Lagrange's identity (boundary value problem)

In general terms, Lagrange's identity for any pair of functions u and v in function space C² (that is, twice differentiable) in n dimensions is:^[1] $${\displaystyle vL[u]-uL^{*}[v]=\nabla \cdot {\boldsymbol {M}},}$$ where: $${\displaystyle M_{i}=\sum _{j=1}^{n}a_{ij}\left(v{\frac {\partial u}{\partial x_{j}}}-u{\frac {\partial v}{\partial x_{j}}}\right)+uv\left(b_{i}-\sum _{j=1}^{n}{\frac {\partial a_{ij}}{\partial x_{j}}}\right),}$$ and $${\displaystyle \nabla \cdot {\boldsymbol {M}}=\sum _{i=1}^{n}{\frac {\partial }{\partial x_{i}}}M_{i},}$$

The operator L and its adjoint operator L^* are given by: $${\displaystyle L[u]=\sum _{i,\ j=1}^{n}a_{i,j}{\frac {\partial ^{2}u}{\partial x_{i}\partial x_{j}}}+\sum _{i=1}^{n}b_{i}{\frac {\partial u}{\partial x_{i}}}+cu}$$ and $${\displaystyle L^{*}[v]=\sum _{i,\ j=1}^{n}{\frac {\partial ^{2}(a_{i,j}v)}{\partial x_{i}\partial x_{j}}}-\sum _{i=1}^{n}{\frac {\partial (b_{i}v)}{\partial x_{i}}}+cv.}$$

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

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

Any second order ordinary differential equation of the form: $${\displaystyle a(x){\frac {d^{2}y}{dx^{2}}}+b(x){\frac {dy}{dx}}+c(x)y+\lambda w(x)y=0,}$$ can be put in the form:^[2] $${\displaystyle {\frac {d}{dx}}\left(p(x){\frac {dy}{dx}}\right)+\left(q(x)+\lambda w(x)\right)y(x)=0.}$$

This general form motivates introduction of the Sturm–Liouville operator L, defined as an operation upon a function f such that: $${\displaystyle Lf={\frac {d}{dx}}\left(p(x){\frac {df}{dx}}\right)+q(x)f.}$$

It can be shown that for any u and v for which the various derivatives exist, Lagrange's identity for ordinary differential equations holds:^[2] $${\displaystyle uLv-vLu=-{\frac {d}{dx}}\left[p(x)\left(v{\frac {du}{dx}}-u{\frac {dv}{dx}}\right)\right].}$$

For ordinary differential equations defined in the interval [0, 1], Lagrange's identity can be integrated to obtain an integral form (also known as Green's formula):^[3][4][5][6] $${\displaystyle \int _{0}^{1}dx\ (uLv-vLu)=\left[p(x)\left(u{\frac {dv}{dx}}-v{\frac {du}{dx}}\right)\right]_{0}^{1},}$$

where ${\displaystyle p=P(x)}$, ${\displaystyle q=Q(x)}$, ${\displaystyle u=U(x)}$ and ${\displaystyle v=V(x)}$ are functions of ${\displaystyle x}$. ${\displaystyle u}$ and ${\displaystyle v}$ having continuous second derivatives on the interval ${\displaystyle [0,1]}$.

We have: $${\displaystyle uLv=u\left[{\frac {d}{dx}}\left(p(x){\frac {dv}{dx}}\right)+q(x)v\right],}$$ and $${\displaystyle vLu=v\left[{\frac {d}{dx}}\left(p(x){\frac {du}{dx}}\right)+q(x)u\right].}$$

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

The leading multiplied u and v can be moved inside the differentiation, because the extra differentiated terms in u and v are the same in the two subtracted terms and simply cancel each other. Thus, $${\displaystyle {\begin{aligned}uLv-vLu&={\frac {d}{dx}}\left(p(x)u{\frac {dv}{dx}}\right)-{\frac {d}{dx}}\left(vp(x){\frac {du}{dx}}\right),\\&={\frac {d}{dx}}\left[p(x)\left(u{\frac {dv}{dx}}-v{\frac {du}{dx}}\right)\right],\end{aligned}}}$$ which is Lagrange's identity. Integrating from zero to one: $${\displaystyle \int _{0}^{1}dx\ (uLv-vLu)=\left[p(x)\left(u{\frac {dv}{dx}}-v{\frac {du}{dx}}\right)\right]_{0}^{1},}$$ as was to be shown.