Lagrange's identity (boundary value problem)

Lagrange's identity and calculus

In terms of Sturm-Liouville theory, Lagrange's identity can be written:

(4)   $\ \int _{0}^{1}(Lu)v-u(Lv)\,dx=-p(u'v-uv'){\bigg |}_{0}^{1}\ ,$

where $\ p=P(x)$, $\ q=Q(x)$, $\ u=U(x)$ and $\ v=V(x)$ are functions of $\ x$. $\ u$ and $\ v$ having continuous second derivatives on the interval $\ [0,1]$. $\ L$ is the Sturm-Liouville differential operator defined by:

(5)   $\ Lu=-(pu')'+qu\ .$

Proof of calculus form

Replace $\ f(x)=pu'$, $\ g(x)=v$, $\ a=0$ and $\ b=1$ into the rule of integration by parts

(6)   $\int _{a}^{b}f'(x)\,g(x)\,dx=f(x)\,g(x){\bigg |}_{a}^{b}-\int _{a}^{b}f(x)\,g'(x)\,dx\ ,$

we have:

(7)   $\int _{0}^{1}(pu')'v\,dx=(pu')v{\bigg |}_{0}^{1}-\int _{0}^{1}(pu')v'\,dx=p(u'v){\bigg |}_{0}^{1}-\int _{0}^{1}(pu')v'\,dx\ .$

Replace $\ f(x)=u$, $\ g(x)=pv'$, $\ a=0$ and $\ b=1$ into the rule ('6') again, we have:

$\ \int _{0}^{1}u'(pv')\,dx=u(pv'){\bigg |}_{0}^{1}-\int _{0}^{1}u(pv')'\,dx$

(8)   $\ -\int _{0}^{1}(pu')v'\,dx=-p(uv'){\bigg |}_{0}^{1}+\int _{0}^{1}u(pv')'\,dx\ .$

Replace ('8') into ('7'), we get:

$\ \int _{0}^{1}(pu')'v\,dx=p(u'v){\bigg |}_{0}^{1}-p(uv'){\bigg |}_{0}^{1}+\int _{0}^{1}u(pv')'\,dx$

(9)   $\ -\int _{0}^{1}(pu')'v\,dx=-p(u'v-uv'){\bigg |}_{0}^{1}-\int _{0}^{1}u(pv')'\,dx\ .$

From the definition ('2'), we can get:

(10)   $\ \int _{0}^{1}(Lu)v\,dx=\int _{0}^{1}[-(pu')'+qu]v\,dx=-\int _{0}^{1}(pu')'v\,dx+\int _{0}^{1}uqv\,dx\ .$

Replace ('9') into ('10'), we have:

$\ \int _{0}^{1}(Lu)v\,dx=-p(u'v-uv'){\bigg |}_{0}^{1}-\int _{0}^{1}u(pv')'\,dx+\int _{0}^{1}uqv\,dx$

(11)   $\ \int _{0}^{1}(Lu)v\,dx=-p(u'v-uv'){\bigg |}_{0}^{1}+\int _{0}^{1}u(Lv)\,dx\ .$

Rearrange terms of ('11') then ('1') is obtained. Q.E.D..