Powers D.L. Boundary Value Problems and Partial Differential Equations
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428 Chapter 7 Numerical Methods
11.
Approximate the solution of the wave equation in a semi-infinite strip 3
units wide. Assume u = 0 on all boundaries, zero initial velocity, and an
initial value for u that is 1 in a corner and 0 elsewhere. Take x =y = 1
and ρ
2
=1/2.
7.6 Comments and References
Our objective in this chapter has been to survey some elementary numerical
methods for problems like those we attacked analytically in earlier chapters.
We have only had enough space to touch on the central topics: obtaining re-
placement equations, solving linear systems of equations by direct and iterative
methods, numerical stability, and order of error.
The methods we have introduced are satisfactory for a first introduction and
for learning something about partial differential equations, but they are not
adequate for any serious problem solving. New techniques for these problems
are superior in speed, accuracy and stability but are also more complicated. Of
the many texts available, two excellent ones are Numerical Analysis by Burden
and Faires, for general methods, and Numerical Solution of Partial Differential
Equations by Smith. (See the Bibliography.)
Almost all numerical methods for linear partial differential equations rely
on the symbolism and theory of matrices. Two outstanding texts on matrix
theory are Applied Linear Algebra, 3rd ed., by Noble and Daniel, and Matrices
by Barnett.
Miscellaneous Exercises
1. Set up and solve replacement equations for this boundary value problem.
Use x = 1/3.
d
2
u
dx
2
−
√
24xu= 0, 0 < x < 1,
du
dx
(0) = 1, u(1) = 1.
2. Use the change of variables x = (r −a)/(b − a) and v(r) = u(x) to con-
vert the equation
1
r
d
dr
r
dv
dr
−q(r)v = f (r), a < r < b,
to an equation in u on the interval 0 < x < 1.
Miscellaneous Exercises 429
3. By means of the transformation mentioned in Exercise 2, a heat problem
on an annular ring is converted to
d
2
u
dx
2
+
1
1 +x
du
dx
=−(1 +x), 0 < x < 1,
u(0) =1, u(1) = 0.
Set up and solve replacement equations for this problem using x =
1/4.
4. The boundary value problem
1
r
d
dr
r
dv
dr
−γ
2
v = 0, a < r < b,
v(a) = 1,v(b) =0,
can be transformed into the problem
d
2
u
dx
2
+
1
α +x
du
dx
−γ
2
L
2
u =0, 0 < x < 1,
u(0) =1, u(1) = 0,
where L = b − a and α = a/L. Set up and solve replacement equations
using x =1/4, α = 1, γ L = 1.
5. Set up replacement equations for the heat problem in the following, and
solve for t up to 1/4, using x =1/4, t = 1/32.
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = u(1, t) = 1 −e
−t
, 0 < t,
u(x, 0) = 0, 0 < x < 1.
6. Same as Exercise 5, but use u(0, t) = u(1, t) = 1 − e
−32(ln 2)t
so that
u(0, t
m
) = 1 −(0.5)
m
.
7. Compare the numerical solution of the problem
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = 0, 0 < t,
u(x, 0) = 1, 0 < x < 1,
430 Chapter 7 Numerical Methods
with the solution of the problem consisting of the equation
∂
2
u
∂x
2
−16u =
∂u
∂t
, 0 < x < 1, 0 < t,
with the same initial and boundary conditions. Use x = 1/4, t =
1/48 in both cases.
8. In Exercise 7, what is the longest stable time step for each of the two
problems?
9. Solve for several time levels using x =1/5andr = 1/2. What is t?
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = 25t, u(1, t) = 0, 0 < t,
u(x, 0) = 0, 0 < x < 1.
10. Same as Exercise 9, except the second boundary condition is
∂u
∂x
(1, t) = 0.
11. This problem describes the displacement of a string whose end is jerked:
∂
2
u
∂x
2
=
∂
2
u
∂t
2
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = 1, 0 < t,
u(x, 0) = 0,
∂u
∂t
(x, 0) = 0, 0 < x < 1.
Solve numerically through one period (until t = 2) with x = t =
1/4.
12. Same problems as Exercise 11, except the right-hand boundary condi-
tion is u(1, t) = h(t),0< t,where
h(t) =
1, 0 < t ≤1,
0, 1 < t ≤2,
and h(t +2) = h(t). Solve numerically with x = t = 1/4 for enough
values of t so that resonance becomes noticeable.
13. Using x = y = 1/4, find a numerical solution of this problem:
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0, 0 < x < 1, 0 < y < 1,
u(x, 0) = 0, u(x, 1) =
2
π
tan
−1
1
x
, 0 < x < 1,
u(0, y) = 1, u(1, y) =
2
π
tan
−1
(y), 0 < y < 1.
Miscellaneous Exercises 431
14. The analytical solution of the problem in Exercise 13 is u(x, y) =
(2/π) tan
−1
(y/x). Compare your numerical results with the exact so-
lution.
15. Using x = y = 1/4andr = 1/4, find a numerical solution for this
problem:
∇
2
u =
∂u
∂t
, 0 < x < 1, 0 < y < 1, 0 < t,
u(x, 0, t) = u(x, 1, t) = 0, 0 < x < 1, 0 < t,
u(0, y, t) = u(1, y, t) = 0, 0 < y < 1, 0 < t,
u(x, y, 0) = 1, 0 < x < 1, 0 < y < 1.
16. TheanalyticalsolutionoftheprobleminExercise15is
u(x, y, t) =
∞
n=1
∞
n=1
4(1 −cos(nπ ))(1 −cos(mπ))
π
2
mn
×sin(nπ x) sin(nπy )e
−(m
2
+n
2
)π
2
t
.
Using just the term m = n = 1 of this solution, compare the ratio
R =
u
1
2
,
1
2
, t
m+1
u
1
2
,
1
2
, t
m
and the ratio of the corresponding u’s computed in Exercise 15.
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Bibliography
Abramowitz, M., and I. Stegun (eds). Handbook of Mathematical Functions,
10th ed. Washington, DC, National Bureau of Standards, 1972. (Reprinted
by Dover, 1974.)
Andrews, L.C. Special Functions of Mathematics for Engineers,2nded.Belling-
ham, WA, SPIE — The International Society for Optical Engineering, 1997.
Barnett, S. Matrices: Methods and Applications. New York, Oxford University
Press, 1990.
Burden, R.L., and J.D. Faires. Numerical Analysis, 7th ed. Belmont, CA,
Brooks/Cole, 2000.
Carslaw, H.S., and J.C. Jaeger. Conduction of Heat in Solids,2nded.NewYork,
Oxford University Press, 1986.
Churchill, R.V., and J.W. Brown. Fourier Series and Boundary Value Problems,
6th ed. New York, McGraw-Hill, 2000.
Churchill, R.V. Operational Mathematics,3rded.NewYork,McGraw-Hill,
1972.
Courant, R., and D. Hilbert. Methods of Mathematical Physics, Vol. 1. New
York, Wiley-Interscience, 1953/1989.
Crank, J. The Mathematics of Diffusion,2nded.NewYork,OxfordUniversity
Press, 1980.
Davis, P.J., and R. Hersh. The Mathematical Experience.Boston,Houghton
Mifflin, 1999.
Erdelyi, A., W. Magnus, F. Oberhettinger, and F. Tricomi. Ta b l e s of Inte g ral
Transforms, Vols. 1 and 2. New York, McGraw-Hill, 1954.
Feller, W. Introduction to Probability Theory and Its Applications, Vol. 1, 3rd ed.
New York, Wiley, 1968.
433
434 Bibliography
Fletcher, N.H., and T.D. Rossing. The Physics of Musical Instruments,2nded.
New York, Springer-Verlag, 1998.
Isenberg, C. The Science of Soap Films and Soap Bubbles. Avon, UK, Tieto Ltd,
1978. (Reprinted by Dover, 1992.)
Jerri, A.J. Integral and Discrete Transforms with Applications and Error Analysis.
New York, Marcel Dekker, 1991.
Jones, D.S., and B.D. Sleeman. Differential Equations and Mathematical Biol-
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Kirchhoff, R.H. Potential Flows: Computer Graphic Solutions.NewYork,Marcel
Dekker, 1998.
Lamb, H. Hydrodynamics, 6th ed. Cambridge, UK, Cambridge University
Press, 1971.
Love, A.E.H. A Treatise on the Mathematical Theory of Elasticity,4thed.New
York, Dover, 1944.
Main, I.G. Vibrations and Waves in Physics,3rded.Cambridge,UK,Cambridge
University Press, 1993.
Morse, P.M., and H. Feshbach. Methods of Theoretical Physics, Part I.NewYork,
McGraw-Hill, 1953.
Murray, J.D. Mathematical Biology. New York, Springer-Verlag, 1993.
Noble, B., and J.W. Daniel. Applied Linear Algebra, 3rd ed. Englewood Cliffs,
NJ, Prentice-Hall, 1988.
Sagan, H. Boundary and Eigenvalue Problems in Mathematical P hysics.New
York, Wiley, 1966. (Reprinted by Dover, 1989.)
Smith, G.D. Numerical Solution of Partial Differential Equations,3rded.New
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McGraw-Hill, 1951.
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(Reprinted by Dover, 1976.)
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1996.
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Appendix:
Mathematical
References
Trigonometric Functions
sin(A ±B) = sin(A) cos(B) ± cos(A) sin(B)
cos(A ±B) = cos(A) cos(B) ∓ sin(A) sin(B)
sin(A) +sin(B) = 2sin
A +B
2
cos
A −B
2
sin(A) −sin(B) = 2cos
A +B
2
sin
A −B
2
cos(A) +cos(B) = 2cos
A +B
2
cos
A −B
2
cos(A) −cos(B) = 2sin
A +B
2
sin
A −B
2
sin(A) sin(B) =
1
2
cos(A −B) −cos(A +B)
sin(A) cos(B) =
1
2
sin(A −B) +sin(A +B)
cos(A) cos(B) =
1
2
cos(A −B) +cos(A +B)
cos(A) =
1
2
e
iA
+e
−iA
, sin(A) =
1
2i
e
iA
−e
−iA
cos
2
(A) +sin
2
(A) = 1, 1 + tan
2
(A) =sec
2
(A)
435
436 Appendix: Mathematical References
Hyperbolic Functions
cosh(A) =
1
2
e
A
+e
−A
, sinh(A) =
1
2
e
A
−e
−A
d cosh(u) = sinh(u)du, d sinh(u) = cosh(u)du
sinh(A ±B) = sinh(A) cosh(B) ± cosh(A) sinh(B)
cosh(A ± B) = cosh(A) cosh(B) ± sinh(A) sinh(B)
sinh(A) +sinh(B) = 2sinh
A +B
2
cosh
A −B
2
sinh(A) −sinh(B) = 2cosh
A +B
2
sinh
A −B
2
cosh(A) + cosh(B) = 2cosh
A +B
2
cosh
A −B
2
cosh(A) − cosh(B) = 2sinh
A +B
2
sinh
A −B
2
sinh(A) sinh(B) =
1
2
cosh(A + B) −cosh(A − B)
sinh(A) cosh(B) =
1
2
sinh(A +B) +sinh(A −B)
cosh(A) cosh(B) =
1
2
sinh(A +B) +cosh(A −B)
cosh
2
(A) −sinh
2
(A) = 1, 1 − tanh
2
(A) =sech
2
(A)
Calculus
1. Derivative of a product
(uv)
= u
v +uv
(uv)
= u
v +2u
v
+uv
(uv)
(n)
= u
(n)
v +
n
1
u
(n−1)
v
+···+
n
n −1
uv
(n−1)
+uv
(n)
In this formula,
n
k
=
n!
(n−k)!k!
is a binomial coefficient.
2. Rules of integration
a.
b
a
c
1
f
1
(x) +c
2
f
2
(x)
dx = c
1
b
a
f
1
(x) dx + c
2
b
a
f
2
(x) dx
Appendix: Mathematical References 437
b.
a
a
f (x) dx = 0
c.
b
a
f (x) dx =−
a
b
f (x) dx
d.
b
a
f (x) dx =
c
a
f (x) dx +
b
c
f (x ) dx
3. Derivatives of integrals
a.
d
dt
b
a
f (x , t) dx =
b
a
∂f
∂t
(x, t) dx
b.
d
dt
t
a
f (x) dx = f (t) (Fundamental theorem of calculus;
a is constant)
c.
d
dt
v(t)
u(t)
f (x , t) dx = f
v(t), t
v
(t) −f
u(t), t
u
(t)
+
v(t)
u(t)
∂f
∂t
(x, t) dx (Leibniz’s rule)
4. Integration by parts
a.
uv
dx = uv −
vu
dx
b.
uv
dx = v
u −vu
+
vu
dx
5. Functions defined by integrals
a. Natural logarithm
ln(x) =
x
1
dz
z
b. Sine-integral function
Si(x) =
x
0
sin(z)
z
dz
c. Normal probability distribution function
(x) =
1
√
2π
x
−∞
e
−z
2
/2
dz
d. Error function
erf(x) =
2
√
π
x
0
e
−z
2
dz
Note: erf(x) = 2
√
2x
−1