7.5 Nonlinear problems

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In a constrained minimization problem

J(u) → min A(u) = 0 constraints (7.154)

the Lagrangian functional

L(u, λ) = J(u)+ < λ,A(u)> <.,.>“duality pairing” (7.155)

is stationary at the minimum point u0, see e.g. [202], i.e.,

δJ(u0) + λd(u0) = 0. (7.156)

This technique is now adopted1 to estimate the error J(u)−J(uh) in nonlinear

problems by introducing a “dual” variable z.

As a model problem we choose the linear Poisson equation on V = {u ∈

H1(Ω) | u = 0 on Γ}

A(u) := −Δu − p = 0. (7.157)

We let

A(u)(ψ) := a(u, ψ) − (p, ψ) a(u, ψ) := (∇u,∇ψ) (7.158)

the weak form and we intend to evaluate the solution at a point x so that

J(u) = (δ0, u) = u(x) . (7.159)

The Gateaux derivatives of these functionals are

J

_(u)(ϕ) :=

_

d

dε

J(u + εϕ)

_

ε=0

= (δ0, ϕ) = J(ϕ) (7.160)

and

A

_(u)(ϕ, z) :=

_

d

d ε

A(u + εϕ)(z)

_

ε=0

= a(ϕ, z) . (7.161)

1 The following is based on Sect. 6.1 in [22]. Added in proof: a good summary can

also be found in Ern A, Guermond J-L (2004) Theory and Practice of Finite

Elements, Springer-Verlag

7.5 Nonlinear problems 527

Next we define the Lagrangian functional

L(u, z) = J(u) − A(u)(z) (7.162)

and we seek for a stationary point {u, z} ∈ V × V of L(., .), i.e.,

L

_(u, z)(ϕ, ψ) =

_

J_(uh)(ϕh) − A_(uh)(ϕh, zh)

−A(u)(ψ)

_

= 0 (7.163)

for all {ϕh, ψh} ∈ V × V .

Evidently in the linear case these two equations are identical to the standard

approach

J(ϕh) − a(ϕh, zh) = 0 for all ϕh ∈ Vh → zh (7.164)

a(uh, ψh) − p(ψh) = 0 for all ψh ∈ Vh → uh . (7.165)

Under appropriate assumptions we have the error representation

J(u) − J(uh) =

1

2ρ(uh)(z − zh) +

1

2ρ

∗(uh, zh)(u − uh) + R(3)

h (7.166)

where

ρ(uh)(z − zh) : = −A(uh)(z − zh) (7.167)

ρ

∗

h(uh, zh)(u − uh) : = J

_(uh) − A

_(uh)(u − uh, zh) (7.168)

and where the remainder term R(3)

h is cubic in the “primal” and “dual” errors

e := u − uh and e∗ := z − zh and involves second and third order Gateaux

derivatives of J(.) and A(.).

In the case of the linear model problem (7.157) we have

− A(uh)(z − zh) = −a(uh, z − zh) + p(z − zh) = −a(uh, z) + p(z)

= −ph(z) + p(z) = −uh(x) + u(x) (7.169)

and

J

_(u − uh) − A

_(uh)(u − uh, zh) = (δ0, u − uh) − a(u − uh, zh)

= (δ0, u) − a(u, zh) = (δ0, u) − (δh

0 , u) = u(x) − uh(x) (7.170)

and of course R(3) is zero in this case so that indeed

J(u) − J(uh) =

1

2

(u(x) − uh(x)) +

1

2

(u(x) − uh(x)) . (7.171)

In the case of a nonlinear equation such as

A(u) := −Δu − u3 − p = 0 (7.172)

the weak form is

528 7 Theoretical details

A(u)(ψ) := (∇u,∇ψ) − (u3, ψ) − (p, ψ) (7.173)

and

A

_(u)(ϕ, z) := (∇ϕ,∇z) − (3u2 ϕ, z) (7.174)

so that the generalized Green’s function z is the solution of

(∇ϕ,∇z) − (3u2 ϕ, z) = J

_(ϕ) for all ϕ ∈ V (7.175)

and the FE approximation zh solves the variational problem

aT (u, zh,ϕh) = J

_(ϕh) for all ϕh

∈ Vh (7.176)

where we have written aT (., ., .) for the Gateaux derivative, i.e., the left-hand

side of (7.175). If J is linear then J_ = J so that

KT (u) z = j ji = J(ϕi) (7.177)

where KT is the tangential stiffness matrix and z is the vector of nodal values

of the field z.

A more pedestrian, engineering approach would go like this: let a(u, v) =

p(v) the nonlinear equation and let u = uh + e then

a(uh + e, v) = p(v) for all v ∈ V (7.178)

or if we do a “Taylor expansion”

a(uh, v) + aT (uh; e, v) + . . . = p(v) (7.179)

and neglect the higher order terms (. . .)

aT (uh; e, v) = p(v) − a(uh, v) = p(v) − ph(v)

=

_

i

__

Ωi

r • v dΩ +

_

Γi

j • v ds

_

:= r(v) (7.180)

where r and j are defined as in (1.425) p. 150.

Next, let a∗

T (uh; e, v) the dual bilinear form defined by switching the last

two arguments in aT

a

∗

T (uh; e, v) := aT (uh; v, e) . (7.181)

If aT is symmetric in the last two arguments—as in hyperelasticity—then

a∗

T = aT. Now let J(v) a linear functional on V and z the solution of

a

∗

T (uh; z, v) = J(v) (7.182)

then

7.6 The derivation of influence functions 529

J(e) = a

∗

T (uh; z, e) = aT (uh; e, z) = r(z) . (7.183)

If zh is the FE solution of (7.182), we can invoke the Galerkin orthogonality

0 = a

∗

T (uh; z − zh, v) = aT (uh; e, z − zh) = r(z − zh) (7.184)

and so we arrive at

J(e) =

_

i

__

Ωi

r • (z − zh) dΩ +

_

Γi

j • (z − zh) ds

_

. (7.185)

Remark 7.1. In the mathematical literature the Gateaux derivative aT is often

replaced by a secant form

aS(u,uh; e, v) :=

_ 1

0

aT (uh + s e; e, v) ds , (7.186)

which can be interpreted as the average Fr´echet derivative of a(u, v). In this

case the error e = u − uh is the solution of the linear variational problem

aS(u,uh; e, v) = a(u, v) − a(uh, v) = r(v) v ∈ V . (7.187)

Often this approach leads to identical formulation—because the exact solution

u is unknown and therefore compromises must be made—though in specific

circumstances this formulation can be advantageous [153].

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