbitrary testfunction $$v$$ with compact support:

$$\int_\Omega (-(u_0 + u_\varphi)'' + u_0 + u_\varphi)v dx= \int_\Omega fvdx$$ yielding $$\int_\Omega (-u_0''+u_0)v dx=\int_\Omega \nabla u_0 \nabla v dx + \int_\Omega u_0v dx = \int_\Omega (f +u_\varphi'' - u_\varphi)vdx$$ which we usually write compactly as $$a(u_0,v) + \int_\Omega u_0v dx = F(v).$$ This integral in the last equation bothers me, since I cannot get rid of it. Is there a way to do so?

Thanks!

• Why do you want to get rid of this integral? – gerw Jan 11 at 8:18
• I assumed, this stands in the way of solving the problem. But the left side can be written as $a^*(u_0,v)$ with $a^*$ being a bilinear form. – dba Jan 11 at 14:12

The problem at hand can be reduced to a (somewhat more general) normed problem: $$\frac{d^2 T}{d\xi^2} - p^2 T(\xi) = F(\xi)$$ The left hand side of this normed problem is handled with help of the following references:
The second reference shows that vertex integration is the most stable one. If we employ this for the right hand side, then the integral $$\int_0^1 F(\xi)f(\xi)\,d\xi$$ results in a load vector $$\vec{F}$$ instead of $$0$$ . Giving for the system of equations as a whole (read the first reference): $$\begin{bmatrix} E_{0,0}^{(1)} & E_{0,1}^{(1)} & 0 & 0 & 0 & \cdots \\ E_{1,0}^{(1)} & E_{1,1}^{(1)}+E_{0,0}^{(2)} & E_{0,1}^{(2)} & 0 & 0 & \cdots \\ 0 & E_{1,0}^{(2)} & E_{1,1}^{(2)}+E_{0,0}^{(3)} & E_{0,1}^{(3)} & 0 & \cdots \\ 0 & 0 & E_{1,0}^{(3)} & E_{1,1}^{(3)}+E_{0,0}^{(4)} & E_{0,1}^{(4)} & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \end{bmatrix} \begin{bmatrix} T_1 \\ T_2 \\ T_3 \\ T_4 \\ T_5 \\ \cdots \end{bmatrix} = \begin{bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ \cdots \end{bmatrix}$$ with the boundary conditions properly imposed.
The original problem - with $$x$$ and $$u$$ instead of $$\xi$$ and $$T$$ - is recovered by employing the following transformations. Herewith: $$\xi_k \;\rightarrow\; x_k$$ and $$T_k \;\rightarrow\; u_k$$ : $$x = (b-a)\xi+a \quad \Longrightarrow \quad \begin{cases} x = a \;\leftrightarrow\; \xi = 0 \\ x = b \;\leftrightarrow\; \xi = 1 \end{cases} \\ u = (\beta-\alpha)T+\alpha \quad \Longrightarrow \quad \begin{cases} u = \alpha \;\leftrightarrow\; T = 0 \\ u = \beta \;\leftrightarrow\; T = 1 \end{cases}$$ Note. Variational formulation and Galerkin method are the same in this case.