ch 3. boundary value problems

Ch 3. Boundary Value Problems

Yoon Mo Jung

Computational Science and Engineering

Yonsei University

Ch 3. Boundary Value Problems 2

§3.1 Differential Equations and Finite Elements

We want to solve

−d

dx

(c(x)

du

dx

)= f(x) : 2nd order DE.

If A =d

dx, then AT = −

d

dx, so ATCAu = f.

A = − (backward difference) AT = −+(−forward difference)

− =

1

−1 1

−1 1

−1 1

+ =

−1 1

−1 1

−1 1

−1

Un+1 − Un Un − Un−1

Consider

Boundary Value Problem −d

dx

(c(x)

du

dx

)= f(x)

Fixed : u(0) = 0

Free : w(1) = 0

where w(x) = c(x)dudx.

−∫ 1

0

d

dx

(c(x)

du

dx

)v(x)dx =

∫ 1

0

f(x)v(x)dx,

for every admissible v.

Applying integration by parts,

−∫ 1

0

d

dx

(c(x)

du

dx

)vdx =

∫ 1

0

c(x)du

dx

dv

dxdx − c(x)

du

dxv∣∣∣10,

c(x)du

dx(x)v(x)

∣∣∣10= c(1)

du

dx(1)︸︷︷︸

w(1)=0

v(1) − c(0)du

dx(0) v(0)︸︷︷︸

impose 0

= 0.

Computational Science & Engineering (CSE) Yoon Mo Jung


Figure 1: Hanging bar.

Weak form :

∫ 1

0

c(x)du

dx

dv

dxdx =

∫ 1

0

f(x)v(x)dx for every admissible v.

: Finite Element Method or Galerkin Method

• The Framework ATCA for a Hanging Bar

The bar stretches like a line of a spring. i.e. One may treat it as a continuum

limit of line of springs. See Figure 1. The point that begins at distance x

along the unstretched bar moves down to x + u(x). We want to find the

displacement u(x).

The internal force at x = 1 must be zero because no external force will

balance it. Thus w(1) = 0.

Note that the bottom end actually move down to x = 1 + u(1). Nev-

ertheless the boundary condition is written at x = 1. The reason is that

displacements u(x) are assumed small.

e(x): the elongation per unit length, called strain.

Step 1 Stretching e(x) = Au(x) = dudx

with u(0) = 0.

cf) ei = ui−ui−1 in the mass-spring system. The difference becomes

derivative.

Step 2 Hooke’s Law w(x) = c(x)e(x) = c(x)dudx.

At a point where the material of the bar suddenly changes, c(x) will



have a jump. The product w(x) = c(x)u′(x) will not jump. The

force balance at the point demands thatw(x) must be continuous. The

jump in c(x) must compensated by a jump in e(x) = dudx.

Step 3 Force balance ATw = −dwdx

= f(x) with w(1) = 0.

See the small piece of bar with thickness x in Figure 1. The internal

forces are w(x) upward and w(x + x) downwards. In equilibrium,

w(x) = w(x+x)+fx ⇒ −w(x + x) − w(x)

x= f(x).

Note that AT = − ddx

is the “transpose” of A = ddx.

displacement u(x)

u(0) = 0→ strain-displacement

e(x) = Au(x) = dudx

→ strain

elongatione(x)

→stress-strain law

(Hooke’s law)

w(x) = c(x)e(x) = c(x)dudx

→ internal stress w(x)

w(1) = 0

→ force balance

f(x) = ATw(x) = −dwdx

→ external force

per unit lengthf(x)

Again we have:

Ku := ATCAu = f .

• General Solution and Examples

Since −(cu′)′ = f and cu′ = w,

c(x)du

dx(x) − c(0)

du

dx(0) = w(x) − w(0) = −

∫ x

0

f(x)dx.

Since w(1) = 0, −w(0) = w(1) − w(0) = −∫ 1

0 f(x)dx.

Internal force (stress) w(x) =

∫ 1

0

f(x)dx −∫ x

0

f(x)dx =

∫ 1

x

f(x)dx.



It means that the internal force W at the point x balances the total applied

force f below that point.

From cu′ = w and integrating,

Displacementdu

dx=

w(x)

c(x)and

∫ x

0

w(x)

c(x)dx = u(x)

by noticing u(0) = 0.

Example 1 A uniform bar is hanging under its own weight. The force per

unit length is f0 = ρg.

Sol) w(x) =∫ 1

x f0dx = (1 − x)f0

and

u(x) =∫ x

0(1−x)f0

c(x)dx =

(x − 1

2x2)

f0c.

Example 2 A bar is hanging under its own weight. In this case, c(x) = 1

if 0 ≤ x ≤ 1/2 and 2 if 1/2 < x ≤ 1. The force per unit length is

f0 = ρg.

Sol) w(x) =∫ 1

x f0dx = (1 − x)f0, which is continuous.

e(x) = dudx

= w(x)c(x)

= (1−x)f0c(x)

, which is discontinuous at x = 1/2.

u(x) =∫ x

0(1−x)f0

c(x)dx =

(x − 1

2x2)f0 if 0 ≤ x ≤ 1/2,

38f0 +

∫ x

1/2(1−x)f0

2dx if 1/2 ≤ x ≤ 1.

=(x − 1

2x2)

f02+ 3

16f0

• The Transpose of A = ddx

Why is the transpose of A = ddx

given by AT = A∗ = − ddx?

What are correct boundary conditions on A and AT ?

The underlying physics is the connection of energy eTw to work (force

·displacement) uTf . The mathematical reason is this requirement for inner

products:

AT is defined by (Au︸︷︷︸e

)Tw = uT (ATw︸︷︷︸f

) for every u and w.



Inner product of functions (e, w) =

∫ 1

0

e(x)w(x)dx

Notice that∫ 1

0 e(x)w(x)dx = limn→∞∑n

i=1 e(xi)w(xi)x.

Letting A = ddx, formally,

(Au,w) =

∫ 1

0

du

dxw(x)dx =

∫ 1

0

u(x)

(d

dx

)T

w(x)dx = (u,ATw)

for every u and w.

By taking integration by parts,∫ 1

0

du

dxw(x)dx =

∫ 1

0

u(x)

(−dw

dx

)dx +

[u(x)w(x)

]x=1

x=0.

For − ddx

to be AT ,[u(x)w(x)

]x=1

x=0= 0 is needed.

Boundary condition

• If A has boundary condition u(0) = 0, AT must have w(1) = 0.

• If A has boundary condition u(0) = u(1) = 0, AT has no condition.

• If A has no condition, AT has two conditions w(0) = 0 and w(1) =

0.

cf) ATCAu = f ⇒ (ATCAu)Tv = fTv for all v

(CAu)TAv = fTv for all v

or ⟨CAu,Av⟩ = ⟨f, v⟩ for all v

• Galerkin’s Method (Finite Element Method, FEM)

Consider the fixed-free case:

Boundary Value Problem −d

dx

(c(x)

du

dx

)= f(x)

Fixed : u(0) = 0

Free : w(1) = 0



where w(x) = c(x)dudx.

−∫ 1

0

d

dx

(c(x)

du

dx

)v(x)dx =

∫ 1

0

f(x)v(x)dx,

for every admissible v.

Applying integration by parts,

−∫ 1

0

d

dx

(c(x)

du

dx

)vdx =

∫ 1

0

c(x)du

dx

dv

dxdx − c(x)

du

dxv∣∣∣10,

c(x)du

dx(x)v(x)

∣∣∣10= c(1)

du

dx(1)︸︷︷︸

w(1)=0

v(1) − c(0)du

dx(0) v(0)︸︷︷︸

=0 :required

= 0.

Weak form :

∫ 1

0

c(x)du

dx

dv

dxdx =

∫ 1

0

f(x)v(x)dx for every admissible v.

Admissibility condition: v(0) = 0.

Finite Element Method is based on the weak form with test functions v(x).

Galerkin Method: Choose n trial functions ϕ1(x), ϕ2(x), · · ·ϕn(x).

U(x) = U1ϕ1(x) + U2ϕ2(x) + · · · + Unϕn(x)

Choose n admissible test functions V1(x), V2(x), · · ·Vn(x).∫ 1

0

c(x)

n∑j=1

Uj

dϕj

dx

dVi

dxdx =

∫ 1

0

f(x)Vi(x)dx

for i = 1, · · ·n: n equations.

KU = F Kij =

∫ 1

0

c(x)dVi

dx

dϕj

dxdx and Fi =

∫ 1

0

f(x)Vi(x)dx

(1)

Construction of the Finite Element Method



Figure 2: blue lines: hat functions ϕi and a combination (piecewise linear): red line.

1. Choose the ϕi and Vi: one unknowns for each ϕ, one equation for each

V .

2. Compute exact or approximate integrals (1). If all ϕi = Vi, then

Kij = Kji.

3. The weak form becomes KU = F . The FEM approximation is∑nj=1 Uiϕi(x).

• Linear Finite Elements

There are three hat functions ϕ1(x), ϕ2(x), ϕ3(x) in Figure 2 and they

are also the test function V1, V2, V3. The functions are local-they are zero

outside intervals of length 2h, 2h, h. There is no overlap between ϕ1 and

the half-hat ϕ3, which makes the integral zero: K13 = K31 = 0. The

stiffness matrix K is tridiagonal.

Example 3 Consider the fixed-free case:

−u′′ = 1, u(0) = 0, w(1) = 0

The true solution is u(x) = x − 12x2.

Since Fi =∫ 1

0 f(x)Vi(x)dx,

F1 =

∫ 1

0

V1(x)dx︸︷︷︸area

= 2 · 12· 13· 1 = 1

3, F2 = 1

3, F3 = 1

213.



Since Kij =∫ 1

0 c(x)dVidx

dϕj

dxdx,

K11 =∫ 1

0

(dV1dx

)2dx = 32 · 1

3+ (−3)2 · 1

3= 6,

K22 = 6, K33 = 32 · 13= 3.

K12 =∫ 1

0 c(x)dV1dx

dϕ2dx

dx = −3 · 3 · 13= −3,

K23 = −3, K13 = 0 (No overlap).

Finally we have KU = F : 6 −3 0

−3 6 −3

0 −3 3

U1

U2

U3

=

1/3

1/3

1/6

gives

U1

U2

U3

=

5/18

4/9

1/2

.

All three values U1, U2, U3 are agree exactly with u(x) = x− 12x2 since

it is a second order scheme: O(h2).

• Comparison with Finite Differences

KU = F in Example 3 is like 1h(−2U) = hf .

cf) Finite difference method: from 1h2(−2U) = f ,

32

2 −1 0

−1 2 −1

0 −1 1

U1

U2

U3

=

1

1

1

.

Notice that it is the first order accurate: O(h).

Example 4 Consider the following DE:

−d

dx

(c(x)

du

dx

)= f(x), f(x) = δ(x − 1/2),

c(x) =

2, 0 ≤ x ≤ 1

3

4, 13≤ x ≤ 1

, u(0) = 0, and w(1) = 0.



First we find the exact solution:

ATCAu = f ⇒

ATw = f,

CAu = w.

ATw = f : −dw

dx= δ(x − 1/2), with w(1) = 0.

⇒ w(x) =

1, x < 1/2,

0, x > 1/2.

CAu = w : c(x)du

dx= w(x) with u(0) = 0.

⇒ u(x) =

∫ x

0

w(x)

c(x)dx =

∫ x

012dx = 1

2x, 0 ≤ x ≤ 1

3,

16+∫ x

1/314dx = 1

4x + 1

12, 1

3≤ x ≤ 1

2,

112

+ 18+∫ x

1/2 0dx = 524, 1

2≤ x ≤ 1.

Now we use finite elements:

F1 =∫ 1

0 δ(x − 12)V1(x)dx = V1(

12) = 1

2,

F2 =∫ 1

0 δ(x − 12)V2(x)dx = 1

2, and F3 = 0.

K11 =∫ 1

0 c(x)(V ′1 )

2dx = 13· 2 · 32 + 1

3· 4 · 32 = 18,

K12 =∫ 1

0 c(x)V ′1V

′2dx = 1

3· 4 · (−3) · 3 = −12, and K13 = 0.

K22 = 24, K23 = −12, K33 = 12.

KU = F :

18 −12 0

−12 24 −12

0 −12 12

U1

U2

U3

=

1/2

1/2

−0

.

Example 5 Allow any c(x) and f(x), using numerical integration.



If we apply the midpoint rule∫ b

a f(x)dx = (b − a)f(a+b2),

F1 =

∫ 1/3

0

f(x)V1(x)dx +

∫ 2/3

1/3

f(x)V1(x)dx

≈1

3· f(1/6) ·

1

2+

1

3· f(3/6) ·

1

2.

Similarly,

K11 =

∫ 1/3

0

c(x)(V ′1 )

2dx +

∫ 2/3

1/3

c(x)(V ′1 )

2dx

≈1

3· c(1/6) · 32 +

1

3· c(3/6) · (−3)2.

Finite differences become painful when the mesh spacing changes since sec-

ond differences need a new formula.

Finite elements have geometric flexibility in 2 and 3 dims.

• More accurate Finite Elements

To increase accuracy, one may

1. refine mesh.

2. increase the degree of finite elements.

ex) quadratic elements: O(h3).



§3.2 Cubic Splines and Forth-Order Equations

Goal - want to solve the following forth order equation:d4udx4 = f(x): bending cf) d2u

dx2 = f(x): stretching

U(x) = A + Bx + Cx2 + Dx3 U(x) = A + Bx

piecewise cubic piecewise linear

:cubic finite elements :linear finite elements

U and U ′ are continuous. U is continuous.

U ′′ and U ′′′ can jump. U ′ can jump.

cf) cubic splines - approximation theory

U , U ′ and U ′′ are continuous. U ′′′ can jump.

U ′′: curvature, bending (concave, convex).

• Forth-Order Equations: Beam Bending

A beam is a horizontal structural element that is capable of withstanding load

primarily by resisting bending. The bending produces internal forces that try

to straight it. The bending force induced into the material of the beam as a

result of the external loads, own weight, span and external reactions to these

loads is called a bending moment.

The displacement u and load f , previously in the direction of the rod, are

now perpendicular to the beam. See Figure 3 (a).

(a) Simply supported beam (b) Cantilever Beam

Figure 3: Beam examples



beam cf) elastic rod

Restoring force: curvature stretching

Au = d2udx2 Au = du

dxIf it is zero, the displacement u is linear and the beam is straight.

Bending moment: M = CAu = cu′′, c: bending stiffness.

Force balance: M ′′ = f(x), equation of equilibrium (between the restoring

force and the applied force).

ATCA for a beam:

displacement u

curvature e = Au = u′′

bending moment M = ce

force balance f = ATM = M ′′

ATCAu = f ord2

dx2

(cd2u

dx2

)= f

(d2

dx2

)T= d2

dx2 : symmetric.

cf)(

ddx

)T= − d

dx: skewsymmetric or asymmetric.

(Au,M) = (u,ATM) =︸︷︷︸want to have

(u,AM).

∫ 1

0

u′′Mdx = −∫ 1

0

u′M ′dx + u′M |10

=

∫ 1

0

uM ′′dx + [u′M − uM ′]10︸︷︷︸impose 0

.

[u′M−uM ′]10 = u′(1)M(1)−u(1)M ′(1)−u′(0)M(0)+u(0)M ′(0).

1. simply supported end: u = 0, M = 0. See Figure 3(a).

2. clamped (fixed) end: u = 0, u′ = 0. See the left end in Figure 4.



Figure 4: Bending a cantilever beam by perpendicular force: (cu′′)′′ = f(x).

3. free end: M = 0, M ′ = 0. See the right end in Figure 4.

4. sliding clamped end: u′ = 0, M ′ = 0. (rare)

Bendingd2

dx2

(cd2u

dx2

)= f or M = c

d2u

dx2and

d2M

dx2= f

M = cd2udx2 = Au requires two conditions on u.

d2Mdx2 = AM = ATM = f requires two conditions on M .

d2

dx2

(cd2u

dx2

)= f ⇒

∫ 1

0

d2

dx2

(cd2u

dx2

)v(x)dx =

∫ 1

0

f(x)v(x)dx

By integration by parts twice,

Weak form:

∫ 1

0

cd2u

dx2

d2v

dx2dx =

∫ 1

0

f(x)v(x)dx

Finite elements methods: U(x) =∑

Uiϕ(x) (trial functions)

Vi(x)(= ϕi(x)) (test functions)will give us discrete system KU = F .

• Cubic Finite Elements

The four coefficients of A + Bx + Cx2 + Dx3 are determined by the

endpoint displacement Udi , U

di+1 and the endpoint slope U s

i , Usi+1. Since

the neighboring intervals end with the same height and slope, the overall

piecewise cubic will have continuous height and slope.



Figure 5: Displacement trial function ϕdi and slope function ϕs

i .

Displacement trial function Slope trial function

ϕd(X) = (|X| − 1)2(2|X| + 1) ϕs(X) = |X(|X| − 1)2

ϕd(0) = 1 dϕs

dX(0) = 1

ϕd(−1) = ϕdi (1) = 0 dϕs

dX(−1) = dϕs

dX(1) = 0

dϕd

dX(−1) = dϕd

dX(0) = dϕd

dX(1) = 0 ϕs(−1) = ϕs(0) = ϕs(1) = 0

See Figure 5.

The finite element approximation U(x) is a combination of these 2N trial

functions (called Hermite cubics):

U(x) = Ud1ϕ

d1(x) + U s

1ϕs1(x) + · · · + Ud

Nϕd1(x) + U s

Nϕs1(x)

We need 2N test functions. Normally these V ’s will be the same as the

ϕ’s to preserve symmetry of the matrix K.

X = x−ihh

: This connect the global coordinate x to the local coordinate

X.

We compute integrals on that standard interval of length 1, and rescale to

the actual x-interval of length h. The local-global mapping is a basic steps

in finite element code.

ϕdi (x) = ϕd

(x − ih

h

), ϕs

i(x) = ϕs

(x − ih

h

).

1. Hat functions could not be used in bending problems, because then

we are multiplying delta functions in U ′′ and V ′′. The integral of δ2



Figure 6: Assembled stiffness matrix

is infinite (∫∞−∞ δ(x)dx = 1 but

∫∞−∞ δ2(x)dx = ∞). Another

reason is that integration by parts twice doesn’t work for hat functions.

2. The error u(x)−U(x) is of order h4, because the trial space include

cubics. U(x) could match the Taylor series of u(x) in each interval,

until the x4 term produces an h4 error. The actual U(x) from finite

elements does not match the Taylor series, because it is based on in-

tegrals (the weak form). U(x) gives the smallest possible error in the

energy∫ 1

0 c(x)(d2u

dx2 − d2Udx2 )

2dx.

3. In practice, the integrals are computed separately over each interval

[ih, (i+ 1)h]. On that interval only four trial functions are non zero:

ϕdi , ϕ

si , ϕ

di+1, ϕ

si+1. Similarly only four test functions are nonzero. This

produces 16 integrals in∫cU ′′V ′′, which go into the 4 × 4 element

stiffness matrix.

How those element matrices assemble into the global stiffness matrixK,

see Figure 6. Since ϕdi and ϕs

i also contribute to the previous element

matrix Ki−1, there will be a 2 by 2 overlap in K and 6 nonzero per

low.

• Finite Differences for (cu′′)′′ = f(x)

(2u)i = ui+1 − 2ui + ui−1.

(4u)i = (22u)i = ui+2 − 4ui+1 + 6ui − 4ui−1 + ui−2.



§3.3 Gradient and Divergence

So far we considered 1-d problems:

In ATCA system, A = ddx, then AT = − d

dxby integration by parts.

In 2-d or 3-d, A = ∇ = grad, then AT = −∇· = −div by Gauss-Green

formula.

∇u(x, y) = grad u =

[∂u∂x∂u∂y

], ∇ = grad =

[∂∂x∂∂y

].

∇·[w1

w2

]= div

[w1

w2

]=

∂w1

∂x+

∂w2

∂y, ∇· = div =

[∂

∂x,

∂

∂y

].

potential

u(x)

A=grad−−−−−−−−−−−−→boundary cond u=u0(x,y)

velocity

v(x, y) = (v1, v2)

C−−−→w=cv

flow rate

w(x, y) = (w1, w2)

AT=−div−−−−−−−−−−−→bd cond w·n=F0(x,y)

source

f(x, y) = −div w

Ku := ATCAu = f

Poisson equation: −div (c grad )u = −∇ · (c∇u) = f

Laplace equation: −div grad u = ∇ · ∇u = ∇2u = 0

• Gradient

Definition 6 Gradient

f : R2,R3 or Rn → R

∇f(x, y) = grad f =

[∂f∂x∂f∂y

]=

[∂∂x∂∂y

]f



The gradient of a scalar field is a vector field that points in the direction of

the greatest rate of increase of the scalar field, and whose magnitude is the

greatest rate of change.

|grad f |: maximum slope (steepest slope).grad f|grad f |: direction of maximum increase.

Definition 7 Directional derivative

Dbf(x)def= lim

h→0

f(x + hb) − f(x)

h

chain rule= ∇f · b with |b| = 1.

The directional derivative of a multivariate differentiable function along a

given vector b at a given point x intuitively represents the instantaneous

rate of change of the function, moving through x in the direction of b.

• Divergence

Definition 8 Divergence

v : R2 → R2, or R3 → R3

(x, y) 7→ (v1(x, y), v2(x, y))

∇ · v : R2 or R3 → R

∇ · v = div v =∂v1

∂x+

∂v2

∂y=

[∂

∂x,∂

∂y

]· [v1, v2]

Divergence is a vector operator that measures the magnitude of a vector

field’s source or sink at a given point, in terms of a signed scalar. More

technically, the divergence represents the volume density of the outward flux

of a vector field from an infinitesimal volume around a given point.

Physicist defines as follows:

∇ · v = limV →0

∮S v · ndAV

(= lim

volume→0

net outward flux

volume

)Computational Science & Engineering (CSE) Yoon Mo Jung


Theorem 9 Divergence Theorem of Gauss:∫V

∇ · vdV =

∫∂V

v · ndA.

In words, the net divergence in the body is the net flux through the boundary

surface.

Theorem 10 Green’s Formula:

First identity :∫V (f∇

2g + ∇f · ∇g)dV =∫∂V f ∂g

∂ndA

Second identity :∫V (f∇

2g − g∇2f)dV =∫∂V

(f ∂g

∂n− g∂f

∂n

)dA

cf) Integration by part:∫ b

a fg′dx = fg∣∣ba−∫ b

a f ′gdx.

Now,∫ b

a fg′′dx = fg′∣∣ba−∫ b

a f ′g′dx.

• Curl

Definition 11 curl

v : R3 → R3

(x, y, z) 7→ (v1(x, y, z), v2(x, y, z), v3(x, y, z))

∇× : R2 or R3 → R2 or R3

curl v =∇ × v

=

(∂v3

∂y−

∂v2

∂z

)i −

(∂v3

∂x−

∂v1

∂z

)j +

(∂v2

∂x−

∂v1

∂y

)k

=

∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

v1 v2 v3

∣∣∣∣∣∣∣Computational Science & Engineering (CSE) Yoon Mo Jung


Curl is a vector operator that describes the infinitesimal rotation of a vector

field. At every point in the field, the curl is represented by a vector. The

attributes of this vector (length and direction) characterize the rotation at

that point. The direction of the curl is the axis of rotation, as determined

by the right-hand rule, and the magnitude of the curl is the magnitude of

rotation. If the vector field represents the flow velocity of a moving fluid,

then the curl is the circulation density of the fluid. A vector field whose curl

is zero is called irrotational.

Physicist defines as follows:

∇×v = maxa

a limA→0

∮C F · r′(s)ds

A

(= lim

area→0

circulation along curve

enclosed area

)Theorem 12 Stokes’ Theorem:∫

S

∇ × F · ndA =

∮C

F · r′(s)ds.

In words, the net circulation through normal direction is the circulation along

the boundary.

Theorem 13 Green’s Theorem:∫∫R

∂F2

∂x−

∂F1

∂ydxdy =

∮C

F1dx + F2dy.

It is just the 2-dim version of Stokes’ Theorem.

We may derive 2d divergence theorem using Green’s theorem:∫∫R

∂v2

∂x−

∂v1

∂ydxdy =

∫B

v1dx + v2dy.

Let v2 = w1 and v1 = −w2.∫∫R

∂w1

∂x+

∂w2

∂ydxdy =

∫B

−w2dx+w1dy =

∮B

w1dy+w2dx.



Figure 7: The integral of w · nds = (w1, w2) · (dy,−dx) = w1dy − w2dx is the flux. In casew2 = 0, this

∮w1dy equals

∫∫∂w1/∂xdxdy.

Figure 7 shows the flux w · n︸︷︷︸“flow rate out”

ds = w1dy−w2dx through a bound-

ary segment of length ds, which we integrate around the boundary B.

If we consider the horizontal flow (w1, 0), the divergence theorem becomes∫∫R

∂w1

∂xdxdy =

∫B

w1dy.

cf) Fundamental Theorem of Calculus:∫ b

adwdx

dx = w(b) − w(a).

It is the divergence theorem in 1D.

Remark Two important identities

1. curl (grad u) = 0: The gradient field is irrotational.

2. div (curl w) = 0.

Question:

1. If v = grad u, then curl v = 0.

If curl v = 0, v = grad u for some u?

2. If s = curl w, then div s = 0.

If div s = 0, s = curl w for some w?

Answer: Yes (Under some condition).

• Converse of the Two Identities



Kirchhoff Voltage Law (KVL): The sum of potential differences

around any loop must be zero. - No circulation.

eTw = 0 for all loops w ⇒ e = −Au.

We change concept: voltage potential and potential difference e velocity v. Then v = −Au.

Continuous form of eTw = 0 is

Zero circulation :

∫C

v1dx + v2dy = 0 for any loop C.

By Green’s theorem, ∫∫R

∂v2

∂x−

∂v1

∂ydxdy.

Since C is arbitrary, so is the enclosed region R, we have

Zero vorticity :∂v2

∂x−

∂v1

∂y= 0 for any point.

Zero vorticity is “no rotation”.

Example 14 v(x, y) = (2x, 2y) and V (x, y) = (2x,−2y) are gra-

dient fields.

curl v = ∂v2∂x

− ∂v1∂y

= 0.∂u∂x

= v1 = 2x ⇒ u(x, y) = x2 + F (y).∂u∂y

= F ′(y) = v2 = 2y ⇒ F (y) = y2 + C.

Hence u(x, y) = x2 + y2 + C.

Similarly, U(x, y) = x2 − y2 + C.

Example 15 w(x, y) = (2y,−2x) is not gradient field.

curl w = ∂w2∂x

− ∂w1∂y

= −4 = 0.

If w = grad u was true,∂u∂x

= w1 = 2y ⇒ u(x, y) = 2xy + F (y).∂u∂y

= 2x + F ′(y) = w2 = −2x ⇒ F ′(y) = −4x ????



Figure 8: The velocity fields v(x, y) = (2x, 2y) and V (x, y) = (2x,−2y) are perpendicular tothe equipotential curves u(x, y) = x2+y2+C (circles) and U(x, y) = x2−y2+C (hyperbolas).

Figure 9: w(x, y) = (2y,−2x) is tangent to the circles: rotation field, not a gradient.

Note that [x, y]T [2y,−2x] = 0.

2

[cos(−π/2) − sin(−π/2)

sin(−π/2) cos(−π/2)

] [x

y

]= 2

[0 1

−1 0

] [x

y

]=

[2y

−2x

]div w = 2 − 2 = 0. w = curl s := curl (0, 0, s).

Kirchhoff Current Law (KCL): Flow-in equals Flow-out. ATw = 0

in network - conservation of mass, source free.

Continuous form of ATw = 0 is

Solenoidal flow : div w = 0.

Example 16 Find the divergence of v(x, y) = (2x, 2y), V (x, y) =

(2x,−2y), and w(x, y) = (2y,−2x).

div v = 4 = source: Mass is not preserved.

div V = 0 and div w = 0: conservation of mass, source-free.



Definition 17

Stream function s(x, y) : w1 =∂s

∂yand w2 = −

∂s

∂x.

If div w = 0, then w = curl s for some s.

Note that if w1 = ∂s∂y

and w2 = −∂s∂x

,

div w =∂w1

∂x+

∂w2

∂y=

∂2s

∂x∂y−

∂2s

∂y∂x= 0.

curl s = curl (0, 0, s(x, y)) =

(∂s

∂y,−

∂s

∂x, 0

).

Physically, the curves s(x, y) = c are the streamlines. The flow travels

along those streamlines.

Example 18 Find stream functions of V (x, y) = (2x,−2y), and

w(x, y) = (2y,−2x).∂s∂y

= 2y,

−∂s∂x

= −2y.⇒ s(x, y) = 2xy + F (x)

∂s∂x

= 2y + F ′(x) = 2y.⇒ s(x, y) = 2xy.

Similarly, s(x, y) = x2 + y2 for w.

Consider V (x, y) = (2x,−2y).

V = grad u, u(x, y) = x2 − y2. Note that curl V = 0.

V = curl s, s(x, y) = 2xy. div V = 0.

div (grad u) = ∇2u = u = 0 :Laplace equation

u is a harmonic function.

What is the relation between u(x, y) and s(x, y)?

z2 = (x + iy)2 = x2 − y2︸︷︷︸u(x,y)

+ i2xy︸︷︷︸s(x,y)

.



§3.4 Laplace Equation

Laplace equation (in 2D): ∇2u = u = uxx + uyy = 0.

u(x, y) = x2 − y2 : potential function

s(x, y) = 2xy : stream function

One interpretation: When the temperature is u, heat flows along the stream-

lines s = constant.

z2 = (x + iy)2 = x2 − y2︸︷︷︸u(x,y)

+ i2xy︸︷︷︸s(x,y)

.

Furthermore, u = 0 and v = 0.

The case zn = (x + iy)n

By chain rule,

(x + iy)n =∂2

∂x2(x + iy)n +

∂2

∂y2(x + iy)n

=n(n − 1)(x + iy)n−2 − n(n − 1)(x + iy)n−2 = 0.

un = Re (x + iy)n and vn = Im (x + iy)n

General case: f(x + iy)

f(x + iy) =∂2

∂x2f(x + iy) +

∂2

∂y2f(x + iy)

=f ′′(x + iy) − f ′′(x + iy) = 0.

Let f(x + iy) = u(x, y) + iv(x, y).

∂2

∂x2f = ∂2u∂x2 + i∂

2v∂x2

∂2

∂y2f = ∂2u

∂y2+ i∂

2v∂y2

f = u + iv

f = 0 ⇔ u = 0 and v = 0.



Remark This magic doesn’t work in 3 dimensions, for ∇2u = u =

uxx + uyy + uzz.

Definition 19 Analytic function:

f(x + iy) = f(z) = c0 + c1z + c2z2 + · · · =

∞∑n=0

anzn.

Example 20

∞∑n=0

zn

n!= ez = ex+iy = exeiy = ex cos y︸︷︷︸

u(x,y)

+i ex sin y︸︷︷︸v(x,y)

.

Harmonic functions : u(x, y) = Re [f(x + iy)],

v(x, y) = Im [f(x + iy)].

• The Cauchy-Riemann Equations

The potential u and stream function s are closely linked. Physically, the

flow velocity v = grad u is along the streamlines s = constant. Ge-

ometrically, the equipotentials u = constant are perpendicular to those

streamlines. Mathematically, u and s both come from the same analytic

function f(x + iy) = u(x, y) + iv(x, y) (This magic doesn’t work in

3 dimensions).

Notice that we we write f(x + iy) = u(x, y) + iv(x, y),

f : C → C but u, v : R2 → R2.

One must distinguish R2 and C:R2 is a vector space, which has no multiplication, so no division.

C is a field, which has multiplication, so division by nonzero is possible.

As a consequence, differentiation is possible on complex valued function but

only partial derivatives are allowed in R2.



Definition 21

f ′(z) = limz→0

f(z + z) − f(z)

z.

Remark Note that z is of the form x + iy, including all possible

directions.

z → 0 ⇔ x → 0 and y → 0.

Thus, to consider the existence of f ′(z), one must consider all possible

combinations of x and y which x → 0 and y → 0.

First we consider z = x, the case y ≡ 0.

limx→0

f(z + x) − f(z)

x

= limx→0

u(x + x, y) + iv(x + x, y) − (u(x, y) + iv(x, y))

x

= limx→0

u(x + x, y) − u(x, y)

x+ i lim

x→0

v(x + x, y) − v(x, y)

x.

Now we invoke multivariable calculus,

limx→0

f(z + x) − f(z)

x= ux + ivx.

Now we consider the case z = iy, x ≡ 0. A similar calculation

leads:

limy→0

f(z + iy) − f(z)

iy

= limy→0

u(x, y + y) − u(x, y)

iy+ i lim

x→0

v(x, y + y) − v(x, y)

iy

= − iuy + vy.

If derivative exists, those two limits must be identical. Hence if f is analytic

(differentiable),

Cauchy-Riemann Equations: ux = vy, uy = −vx.



Remark Cauchy-Riemann equations reveals the relation between complex

differentiation and multivariable calculus on R2.

Example 22 f(z) = z, f(x + iy) = x − iy.

u(x, y) = x, v(x, y) = −y.

ux = 1 = −1 = vy: z is not analytic (not differentiable).

Its analog in R2 is g(x, y) = (x,−y) := (g1, g2).

Jg =

[∂g1∂x

∂g1∂y

∂g2∂x

∂g2∂y

]=

[1 0

0 −1

].

Partial derivatives can be taken as many times as we want, since it consider

the instantaneous change along x or y direction.

From the Cauchy-Riemann equations,

u = uxx + uyy = (vy)x + (−vx)y = 0.

Similarly v = 0.

Furthermore,

grad u·grad v =

[ux

uy

]·[vx

vy

]= uxvx+uyvy = vyvx+(−vx)vy = 0.

grad u⊥grad v: The equipotentials u = constant are perpendicular to the

streamlines s = constant.

Remark

f(x + iy) = u(x, y) + iv(x, y) is analytic

⇒⇐

+ some conditions

u, v satisfy Cauchy-Riemann equations.

• Polar Coordinates: Laplace Equation in a Circle

Polar coordinates: z = x + iy = reiθ = r(cos θ + i sin θ)

Power of z: zn = (x + iy)n = rneinθ = rn cosnθ︸︷︷︸harmonic

+i rn sinnθ︸︷︷︸harmonic



Cartesian coordinates Polar coordinates

x = r cos θ r =√x2 + y2

y = r sin θ arctan yx

∂x∂r

= cos θ ∂r∂x

= x√x2+y2

∂x∂θ

= −r sin θ ∂r∂y

= y√x2+y2

... ...

u = uxx + uyy u = ∂2u∂r2

+ 1r∂u∂r

+ 1r2

∂2u∂θ2

Laplace equation in r, θ :∂2u

∂r2+

1

r

∂u

∂r+

1

r2

∂2u

∂θ2= 0

Example 23 log z = log r + iθ where z = reiθ.

Note that log r = 0 and θ = 0 except r = 0 (origin).

Radial flow u = log r and s = θ

The equipotentials log r = c are circles around the origin and the stream-

lines are the ray θ = c going out from the origin, perpendicular to the

circles. The flow is leaving from a point source δ(x, y) at the center.

The magnitude of that source is the flux through any circle of radius R.

The flux must be the same for all circles (all R) because there are no other

sources:

Flux =

∫r=R

w · nds =

∫ 2π

0

1

RRdθ = 2π.

Sol) u = log r = log√

x2 + y2.

w = grad u = 1r

(∂r∂x

, ∂r∂y

)= 1√

x2+y2

(x√

x2 + y2,

y√x2 + y2

)︸︷︷︸

x|x|=

(x,y)|(x,y)|

n = x|x| =

(x,y)|(x,y)| on the circle C : r = R.



w · n = 1r

x|x| ·

x|x| =

1ron C.

C : (R cos θ,R sin θ), 0 ≤ θ ≤ 2π.dsdθ

=

√dxdθ

2+ dy

dθ

2=√

(−R sin θ)2 + (R cos θ)2 = R

Thus, ds = Rdθ.

If u = log r,∫∫D

udxdy =

∫∫D

∇ · ∇dxdy“ = ”div thm

∫C

∇ · nds = 2π.

Remark We cannot apply the divergence theorem directly. We need some

limiting procedure.

Since R is arbitrary,∫∫udxdy = 2π and u = 0 except r = 0.

Definition 24 2D Dirac delta:∫∫δ(x, y)dxdy = 1 and δ(x, y) = 0 except (x, y) = (0, 0).∫∫

F (x, y)δ(x, y)dxdy = F (0, 0).

(

1

2πlog r

)= δ(x, y)

Remark 12π

log r is called the fundamental solution for Laplace equation.

cf) 1D:

u′′ = δ, u(x) =

x if x ≥ 0,

0 if x < 0.

3D:

u = δ, u =1

4π

1

r.



Example 25 Dirichlet problem on unit circle:u = 0,

u(1, θ) = u0(θ).

Let

u(r, θ) =∞∑

n=0

(anrn cosnθ + bnr

n sinnθ).

u(1, θ) =∑∞

n=0(an cosnθ + bn sinnθ) = u0(θ)

: Fourier series of u0(θ).

Remark The separation of variables u(r, θ) = R(r)Θ(θ) will give such

a representation.

• Poisson’s Equation in a Square

Let Ω be the unit square in x, y plane. Consider the following Poisson

equation: − u = f(x, y) in Ω,

u = 0 on ∂Ω.

Eigen function: − u = λu.

Let

umn = sinmπx sinnπy .

−(uxx + uyy) =(m2 + n2)π2 sinmπx sinnπy

=(m2 + n2)π2u := λmnu

and umn = 0 on ∂Ω.

Let

u(x, y) =∞∑

n=1

∞∑m=1

bmnumn =∞∑

n=1

∞∑m=1

bmn sinmπx sinnπy.



−u =∑

bmnλmnumn = f(x, y)

=∞∑

n=1

∞∑m=1

bmn(m2 + n2)π2︸︷︷︸bmn

sinmπx sinnπy = f(x, y)

2D Fourier sine series of f(x, y).

Remark The separation of variables u(x, y) = X(x)Y (y) will give such

a representation.



Figure 10: 5-point stencil at inside nodes (fewer −1’s next to boundary).

§3.5 Finite Differences and Fast Poisson Solvers

It is extremely unusual to use eigenvectors to solve a linear system KU =

F . Let the eigenvalue decomposition of K be SΛS−1. Especially, S must

be especially fast to work with since both S and S−1 are required.

Consider the following the following Poisson equation in a square Ω:− u = f(x, y) in Ω,

u = 0 on ∂Ω.

We replace the derivatives by second differences under uniform discretization:

uxx ≈ −Ui+1,j−2Ui,j+Ui−1,j

h2 and uyy ≈ −Ui,j+1−2Ui,j+Ui,j−1

h2 .

(K2D)U = F :

4Ui,j − Ui+1,j − Ui−1,j − Ui,j+1 − Ui,j−1 = h2f(ih, jh)(2)

The inside rows of K2D have five nonzero entries 4,−1,−1,−1,−1.

When (i, j) is next to a boundary point of the square, the know value u0 at

the neighboring boundary point moves to the right side of (2). It becomes

part of the vector F , and an entry −1 drops out of the corresponding row

of K. So K2D have five nonzeros on inside rows, and fewer nonzeros on

next-to-boundary rows. See Figure 10.



Using blocks of size N , we can create the 2D matrix form the familiar N by

N second difference matrix K. We number the nodes of the square a row

at a time. Then the −1’s for the neighbors above and below are N position

away from the main diagonal of K2D. The 2D matrix is block tridiagonal

with tridiagonal blocks, and sparse:

KN×N =

2 −1

−1 2 −1

· · ·−1 2

K2DN2×N2 =

K + 2I −I

−I K + 2I −I

· · ·−I K + 2I

Its bandwidth w = N is the distance from the main diagonal to the nonze-

ros in −I.

Definition 26 Let A be an m × n matrix and B be a p × q matrix.

Then the Kronecker product A⊗

B is the mp ⊗ nq block matrix:

A ⊗ B =

a11B · · · a1nB... . . . ...

am1B amnB

i.e. aij is replaced by aijB.

Using the Kronecker product,

K2D = I⊗K+K⊗I =

K

. . .. . .

K

+

2I −I

−I 2I −I. . .

−I 2I

.



Figure 11: Typical rows of K2D have 5 nonzeros. Elimination fills in the band.

• Elimination and Fill-in

It is sparse and banded, but the bandwidth is not so small. (Meshpoints

cannot be numbered so that all five neighbors in the stencil receives near by

numbers.)

Elimination fills in the zeros inside the band. We add row 1 (times 14) to

row 2 to eliminate the −1 in position (2, 1). But the last −1 in row 1

produces a new −14in row 2. A zero inside the band has disappeared. As

elimination continues, virtually the whole band of multipliers in L is filled in.

See Figure 11.

• Solver Using Eigenvalues

Note that the eigenvectors and eigenvalues of Kn are

yk = (sin kπh, · · · , sin kπh), h =1

n + 1,

λk = 2 − 2 cos kπh, k = 1, · · · , n.There will be similar pattern for K2D.

To solve KU = F , let U = α1y1 + · · · + αnyn.

We have to find α1, · · · , αn.

Expand F w.r.t. y1, · · · , yn: F = a1y1 + · · · anyn.

KU = α1λ1y1 + · · · + αnλnyn = F = a1y1 + · · · anyn.

Hence α1 = a1/λ1, · · · , αn = an/λn.



We describe the method in the following 3 steps:

Step 1 Expand F as a combination F = a1y1 + · · · anyn of the

eigenvectors.

Sa = F

y1 · · · yn

a1

...

an

= a1y1 + · · · anyn = F

Step 2 Divide each ak by λk.

Step 3 Recombine eigenvectors into U = a1/λ1y1 + · · · +

an/λnyn.

U =

y1 · · · yn

a1/λ1

...

an/λn

= SΛ−1a = SΛ−1S−1F.

Step 2 is fast. The success of the method depends on the speed of step 1

and step 3: multiplying quickly by S−1 and S. Normally they both need

n2 operations since they are full matrices. But the sine eigenvectors in S

give the Discrete Sine Transform, and the Fast Fourier Transform executes

S and S−1 in n log2 n steps.

Remark In one dimension, it is slower than cn from tridiagonal elimination.

But in two dimensions n2 log2 n2 easily wins over (n2)2.

• Fast Poisson Solvers

Eigenvectors ykl (i, j) component: sinikπ

N + 1sin

jlπ

N + 1



Eigenvalues λkl K2Dykl = λklykl

λkl = 2 − coskπ

N + 1+ 2 − cos

lπ

N + 1

By 2-dim sine transform,

Fi,j =N∑k=1

N∑l=1

akl sinikπ

N + 1sin

jlπ

N + 1,

Ui,j =N∑k=1

N∑l=1

akl

λkl

sinikπ

N + 1sin

jlπ

N + 1.



§3.6 The Finite Element Method

The real success of finite element method is in two and three dimensions,

where finite differences mostly hope for a square or rectangle or cube. The

boundaries can be curved and the mesh can be unstructured. See Figure 12.

(a) Triangularization of 2d domain. (b) Pyramid Function which is piecewise linear.

Figure 12: Triangularization and pyramid function examples.

Procedure

1. Write the equation in its weak form, integrated with test function v(x, y).

2. Subdivide the region into triangle or quadrilaterals.

3. Choose N simple trial functions ϕj(x, y) and look for U = U1ϕi +

· · · + UNϕN . The 1D hat function ϕ(x) can change to 2D pyramid

function ϕ(x, y).

4. Produce N equation KU = F from test functions V1, · · · , VN

(often Vj = ϕj).

5. Assemble the stiffness matrix K and the load vector F . Solve KU =

F .

Laplace equation: − ∇2u = −(uxx + uyy) = 0.

Multiply the strong form by a test function and integrate,∫∫Ω

−(uxx + uyy)vdxdy = 0.



By Gauss-Green’s formula,

Weak form:

∫∫Ω

(uxvx + uyvy)dxdy =

∫∂Ω

∂u

∂nvds.

Boundary condition on u and v:

Essential condition: u = u0(x, y) on ∂Ω.

v = 0 is required at ∂Ω.

Natural condition: ∂u∂n

= F0(x, y) on ∂Ω.

No condition is imposed on v.

With the essential boundary condition,

Weak form:

∫∫Ω

(uxvx + uyvy)dxdy = 0 for all admissible v.

The beauty of this weak form is that it involves only first derivatives. It

has Au and Av instead of ATAu. We can create trial and test functions

much more easily when only the first derivatives are required.

Poisson equation −∇2u = −(uxx + uyy) = f(x, y) has the weak

form: ∫∫Ω

(uxvx + uyvy)dxdy =

∫∫Ω

fvdxdy.

Remark (ATCAu)Tv = fTv becomes the weak form: (CAu)TAv =

fTv.

• Trial and Test Functions: Galerkin’s Method

Approximation U(x, y) =N∑j=1

Ujϕj(x, y) (plus UB(x, y) if needed).



The extra functionUB(x, y) satisfies any nonzero essential conditionUB =

u0 at boundary mesh points. The trial functions ϕ1, · · · , ϕN equal zero

on the boundary.

We have N unknown coefficients U1, · · · , UN , and we need

N equations. Choose test functions Vi, · · · , VN .

Replace u in the weak form by∑N

j=1 Ujϕj and v by Vi:∫∫Ω

N∑j=1

Uj

∂ϕj

∂x

∂Vi

∂x+

N∑j=1

Uj

∂ϕj

∂y

∂Vi

∂y

dxdy

=

∫∫Ω

f(x, y)Vi(x, y)dxdy.

Stiffness matrix Kij =

∫∫Ω

[∂ϕj

∂x

∂Vi

∂x+

∂ϕj

∂y

∂Vi

∂y

]dxdy

and we have KU = F .

If Vi = ϕi, the matrix K is symmetric and positive definite.

If ϕj are localized, K is sparse (Kij = 0 when the graphs of ϕj and Vi

don’t overlap).

The right hand side F comes from f(x, y) and also from nonzero boundary

values u = u0(x, y) (in UB), or∂u∂n

= F0(x, y) .

• Pyramid Functions

We consider Laplace equation −∇2u = −(uxx + uyy) = 0 inside a

square. The boundary values are given around the whole square of side 2h.

See Figure 13.

We choose approximation U(x, y) as linear inside each of the eight trian-

gles. Thus U = a + bx + cy in each triangle. For T1, those numbers



Figure 13: Approximation U(x, y) and test pyramid V = ϕ, linear in eight triangles.

a, b, c are determined by the values U,UE, UN at the corners. UE, UN

are known from u0(x, y). The interior value U is what we want to find.

Only one test function V is required because we have one unknown U .

Test function V (x, y) = 1 − xh− y

hin T1

V (x, y) = 1 − xh

in T2

U(x, y) adjusts a+ bx+ cy to match U,UE, UN at the corners of T1:

U + xh(UE − U) + y

h(UN − U) in T1,

U + xh(UE − U) + y

h(UE − USE) in T2.

The integral of UxVx +UyVy in the weak form is easy, because all deriva-

tives are constants. Just multiply by the area 12h2.

∫∫T1

UxVx + UyVydxdy

=

∫∫T1

[(UE − U

h

)(−1

h

)+

(UN − U

h

)(−1

h

)]dxdy

=1

2(−UE + 2U − UN).∫∫

T2

[(UE − U

h

)(−1

h

)+

(UE − USE

h

)(0)

]dxdy

=1

2(U − UE).



Over T3:12(U − US).

Over T4:12(−UW + 2U − US).

Over T5:12(U − UW ).

Over T6:12(U − UN).

⇒ 4U − UE − UW − UN − US = 0

KU = F 4U = UE + UW + UN + US

U is the average of those boundary values: Discrete mean value theorem.

Remark 1. Mean value theorem for harmonic function:

u =1

|C|

∮C

uds =1

|R|

∫∫R

udxdy.

2. It is exactly the five-point Laplacian in FDM.

• Element Matrices and Element Vectors

Within a single interval in 1D, or one triangle in 2D, the approximation

solution U only involves two or three of the trial functions ϕi (the others

are zero in that element). The contribution to the global matrix K will

be a 2 by 2 or 3 by 3 element matrix Ke. When assembled into K,

this matrix will overlap other element matrices from triangles that share a

meshpoint.

(Ke)ij =

∫∫e

(∂ϕi

∂x

∂ϕj

∂x+

∂ϕi

∂y

∂ϕj

∂y

)dxdy, i, j = 1, 2, 3.



With U =∑N

j=1 Ujϕj and Vi = ϕi,∫∫Ω

U2x + U2

ydxdy =N∑i=1

N∑j=1

UiUj

∫∫Ω

(∂ϕi

∂x

∂ϕj

∂x+

∂ϕi

∂y

∂ϕj

∂y

)dxdy︸︷︷︸

Kij

=N∑i=1

Ui

N∑j=1

KijUj

=UTKU : N × N matrix.∫∫e

U2x + U2

ydxdy = UTKeU : 3 × 3 matrix.

We have an N × N matrix, but its only nonzero entries are in the 3 × 3

matrix KT . At the end, K is an assembly of Ke from all triangles.

In a standard right triangle T , the three unknowns Ui, Ui+1, Ui+2 are the

corners of T . See Figure 14.

∫∫T

U2x + U2

ydxdy =

[(Ui+1 − Ui

h

)2

+

(Ui+2 − Ui

h

)2]

h2

2︸︷︷︸area

.

Element

matrix KT

UTKTU = [Ui Ui+1 Ui+2]

1 −12

−12

−12

12

0

−12

0 12

Ui

Ui+1

Ui+2

The general element matrixKe depends on the tangents of the corner angles.

See Figure 14.

Element

matrixKe =

C2 + C3 −C3 −C2

−C3 C1 + C3 −C1

−C2 −C1 C1 + C3

with ci =1

2 tan θi



Figure 14: The standard triangle T (lengths 1 or h) and an arbitrary triangle e.

One can derive the matrix above using the following observation:

With U = U1ϕ1 + U2ϕ2 + U3ϕ3,∫e ∇U · ∇ϕ1∫e ∇U · ∇ϕ2∫e ∇U · ∇ϕ3

=

∫e |∇ϕ1|2

∫e ∇ϕ1 · ∇ϕ2

∫e ∇ϕ1 · ∇ϕ3∫

e ∇ϕ2 · ∇ϕ1

∫e |∇ϕ2|2

∫e ∇ϕ2 · ∇ϕ3∫

e ∇ϕ3 · ∇ϕ1

∫e ∇ϕ3 · ∇ϕ2

∫e |∇ϕ3|2

U1

U2

U3

.

Remark Ke is only semidefinite. The constant vector [1, 1, 1] is in the null

space. U = 1 at all three corners gives U = constant and grad U = 0 in

the triangle. A fixed boundary U = u0 will make K positive definite.

How do boundary values of U enter the equation KU = F ?

If U is known at a meshpoint, it moves to the right side (into F ). This is

equivalent to the extra termUB(x, y) that matches the boundary condition.

The load vector F comes from assembling integrals of f(x, y) times U :

Element load

vector Fe

∫∫e

f(∑

Uiϕi)dxdy =∑

Ui

∫∫e

fϕidxdy = UTFe

Each triangle produces a vector Fe with 3 nonzeros. On assembly, those

Fe overlap.



As an approximation, we could look at the centroid of the triangle P . For

linear elements, U(P ) is the average:

U(P ) =Ui + Ui+1 + Ui+2

3.

One-point

integration

∫∫e

f(x, y)U(x, y)dxdy

≈ f(P )

(Ui + Ui+1 + Ui+2

3

)(area of e)

• Global Matrix K and F from Ke and Fe

1. K and F are assembled from element matrices Ke and load vectors

Fe.

2. The intervals and triangles (or quads) can have different lengths and

shapes.

3. First, K and F can be constructed with no boundary conditions. Then

for boundary nodes put I and u0 into F .

To ease understanding, we consider the following 1D problem:− u′′ = 1

u(0) = 0, u′(1) = 0.

Its weak form is∫uxvxdx =

∫1vdx.

Using the discretization in Figure 15,

U = U0ϕ0 + U1ϕ1 + U2ϕ2.

Element matrices Kh and KH .

Assemble Kh and KH into K.



Figure 15: A hat and two half-hats: K will be 3 by 3 and singular.

In the first interval,∫ h

0

(U ′)2dx =

∫ h

0

(−U0

h+

U1

h

)2

dx =1

h(U1 − U0)

2

=[U0 U1]1

h

[1 −1

−1 1

]︸︷︷︸

Kh

[U0

U1

].

Element

matrix Kh

[ ∫ h

0 (ϕ′0)

2dx∫ h

0 ϕ′0 · ϕ′

1dx∫ h

0 ϕ′1 · ϕ′

0dx∫ h

0 (ϕ′1)

2dx

]=

1

h

[1 −1

−1 1

]

Similarly,

KH =

[ ∫ h+H

h (ϕ′1)

2dx∫ h+H

h ϕ′1 · ϕ′

2dx∫ h+H

h ϕ′2 · ϕ′

1dx∫ h+H

h (ϕ′2)

2dx

]=

1

H

[1 −1

−1 1

].

Now assemble Kh and KH into their right places in K. They overlap:

Global matrix K =

1h

−1h

0

−1h

1h+ 1

H− 1

H

0 − 1H

1H

This matrix has the vector [1 1 1]T in its nullspace. If U0 = U1 = U2

then U = constant and dUdx

= 0.



Consider the middle row:[−

1

h

1

h+

1

H−

1

H

] U0

U1

U2

=U1 − U0

h+

U2 − U1

H.

This is a difference of first differences. Finite element automatically adjust

for changes in meshsize.

For −(c(x)u′(x))′ = f(x),[−c1

h

c1

h+

c2

H−

c2

H

]lead to

c1(U1 − U0)

h+

c2(U2 − U1)

H.

Now impose an essential boundary condition U0 = 0, corre-

sponding to u(0) = 0. In the boundary row and column, insert the identity

matrix:

Change K to Kb Kb =

1 0 0

0 1h+ 1

H− 1

H

0 − 1H

1H

When we put F0 = 0 on the right side of the equation, the solution will have

U0 = 0. In the lower 2 by 2 submatrix is the usual fixed-free matrix[2 −1

−1 1

]when h = H .

∫ h

0

f(x)︸︷︷︸=1

U(x)dx =

∫ h

0

[U0 + (U1 − U0)

x

h

]dx

=x

h(U0 − U1) = [U0 U1]

h

2

[1

1

].

Fh =

[ ∫ h

0 f(x)ϕ0(x)dx∫ h

0 f(x)ϕ1(x)dx

]=

h

2

[1

1

]Computational Science & Engineering (CSE) Yoon Mo Jung


and

FH =

[ ∫ h+H

h f(x)ϕ1(x)dx∫ h+H

h f(x)ϕ2(x)dx

]=

H

2

[1

1

].

Assemble load vector F

before boundary conditionF =

[Fh

−

]+

[−FH

]=

1

2

h

h + H

H

Note that h/2, (h+H)/2,H/2 are the areas under the graphϕ0, ϕ0, ϕ0.

To impose U0 = 0 at the left end point, for a fixed-free bar, change the

first component of F to zero in Fb. Finally we have:

KbFb = Fb

1 0 0

0 1h+ 1

H− 1

H

0 − 1H

1H

U0

U1

U2

=1

2

0

h + H

H

It gives U0

U1

U2

=1

2

0

h + hH

1

.

If the essential boundary condition U0 = u(0) = A, that number is

multiplying −1/h in row 2 of K, so −A/h must move into the load

vector Fb as A/h. 1 0 0

0 1h+ 1

H− 1

H

0 − 1H

1H

U0

U1

U2

=

A12(h + H) + A

h12H

.

• Element Matrix in Two Dimension

The trial function have ϕ = a+bx+cy inside each triangle. Those coef-



Figure 16: Standard triangle T .

ficient a, b, c are determined by the three values 1, 0, 0 at the corners. We

want to construct the three local basis function ϕ1, ϕ2, ϕ3 on the standard

triangle given in Figure 16 satisfying

ϕ1 ϕ2 ϕ3

(0, 0) 1 0 0

(1, 0) 0 1 0

(0, 1) 0 0 1

.

We define the position matrix P :

Position matrix P

Standard triangle

1 0 0

1 1 0

1 0 1

Node 1 at (0, 0)

Node 2 at (1, 0)

Node 3 at (0, 1)

Finding the coefficient ai + bix + ciy for ϕ1, ϕ2, ϕ3 lead PC = I:

C = P−1 give the

coefficients a, b, c

in the three ϕ′s

1 x1 y1

1 x2 y2

1 x3 y3

a1 a2 a3

b1 b2 b3c1 c2 c3

=

1 0 0

0 1 0

0 0 1

For the standard triangle,

Standard

triangle

1 0 0

1 1 0

1 0 1

1 0 0

−1 1 0

−1 0 1

=

1 0 0

0 1 0

0 0 1

ϕ1 =1 − x − y

ϕ2 =x

ϕ3 =y



The derivatives of a+ bx+ cy are b and c. Those b’s and c’s are in rows

2 and 3 of C, and they enter the integrals of r the 3 by 3 element matrix

Ke:

Stiffness matrix for element e

(Ke)ij =

∫∫e

(∂ϕi

∂x

∂ϕj

∂x+

∂ϕi

∂y

∂ϕj

∂y

)dxdy = (area)(bibj + cicj).

The area of the triangle is | detP |/2. For the standard triangle,

Standard

element

matrix

KT = (area)(bibj + cicj) =1

2

2 −1 −1

−1 1 0

−1 0 1

• Quadrilateral Elements

In one dimension, we moved from linear element a+ bx to quadratics and

cubics. In two dimension it is often enough to allow an xy term (Q1 element

for rectangles). Even better are the x2, xy, y2 term that produces P2 for

triangles.

Bilinear Q1 elements: U = a + bx + cy + dxy on rectangle. Ele-

ment matrix is 4 by 4.

Quadratic P2 elements: U = a + bx + cy + dxy + ex2 + fy2

on triangle. Element matrix is 6 by 6.


ch 3. boundary value problems

Documents