Math 257/316 Assignment 6 Solutions
1. For the function
f (x) =
⇢
1
1
2x 0  x < 1/2
1/2  x  1
defined on [0, 1], sketch (several periods of) its even and odd 2-periodic extensions,
and for each x 2 [0, 1] determine the value to which its Fourier sine and cosine series
converge.
The even and odd 2-periodic extensions look like (over 3 periods):
1.2
1
1
0.8
0.5
0.6
0
0.4
0.2
-0.5
0
-1
-0.2
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
Then in light of the Fourier convergence theorem, for x 2 [0, 1], we have
8
⇢
6 {0, 12 , 1}
< f (x) x 2
1
f (x) x 6= 2
1
x = 12
F.C.S. of f =
F.S.S. of f =
1
1 ,
2
x
=
:
2
2
0
x = 0, 1
2
2. For the following heat conduction problem with homogeneous Dirichlet boundary
conditions:
8
ut = 5uxx , (0 < x < 1.5, t > 0)
>
>
<
u(0, t) = u(1.5,
t) = 0 ,
⇢
0
0  x  .75
>
>
.
: u(x, 0) =
1
.75 < x  1.5
1
3
(a) Find the solution (in series form).
We have L = 3/2, ↵2 = 5, so
u(x, t) =
1
X
bn sin(2n⇡x/3)e
20n2 ⇡ 2 t/9
n=1
with
bn =
=
2
3/2
Z
3/2
sin(2n⇡x/3)u(x, 0)dx =
0
2
(cos(n⇡/2)
n⇡
cos(n⇡)) =
8
<
:
4
3
2
n⇡
0
4
n⇡
Z
3/2
sin(2n⇡x/3)dx =
3/4
2
3/2
cos(2n⇡x/3)|3/4
n⇡
n odd
n = 4k
,
n = 4k + 2
so (there are several ways to write this)
1
2X 1
u(x, t) =
sin(2(2k + 1)⇡x/3)e
⇡
2k + 1
k=0
1
X
4
⇡
k=0
20(2k+1)2 ⇡ 2 t/9
1
sin(2(4k + 2)⇡x/3)e
4k + 2
20(4k+2)2 ⇡ 2 t/9
.
(b) (Excel) Use the finite di↵erence method discussed in class to numerically approximate the solution u(x, t). Use the discretization parameters x = 0.06,
t = 0.0003. (You are free to use one of the posted spreadsheet examples as a
template.) Hand in the approximate value you obtain at x = 0.42, t = 0.015.
Compare this value with the value given by the first non-zero term of your series
solution from part (a). Do the same for the first two non-zero terms of your
series solution from part (a).
The computed value is u(0.42, 0.015) ⇡ 0.192. The first term of the series solution gives
2
sin(2⇡(0.42)/3)e
⇡
while the first two terms give
u(0.42, 0.015) ⇡
20⇡ 2 (0.015)/9
⇡ 0.353
2
2⇡(0.42)
2
4⇡(0.42)
2
2
sin(
)e 20⇡ (0.015)/9
sin(
)e 20(4)⇡ (0.015)/9 ⇡ 0.185
⇡
3
⇡
3
(much closer to the numerically computed value!).
(c) (Excel) Change the boundary conditions in the problem to u(0, t) = 0, u(1.5, t) =
1. Use the numerical computation to compute the solution up to time t = 0.03
and verify that it approaches a steady state. Hand in a plot showing the approximate solution at times t = 0, t = 0.003, t = 0.01, and t = 0.03.
See attached page.
u(0.42, 0.015) ⇡
2
Sheet1
Heat equation
Dx
Dt
Alpha^2
K
0.060
0.000
5.000
0.417
Legend:
Boundary conditions
Initial condition
Interval of x
Interval of t
x
t
0.0000
0.0003
0.0006
0.0009
0.0012
0.0015
0.0018
0.0021
0.0024
0.0027
0.0030
0.0033
0.0036
0.0039
0.0042
0.0045
0.0048
0.0051
0.0054
0.0057
0.0060
0.0063
0.0066
0.0069
0.0072
0.0075
0.0078
0.0081
0.0084
0.0087
0.0090
0.0093
0.0096
0.0099
0.0102
0.0105
0.0108
0.0111
0.0114
0 0.06 0.12 0.18 0.24
0.3 0.36 0.42 0.48 0.54
1.2 1.26 1.32 1.38 1.44
1.5
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0 0.417 0.583
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0 0.174 0.313 0.688 0.826
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0 0.072 0.159 0.411 0.589 0.841 0.928
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0 0.03 0.078 0.228 0.38 0.62 0.772 0.922 0.97
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0 0.013 0.038 0.121 0.229 0.417
0.583
0.771 0.879 0.962 0.987
1
1
1
1
1
1
1
Chart
Title
0
0
0
0
0
0
0 0.005 0.018 0.062 0.131 0.262 0.408 0.592 0.738 0.869 0.938 0.982 0.995
1
1
1
1
1
1
0
0
0
0
0
0 0.002 0.008 0.031 0.072 0.157 0.268 0.424 0.576 0.732 0.843 0.928 0.969 0.992
1
1
1
1
1
1
0
0
0
0
0 0.001 0.004 0.015 0.039 0.09 0.168 0.287 0.423 0.577 0.713 0.832 0.91 0.961 0.985
1
1
1
1
1
1
0
0
0
0
0 0.002 0.007 0.02 0.05 0.101 0.185 0.294 0.43 0.57 0.706 0.815 0.899 0.95 0.98 0.993
1
1
1
1
1
0
0
0
0 0.001 0.004 0.01 0.027 0.059 0.115 0.195 0.305 0.431 0.569 0.695 0.805 0.885 0.941 0.973 0.99
1
1
1
1
1
0
0 ###
0 0.002 0.005 0.015 0.033 0.069 0.125 0.208 0.312 0.436 0.564 0.688 0.792 0.875 0.931 0.967 0.985 0.995
1
1
1
1
0 ###
0 0.001 0.003 0.008 0.019 0.04 0.078 0.136 0.217 0.32 0.438 0.562 0.68 0.783 0.864 0.922 0.96 0.981 0.992
1
1
1
1
0 ###
0 0.001 0.004 0.01 0.023 0.047 0.087 0.145 0.226 0.326 0.441 0.559 0.674 0.774 0.855 0.913 0.953 0.977 0.99
1
1
1
1
u(x,0)
0
0 0.001 0.002 0.005 0.013 0.028 0.054 0.095 0.155 0.234 0.332 0.442 0.558 0.668 0.766 0.845 0.905 0.946 0.972 0.987 0.995
1
1
1
0
0 0.001 0.003 0.007 0.016 0.032 0.06 0.102 0.163 0.242 0.337 0.445 0.555 0.663 0.758 0.837 0.898 0.94 0.968 0.984 0.993
u(x,0.003)1
1
1
0
0 0.001 0.004 0.009 0.019 0.037 0.066 0.11 0.171 0.249 0.342 0.446 0.554 0.658 0.751 0.829 0.89 0.934 0.963 0.981 0.991
1
1
1
u(x,0.0099)
0 0.001 0.002 0.005 0.011 0.022 0.042 0.072 0.117 0.178 0.255 0.347 0.448 0.552 0.653 0.745 0.822 0.883 0.928 0.958 0.978 0.989
0.995
1
1
u(x,0.030)
0 0.001 0.003 0.006 0.013 0.026 0.046 0.078 0.124 0.185 0.261 0.351 0.449 0.551 0.649 0.739 0.815 0.876 0.922 0.954 0.974 0.987 0.994
1
1
0 0.001 0.003 0.008 0.016 0.029 0.051 0.084 0.13 0.191 0.267 0.354 0.45 0.55 0.646 0.733 0.809 0.87 0.916 0.949 0.971 0.984 0.992
1
1
0 0.002 0.004 0.009 0.018 0.033 0.056 0.089 0.136 0.197 0.272 0.358 0.452 0.548 0.642 0.728 0.803 0.864 0.911 0.944 0.967 0.982 0.991
1
1
0 0.002 0.005 0.011 0.02 0.036 0.06 0.095 0.142 0.203 0.276 0.361 0.453 0.547 0.639 0.724 0.797 0.858 0.905 0.94 0.964 0.98 0.989 0.995
1
0 0.003 0.006 0.012 0.023 0.04 0.065 0.1 0.148 0.208 0.281 0.364 0.454 0.546 0.636 0.719 0.792 0.852 0.9 0.935 0.96 0.977 0.988 0.994
1
0 0.003 0.007 0.014 0.025 0.043 0.069 0.105 0.153 0.213 0.285 0.367 0.455 0.545 0.633 0.715 0.787 0.847 0.895 0.931 0.957 0.975 0.986 0.993
1
0 0.004 0.008 0.016 0.028 0.047 0.073 0.11 0.158 0.218 0.289 0.37 0.456 0.544 0.63 0.711 0.782 0.842 0.89 0.927 0.953 0.972 0.984 0.992
1
0 0.004 0.01 0.018 0.031 0.05 0.077 0.115 0.163 0.223 0.293 0.372 0.457 0.543 0.628 0.707 0.777 0.837 0.885 0.923 0.95 0.969 0.982 0.99
1
0 0.005 0.011 0.02 0.033 0.053 0.082 0.119 0.168 0.227 0.297 0.374 0.458 0.542 0.626 0.703 0.773 0.832 0.881 0.918 0.947 0.967 0.98 0.989
1
0 0.005 0.012 0.022 0.036 0.057 0.086 0.124 0.172 0.231 0.3 0.377 0.458 0.542 0.623 0.7 0.769 0.828 0.876 0.914 0.943 0.964 0.978 0.988 0.995
0 0.006 0.013 0.024 0.039 0.06 0.089 0.128 0.177 0.235 0.303 0.379 0.459 0.541 0.621 0.697 0.765 0.823 0.872 0.911 0.94 0.961 0.976 0.987 0.994
0 0.006 0.014 0.026 0.041 0.063 0.093 0.132 0.181 0.239 0.307 0.381 0.46 0.54 0.619 0.693 0.761 0.819 0.868 0.907 0.937 0.959 0.974 0.986 0.994
0 0.007 0.016 0.028 0.044 0.067 0.097 0.136 0.185 0.243 0.309 0.383 0.46 0.54 0.617 0.691 0.757 0.815 0.864 0.903 0.933 0.956 0.972 0.984 0.993
0 0.008 0.017 0.029 0.047 0.07 0.101 0.14 0.189 0.247 0.312 0.385 0.461 0.539 0.615 0.688 0.753 0.811 0.86 0.899 0.93 0.953 0.971 0.983 0.992
0 0.008 0.018 0.031 0.049 0.073 0.104 0.144 0.193 0.25 0.315 0.386 0.462 0.538 0.614 0.685 0.75 0.807 0.856 0.896 0.927 0.951 0.969 0.982 0.992
0 0.009 0.02 0.033 0.052 0.076 0.108 0.148 0.196 0.253 0.318 0.388 0.462 0.538 0.612 0.682 0.747 0.804 0.852 0.892 0.924 0.948 0.967 0.98 0.991
0 0.01 0.021 0.035 0.054 0.079 0.111 0.151 0.2 0.256 0.32 0.39 0.463 0.537 0.61 0.68 0.744 0.8 0.849 0.889 0.921 0.946 0.965 0.979 0.99
0 0.01 0.022 0.037 0.057 0.082 0.115 0.155 0.203 0.259 0.322 0.391 0.463 0.537 0.609 0.678 0.741 0.797 0.845 0.885 0.918 0.943 0.963 0.978 0.99
0 0.011 0.024 0.039 0.059 0.085 0.118 0.158 0.206 0.262 0.325 0.393 0.464 0.536 0.607 0.675 0.738 0.794 0.842 0.882 0.915 0.941 0.961 0.976 0.989
0 0.012 0.025 0.041 0.062 0.088 0.121 0.161 0.21 0.265 0.327 0.394 0.464 0.536 0.606 0.673 0.735 0.79 0.839 0.879 0.912 0.938 0.959 0.975 0.988
0 0.012 0.026 0.043 0.064 0.091 0.124 0.165 0.213 0.268 0.329 0.395 0.465 0.535 0.605 0.671 0.732 0.787 0.835 0.876 0.909 0.936 0.957 0.974 0.988
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Axis Title
Parameters:
Finite difference scheme
Page 1
0.6 0.66 0.72 0.78 0.84
0.9 0.96 1.02 1.08 1.14
(d) (Excel) In your computation, change t from t = 0.0003 to t = 0.0004
and see what happens. Does your computation make any sense? What is the
explanation?
Increasing the time step t to 0.004 makes the computation go wild: enormous
numbers appear very quickly. That is because the value of K = ↵2 ( t)/( x)
becomes K = 0.556, and the stability condition K < 1/2 is violated, producing
spurious, exponentially growing solutions.
3. (Excel) Consider the following heat conduction problem with “insulating” BCs:
ut = uxx ,
0<x<1
ux (0, t) = 0
and ux (1, t) = 0
⇢
0 if 0  x  1/2,
u(x, 0) = f (x) =
1 if 1/2 < x  1,
The solution is given by
u(x, t) =
1
a0 X
+
an cos (n⇡x) e
2
n2 ⇡ 2 t
n=1
where
an = 2
Z
1
f (x) cos (n⇡x) dx =
0
(
1
2
n⇡
sin
n = 0,
n⇡
2
n > 0.
[You should be able to find this solution yourself by separating variables - if you feel
you need more practice, try to check it.]
(a) Describe, with the aid of a sketch, how the temperature varies over time at the
following points; (a) x = 0, (b) x = 1/2, (c) x = 1. To what value does the
solution tend as t ! 1?
In light of the Fourier convergence theorem (recall the Fourier cosine series
describing u(x, 0) is Fourier series of the even periodic extension of f (x)), we
have
1
1 1
1
1
1
u(0, 0) = f (0) = 0, u( , 0) = [f ( )+f ( +)] = [0+1] = , u(1, 0) = f (1) = 1.
2
2 2
2
2
2
Moreover, for every x, limt!1 u(x, t) = 1/2. For x = 1/2, in fact, we have
u(1/2, t) = 1/2 for all t, by direct substitution of x = 1/2 into the series – all
the terms except a0 /2 vanish. Based on these values, one can make a reasonable
qualitative sketch of u(0, t), u(1/2, t), and u(1, t) as functions of t:
3
u(0,t)
u(1/2,t)
u(1,t)
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
(b) Now use an Excel spreadsheet and a finite di↵erence approximation to solve
the problem numerically, taking N = 10 ( x = 0.1) and t = 0.004 (you can
use the template on the website, or your spreadsheet from assignment 5, but
remember to account for the derivative boundary conditions). To what value
does the solution appear to be converging as t ! 1? Why is this? How could
you improve the numerical solution to get closer to the correct answer?
A spreadsheet showing this computation is attached, with a plot of the numerical
solution at several times. The solution appears to be tending to a constant value
of (about) 0.45, whereas we know the true solution tends to 0.5. This is because
in representing the initial data we chose the value 0 at the grid point x = 1/2,
and so e↵ectively have represented a function with average value (integral) less
than 0.5 (i.e. about 0.45) where the true initial data has average value 0.5 (and
the solution of the heat equation with insulating BCs tends to the average value
of the initial data as t ! 1).
4
dx =
Time \x ->
0
0
0.01
0.01
0.02
0.02
0.02
0.03
0.03
0.04
0.04
0.04
0.05
0.05
0.06
0.06
0.06
0.07
0.07
0.08
0.08
0.08
0.09
0.09
0.1
0.1
0.1
0.11
0.11
0.12
0.12
0.12
0.13
0.13
0.14
0.14
0.14
0.15
0.15
0.16
0.16
0.16
0.17
0.17
0.18
0.18
0.18
0.19
0.1 dt =
1
0
0.16
0.06
0.1
0.08
0.09
0.09
0.1
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
0.21
0.22
0.23
0.24
0.24
0.25
0.26
0.27
0.27
0.28
0.29
0.29
0.3
0.31
0.31
0.32
0.32
0.33
0.33
0.34
0.34
0.34
0.35
0.35
0.36
0.36
0.36
0.37
0.37
0
0
0.4
0.08
0.14
0.08
0.09
0.08
0.09
0.09
0.1
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
0.21
0.22
0.23
0.23
0.24
0.25
0.26
0.26
0.27
0.28
0.29
0.29
0.3
0.3
0.31
0.31
0.32
0.33
0.33
0.33
0.34
0.34
0.35
0.35
0.36
0.36
0.36
0.37
0 alpha^2
1 k=a dt/dx^2 0.4 pi
0.1
0.2
0.3
0.4
0.5
0
0
0
0
0
0
0
0
0
0.4
0.16
0
0
0.16
0.32
0.06
0.06
0.06
0.16
0.4
0.1
0.06
0.1
0.22
0.38
0.08
0.09
0.13
0.24
0.41
0.09
0.1
0.16
0.26
0.41
0.09
0.12
0.18
0.28
0.42
0.1
0.13
0.2
0.3
0.42
0.11
0.15
0.21
0.31
0.43
0.12
0.16
0.22
0.32
0.43
0.13
0.17
0.23
0.32
0.43
0.14
0.18
0.24
0.33
0.44
10.15
0.19
0.25
0.34
0.44
0.16
0.2
0.26
0.34
0.44
0.9
0.17
0.21
0.27
0.35
0.44
0.18
0.22
0.28
0.35
0.44
0.8
0.19
0.23
0.28
0.36
0.44
0.7 0.2
0.24
0.29
0.36
0.44
0.21
0.24
0.3
0.37
0.45
0.6
0.22
0.25
0.3
0.37
0.45
0.50.23
0.26
0.31
0.37
0.45
0.24
0.27
0.32
0.38
0.45
0.4
0.24
0.27
0.32
0.38
0.45
0.25
0.28
0.33
0.38
0.45
0.3
0.26
0.29
0.33
0.39
0.45
0.20.27
0.29
0.34
0.39
0.45
0.27
0.3
0.34
0.39
0.45
0.1
0.28
0.31
0.34
0.39
0.45
0.31
0.35
0.4
0.45
00.29
0.29 Col
0.32 Col0.35 Col 0.4Col 0.45
Col
Col
umn
umn
0.3 umn
0.32 umn
0.36 umn 0.4umn 0.45
C
H
0.31 D0.33 E 0.36 F
0.4G 0.45
0.31
0.33
0.36
0.4
0.45
0.32
0.34
0.37
0.41
0.45
0.32
0.34
0.37
0.41
0.45
0.33
0.34
0.37
0.41
0.45
0.33
0.35
0.38
0.41
0.45
0.34
0.35
0.38
0.41
0.45
0.34
0.36
0.38
0.41
0.45
0.34
0.36
0.38
0.42
0.45
0.35
0.36
0.39
0.42
0.45
0.35
0.37
0.39
0.42
0.45
0.36
0.37
0.39
0.42
0.45
0.36
0.37
0.39
0.42
0.45
0.36
0.38
0.4
0.42
0.45
0.37
0.38
0.4
0.42
0.45
0.37
0.38
0.4
0.42
0.45
3.14
0.6
0.7
0.8
0.9
1
1
1
1
1
1
0.6
1
1
1
0.6
0.68
0.84
1
0.84
0.92
0.6
0.84
0.87
0.94
0.86
0.62
0.76
0.88
0.88
0.92
0.58
0.75
0.83
0.9
0.89
0.58
0.71
0.83
0.87
0.9
0.57
0.71
0.8
0.86
0.87
0.56
0.69
0.79
0.84
0.86
0.56
0.68
0.77
0.83
0.85
0.55
0.67
0.76
0.81
0.83
0.55
0.66
0.74
0.8
0.82
0.55
0.65
0.73
0.78
0.8
0.54
0.64
0.72
0.77
0.79
0.54
0.63
0.71
0.76
0.77
0.54
0.63
0.7
0.74
0.76
0.53
0.62
0.69
0.73
0.75
0.53
0.61
0.68
0.72
0.73
0.53
0.61
0.67
0.71
0.72
0.53
0.6
0.66
0.7
0.71
0.52
0.59
0.65
0.69
0.7
0.52
0.59
0.64
0.68
0.69
0.52
0.58
0.63
0.67
0.68
0.52
0.58
0.63
0.66
0.67
0.51
0.57
0.62
0.65
0.66
0.51
0.57
0.61
0.64
0.65
0.51
0.56
0.61
0.64
0.65
0.51
0.56
0.6
0.63
0.64
0.5
0.55
0.6
0.62
0.63
0.5
0.55
0.59
0.61
0.62
0.5
0.55 Col
0.58 Col0.61Col 0.62
ColCol
umn
0.5 I umn
0.54 umn
0.58 umn0.6umn0.61
J0.54 K0.57 L 0.6M
0.5
0.6
0.49
0.54
0.57
0.59
0.6
0.49
0.53
0.56
0.58
0.59
0.49
0.53
0.56
0.58
0.59
0.49
0.53
0.56
0.57
0.58
0.49
0.52
0.55
0.57
0.58
0.49
0.52
0.55
0.56
0.57
0.49
0.52
0.54
0.56
0.57
0.48
0.52
0.54
0.56
0.56
0.48
0.51
0.54
0.55
0.56
0.48
0.51
0.53
0.55
0.55
0.48
0.51
0.53
0.54
0.55
0.48
0.51
0.53
0.54
0.54
0.48
0.5
0.52
0.54
0.54
0.48
0.5
0.52
0.53
0.54
0.48
0.5
0.52
0.53
0.53
1
0.84
0.94
0.88
0.9
0.87
0.86
0.84
0.83
0.81
0.8
0.78
0.77
0.76
0.74
0.73
0.72
0.71
0.7
0.69
0.68
0.67
0.66
0.65
0.64
0.64
0.63
0.62
0.61
0.61
0.6
0.6
0.59
0.58
0.58
0.57
0.57
0.56
0.56
0.56
0.55
0.55
0.54
0.54
0.54
0.53
0.53