Engineering Mathematics | CHEN30101
solutions to sheet 3
1. Consider the initial value problem: find y(t) such that
y ′ = y 2 t,
y(0) = 1.
Take a step size h = 0.1 and verify that the forward Euler approximation
to y(0.1) is y1 = 1.0 and that the approximations to subsequent points
are y2 = 1.01, y3 = 1.0304 and y4 = 1.0623. Compare these estimates
with the exact values y(0.1), y(0.2), y(0.3) and y(0.4) obtained by solving
the ODE algebraically.
Running the forward Euler (f e) process the estimates yj ≈ y(tj ) are
tabulated below. The results are rounded to four decimal places.
h
tj
yj
f (yj , tj )
y(tj )
0.1
0.1
0.1
0.1
0.1
0
0.1
0.2
0.3
0.4
1.0000
1.0000
1.0100
1.0304
1.0623
0.0000
0.1000
0.2040
0.3185
0.4514
1.0000
1.0050
1.0204
1.0471
1.0870
Note that the exact solution is the function y(t) =
2
.
2 − t2
2. By taking a step size of h = 0.5, compute the forward Euler approximation
of z(5), where z(t) satisfies the initial value problem:
z′ =
1
z,
50(1 + t)
z(0) = 1.
Running (f e) the approximate solution zj is tabulated below. The results
are rounded to four decimal places.
h
tj
zj
f (zj , tj )
z(tj )
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
1.0000
1.0100
1.0167
1.0218
1.0259
1.0293
1.0323
1.0348
1.0371
1.0392
1.0411
0.0200
0.0135
0.0102
0.0082
0.0068
0.0059
0.0052
0.0046
0.0041
0.0038
0.0035
1.0000
1.0081
1.0140
1.0185
1.0222
1.0254
1.0281
1.0305
1.0327
1.0347
1.0365
Note that the exact solution is the function z(t) = (1 + t)1/50 .
1
3. Consider the initial value problem: find y(t) such that
p
y ′ = π 1 − y 2 , y(0) = 0.
Take a step size h = 0.1 and show that computing the forward Euler
approximation to y(1) is problematic.
Running the forward Euler process gives ...
If y = 1.0207, y ′ = π
p
h
tj
yj
f (yj , tj )
0.1
0.1
0.1
0.1
0.1
0
0.1
0.2
0.3
0.4
0.0000
0.3142
0.6124
0.8608
1.0207
3.1416
2.9825
2.4835
1.5991
1 − y 2 cannot be calculated and the method stops.
4. Consider the initial value problem: find y(x) such that
1 x−y
e , y(0) = 0.
2
Compute the forward Euler approximation of y(1) using (a) a single step
of size h = 1, (b) two steps each of size h = 0.5, (c) four steps each of size
h = 0.25, and (d) ten steps each of size h = 0.1. Compare your estimates
with the exact value.
y′ =
(a) Running (f e) for h = 1 gives
h xj
1
1
0
1
yj
f (yj , xj )
0.0000
0.5000
0.5000
0.8244
so y(1) ≈ 0.5.
(b) Running (f e) for h = 0.5 gives
h
xj
yj
0.5 0 0.0000
0.5 0.5 0.2500
0.5 1 0.5710
f (yj , xj )
0.5000
0.6420
0.7679
so y(1) ≈ 0.5710.
(c) Running (f e) for h = 0.25 gives
h
xj
yj
0.25
0
0.0000
0.25 0.25 0.1250
0.25 0.5 0.2666
0.25 0.75 0.4245
0.25
1
0.5976
so y(1) ≈ 0.5976.
2
f (yj , xj )
0.5000
0.5666
0.6314
0.6924
0.7477
(d) Running (f e) for h = 0.1 gives
h
xj
yj
f (yj , xj )
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.0000
0.0500
0.1026
0.1577
0.2153
0.2755
0.3381
0.4030
0.4703
0.5398
0.6115
0.5000
0.5256
0.5512
0.5765
0.6014
0.6259
0.6497
0.6729
0.6953
0.7168
0.7374
so y(1) ≈ 0.6115.
x
The solution is y(x) = ln 1+e
so y(1) = 0.6201 to four decimal places.
2
The accuracy of the forward Euler approximation improves when the step
size h is reduced.
5. By taking a step size of h = 0.5, compute the modified Euler approximation of z(2), where z(t) satisfies the initial value problem:
z′ =
1
z,
50(1 + t)
z(0) = 1.
Running the modified Euler (me) process the approximate solution yj is
tabulated below. Note that tm = tj + h2 and zm = zj + h2 f (zj , tj ).
h
tj
zj
0.5 0 1.0000
0.5 0.5 1.0080
0.5 1 1.0138
0.5 1.5 1.0183
0.5 2 1.0220
f (zj , tj )
tm
zm
f (zm , tm )
z(tj )
0.0200
0.0134
0.0101
0.0081
0.0068
0.2500
0.7500
1.2500
1.7500
2.2500
1.0050
1.0114
1.0164
1.0204
1.0238
0.0161
0.0116
0.0090
0.0074
0.0063
1.0000
1.0081
1.0140
1.0185
1.0222
These results are clearly much more accurate than those obtained for the
forward Euler approximation scheme (cf. question 2).
6. Consider the initial value problem: find y(x) such that
y ′ = 17 − 12x − 3y,
y(0) = 9.
Using four steps of size h = 0.25, estimate y(1) using (i) the forward
Euler method, (ii) the modified Euler method. Compare your estimates
with the exact value.
3
The exact solution is the function y(x) = 2e−3x + 7 − 4x. Running the
forward Euler process gives ...
h
xj
yj
f (yj , xj )
0.25
0
9.0000 −10.0000
0.25 0.25 6.5000 −5.5000
0.25 0.5 5.1250 −4.3750
0.25 0.75 4.0313 −4.0938
0.25
1
3.0078
Running the modified Euler process gives ...
h
xj
yj
f (yj , xj )
xm
ym
0.25
0
9.0000 −10.0000 0.1250 7.7500
0.25 0.25 7.0625 −7.1875 0.3750 6.1641
0.25 0.5 5.5645 −5.6934 0.6250 4.8528
0.25 0.75 4.2999 −4.8996 0.8750 3.6874
0.25
1
3.1593 −4.4779 1.1250 2.5996
f (ym , xm )
y(xj )
−7.7500
−5.9922
−5.0583
−4.5622
−4.2987
9.0000
6.9447
5.4463
4.2108
3.0996
Neither solution is particularly accurate but the modified Euler method
gives more accurate results than the forward Euler method.
7. Heun’s method for approximating the solution of an initial value problem
is given by
K1 = f (xn , yn ), K2 = f (xn + h, yn + hK1 )
h
yn+1 = yn + (K1 + K2 ).
2
Carry out two steps of the method (with step size h = 0.5) to compute
an estimate of the solution y(x) of the problem
y ′ = y − x,
y(0) = 20.
Running the process with x∗ = xn + h and y∗ = yn + hK1 gives ...
h
xn
yn
K1
x∗
y∗
K2
y(tn )
0.5 0.0 20.0000 20.0000 0.5 30.0000 29.5000 20.0000
0.5 0.5 32.3750 31.8750 1.0 48.3125 47.3125 32.3257
0.5 1.0 52.1719 51.1719 1.5 77.7578 76.2578 52.6474
8. Consider the initial value problem: find y(t) such that
y ′ = 1 + ln y,
y(0) = 1.
Carry out a step of the 4th-order Runge–Kutta method (with step size
h = 0.1) and verify that K1 = 1, K2 = 1.048790, K3 = 1.051111 and
K4 = 1.099946 so that the approximation to y(0.1) is given by y1 =
1.104996.
4
For the next step of the method, show that K1 = 1.099842, K2 =
1.148410, K3 = 1.150501 and K4 = 1.198888 so that y2 = 1.219938.
Continue for one more step and verify that y3 = 1.344702.
Running the Runge–Kutta (rk4) process gives ...
h = 0.1
first temporary point
second temporary point
third temporary point
new point
t = 0.00 y0 = 1.000000
t = 0.05 ym = 1.050000
t = 0.05 ym = 1.052440
t = 0.10 y∗ = 1.105111
t = 0.10 y1 = 1.104996
K1
K2
K3
K4
= 1.000000
= 1.048790
= 1.051111
= 1.099946
h = 0.1
first temporary point
second temporary point
third temporary point
new point
t = 0.10 y1 = 1.104996
t = 0.15 ym = 1.159988
t = 0.15 ym = 1.162416
t = 0.20 y∗ = 1.220046
t = 0.20 y2 = 1.219938
K1
K2
K3
K4
= 1.099842
= 1.148410
= 1.150501
= 1.198888
h = 0.1
first temporary point
second temporary point
third temporary point
new point
t = 0.20 y2 = 1.219938
t = 0.25 ym = 1.279878
t = 0.25 ym = 1.282277
t = 0.30 y∗ = 1.344802
t = 0.30 y3 = 1.344702
K1
K2
K3
K4
= 1.198800
= 1.246765
= 1.248637
= 1.296247
√
9. Consider fluid draining out of a tank according to the equation y ′ = −k y
with k = 0.1 and y = 20 at t = 0. Use the 4th-order Runge–Kutta method
with step size h = 10 to estimate the values of y(t) (the height, in metres,
of fluid in the tank) for t = 10, 20 and 30 measured in seconds.
Running the (rk4) process gives ...
first step
h = 10
t = 0 y0 = 20.0000
first temporary point
t = 5 ym = 17.7639
second temporary point t = 5 ym = 17.8926
third temporary point
t = 10 y∗ = 15.7700
new point
t = 10 y1 = 15.7779
K1
K2
K3
K4
= −0.4472
= −0.4215
= −0.4230
= −0.3971
h = 10
first temporary point
second temporary point
third temporary point
new point
t = 10 y1 = 15.7779
t = 15 ym = 13.7918
t = 15 ym = 13.9210
t = 20 y∗ = 12.0468
t = 20 y2 = 12.0558
K1
K2
K3
K4
= −0.3972
= −0.3714
= −0.3731
= −0.3471
h = 10
first temporary point
second temporary point
third temporary point
new point
t = 20 y2 = 12.0558
t = 25 ym = 10.3197
t = 25 ym = 10.4496
t = 30 y∗ = 8.8232
t = 30 y3 = 8.8337
K1
K2
K3
K4
= −0.3472
= −0.3212
= −0.3233
= −0.2970
5
10. Consider the initial value problem: find y(x) such that
y ′ = −2xy 2
y(0) = 1.
Carry out two steps of each of the following methods for solving the
equation. (a) forward Euler with h = 1 (b) forward Euler with h = 0.5
(c) modified Euler with h = 1 (d) modified Euler with h = 0.5 (e) Runge–
Kutta with h = 1 (f) Runge–Kutta with h = 0.5. Compare your results
with the exact solution.
(a) Running (f e) with h = 1 ...
h xj
yj f (yj , xj ) y(xj )
1 0.0
1.0000
0.0000 1.0000
1 1.0
1.0000 −2.0000 0.5000
1 2.0 −1.0000
0.2000
Unrealistic results.
(b) Running (f e) with h = 0.5 ...
h xj
yj f (yj , xj ) y(xj )
0.5 0.0 1.0000
0.0000 1.0000
0.5 0.5 1.0000 −1.0000 0.8000
0.5 1.0 0.5000
0.5000
The final value is correct (by coincidence!).
(c) Running (me) with h = 1 ...
h xj
yj f (yj , xj )
xm
ym f (ym , xm ) y(xj )
1 0 1.0000
0.0000 0.5000 1.0000
−1.0000 1.0000
1 1 0.0000
0.0000 1.5000 0.0000
0.0000 0.5000
1 2 0.0000
0.0000
0.2000
All subsequent values will be zero. Not a realistic solution.
(d) Running (me) with h = 0.5 ...
h xj
yj f (yj , xj )
xm
ym f (ym , xm ) y(xj )
0.5 0.0 1.0000
0.0000 0.2500 1.0000
−0.5000 1.0000
0.5 0.5 0.7500 −0.5625 0.7500 0.6094
−0.5570 0.8000
0.5 1.0 0.4715 −0.4446 1.2500 0.3603
−0.3246 0.5000
These values are not particularly accurate but are pointing in the
right direction.
6
(e) Running (rk4) with h = 1 ...
h = 1.0
first temporary point
second temporary point
third temporary point
new point
x = 0.00 y0
x = 0.50 ym
x = 0.50 ym
x = 1.00 y∗
x = 1.00 y1
= 1.0000
K1 = 0.0000
= 1.0000 K2 = −1.0000
= 0.5000 K3 = −0.2500
= 0.7500 K4 = −1.1250
= 0.3958
h = 1.0
first temporary point
second temporary point
third temporary point
new point
x = 1.00 y1
x = 1.50 ym
x = 1.50 ym
x = 2.00 y∗
x = 2.00
y
= 0.3958
= 0.2391
= 0.3100
= 0.1075
= 0.1826
K1
K2
K3
K4
= −0.3134
= −0.1716
= −0.2884
= −0.0462
These values are not particularly accurate but are again pointing in
the correct direction.
(f) Running (rk4) with h = 0.5 ...
h = 0.5
first temporary point
second temporary point
third temporary point
new point
x = 0.00 y0
x = 0.25 ym
x = 0.25 ym
x = 0.50 y∗
x = 0.50 y1
= 1.0000
K1 = 0.0000
= 1.0000 K2 = −0.5000
= 0.8750 K3 = −0.3828
= 0.8086 K4 = −0.6538
= 0.7984
h = 0.5
first temporary point
second temporary point
third temporary point
new point
x = 0.50 y1
x = 0.75 ym
x = 0.75 ym
x = 1.00 y∗
x = 1.00 y2
= 0.7984
= 0.6390
= 0.6452
= 0.4861
= 0.4997
K1
K2
K3
K4
= −0.6374
= −0.6125
= −0.6245
= −0.4726
These values are much more accurate.
11. Express the following high-order equations as systems of first-order equations.
(a)
(b)
(c)
(d)
(e)
dy
d2 y
+ 12y = 0; y(0) = 1,
(0) = 0.
2
dx
dx
d2 y
dy
dy
+2
− 4y = x; y(0) = 0,
(0) = 0.
dx2
dx
dx
dy
dy
d2 y
d3 y
3
−
+
y
=
x
,
y(0)
=
1,
(0)
=
0,
(0) = 0.
dx3 dx
dx
dx2
d2 p
dp
1
dp
=
; p(0) = 1,
(0) = 2.
2
dt
(p + t) dt
dt
the coupled system
dz
d2 y
−2
= y + z;
2
dt
dt
d2 z
dy
+2
= y − z;
2
dt
dt
7
y(t0 ) = y0 ,
z(t0 ) = z0 ,
dy
(t0 ) = vA ,
dt
dz
(t0 ) = vB .
dt
(a) Introducing z =
dy
dz
d2 y
so
= 2 gives the following coupled system.
dx
dx
dx
dy
= z, y(0) = 1,
dx
dz
•
= −12y, z(0) = 0.
dx
dz
d2 y
dy
so
= 2 gives the following coupled system.
Introducing z =
dx
dx
dx
dy
•
= z, y(0) = 0,
dx
dz
= x + 4y − 2z, z(0) = 0.
•
dx
dy
dz
d2 y
Introducing z =
and w =
= 2 gives the following system.
dx
dx
dx
dy
•
= z, y(0) = 1,
dx
dz
= w, z(0) = 0,
•
dx
dw
•
= x3 + z − y, w(0) = 0.
dx
dp
dq
d2 p
Introducing q =
so
= 2 gives the following system.
dt
dt
dt
dp
= q, p(0) = 1,
•
dt
q
dq
=
, q(0) = 2.
•
dt
p+t
dy
dz
Finally, introducing x =
and w =
gives the following system
dt
dt
of four equations with associated initial conditions.
dy
•
= x, y(t0 ) = y0 ,
dt
dx
= y + z + 2w, x(t0 ) = vA ,
•
dt
dz
•
= w, z(t0 ) = z0 ,
dt
dw
•
= y − z − 2x, w(t0 ) = vB .
dt
•
(b)
(c)
(d)
(e)
8
12. Consider the coupled initial value problem
dx
= −3y,
dt
dy
= 3x,
dt
x(0) = 1,
y(0) = 0.
Approximate the solution of this system using the Runge–Kutta method
with step size h = 0.1. Show that at the first step, the K values are (0, 3),
(−0.45, 3), (−0.45, 2.9325), (−0.87975, 2.865) giving new values x1 and y1
of 0.955338 and 0.2955. Show that the second step gives values x2 and y2
of 0.825349 and 0.564604. Compare these results with the exact values
given by x = cos 3t, y = sin 3t.
Running (rk4) with h = 0.1 ...
h = 0.1
first temporary point
second temporary point
third temporary point
new point
t
t
t
t
t
=
=
=
=
=
0.00
0.05
0.05
0.10
0.10
x0
xm
xm
x∗
x1
=
=
=
=
=
1.000000
1.000000
0.977500
0.955000
0.955338
y0
ym
ym
y∗
y1
=
=
=
=
=
0.000000
0.150000
0.150000
0.293250
0.295500
Kf 1 = 0.000000
Kf 2 = -0.450000
Kf 3 = -0.450000
Kf 4 = -0.879750
h = 0.1
first temporary point
second temporary point
third temporary point
new point
t
t
t
t
t
=
=
=
=
=
0.10
0.15
0.15
0.20
0.20
x1
xm
xm
x∗
x2
=
=
=
=
=
0.955338
0.911013
0.889517
0.825692
0.825349
y1
ym
ym
y∗
y2
=
=
=
=
=
0.295500
0.438801
0.432152
0.562355
0.564604
Kf 1
Kf 2
Kf 3
Kf 4
=
=
=
=
-0.886500
-1.316402
-1.296456
-1.687066
The exact solutions are the functions x(t) = cos 3t, y(t) = sin 3t. Thus
when when t = 0.1, we have x = 0.955336 and y = 0.295520 and when t =
0.2 we have x = 0.825336 and y = 0.564642. We see that the numerical
approximations are correct to 4 but not 5 decimal places.
13. Consider the coupled initial value problem
dx
1
= y,
dt
3
dy
= 6x3 ,
dt
x(0) = 1;
y(0) = −3.
Compute an approximation to the solution vector (x, y) at the final time
t = 0.5 using the Runge–Kutta method with step size h = 0.1. Then
repeat the calculation for h = 0.05. Compare the numerical results with
the exact values given by x = 1/(t + 1), y = −3/(t + 1)2 .
Editing rhs and running rk4 gives the following output:
>> [time, zzrk4, nfeval] = rk4([1;-3],0.5,5);
RK4 iteration in 2 dimensions ...
t
x
y
0.00
1.000000
-3.000000
0.00
1.000000
-3.000000
0.10
0.909095
-2.479325
0.20
0.833340
-2.083307
0.30
0.769239
-1.775110
0.40
0.714296
-1.530563
0.50
0.666680
-1.333273
all done
9
Kg1
Kg2
Kg3
Kg4
=
=
=
=
3.000000
3.000000
2.932500
2.865000
Kg1
Kg2
Kg3
Kg4
=
=
=
=
2.866013
2.733038
2.668552
2.477076
>> [time, zzrk4, nfeval] = rk4([1;-3],0.5,10);
RK4 iteration in 2 dimensions ...
t
x
y
0.00
1.000000
-3.000000
0.00
1.000000
-3.000000
0.05
0.952381
-2.721088
0.10
0.909091
-2.479338
0.15
0.869566
-2.268430
0.20
0.833334
-2.083332
0.25
0.800000
-1.919998
0.30
0.769231
-1.775146
0.35
0.740741
-1.646088
0.40
0.714286
-1.530609
0.45
0.689656
-1.426869
0.50
0.666667
-1.333330
all done
The exact values are x(0.5) = 32 ≈ 0.666667 and y(0.5) = − 34 ≈ −1.333333.
The approximation errors for h = 0.1 are |x(0.5) − x5 | = 0.000013 and
|y(0.5) − y5 | = 0.000060. For the smaller value h = 0.05, we have
|x(0.5) − x10 | = 0.000000 and |y(0.5) − y10 | = 0.000003.
The numerical results are much more accurate for the smaller value of h.
14. Consider the second-order initial value problem
d2 y
+ y + ǫy 3 = 0;
dt2
y(0) = 1,
dy
(0) = 0
dt
with the value ǫ = 0.1. (This is an example of a Duffing oscillator.)
Compute three steps of Runge–Kutta with a step size h = 0.1.
The first step is to express the second-order problem as a coupled firstorder system.
dy
= v;
y(0) = 1,
dt
dv
= −y − 0.1y 3 ;
v(0) = 0.
•
dt
•
Editing rhs and running rk4 gives the following output:
>> [time, zzrk4, nfeval] = rk4([1;0],0.3,3);
RK4 iteration in 2 dimensions ...
t
x
y
0.00
1.000000
0.000000
0.10
0.994506
-0.109762
0.20
0.978095
-0.218104
0.30
0.950979
-0.323645
all done
10
15. A sphere floating in water (in the absence of friction and with its centre y
above the water level) experiences a force due to gravity of − 34 πr 3 ρg and
a force due to buoyancy of 13 ρwater π 2r 3 + y(y 2 − 3r 2 ) g. It’s motion is
determined by the ODE
ρwater
y
y3
d2 y
= −g + g
2+ 3 −3
dt2
4ρ
r
r
A 0.4 metre radius sphere with density 0.7 of that of water is initially at
rest with its centre 25 cm below the level of the water. You may assume
that g = 10 ms−2 .
(a) Express the equation of motion as a system of two first-order equations and state the appropriate initial conditions.
(b) Take a step size of h = 0.1 and compute two steps of Runge–Kutta
to estimate the position of the centre of the sphere when t = 0.1 and
when t = 0.2.
The first step is to express the second-order problem as a coupled firstorder system:
dy
= v with y(0) = −0.25
dt
ρwater
y
y3
dv
= −g + g
2+ 3 −3
with v(0) = 0
•
dt
4ρ
r
r
•
Next, we substitute the given values for the parameters to give
dy
= v with y(0) = −0.25
dt
25 20 3125 3 375
dv
= −10 +
y −
y
2 + 15.625y 3 − 7.5y = − +
•
dt
7
7
56
14
with v(0) = 0
•
Finally, editing rhs and running rk4 gives the following output:
>> [time, zzrk4, nfeval] = rk4([-0.25;0],0.2,2);
RK4 iteration in 2 dimensions ...
t
x
y
0.00
-0.250000
0.000000
0.10
-0.235369
0.288436
0.20
-0.194009
0.525712
all done
The sphere is quickly floating back to the surface! Running rk4 for longer
times suggests that the sphere tends to bob up and down in the water.
This is illustrated by the following output.
11
>> [time, zzrk4, nfeval] = rk4([-0.25;0],10,100);
RK4 iteration in 2 dimensions ...
t
x
y
0.00
-0.250000
0.000000
0.00
-0.250000
0.000000
0.10
-0.235369
0.288436
0.20
-0.194009
0.525712
0.30
-0.133823
0.656006
0.40
-0.067990
0.633737
0.50
-0.012547
0.451335
0.60
0.018247
0.151392
0.70
0.016357
-0.188037
0.80
-0.017715
-0.478234
0.90
-0.075068
-0.643951
1.00
-0.140999
-0.647951
1.10
-0.199604
-0.503193
1.20
-0.238239
-0.257729
1.30
-0.249634
0.032871
1.40
-0.231837
0.317951
1.50
-0.187943
0.545862
1.60
-0.126467
0.660830
1.70
-0.061066
0.620233
1.80
-0.007808
0.422030
1.90
0.019567
0.113801
2.00
0.013912
-0.224085
2.10
-0.023284
-0.503314
.
.
.
9.80
-0.009216
-0.422804
9.90
-0.062376
-0.617180
10.00
-0.127317
-0.654832
all done
>> plot(time,zzrk4(1,:),’-kx’), axis square
16. Compute a centered difference approximation to the BVP
x2
dy
d2 y
− 4x
+ 4y = x3 ,
2
dx
dx
x ∈ (1, 2);
y(1) = 1, y(2) = 6,
using a grid of size h = 0.25. Compare your result with the exact solution
values y(1.25) = 4.771, y(1.5) = 5.9834, y(1.75) = 6.1867.
We start by dividing the interval into 4 pieces so that
x0 = 1,
x1 = 1.25,
x2 = 1.5,
x3 = 1.75,
x4 = 2,
and then we define the grid values
y0 = y(x0 ) = 1;
y1 ≈ y(x1 ),
y2 ≈ y(x2 ),
12
y3 ≈ y(x3 ),
y4 = y(x4 ) = 6.
We also define the two finite difference approximations
1
1
(yj−1 − 2yj + yj+1 ),
y ′ (xj ) =
(yj+1 − yj−1 ).
2
h
2h
To simplify notation define y1 = A, y2 = B and y3 = C.
• y ′′ (xj ) =
The difference approximations at the point x1 = 1.25 are y ′ (xj ) ≈ B−1
0.5 =
=
−32A
+
16B
+
16.
Substituting
these
2B − 2 and y ′′ (xj ) ≈ B−2A+1
0.252
expressions into the ODE gives
1.252 (−32A + 16B + 16) + 4 × 1.25 (2B − 2) + 4A = 1.253
that is, −46A + 35B = −13.046875.
The difference approximations at the point x2 = 1.5 are y ′ (xj ) ≈ C−A
0.5 =
=
16A
−
32B
+
16C.
Substituting
these
2C − 2A and y ′′ (xj ) ≈ C−2B+A
0.252
expressions into the ODE gives
1.52 (16A − 32B + 16C) + 4 × 1.5 (2C − 2A) + 4B = 1.53
that is, 24A − 68B + 48C = 3.375.
The difference approximations at the point x3 = 1.75 are y ′ (xj ) ≈ 6−B
0.5 =
12 − 2B and y ′′ (xj ) ≈ 6−2C+B
=
96
−
32C
+
16B.
Substituting
these
0.252
expressions into the ODE gives
1.752 (96 − 32C + 16B) + 4 × 1.75 (12 − 2B) + 4C = 1.753
that is 35B − 94C = −372.640625.
Solving the 3 linear equations gives the results A = 4.90892(= y1 ), B =
6.07895(= y2 ) and C = 6.2277(= y3 ). Comparing with the exact values
we see that the errors are of order of 0.1.
17. Compute a centered difference approximation to the BVP
π2
d2 y
+
y = ǫy 3 , x ∈ (0, 1); y(0) = 1, y(1) = 0,
dx2
4
for the value ǫ = 0.1, using a grid of size h = 0.5. Compare your computed
result with the exact solution at the interior grid point.
Dividing the interval into 2 pieces we introduce the unknown grid point
value y1 ≈ y(0.5). To simplify notation we let A = y1 .
0−1
The difference approximations at the point x1 = 0.5 are y ′ (xj ) ≈ 2×0.5
=
0−2A+1
−1 and y ′′ (xj ) ≈ 0.52 = 4 − 8A. Substituting these expressions into
the ODE gives
4 − 8A + (π 2 /4)A = 0.1A3 .
This is a nonlinear equation. To solve it we set up a Newton iteration
with f (A) = 4 + (π 2 /4 − 8)A − 0.1A3 so that f ′ (A) = π 2 /4 − 8 − 0.3A2 .
The iteration converges to the solution A = 0.7163 (= y1 ).
A |f (A)|
f ′ (A)
0.0000 4.0000 −5.5326
0.7230 0.0377 −5.6894
0.7163 0.0000 −5.6865
13
The exact solution solution to the BVP when ǫ = 0 is the function
y(x) = cos( πx
2 ) so that y(0.5) = 0.7071. The difference approximation
is surprisingly accurate.
18. An iron bar is kept at temperatures T0 and T1 at its two ends. The
radiation of heat at point x is proportional its temperature −AT (x) and
2
this effect is balanced by thermal diffusion, that is, −κ ddxT2 . Thus, when
the temperature T (x) has reached equilibrium, it satisfies the BVP
−κ
d2 T
= −AT,
dx2
x ∈ (0, 1);
T (0) = T0 , T (1) = T1 .
Write down a centered difference approximation to T (x) for values of x of
0.25, 0.5 and 0.75 and solve the resulting linear system in the case where
A = 1, κ = 1, T0 = 1 and T1 = 2.
Multiplying by minus one and substituting parameter values gives
d2 T
−T =0
dx2
with T (0) = 1 and T (1) = 2.
Let’s define grid values so that T (0.25) ≈ P , T (0.5) ≈ Q and T (0.75) ≈ R.
+1
=
The difference approximation at the point x1 = 0.25 is T ′′ (xj ) ≈ Q−2P
0.252
16Q − 32P + 16. Substituting this expression into the ODE gives
16Q − 33P + 16 = 0.
(a)
The difference approximation at the point x2 = 0.5 is T ′′ (xj ) ≈ R−2Q+P
=
0.252
16R − 32Q + 16P . Substituting this expression into the ODE gives
16P − 33Q + 16R = 0.
(b)
The difference approximation at the point x3 = 0.75 is T ′′ (xj ) ≈ 2−2R+Q
=
0.252
32 − 32R + 16Q. Substituting this expression into the ODE gives
16Q − 33R + 32 = 0.
(c)
Solving the 3 linear equations (a)–(c) gives the results P = 1.13019, Q =
1.33102 and R = 1.61504.
Note that the exact solution to the BVP is
T =
2e − 1 x e2 − 2e −x
e + 2
e .
e2 − 1
e −1
Thus the exact values are T (0.25) = 1.12963, T (0.5) = 1.33023, T (0.75) =
1.61440 so the numerical values are in good agreement.
14
19. Imagine two metal bars, one with ends maintained at temperature T =
600 and the other with a temperature maintained at T = 400. As well
as thermal diffusion and general radiation, there is heat transferred from
one bar to the other. The bars thus satisfy a coupled system of ODEs
d2 T1
+ 0.01 T1 + 0.005 (T1 − T2 ) = 0,
dx2
d2 T2
−0.01
+ 0.01 T2 + 0.005 (T2 − T1 ) = 0,
dx2
−0.01
subject to T1 (0) = T1 (1) = 600 and T2 (0) = T2 (1) = 400.
Take a grid with points x0 = 0, x1 = 0.25, x2 = 0.5, x3 = 0.75 and x4 = 1
and compute a centered difference approximation to T1 and T2 .
First we introduce new functions y = 0.01T1 and z = 0.01T2 and substitute into the ODEs. This gives the rescaled equations:
y ′′ − 1.5y + 0.5z = 0
z ′′ + 0.5y − 1.5z = 0
subject to y(0) = y(1) = 6 and z(0) = z(1) = 4.
Let’s define grid values so that A ≈ y(0.25), B ≈ y(0.5), C ≈ y(0.75),
D ≈ z(0.25), E ≈ z(0.5) and F ≈ z(0.75).
Substituting the second derivative difference approximation into the ODEs
evaluated at the grid points gives a system of 6 algebraic equations:
6 − 2A + B
0.252
C − 2B + A
0.252
B − 2C + 6
0.252
4 − 2D + E
0.252
D − 2E + F
0.252
E − 2F + 4
0.252
− 1.5A + 0.5D = 0 ⇐⇒ −67A + 32B + D + 192 = 0
− 1.5B + 0.5E = 0 ⇐⇒ 32A − 67B + 32C + E = 0
− 1.5C + 0.5F = 0 ⇐⇒ 32B − 67C + F + 192 = 0
− 1.5D + 0.5A = 0 ⇐⇒ A − 67D + 32E + 128 = 0
− 1.5E + 0.5B = 0 ⇐⇒ B + 32D − 67E + 32F = 0
− 1.5F + 0.5C = 0 ⇐⇒ C + 32E − 67F + 128 = 0
Solving these equations gives A = 5.42, B = 5.23, C = 5.42, D = 3.73,
E = 3.64, F = 3.73. The corresponding temperature estimates are
x = 0 x = 0.25 x = 0.5 x = 0.75 x = 1
Bar 1 600
542
523
542
600
Bar 2 400
373
364
373
400
15