Institute of Scientific Computing
Technische Universität Braunschweig
Dr. Bojana Rosic
Dr. Thilo Moshagen
Winter Term 2015
December 3, 2015
Introduction to Scientific Computing:
SOLUTIONS: BANACHS F IXED P OINT T HEOREM
Exercise 1: Contractive domain
Consider
(4 points)
3
2 xn − x2n .
xn+1 = T (xn ) =
4
2
2
2
Show that T is a contraction on the interval , 1 , while T on , 1 is no contraction.
5
5
SOLUTION 1: For showing that T is a contraction it is sufficient to show that DT < 1 for all x ∈ 25 , 1 .
So examine the Lipshitz constant. Obviously, as
DT (xn ) =
3
(1 − xn ) ,
2
DT takes its highest absolute value at the left interval boundary, where it is 34 −2 · 25 + 2 =
So T is contractive there.
For showing that T 2 is not a contraction it is sufficient to show DT 2 > 1 for one x ∈ 25 , 1 .
First, for simplification and for getting the idea, T̃ (x) = 2x − x2 = −x(x − 2) and thus
36
45
=
18
20 .
T̃ 2 (x) = (x(x − 2))2 = x2 (x − 2)2 ,
so has double zeros at 0 and 2. The derivative is
DT̃ 2 = 2x(x − 2)2 + 2x2 (x − 2) = 2x(x − 2)(x − 2 + x) = 4x(x − 2)(x − 1),
so it has one more zero at 1, so DT has zeros at 0 and 1 and is positive between them, so also inside the
interval of interest. Further,
⇒
D2 T̃ 2 =4 [(x − 2)(x − 1) + x(x − 1) + x(x − 2)]
=4 x2 − 3x + 2 + x2 − x + x2 − 2x
=4 3x2 − 6x + 2
2
4
= x2 − 2x +
3
3
r
1
x1,2 =1 ±
.
3
(1)
(2)
(3)
(4)
(5)
p
It is 1 − 1/ (3) = 0.42265 > 0.4 = 2/5, so the maximum of DT 2 is inside the interval. At this
point, derivativeDT̃ 2 evaluates to 1.5396 (matlab or calculator), which is bigger than 3/2. But already
33
9
2
4 2 = 4 > 1. So T is not contractive.
Remark Mind that neither T 2 nor T map from 25 , 1 −→ 25 , 1 , so Fixed Point iteration cannot be
applied. The question was only about contractivity.
Exercise 2: Fixed Point Iteration and Error Estimates
(12 points)
A fixed-point iteration xn+1 = φ(xn ) is defined by
φ(x) = e−x .
(a) Verify that φ(x) meets the conditions of Banach’s fixed point theorem for the interval [0.5,0.69].
Determine the constant of contraction ρ.
(4 points)
(b) Given x0 = 0.55. Calculate x1 and x2 and determine the a priori and a posterior error bounds.
How many iterations are necessary if you look at the a priori estimation to achieve an accuracy of
10−15 ?
(4 points)
(c) With the help of computer make first 100 iterations. In which iteration do you achieve accuracy
of 10−15 ? Compare the a priori and a posterior error bounds.
(4 points)
SOLUTION 2:
• First, we state (by evaluation of interval boundaries and by the monotony of φ) that the image
of [0.5 0.69] under φ is [0.60653 0.50158] and so a subset of [0.5 0.69], so φ : [0.50.69] −→
[0.50.69]. Second, obviously Dφ(x) = −e−x = −φ(x), so kDφ(x)k = e−x which is < 1 for
x < 0.
Left interval boundary has lowest derivative, and it evaluates to 0.60653 (matlab), which is the
Lipshitz constant valid on the given interval, so ρ = 0.60653.
• x1 = ex0 = 5.7695e − 01, x2 = ex1 = 5.6161e − 01 (matlab).
The a priori error bound for first iteration is
kxn − x∗ k ≤ ρn (1 − ρ)−1 kx1 − x0 k = 4.1543e − 02
(6)
The a posteriori error bound is
kxn+1 − x∗ k ≤ ρ(1 − ρ)−1 kxn+1 − xn k = 4.1543e − 02
(7)
too, the expressions are the same for n = 1. For n = 2, a priori bound is 2.5197e-02, a posteriori
is 2.3648e-02.
• One remarks that the a posteriori estimate is lower than the a priori estimate, so more exact. In
iteration 58 the desired precision is achieved. The a priori estimation guarantees this precision
only after iteration 65.
The script
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int = [0.5, 0.69]
exp(-int)
%contractivity constant:
rho = ans(1)
%%b.)
x(1)=0.55;
n= 3%
for i= 2:n
x(i) = exp(-x(i-1));
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end
n=100
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for i= 2:n
e_a_priori(i)
if e_a_priori< 1e-15
i
break;
end
end
= abs(rhoˆ(i-1)*(x(2)-x(1))/(1-rho));
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pause;
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%weg damite_a_priori
= abs(rhoˆn*(x(2)-x(1))/(1-rho))
%e_a_posteriori = abs(rho*(x(n)-x(n-1))/(1-rho))
%%c)
for i= 2:n
x(i) = exp(-x(i-1));
e_a_posteriori(i)
= abs(rho*(x(i)-x(i-1))/(1-rho));
if e_a_posteriori< 1e-15
i
break;
end
end
%fuer b.)
plot(e_a_priori(2:end), ’r’)
hold on
plot(e_a_posteriori(2:end))
figure
plot(e_a_priori(20:end), ’r’)
hold on
plot(e_a_posteriori(20:end))
lists the needed commands.
Exercise 3:
(a) Can we obtain the solution of a following nonlinear system
x1 = g1 (x1 , x2 ) =
1
[1 − x2 − sin(x1 + x2 )]
10
x2 = g2 (x1 , x2 ) =
1
[2 + x1 + cos(x1 − x2 )]
10
in R2 by using the fixed point iterations?
(16 points)
(10 points)
(b) Make the Matlab program for the fixed point iteration
(k+1)
x∗
(k)
= F (x∗ )
in which F = (g1 , g2 )T . Compute the solution x∗ = (x1∗ , x2∗ ) by starting from the initial point
(0, 0)T . Plot obtained results.
(6 points)
SOLUTION 2:
• As fixed point iteration on R2 is asked, one has to check if g(R2 ) = R2 , and prove the contractivity
everywhere. The first condition is obviously fulfilled. Second,
!
!
− cos(x1 + x2 ) 1 − cos(x1 + x2 )
∇g1
1
.
(8)
=
Dg = ∇g =
10 1 − sin(x1 − x2 )
sin(x1 − x2 )
∇g2
According to Banachs fixed point theorem, it is sufficient to find one norm in which kDgk < 1. It
P
is k.k∞ = maxi j |aij |, and it is temptative to try this one, as cancellations can be expected:
first line: |− cos(x1 + x2 )| + |1 − cos(x1 + x2 )| = cos(x1 + x2 ) + 1 − cos(x1 + x2 ) = 1
(9)
second line: |1 − sin(x1 − x2 )| + |sin(x1 − x2 )| = 1 − sin(x1 − x2 ) + sin(x1 − x2 ) = 1,
(10)
so this matrix norm is 1/10, and the mapping is contractive. Banach fixed point iteration is applicable.
• The script
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function task3(maxIter)
x(1,:)= [0 0 ];
for i=2:maxIter
x(i,:)= F(x(i-1,:));
end
plot(x(:,1));hold on
plot(x(:,2))
%plot(x)
figure
plot([x(:,1) x(:,2)])
title(’x1 and x2 stepwise’)
legend(’x1’, ’x2’)
figure
plot(x(:,1), x(:,2))
title(’Search path in \matbb R’)
end
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function out=F(x)
out(1)=(1-x(2)-sin(x(1)+x(2)))/10;
out(2)=2+x(1)+cos(x(1)-x(2))
end
solves the task.