Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 5 ppt
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37.9 Hints
Hint 37.1
Hint 37.2
Hint 37.3
Hint 37.4
Hint 37.5
Hint 37.6
Hint 37.7
Impose the boundary conditions
u(0, t) = u(2π, t), u
θ
(0, t) = u
θ
(2π, t).
Hint 37.8
Apply the separation of variables u(x, y) = X(x)Y (y). Solve an eigenvalue problem for X(x).
Hint 37.9
Hint 37.10
1734
Hint 37.11
Hint 37.12
There are two ways to solve the problem. For the first method, expand the solution in a series of the form
u(x, t) =
∞
n=1
a
n
(t) sin
nπx
L
.
Because of the inhomogeneous boundary conditions, the convergence of the series will not be uniform. You can
differentiate the series with respect to t, but not with respect to x. Multiply the partial differential equation by the
eigenfunction sin(nπx/L) and integrate from x = 0 to x = L. Use integration by parts to move derivatives in x from
u to the eigenfunctions. This process will yield a first order, ordinary differential equation for each of the a
n
’s.
For the second method: Make the change of variables v(x, t) = u(x, t) − µ(x), where µ(x) is the equilibrium
temperature distribution to obtain a problem with homogeneous boundary conditions.
Hint 37.13
Hint 37.14
Hint 37.15
Hint 37.16
Hint 37.17
1735
Hint 37.18
Use separation of variable s to find eigen-solutions of the partial differential equation that satisfy the homogeneous
boundary conditions. There will be two eigen-solutions for each eigenvalue. Expand u(x, t) in a series of the eigen-
solutions. Use the two initial conditions to determine the constants.
Hint 37.19
Expand the solution in a series of eigenfunctions in x. D etermine these eigenfunctions by using separation of variables
on the homogeneous partial differential equation. You will find that the answer has the form,
u(x, y) =
∞
n=1
u
n
(y) sin
nπx
a
.
Substitute this series into the partial differential equation to determine ordinary differential equations for each of the
u
n
’s. The boundary conditions on u(x, y) will give you boundary conditions for the u
n
’s. Solve these ordinary differential
equations with Green functions.
Hint 37.20
Solve this problem by expanding the solution in a series of eigen-solutions that satisfy the partial differential equation
and the homogeneous boundary conditions. Use the initial conditions to determine the coefficients in the expansion.
Hint 37.21
Use separation of variables to find eigen-solutions that satisfy the partial differential equation and the homogeneous
boundary conditions. The solution is a linear combination of the eigen-solutions. The whole solution will be exponentially
decaying if each of the eigen-solutions is exponentially decaying.
Hint 37.22
For parts (a), (b) and (c) use separation of variables. For part (b) the eigen-solutions will involve Bessel functions. For
part (c) the eigen-solutions will involve spherical Bessel functions. Part (d) is trivial.
Hint 37.23
The solution is a linear combination of eigen-solutions of the partial differential equation that satisfy the homogeneous
boundary conditions. Determine the coefficients in the expansion with the initial condition.
1736
Hint 37.24
The problem is
u
rr
+
1
r
u
r
+
1
r
2
u
θθ
= 0, 0 < r < 1, 0 < θ < π
u(r, 0) = u(r, π) = 0, u(0, θ) = 0, u(1, θ) = 1
The solution is a linear combination of eigen-solutions that satisfy the partial differential equation an d the three
homogeneous boundary conditions.
Hint 37.25
Hint 37.26
Hint 37.27
Hint 37.28
Hint 37.29
Hint 37.30
Hint 37.31
Hint 37.32
1737
Hint 37.33
Hint 37.34
Hint 37.35
Hint 37.36
Hint 37.37
1738
37.10 Solutions
Solution 37.1
We expand the solution in eigenfunctions in x and y which satify the boundary conditions.
u =
∞
m,n=1
u
mn
(t) sin
mπx
a
sin
nπy
b
We expand the inhomogeneities in the eigenfunctions.
q(x, y, t) =
∞
m,n=1
q
mn
(t) sin
mπx
a
sin
nπy
b
q
mn
(t) =
4
ab
a
0
b
0
q(x, y, t) sin
mπx
a
sin
nπy
b
dy dx
f(x, y) =
∞
m,n=1
f
mn
sin
mπx
a
sin
nπy
b
f
mn
=
4
ab
a
0
b
0
f(x, y) sin
mπx
a
sin
nπy
b
dy dx
1739
We substitute the expansion of the solution into the diffusion equation and the initial condition to determine initial
value problems for the coefficients in the expansion.
u
t
− κ(u
xx
+ u
yy
) = q(x, y, t)
∞
m,n=1
u
mn
(t) + κ
mπ
a
2
+
nπ
b
2
u
mn
(t)
sin
mπx
a
sin
nπy
b
=
∞
m,n=1
q
mn
(t) sin
mπx
a
sin
nπy
b
u
mn
(t) + κ
mπ
a
2
+
nπ
b
2
u
mn
(t) = q
mn
(t)
u(x, y, 0) = f(x, y)
∞
m,n=1
u
mn
(0) sin
mπx
a
sin
nπy
b
=
∞
m,n=1
f
mn
sin
mπx
a
sin
nπy
b
u
mn
(0) = f
mn
We solve the ordinary differential equations for the coefficients u
mn
(t) subject to their initial conditions.
u
mn
(t) =
t
0
exp
−κ
mπ
a
2
+
nπ
b
2
(t − τ)
q
mn
(τ) dτ + f
mn
exp
−κ
mπ
a
2
+
nπ
b
2
t
Solution 37.2
After looking at this problem for a minute or two, it seems like the answer would have the form
u = sin(x)T (t).
This form satisfies the boundary conditions. We substitute it into the heat equation and the initial condition to determine
1740
T
sin(x)T
= −κ sin(x)T + A sin(x), T(0) = 0
T
+ κT = A, T (0) = 0
T =
A
κ
+ c
e
−κt
T =
A
κ
1 −
e
−κt
Now we have the solution of the heat equation.
u =
A
κ
sin(x)
1 −
e
−κt
Solution 37.3
First we write the Laplacian in p olar coordinates.
u
rr
+
1
r
u
r
+
1
r
2
u
θθ
= 0
1. We introduce the separation of variables u(r, θ) = R(r)Θ(θ).
R
Θ +
1
r
R
Θ +
1
r
2
RΘ
= 0
r
2
R
R
+ r
R
R
= −
Θ
Θ
= λ
We have a regular Sturm-Liouville problem for Θ and a differential equation for R.
Θ
+ λΘ = 0, Θ
(0) = Θ(π/2) = 0 (37.8)
r
2
R
+ rR
− λR = 0, R is bounded
1741
First we solve the problem for Θ to determine the eigenvalues and eigenfunctions. The Rayleigh quotient is
λ =
π/2
0
(Θ
)
2
dθ
π/2
0
Θ
2
dθ
Immediately we see that the eigenvalues are non-negative. I f Θ
= 0, then the right boundary condition implies
that Θ = 0. Thus λ = 0 is not an eigenvalue. We find the general solution of Equation 37.8 for positive λ.
Θ = c
1
cos
√
λθ
+ c
2
sin
√
λθ
The solution that satisfies the left boundary condition is
Θ = c cos
√
λθ
.
We apply the right boundary condition to determine the eigenvalues.
cos
√
λ
π
2
= 0
λ
n
= (2n − 1)
2
, Θ
n
= cos ((2n − 1)θ) , n ∈ Z
+
Now we solve the differential equation for R. Since this is an Euler equation, we make the su bstitition R = r
α
.
r
2
R
n
+ rR
n
− (2n − 1)
2
R
n
= 0
α(α − 1) + α −(2n − 1)
2
= 0
α = ±(2n − 1)
R
n
= c
1
r
2n−1
+ c
2
r
1−2n
The solution which is bounded in 0 ≤ r ≤ 1 is
R
n
= r
2n−1
.
1742
The solution of Laplace’s equation is a linear combination of the eigensolutions.
u =
∞
n=1
u
n
r
2n−1
cos ((2n − 1)θ)
We use the boundary condition at r = 1 to determin e the coefficients.
u(1, θ) = f(θ) =
∞
n=1
u
n
cos ((2n − 1)θ)
u
n
=
4
π
π/2
0
f(θ) cos ((2n − 1)θ) dθ
2. We introduce the separation of variables u(r, θ) = R(r)Θ(θ).
R
Θ +
1
r
R
Θ +
1
r
2
RΘ
= 0
r
2
R
R
+ r
R
R
= −
Θ
Θ
= λ
We have a regular Sturm-Liouville problem for Θ and a differential equation for R.
Θ
+ λΘ = 0, Θ
(0) = Θ
(π/2) = 0 (37.9)
r
2
R
+ rR
− λR = 0, R is bounded
First we solve the problem for Θ to determine the eigenvalues and eigenfunctions. We recognize this problem as
the generator of the Fourier cosine series.
λ
n
= (2n)
2
, n ∈ Z
0+
,
Θ
0
=
1
2
, Θ
n
= cos (2nθ) , n ∈ Z
+
1743
Now we solve the differential equation for R. Since this is an Euler equation, we make the su bstitition R = r
α
.
r
2
R
n
+ rR
n
− (2n)
2
R
n
= 0
α(α − 1) + α −(2n)
2
= 0
α = ±2n
R
0
= c
1
+ c
2
ln(r), R
n
= c
1
r
2n
+ c
2
r
−2n
, n ∈ Z
+
The solutions which are bounded in 0 ≤ r ≤ 1 are
R
n
= r
2n
.
The solution of Laplace’s equation is a linear combination of the eigensolutions.
u =
u
0
2
+
∞
n=1
u
n
r
2n
cos (2nθ)
We use the boundary condition at r = 1 to determin e the coefficients.
u
r
(1, θ) =
∞
n=1
2nu
n
cos(2nθ) = g(θ)
Note that the constant term is missing in this cosine series. g(θ) has such a series expansion only if
π/2
0
g(θ) dθ = 0.
This is the condition for the existence of a solution of the problem. If this is satisfied, we can solve for the
coefficients in the expansion. u
0
is arbitrary.
u
n
=
4
π
π/2
0
g(θ) cos (2nθ) dθ, n ∈ Z
+
1744
Solution 37.4
1.
u
t
= ν(u
xx
+ u
yy
)
XY T
= ν(X
Y T + XY
T )
T
νT
=
X
X
+
Y
Y
= −λ
X
X
= −
Y
Y
− λ = −µ
We have boundary value problems for X(x) and Y (y) and a differential equation for T(t).
X
+ µX = 0, X
(0) = X
(1) = 0
Y
+ (λ − µ)Y = 0, Y (0) = Y (1) = 0
T
= −λνT
2. The solutions for X(x) form a cosine series.
µ
m
= m
2
π
2
, m ∈ Z
0+
, X
0
=
1
2
, X
m
= cos(mπx)
The solutions for Y (y) form a sine series.
λ
mn
= (m
2
+ n
2
)π
2
, n ∈ Z
+
, Y
n
= sin(nπx)
We solve the ordinary differential equation for T (t).
T
mn
=
e
−ν(m
2
+n
2
)π
2
t
We expand the solution of the heat equation in a series of the eigensolutions.
u(x, y, t) =
1
2
∞
n=1
u
0n
sin(nπy)
e
−νn
2
π
2
t
+
∞
m=1
∞
n=1
u
mn
cos(mπx) sin(nπy)
e
−ν(m
2
+n
2
)π
2
t
1745
We use the initial condition to determine the coefficients.
u(x, y, 0) = f(x, y) =
1
2
∞
n=1
u
0n
sin(nπy) +
∞
m=1
∞
n=1
u
mn
cos(mπx) sin(nπy)
u
mn
= 4
1
0
1
0
f(x, y) cos(mπx) sin(nπy) dx dy
Solution 37.5
We use the separation of variables u(x, t) = X(x)T (t) to find eigensolutions of the heat equation that satisfy the
boundary conditions at x = 0, π.
u
t
= νu
xx
XT
= νX
T
T
νT
=
X
X
= −λ
The problem for X(x) is
X
+ λX = 0, X
(0) = X
(π) = 0.
The eigenfunctions form the familiar cosine series.
λ
n
= n
2
, n ∈ Z
0+
, X
0
=
1
2
, X
n
= cos(nx)
Next we solve the differential equation for T (t).
T
n
= −νn
2
T
n
T
0
= 1, T
n
=
e
−νn
2
t
We expand the solution of the heat equation in a series of the eigensolutions.
u(x, t) =
1
2
u
0
+
∞
n=1
u
n
cos(nx)
e
−νn
2
t
1746
We use the initial condition to determine the coefficients in the series.
u(x, 0) = x =
1
2
u
0
+
∞
n=1
u
n
cos(nx)
u
0
=
2
π
π
0
x dx = π
u
n
=
2
π
π
0
x cos(nx) dx =
0 even n
−
4
πn
2
odd n
u(x, t) =
π
2
−
∞
n=1
odd n
4
πn
2
cos(nx)
e
−νn
2
t
Solution 37.6
We expand the solution in a Fourier series.
φ =
1
2
a
0
(r) +
∞
n=1
a
n
(r) cos(nθ) +
∞
n=1
b
n
(r) sin(nθ)
We substitute the series into the Laplace’s equation to determine ordinary differential equations for the coefficients.
∂
∂r
r
∂φ
∂r
+
1
r
2
∂
2
φ
∂θ
2
= 0
a
0
+
1
r
a
0
= 0, a
n
+
1
r
a
n
− n
2
a
n
= 0, b
n
+
1
r
b
n
− n
2
b
n
= 0
The solutions that are bounded at r = 0 are, (to within multiplicative constants),
a
0
(r) = 1, a
n
(r) = r
n
, b
n
(r) = r
n
.
Thus φ(r, θ) has the form
φ(r, θ) =
1
2
c
0
+
∞
n=1
c
n
r
n
cos(nθ) +
∞
n=1
d
n
r
n
sin(nθ)
1747
We apply the boundary condition at r = R.
φ(R, θ) =
1
2
c
0
+
∞
n=1
c
n
R
n
cos(nθ) +
∞
n=1
d
n
R
n
sin(nθ)
The coefficients are
c
0
=
1
π
2π
0
φ(R, α) dα, c
n
=
1
πR
n
2π
0
φ(R, α) cos(nα) dα, d
n
=
1
πR
n
2π
0
φ(R, α) sin(nα) dα.
We substitute the coefficients into our series solution.
φ(r, θ) =
1
2π
2π
0
φ(R, α) dα +
1
π
∞
n=1
r
R
n
2π
0
φ(R, α) cos(n(θ −α)) dα
φ(r, θ) =
1
2π
2π
0
φ(R, α) dα +
1
π
2π
0
φ(R, α)
∞
n=1
r
R
n
e
ın(θ−α)
dα
φ(r, θ) =
1
2π
2π
0
φ(R, α) dα +
1
π
2π
0
φ(R, α)
r
R
e
ı(θ−α)
1 −
r
R
e
ı(θ−α)
dα
φ(r, θ) =
1
2π
2π
0
φ(R, α) dα +
1
π
2π
0
φ(R, α)
r
R
e
ı(θ−α)
−
r
R
2
1 − 2
r
R
cos(θ −α) +
r
R
2
dα
φ(r, θ) =
1
2π
2π
0
φ(R, α) dα +
1
π
2π
0
φ(R, α)
Rr cos(θ − α) − r
2
R
2
+ r
2
− 2Rr cos(θ − α)
dα
φ(r, θ) =
1
2π
2π
0
φ(R, α)
R
2
− r
2
R
2
+ r
2
− 2Rr cos(θ − α)
dα
Solution 37.7
In order that the solution is continuously differentiable, (which it must be in order to satisfy the differential equation),
we impose the boundary conditions
u(0, t) = u(2π, t), u
θ
(0, t) = u
θ
(2π, t).
1748
We apply the separation of variables u(θ, t) = Θ(θ)T (t).
u
t
= κu
θθ
ΘT
= κΘ
T
T
κT
=
Θ
Θ
= −λ
We have the self-adjoint eigenvalue problem
Θ
+ λΘ = 0, Θ(0) = Θ(2π), Θ
(0) = Θ
(2π)
which has the eigenvalues and orthonormal eigenfunctions
λ
n
= n
2
, Θ
n
=
1
√
2π
e
ınθ
, n ∈ Z.
Now we solve the problems for T
n
(t) to obtain eigen-solutions of the heat equation.
T
n
= −n
2
κT
n
T
n
=
e
−n
2
κt
The solution is a linear combination of the eigen-solutions.
u(θ, t) =
∞
n=−∞
u
n
1
√
2π
e
ınθ
e
−n
2
κt
We use the initial conditions to determine the coefficients.
u(θ, 0) =
∞
n=−∞
u
n
1
√
2π
e
ınθ
= f(θ)
u
n
=
1
√
2π
2π
0
e
−ınθ
f(θ) dθ
1749
Solution 37.8
Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields
X
X
= −
Y
Y
= −λ.
With the homogeneous boundary conditions, we have the two problems
X
+ λX = 0, X(0) = X(1) = 0,
Y
− λY = 0, Y (1) = 0.
The eigenvalues and orthonormal eigenfunctions for X(x) are
λ
n
= (nπ)
2
, X
n
=
√
2 sin(nπx).
The general solution for Y is
Y
n
= a cosh(nπy) + b sinh(nπy).
The solution for that satisfies the right homogeneous boundary condition, (up to a multiplicative constant), is
Y
n
= sinh(nπ(1 − y))
u(x, y) is a linear combination of the eigen-solutions.
u(x, y) =
∞
n=1
u
n
√
2 sin(nπx) sinh(nπ(1 −y))
We use the inhomogeneous boundary condition to determine coefficients.
u(x, 0) =
∞
n=1
u
n
√
2 sin(nπx) sinh(nπ) = f(x)
u
n
=
√
2
1
0
sin(nπξ)f(ξ) dξ
1750
Solution 37.9
We substitute u(r, θ) = R(r)Θ(θ) into the partial differential equation.
∂
2
u
∂r
2
+
1
r
∂u
∂r
+
1
r
2
∂
2
u
∂θ
2
= 0
R
Θ +
1
r
R
Θ +
1
r
2
RΘ
= 0
r
2
R
R
+ r
R
R
= −
Θ
Θ
= λ
r
2
R
+ rR
− λR = 0, Θ
+ λΘ = 0
We assume that u is a strong solution of the partial differential equation and is thus twice continuously differentiable,
(u ∈ C
2
). In particular, this implies that R and Θ are bounded and that Θ is continuous and has a continuous first
derivative along θ = 0. This gives us a boundary value problem for Θ and a differential equation for R.
Θ
+ λΘ = 0, Θ(0) = Θ(2π), Θ
(0) = Θ
(2π)
r
2
R
+ rR
− λR = 0, R is bounded
The eigensolutions for Θ form the familiar Fourier series.
λ
n
= n
2
, n ∈ Z
0+
Θ
(1)
0
=
1
2
, Θ
(1)
n
= cos(nθ), n ∈ Z
+
Θ
(2)
n
= sin(nθ), n ∈ Z
+
Now we find the bounded solutions for R. The equation for R is an Euler equation so we use the substitution
R = r
α
.
r
2
R
n
+ rR
n
− λ
n
R
n
= 0
α(α − 1) + α −λ
n
= 0
α = ±
λ
n
1751
First we consider the case λ
0
= 0. The solution is
R = a + b ln r.
Boundedness demand s that b = 0. Thus we have the solution
R = 1.
Now we consider the case λ
n
= n
2
> 0. The solution is
R
n
= ar
n
+ br
−n
.
Boundedness demand s that b = 0. Thus we have the solution
R
n
= r
n
.
The solution for u is a linear combination of the eigensolutions.
u(r, θ) =
a
0
2
+
∞
n=1
(a
n
cos(nθ) + b
n
sin(nθ)) r
n
The boundary condition at r = 1 determines the coefficients in the expansion.
u(1, θ) =
a
0
2
+
∞
n=1
[a
n
cos(nθ) + b
n
sin(nθ)] = f(θ)
a
n
=
1
π
2π
0
f(θ) cos(nθ) dθ, b
n
=
1
π
2π
0
f(θ) sin(nθ) dθ
Solution 37.10
A normal mode of frequency ω is periodic in time.
v(r, θ, t) = u(r, θ)
e
ıωt
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We substitute this form into the wave equation to obtain a Helmholtz equation, (also called a reduced wave equation).
1
r
∂
∂r
r
∂u
∂r
+
1
r
2
∂
2
u
∂θ
2
= −
ω
2
c
2
u, u(1, θ) = 0,
∂
2
u
∂r
2
+
1
r
∂u
∂r
+
1
r
2
∂
2
u
∂θ
2
+ k
2
u = 0, u(1, θ) = 0
Here we have defined k =
ω
c
. We apply the separation of variables u = R(r)Θ(θ) to the Helmholtz equation.
r
2
R
Θ + rR
Θ + RΘ
+ k
2
r
2
RΘ = 0,
r
2
R
R
+ r
R
R
+ k
2
r
2
= −
Θ
Θ
= λ
2
Now we have an ordinary differential equation for R(r) and an ei genvalue problem for Θ(θ).
R
+
1
r
R
+
k
2
−
λ
2
r
2
R = 0, R(0) is bounded, R(1) = 0,
Θ
+ λ
2
Θ = 0, Θ(−π) = Θ(π), Θ
(−π) = Θ
(π).
We compute the eigenvalues and eigenfunctions for Θ.
λ
n
= n, n ∈ Z
0+
Θ
0
=
1
2
, Θ
(1)
n
= cos(nθ), Θ
(2)
n
= sin(nθ), n ∈ Z
+
The differential equations for the R
n
are Bessel equations.
R
n
+
1
r
R
n
+
k
2
−
n
2
r
2
R
n
= 0, R
n
(0) is bounded, R
n
(1) = 0
The general solution is a linear combination of order n Bessel functions of the first and second kind.
R
n
(r) = c
1
J
n
(kr) + c
2
Y
n
(kr)
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Since the Bessel function of the second kind, Y
n
(kr), is unbounded at r = 0, the solution has the form
R
n
(r) = cJ
n
(kr).
Applying the second boundary condition gives us the admissable frequencies.
J
n
(k) = 0
k
nm
= j
nm
, R
nm
= J
n
(j
nm
r), n ∈ Z
0+
, m ∈ Z
+
Here j
nm
is the m
th
positive root of J
n
. We combining the above results to obtain the normal modes of oscillation.
v
0m
=
1
2
J
0
(j
0m
r)
e
ıcj
0m
t
, m ∈ Z
+
v
nm
= cos(nθ + α)J
nm
(j
nm
r)
e
ıcj
nm
t
, n, m ∈ Z
+
Some normal modes are plotted in Figure 37.2. Note that cos(nθ + α) represents a linear combination of cos(nθ) and
sin(nθ). This form is preferrable as it illustrates the circular symmetry of the problem.
Solution 37.11
We will expand the solution in a complete, orthogonal set of functions {X
n
(x)}, where the coefficients are functions
of t.
φ =
n
T
n
(t)X
n
(x)
We will use separation of variables to determine a convenient set {X
n
}. We substitite φ = T (t)X(x) into the diffusion
equation.
φ
t
= a
2
φ
xx
XT
= a
2
X
T
T
a
2
T
=
X
X
= −λ
T
= −a
2
λT, X
+ λX = 0
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Note that in order to satisfy φ(0, t) = φ(l, t) = 0, the X
n
must satisfy the same homogeneous boundary conditions,
X
n
(0) = X
n
(l) = 0. This gives us a Sturm-Liouville problem for X(x).
X
+ λX = 0, X(0) = X(l) = 0
λ
n
=
nπ
l
2
, X
n
= sin
nπx
l
, n ∈ Z
+
Thus we seek a solution of the form
φ =
∞
n=1
T
n
(t) sin
nπx
l
. (37.10)
This solution automatically satisfies the boundary conditions. We will assume that we can diffe rentiate it. We will
substitite this form into the diffusion equation and the initial condition to determine the coefficients in the series , T
n
(t).
First we substitute Equation 37.10 into the partial differential equation for φ to determine ordinary differential equations
for the T
n
.
φ
t
= a
2
φ
xx
∞
n=1
T
n
(t) sin
nπx
l
= −a
2
∞
n=1
nπ
l
2
T
n
(t) sin
nπx
l
T
n
= −
anπ
l
2
T
n
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Now we substitute Equation 37.10 into the initial condition for φ to determine initial conditions for the T
n
.
∞
n=1
T
n
(0) sin
nπx
l
= φ(x, 0)
T
n
(0) =
l
0
sin
nπx
l
φ(x, 0) dx
l
0
sin
2
nπx
l
dx
T
n
(0) =
2
l
l
0
sin
nπx
l
φ(x, 0) dx
T
n
(0) =
2
l
l/2
0
sin
nπx
l
x dx +
2
l
l/2
0
sin
nπx
l
(l −x) dx
T
n
(0) =
4l
n
2
π
2
sin
nπ
2
T
2n−1
(0) = (−1)
n
4l
(2n − 1)
2
π
2
, T
2n
(0) = 0, n ∈ Z
+
We solve the ordinary differential equations for T
n
subject to the initial conditions.
T
2n−1
(t) = (−1)
n
4l
(2n − 1)
2
π
2
exp
−
a(2n − 1)π
l
2
t
, T
2n
(t) = 0, n ∈ Z
+
This determines the series representation of the solution.
φ =
4
l
∞
n=1
(−1)
n
l
(2n − 1)π
2
exp
−
a(2n − 1)π
l
2
t
sin
(2n − 1)πx
l
From the initial condition, we know that the the solution at t = 0 is C
0
. That is, it is continuous, but not
differentiable. The series representation of the solution at t = 0 is
φ =
4
l
∞
n=1
(−1)
n
l
(2n − 1)π
2
sin
(2n − 1)πx
l
.
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That the coefficients decay as 1/n
2
corroborates that φ(x, 0) is C
0
.
The derivatives of φ with respect to x are
∂
2m−1
∂x
2m−1
φ =
4(−1)
m+1
l
∞
n=1
(−1)
n
(2n − 1)π
l
2m−3
exp
−
a(2n − 1)π
l
2
t
cos
(2n − 1)πx
l
∂
2m
∂x
2m
φ =
4(−1)
m
l
∞
n=1
(−1)
n
(2n − 1)π
l
2m−2
exp
−
a(2n − 1)π
l
2
t
sin
(2n − 1)πx
l
For any fixed t > 0, the coefficients in the series for
∂
n
∂x
φ decay exp onen tially. These series are uniformly convergent in
x. Thus for any fixed t > 0, φ is C
∞
in x.
Solution 37.12
u
t
= κu
xx
, 0 < x < L, t > 0
u(0, t) = T
0
, u(L, t) = T
1
, u(x, 0) = f(x),
Method 1. We solve this problem with an eigenfunction expansion in x. To find an appropriate set of eigenfunctions,
we apply the separation of variables, u(x, t) = X(x)T (t) to the partial differential equation with the homogeneous
boundary conditions, u(0, t) = u(L, t) = 0.
(XT)
t
= (XT)
xx
XT
= X
T
T
T
=
X
X
= −λ
2
We have the eigenvalue problem,
X
+ λ
2
X = 0, X(0) = X(L) = 0,
which has the solutions,
λ
n
=
nπx
L
, X
n
= sin
nπx
L
, n ∈ N.
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