Proof of the divergence theorem Proof of the divergence theorem
Partial integration in  Rn  Partial integration in  Rn 
GREEN's formula GREEN's formula
GREEN's formula Index
© 2001-2007 Prof. Dr. Hans Wilhelm Alt, University Bonn, Germany

Partial Integration

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Definition: Let  Ω⊂Rn  be an open bounded set. For  k∈N 
Ck(clos(Ω);Y) := {
f: clos(Ω) → Y;   f is k-times continuously differentiable in Ω is in Ω k-mal stetig differenzierbar und
and all partial derivatives of order ≦k
have a continuous extension on clos(Ω) }.
Here, in general,  Y  can be a BANACH space. However, for the purpose of this lecture it suffices to consider the case  Y=Rl .

Let  u:Ω → Y  be  k -times continuously differentiable. For multiindices  α=(α1,...,αn)  of order  |α|=∑ i=1nαi≦k  (see definition:3-14)

∂αu := ∂1α1...∂nαn u .
Here
∂ilu :=


∂i...∂i
l-mal
u für l∈N,     ∂i0u := u .

Partial integration (Examples)    [sect:6-8]
In the following  Ω⊂Rn  is a bounded GAUSS domain (see definition:6-7), and for simplicity all functions are assumed to be in  C1(clos(Ω);Rn) .
  • [sect:6-8-(1)] For  f ∈C1(clos(Ω);R)  and  ϕ∈C1(clos(Ω);Rn) 
     
    ∫
    ∂Ω
    		 ϕf•νΩ dHn-1 =
     
    ∫
    Ω
    		 ϕ div (f)   dLn +
     
    ∫
    Ω
    		 ∇ϕ•f   dLn
  • [sect:6-8-(2)] For  u,v ∈C1(clos(Ω))  and  i∈{1,...,n} 
     
    ∫
    Ω
    		 u ∂i v dLn =
     
    ∫
    ∂Ω
    		 uv νΩ •ei dHn-1 -
     
    ∫
    Ω
    		 (∂i u) v dLn .
    Spezial case: If  u=0  on  ∂Ω  or  v=0  on  ∂Ω , then
     
    ∫
    Ω
    		 u∂iv dLn = -
     
    ∫
    Ω
    		 ∂iu v dLn .
  • [sect:6-8-(3)] For  f∈C1(clos(Ω);Y) ,  Y=Rl , and  i∈{1,...,n} 
     
    ∫
    Ω
    		 ∂i f dLn =
     
    ∫
    ∂Ω
    		 (νΩ•ei) f dHn-1 .
  • [sect:6-8-(4)] Let  u,v∈Ck(clos(Ω))  with  k≧1 . The following holds: If  ∂αv=0  on  ∂Ω  for all multiindices  α  of order  |α|≦k-1 , then
     
    ∫
    Ω
    		 u∂αv dLn = (-1)|α|
     
    ∫
    Ω
    		 ∂αu v dLn .

Proof sect:6-8-(1).   div (ϕf) = ϕ( div (f)) + (∇ϕ) f , wende dann satz:6-7 an.
Proof sect:6-8-(2). Diese Aussage folgt direkt aus sect:6-8-(1) mit  ϕ= u ,  f = v ei . Dann gilt:   div (f) = ∂i v ,  ∇ϕ•f = (∇u •ei) v = (∂i u) v .
Proof sect:6-8-(3). Wende sect:6-8-(2) an auf  u=1 ,  v  eine Komponente von  f .

Proof sect:6-8-(4). Führe den Beweis induktiv nach  |α| . Ist  |α|>0 , so gibt es ein  i  mit  αi>0 . Bezeichnet  ei  den  i -ten Einheitsvektor, so gilt deshalb  α=ei+β  für einen Multiindex  β  mit  |β|=|α|-1 . Ferner ist
∂αv = ∂i∂βv = ∂β(∂iv)
und  ∂βv=0  auf  ∂Ω . Mit sect:6-8-(2) und der Induktionsvoraussetzung, angewendet auf  ∂βv , folgt
 
∫
Ω
 u·∂αv dLn
=
 
∫
Ω
 u·∂i(∂βv) dLn
=
-
 
∫
Ω
 ∂iu·∂βv dLn
=
-(-1)|β|
 
∫
Ω
 ∂β(∂iu)·v dLn
=
(-1)|α|
 
∫
Ω
 ∂αu·v dLn .

Version 1.7
H.W. Alt - 02.01.2007

Proof of the divergence theorem Proof of the divergence theorem
Partial integration in  Rn  Partial integration in  Rn 
GREEN's formula GREEN's formula
GREEN's formula Index
© 2001-2007 Prof. Dr. Hans Wilhelm Alt, University Bonn, Germany