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

In general terms, Lagrange's identity for any pair of functions u and v  in function space C ² (that is, twice differentiable) in n dimensions is:

$vL[u]-uL^{*}[v]=\nabla \cdot {\boldsymbol {M}}\ ,$

where:

$M_{i}=\Sigma _{j=1}^{n}a_{ij}\left(v{\frac {\partial u}{\partial x_{j}}}-u{\frac {\partial v}{\partial x_{j}}}\right)+uv\left(b_{i}-\Sigma _{j=1}^{n}{\frac {\partial a_{ij}}{\partial x_{j}}}\right)\ ,$

and

$\nabla \cdot {\boldsymbol {M}}=\Sigma _{i=1}^{n}{\frac {\partial }{\partial x_{i}}}M_{i}\ .$

The operator L and and its adjoint operator L^* are given by:

$L[u]=\Sigma _{i,\ j=1}^{n}a_{i,j}{\frac {\partial ^{2}u}{\partial x_{i}\partial x_{j}}}+\Sigma _{i=1}{n}b_{i}{\frac {\partial u}{\partial x_{i}}}+cu$

and

$L^{*}[v]=\Sigma _{i,\ j=1}^{n}{\frac {\partial ^{2}(a_{i,j}v)}{\partial x_{i}\partial x_{j}}}-\Sigma _{i=1}{n}{\frac {\partial (b_{i}v)}{\partial x_{i}}}+cv\ .$

If Lagrange's identity is integrated over a bounded region, then the divergence theorem can be used to form Green's second identity in the form:

$\int _{\Omega }vL[u]\ d\Omega =\int _{\Omega }uL^{*}[v]\ d\Omega +\int _{S}{\boldsymbol {M\cdot n}}\ dS\ ,$

where S is the surface bounding the volume Ω and n is the unit outward normal to the surface S.

Ordinary differential equations

Any second order ordinary differential equation of the form:

$a(x){\frac {d^{2}y}{dx^{2}}}+b(x){\frac {dy}{dx}}+c(x)y+\lambda w(x)y=0\ ,$

can be put in the form:

${\frac {d}{dx}}\left(p(x){\frac {dy}{dx}}\right)+\left(q(x)+\lambda w(x)\right)y(x)=0\ .$

If the operator L is defined as an operation upon a function f as:

$Lf={\frac {d}{dx}}\left(p(x){\frac {df}{dx}}\right)+q(x)f\ ,$

then using the above form for the general second order differential equation, it can be shown that for any u and v for which the various derivatives exist, Lagrange's identity for ordinary differential equations holds:

$uLv-vLu={\frac {d}{dx}}\left[p(x)\left(v{\frac {du}{dx}}-u{\frac {dv}{dx}}\right)\right]\ .$

For ordinary differential equations defined in the interval [0, 1], Lagrange's identity can be integrated to obtain the form:

(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 $\ [0,1]$. The "prime" notation ‘ '  ’ represents differentiation: ${\frac {d}{dx}}$. The symbol L is the Sturm–Liouville differential operator expressed in "prime" notation as:

(2)   $\ Lu=-(pu')'+qu.\$

Proof of form for ordinary differential equations

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..