Integration by parts in Rn\mathbb R^n

  1. Review of the Basic Theorems

Review of the Basic Theorems

In what follows, UU is assumed to be an open, bounded subset of Rn\mathbb R ^n with C1C^1 boundary and uu, vv sufficiently regular functions. Then

∫Uuxidx=∫∂UuνidS \int_U u_{x_i} dx = \int_{\partial U} u\nu_i dS

is the basic identity from which the following are built. Applied to the product uvuv, this gives

∫Uuxivdx=−∫Uuvxidx+∫∂UuvνidS \int_U u_{x_i}v dx = -\int_U uv_{x_i} dx+\int_{\partial U} uv\nu_i dS

And applying (1) to the first-order derivatives uxiu_{x_i} and adding up the equalities we obtain

∫UΔudx=∫∂U∂u∂νdS \int_U \Delta u dx = \int_{\partial U} \frac{\partial u}{\partial \nu} dS

where ∂u∂ν\frac{\partial u}{\partial \nu} is defined to be equal to the inner product between the gradient and the outward unit vector, ⟨gradu,ν⟩\langle \textnormal{grad} u , \nu \rangle. Analogously, applying again (1) to the products uvxiu v_{x_i} and summing over ii one gets

∫U⟨gradu,gradv⟩dx=−∫∂UuΔvdx+∫∂Uu∂v∂νdS \int_U \langle \textnormal{grad} u, \textnormal{grad} v \rangle dx = -\int_{\partial U} u\Delta v dx +\int_{\partial U} u \frac{\partial v}{\partial \nu} dS

Finally, taking advantage of symmetry in the LHS of (4) by expressing it in terms of Δu\Delta u results in

∫U(uΔv−vΔu)dx=∫∂U(u∂v∂ν−v∂u∂ν)dS \int_U (u\Delta v-v\Delta u) dx = \int_{\partial U} \left ( u\frac{\partial v}{\partial \nu}-v\frac{\partial u}{\partial \nu} \right ) dS

Two often overlooked identities are related to the biharmonic operator and fall as corollaries of the former relations are

∫UuΔ2vdx=∫∂Uu∂Δv∂νdS−∫∂UΔv∂u∂νdS+∫UΔuΔvdx \int_U u \Delta^2 v dx= \int_{\partial U} u \frac{ \partial \Delta v}{\partial \nu} dS -\int_{\partial U} \Delta v \frac{\partial u}{\partial \nu} dS +\int_U \Delta u \Delta v dx

and for the symmetrical case

∫U(uΔ2v−vΔ2u)dx=∫∂U(u∂Δv∂ν−v∂Δu∂ν)dS+∫∂U(Δu∂v∂ν−v∂u∂ν)dS \int_U \left ( u \Delta^2 v-v\Delta^2 u \right )dx= \int_{\partial U} \left ( u \frac{\partial \Delta v}{\partial \nu}- v\frac{\partial \Delta u}{\partial \nu} \right ) dS +\int_{\partial U} \left ( \Delta u \frac{\partial v}{\partial \nu}- v\frac{\partial u}{\partial \nu} \right )dS

as an evident consequence of (6). To prove (6), use (4) to see that

∫UuΔ2vdx=−∫U⟨gradu,gradΔv⟩dx+∫∂Uu∂Δv∂νdS \int_U u \Delta^2 v dx = -\int_U \langle \textnormal{grad} u, \textnormal{grad} \Delta v \rangle dx +\int_{\partial U} u \frac{\partial \Delta v}{\partial \nu} dS

and that the integral over UU can in turn be decomposed as

∫U⟨gradu,gradΔv⟩dx=−∫UΔuΔvdx+∫∂UΔv∂u∂νdS \int_U \langle \textnormal{grad} u, \textnormal{grad}\Delta v \rangle dx= -\int_U \Delta u \Delta v dx + \int_{\partial U} \Delta v \frac{\partial u}{\partial \nu} dS
CC BY-SA 4.0 Miguel Bustamante. Last modified: August 08, 2023. Website built with Franklin.jl and the Julia programming language.