Lagrange's identity (boundary value problem)

For other meanings of the term, see Lagrange's identity (disambiguation).

In the study of ordinary differential equations and their associated boundary value problems, Lagrange's identity gives the boundary terms arising from integration by parts of a self-adjoint linear differential operator. Lagrange's identity is fundamental in Sturm–Liouville theory. In more than one independent variable, Lagrange's identity is generalized by Green's second identity.

Statement

Lagrange's identity can be written:

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

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:

(2)   $\ Lu=-(pu')'+qu.\$

Proof

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

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

we have:

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

Replace $\ f(x)=u$, $\ g(x)=pv'$, $\ a=0$ and $\ b=1$ into the rule ('3') again, we have:

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

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

Replace ('5') into ('4'), we get:

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

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

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

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

Replace ('6') into ('7'), we have:

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

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

Rearrange terms of ('8') then ('1') is obtained. Q.E.D..