Computing weak derivative, example 1 Evans PDE.

Question

Consider the following problem: Thus we have $$\int_{0}^{2}u\phi'dx = \int_{0}^{1}x\phi'dx + \int_{1}^{2}\phi'dx$$ $$=-\int_{0}^{1}\phi dx + \underbrace{\phi(2)-\phi(1)}_{\text{why is this }0?} = -\int_{0}^{1}\phi dx = -\int_{0}^{2}v\phi dx.$$ I don't get why the $\int_{1}^{2}\phi'$ is equal to $0.$ Any hints will be much appreciated.

Answer 1

The integration by parts in your second equality should read: \begin{align} \int_0^1 x\phi'(x) dx + \int_1^2 \phi' dx &= [x\phi(x)]_0^1 - \int_0^1 \phi dx +\phi(2) - \phi(1) \\ &= \phi(1) - \int_0^1 \phi dx + \phi(2) - \phi(1)\\ &=- \int_0^1 \phi dx. \end{align} Observe that $\phi(2)=0$ since $\phi$ belongs to the space $C_0^\infty(0,2)$ and therefore has compact support on $(0,2)$.