Partial differential equations and the finite element method 1--PDE

PDE

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Background

• The averaged quantities of many natural processes such as the deformation, density, velocity, pressure,temperature,concentration, or electromagnetic field are governed by partial differential equations (PDEs).
• Most PDEs used in practice only contain the first and second partial derivatives (we call them second-order PDEs).

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Definition

Let  O be an open set in Rd R d . A sufficiently general form of a linear second-order PDE in n independent variables z=(z1,…,zn)T z = ( z 1 , … , z n ) T is

−∑i,j=1n∂∂zi(aij∂u∂zj)+∑i=1n(ci∂u∂zi+∂biu∂zi)+a0u=f(z) − ∑ i , j = 1 n ∂ ∂ z i ( a i j ∂ u ∂ z j ) + ∑ i = 1 n ( c i ∂ u ∂ z i + ∂ b i u ∂ z i ) + a 0 u = f ( z )

where

aij=aij(z),bi=bi(z),ci=ci(z),a0=a0(z). a i j = a i j ( z ) , b i = b i ( z ) , c i = c i ( z ) , a 0 = a 0 ( z ) .

and

u(z)∈C2() u ( z ) ∈ C 2 ( O )

aij(z),bi(z),ci(z)∈C1() a i j ( z ) , b i ( z ) , c i ( z ) ∈ C 1 ( O )

a0(z),f(z)∈C() a 0 ( z ) , f ( z ) ∈ C ( O )

• Assume that the coefficient matrix
  
  A(z)={aij}ni,j=1 A ( z ) = { a i j } i , j = 1 n
  
  is symmetric.

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Definition 1.1 (Elliptic, parabolic and hyperbolic equations) Consider a second-order PDE with a symmetric coefficient matrix A(z).

1. The equation is said to be elliptic at z if A (z) is positive definite.

2. The equation is said to be parabolic at z if A(z) is positive semidefinite, but not positive definite, and the rank of (A(z), b(z), c(z)) is equal to n.

3.The equation is said to be hyperbolic at z if A (z) has one negative and n-1 positive eigenvalues.

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Operator notation

• It is customary to write elliptic PDEs in a compact form
  
  Lu=f L u = f
  
  where

Lu=−∑i,j=1n∂∂zi(aij∂u∂zj)+∑i=1n(ci∂u∂zi+∂biu∂zi)+a0u L u = − ∑ i , j = 1 n ∂ ∂ z i ( a i j ∂ u ∂ z j ) + ∑ i = 1 n ( c i ∂ u ∂ z i + ∂ b i u ∂ z i ) + a 0 u

is a second-order elliptic differential operator.

• In physics, one of the independent variables usually is the time t.

The typical operator form of parabolic equations is

∂u∂t+Lu=f ∂ u ∂ t + L u = f

where L is an elliptic differential operator.

• Typical second-order hyperbolic equation

∂2u∂t2+Lu=f ∂ 2 u ∂ t 2 + L u = f

where L is an elliptic differential operator.

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Examples:

Consider linear PED in 2 variables with constant coefficients:

auxx+buxy+cuyy+dux+euy+fu=g(x,y). a u x x + b u x y + c u y y + d u x + e u y + f u = g ( x , y ) .

a, b, and c cannot all be zero. Then

A=[ab/2b/2c],(1) (1) A = [ a b / 2 b / 2 c ] ,

• If b2–4ac>0 b 2 – 4 a c > 0 , hyperbolic. The wave equation:

α2uxx=utt α 2 u x x = u t t

• If b^2– 4ac = 0, parabolic. The heat conduction equation:

α2uxx=ut α 2 u x x = u t

• If b2–4ac<0 b 2 – 4 a c < 0 , elliptic. The Laplace equation:

  uxx+uyy=0 u x x + u y y = 0

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Definition 1.2 (Hadamard’s well-posedness) A problem is said to he well-posed if
 1. it has a unique solution, (existence and uniqueness)
 2. the solution depends continuously on the given data. (stable)
 Otherwise the problem is ill-posed.

Example: The One-Dimensional Heat Conduction Equation

Consider the one-dimensional version of the heat transfer equation with normalized
coefficients:

uxx=ut,Ω=(0,π),t∈(0,T) u x x = u t , Ω = ( 0 , π ) , t ∈ ( 0 , T )

The initial condition: the initial temperature distribution within the bar

u(x,0)=u0(x) u ( x , 0 ) = u 0 ( x )

such that u0(0)=u0(π)=0 u 0 ( 0 ) = u 0 ( π ) = 0 .

and the boundary condition: fix the temperature at the end points to

u(0,t)=u(π,t)=0 u ( 0 , t ) = u ( π , t ) = 0

and ask about the solution u(x,t) u ( x , t ) .

Note that:
1. If the boundary conditions specify u u , e.g. u(0,t)=f(t) u ( 0 , t ) = f ( t ) and u(π,t)=g(t) u ( π , t ) = g ( t ) , then they are often called Dirichlet conditions.
2. If they specify the (spatial) derivative, e.g. ux(0,t)=f(t) u x ( 0 , t ) = f ( t ) and ux(π,t)=g(t) u x ( π , t ) = g ( t ) , then they are often called Neumann conditions.
3. If the boundary conditions are linear combinations of u u and its derivative, e.g. αu(0,t)+βux(0,t)=f(t) α u ( 0 , t ) + β u x ( 0 , t ) = f ( t ) , then they are called Robin conditions.
If the specified functions in a set of condition are all equal to zero, then they
are homogeneous.

Our current example, is a homogeneous Dirichlet type problem.
The initial condition u0(x) u 0 ( x ) can be expressed by means of the Fourier expansion

u0(x)=1/2a+∑n=1∞bncos(nx)+∑n=1∞cnsin(nx) u 0 ( x ) = 1 / 2 a + ∑ n = 1 ∞ b n c o s ( n x ) + ∑ n = 1 ∞ c n s i n ( n x )

because u0(0)=u0(π)=0 u 0 ( 0 ) = u 0 ( π ) = 0 , that is

0=1/2a+∑n=1∞bn 0 = 1 / 2 a + ∑ n = 1 ∞ b n

and

0=1/2a+∑n=1∞bncos(nπ) 0 = 1 / 2 a + ∑ n = 1 ∞ b n c o s ( n π )

it is easy to get a=0,bn=0 a = 0 , b n = 0 . So we have

u0(x)=∑n=1∞cnsin(nx) u 0 ( x ) = ∑ n = 1 ∞ c n s i n ( n x )

Background: the Fourier series of a function f(x) is given by

f(x)=1/2a+∑n=1∞bncos(nx)+∑n=1∞cnsin(nx) f ( x ) = 1 / 2 a + ∑ n = 1 ∞ b n c o s ( n x ) + ∑ n = 1 ∞ c n s i n ( n x )

where

a=1/π∫π−πf(x)dx a = 1 / π ∫ − π π f ( x ) d x

bn=1/π∫π−πf(x)cos(nx)dx b n = 1 / π ∫ − π π f ( x ) c o s ( n x ) d x

cn=1/π∫π−πf(x)sin(nx)dx c n = 1 / π ∫ − π π f ( x ) s i n ( n x ) d x

Thus it is easy to verify that the exact solution u(x,t) u ( x , t ) has the form

u(x,t)=∑n=1∞cne−n2tsin(nx) u ( x , t ) = ∑ n = 1 ∞ c n e − n 2 t s i n ( n x )

General existence and uniqueness results

In the following we consider a pair of Hilbert spaces V V and W W , and an equation of the form

Lu=f L u = f

where L:D(L)⊂V→W L : D ( L ) ⊂ V → W is a linear operator and f∈W f ∈ W .

Theorem 1.1 (Hahn-Banach) Let U U be a subspace of a (real or complex) normed space V V , and g∈U′ g ∈ U ′ is a linear form over U U . Then there exists an extension h∈V′ h ∈ V ′ moreover satisfying ∥g∥U′=∥h∥V′ ∥ g ∥ U ′ =∥ h ∥ V ′ .

Note: Hahn-Banach THM 允许了定义在某个向量空间上的有界线性算子扩张到整个空间，并说明了存在「足够」的连续线性泛函，定义在每一个赋范向量空间. 该定理为线性泛函的保持范数不变的可延拓定理.

Theorem 1.2 (Basic existence result) Let V V , W W be Hilbert spaces and L:D(L)⊂V→W L : D ( L ) ⊂ V → W a bounded linear operator. Then R(L)=W R ( L ) = W if and only if both R(L) R ( L ) is closed and R(L)⊥=0 R ( L ) ⊥ = 0

Note1:就是说L的值域是W的充分必要条件为L的值域是闭的且值域的正交补是{0}。
Note2:It is easy to show that every continuous operator is closed. However, there are closed operators which are not continuous.

Theorem 1.3 (Basic existence and uniqueness result) Let V,W V , W be Hilbert spaces and
L:D(L)⊂V→W L : D ( L ) ⊂ V → W a closed linear operator. Assume that there exists a constant C>0 C > 0
such that

∥Lv∥W≥C∥v∥V,∀v∈D(L) ∥ L v ∥ W ≥ C ∥ v ∥ V , ∀ v ∈ D ( L )

(this inequality sometimes is called the stability or coercivity estimate). If R(L)⊥=0 R ( L ) ⊥ = 0 ,
then the operator equation Lu=f L u = f has a unique solution.

Note: the condition ∥Lv∥W≥C∥v∥V ∥ L v ∥ W ≥ C ∥ v ∥ V means that if Lv1=w1 L v 1 = w 1 is very close to Lv2=w2 L v 2 = w 2 , then δ>∥w1−w2∥W≥C∥v1−v2∥V δ >∥ w 1 − w 2 ∥ W ≥ C ∥ v 1 − v 2 ∥ V , that is v1 v 1 is very close to v2 v 2 , which can ensure uniqueness.

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ref: 1. Bolín, Pavel. Partial differential equations and the finite element method[M]. Wiley-Interscience, 2006.
2. https://www.math.psu.edu/tseng/class/Math251/Notes-PDE%20pt1.pdf