Articolo G.A. Partial Differential Equations and Boundary Value Problems with Maple V
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518 Chapter 8
The corresponding homogeneous boundary conditions on v(x, t) are
v(0,t)= 0 and v(1,t)= 0
The initial conditions on v(x, t) are
v(x, 0) = f(x) −s(x, 0)
which evaluates to
v(x, 0) = f(x)
and
v
t
(x, 0) = g(x) −s
t
(x, 0)
which evaluates to
v
t
(x, 0) = g(x) +x −1
To solve the v(x, t) equation, we temporarily set q(x, t) = 0, and we assume a solution of the
form
v(x, t) = X(x)T(t)
From the method of separation of variables, the corresponding spatial-dependent equation is
the Euler-type differential equation
d
2
dx
2
X(x)+λX(x)= 0
The corresponding homogeneous boundary conditions are type 1 at x = 0 and type 1 at x = 1:
X(0) = 0 and X(1) = 0
The allowed eigenvalues and corresponding orthonormal eigenfunctions, obtained from
Example 2.5.1, are
λ
n
= n
2
π
2
and
X
n
(x) =
√
2 sin(nπx)
for n = 1, 2, 3,....
The statement of orthonormality with the weight function w(x) = 1 reads
1
0
2 sin(nπx) sin(mπx)dx = δ(n, m)
Nonhomogeneous Partial Differential Equations 519
The time-dependent differential equation reads
d
2
dt
2
T
n
(t) +
λ
n
T
n
(t)
4
= Q
n
(t)
where Q
n
(t) is the inner product of q(x, t) with the spatial eigenfunction—that is,
Q
n
(t) =
1
0
(−x sin(t) +sin(t))
√
2 sin(nπx)dx
Evaluation of this integral yields
Q
n
(t) =
√
2 sin(t)
nπ
Thus, the nonhomogeneous time-dependent differential equation reads
d
2
dt
2
T
n
(t) +
n
2
π
2
T
n
(t)
4
=
√
2 sin(t)
nπ
A set of basis vectors for this differential equation, from Section 1.4, reads
T 1(t) = cos
nπ t
2
and T 2(t) = sin
nπ t
2
Evaluation of the corresponding second-order Green’s function yields
G2(t, τ) =
2 sin
nπ(t−τ)
2
nπ
Thus, the solution to the preceding time-dependent differential equation is
T
n
(t) = C(n) cos
nπ t
2
+D(n) sin
nπ t
2
+
t
0
2
√
2 sin
nπ(t−τ)
2
sin(τ)
n
2
π
2
dτ
for n = 1, 2, 3,....
Finally, the solution for the variable or particular portion of the problem is constructed from the
superposition of the products of the time-dependent solutions and the spatial eigenfunctions
evaluated earlier and this reads
v(x, t) =
∞
n=1
C(n) cos
nπt
2
+D(n) sin
nπt
2
+
4
√
2 sin(t)
nπ
n
2
π
2
−4
−
8
√
2 sin
nπt
2
n
2
π
2
n
2
π
2
−4
√
2 sin(nπx) (8.86)
The coefficients C(n) and D(n) are to be determined from the initial conditions on the problem.
520 Chapter 8
8.6 Initial Condition Considerations for the
Nonhomogeneous Wave Equation
At time t = 0, the two initial conditions on the solution are u(x, 0) = f(x) and u
t
(x, 0) = g(x).
This converts to the equivalent initial conditions on the variable portion of the solution v(x, t)
to read
v(x, 0) = f(x) −s(x, 0)
and
v
t
(x, 0) = g(x) −s
t
(x, 0)
If we set t = 0 in the eigenfunction expansion of the variable portion of the solution for v(x, t)
shown here
v(x, t) =
∞
n=0
⎛
⎝
C(n) cos
(
ω
n
t
)
+D(n) sin
(
ω
n
t
)
+
t
0
sin
(
ω
n
(t −τ)
)
Q
n
(τ)
ω
n
dτ
⎞
⎠
X
n
(x)
and its corresponding time derivative, then the initial conditions on v(x, t) lead us to the
following two series equations
f(x) −s(x, 0) =
∞
n=0
C(n)X
n
(x)
and
g(x) −s
t
(x, 0) =
∞
n=0
D(n)ω
n
X
n
(x)
These are the familiar Fourier series expansions of the left-hand functions in terms of the
orthonormalized eigenfunctions. To evaluate the Fourier coefficients C(n) and D(n), we take
the inner products of both sides with respect to the orthonormal eigenfunctions X
n
(x) and the
weight function w(x) = 1, and we get
a
0
(f(x) −s(x, 0))X
m
(x)dx =
∞
n=0
C(n)
⎛
⎝
a
0
X
n
(x)X
m
(x) dx
⎞
⎠
and
a
0
(
g(x) −s
t
(x, 0)
)
X
m
(x)dx =
∞
n=0
D(n)ω
n
⎛
⎝
a
0
X
n
(x)X
m
(x) dx
⎞
⎠
Nonhomogeneous Partial Differential Equations 521
Taking advantage of the statement of orthonormality, these two series reduce to
a
0
(
f(x) −s(x, 0)
)
X
m
(x)dx =
∞
n=0
C(n)δ(m, n)
and
a
0
(
g(x) −s
t
(x, 0)
)
X
m
(x)dx =
∞
n=0
D(n)ω
n
δ(m, n)
Evaluating these sums and taking advantage of the mathematical character of the Kronecker
delta functions (only the m = n terms survive), the Fourier coefficients C(n) and D(n) are
evaluated from the integrals
C(n) =
a
0
(f(x) −s(x, 0))X
n
(x) dx
and
D(n) =
a
0
(
g(x) −s
t
(x, 0)
)
X
n
(x)
ω
n
dx
Now that we have evaluated the coefficients C(n) and D(n), we can write the final solution to
our original nonhomogeneous partial differential equation as the sum of the variable plus the
linear solutions
u(x, t) =
∞
n=0
⎛
⎝
C(n) cos
(
ω
n
t
)
+D(n) sin
(
ω
n
t
)
+
t
0
sin
(
ω
n
(
t −τ
))
Q
n
(τ)
ω
n
dτ
⎞
⎠
X
n
(x) +s(x, t)
(8.87)
where C(n), D(n), Q
n
(τ), and s(x, t) were all evaluated earlier. We demonstrate the preceding
concepts for the illustrative problem from Section 8.5.
DEMONSTRATION: We consider the example problem done in Section 8.5 for the special
case where the initial conditions are
u(x, 0) = f(x)
where
f(x) = 0
and
u
t
(x, 0) = g(x)
522 Chapter 8
where
g(x) = 0
SOLUTION: We previously evaluated the linear portion of the solution s(x, t) =
−x sin(t) +sin(t) and the natural angular frequency ω
n
= nπ/2. The given initial conditions on
u(x, t) translate to the corresponding initial conditions on the variable portion of the solution
shown here:
v(x, 0) = f(x) −s(x, 0)
v
t
(x, 0) = g(x) −s
t
(x, 0)
Insertion of these values into the integral for C(n)
C(n) =
a
0
(
f(x) −s(x, 0)
)
X
n
(x) dx
yields
C(n) =
1
0
0
√
2 sin(nπx) dx
which evaluates to
C(n) = 0
Similarly, the integral for D(n)
D(n) =
a
0
(
g(x) −s
t
(x, 0)
)
X
n
(x)
ω
n
dx
yields
D(n) = 2
⎛
⎝
1
0
(x −1)
√
2 sin(nπx)
nπ
dx
⎞
⎠
which evaluates to
D(n) =
−2
√
2
n
2
π
2
(8.88)
for n = 1, 2, 3,....
Nonhomogeneous Partial Differential Equations 523
With these values for the Fourier coefficients, the sum of the particular portion plus the linear
portion yields the final solution
u(x, t) =−4
∞
n=1
nπ sin
nπt
2
−2 sin(t)
sin(nπx)
nπ
n
2
π
2
−4
+(1 −x) sin(t)
Note how the solution satisfies the given boundary and initial conditions. The details for the
development of this solution along with the graphics are given later in one of the Maple
worksheet examples.
8.7 Example Nonhomogeneous Problems for the
Wave Equation
EXAMPLE 8.7.1: (The whip-rope problem) We seek the amplitude u(x, t) for transverse wave
motion along a taut rope over the interval I ={x |0 <x<1}. The left end of the rope is
attached to a vertical oscillatory motor, which provides a sinusoidal motion to the end of the
rope, and the right end is held fixed. There is no damping in the system. There is no external
force acting on the rope, and the rope has an initial displacement distribution u(x, 0) = f(x)
and an initial speed distribution u
t
(x, 0) = g(x) given as follows. The wave speed is c = 1/2.
SOLUTION: The nonhomogeneous wave equation is
∂
2
∂t
2
u(x, t) = c
2
∂
2
∂x
2
u(x, t)
+h(x, t)
The boundary conditions are nonhomogeneous type 1 at the left and homogeneous type 1 at the
right:
u(0,t)= sin(t) and u(1,t)= 0
The initial displacement distribution is
u(x, 0) = 0
and the initial speed distribution is
u
t
(x, 0) = 0
The external applied force is
h(x, t) = 0
The solution is u(x, t) = v(x, t) +s(x, t), where s(x, t) is the linear portion of the solution and
v(x, t) is the variable portion of the solution that satisfies the partial differential equation
∂
2
∂t
2
v(x, t) = c
2
∂
2
∂x
2
v(x, t)
+q(x, t)
524 Chapter 8
where
q(x, t) = h(x, t) −
∂
2
∂t
2
s(x, t)
Assignment of system parameters
> restart: with(plots):a:=1:c:=1/2:h(x,t):=0:
> A(t):=sin(t):B(t):=0:kappa[1]:=1:kappa[2]:=0:kappa[3]:=1:kappa[4]:=0:
> b(t):=(kappa[3]*a*A(t)+A(t)*kappa[4]−B(t)*kappa[2])/(kappa[1]*kappa[3]*a+
kappa[1]*kappa[4]−kappa[2]*kappa[3]);
b(t) := sin(t) (8.89)
> m(t):=(kappa[1]*B(t)−A(t)*kappa[3])/(kappa[1]*kappa[3]*a+kappa[1]*kappa[4]−
kappa[2]*kappa[3]);
m(t) := −sin(t) (8.90)
Linear portion of solution
> s(x,t):=m(t)*x+b(t);s(x,0):=eval(subs(t=0,s(x,t))):s[t](x,0):=eval(subs(t=0,diff(s(x,t),t))):
s(x, t) := −sin(t)x +sin(t) (8.91)
By the method of separation of variables, the variable solution is
v(x, t) =
∞
n=0
X
n
(x)T
n
(t)
where T
n
(t) is the solution to the time-dependent differential equation
d
2
dt
2
T
n
(t) +c
2
λ
n
T
n
(t) = Q
n
(t)
and X
n
(x) is the solution to the spatial-dependent eigenvalue equation
d
2
dx
2
X
n
(x) +λ
n
X
n
(x) = 0
with boundary conditions
X(0) = 0 and X(1) = 0
The corresponding homogeneous eigenfunction problem consists of the Euler equation, along
with the homogeneous boundary conditions that are type 1 at the left and type 1 at the right.
Nonhomogeneous Partial Differential Equations 525
The allowed eigenvalues and corresponding orthonormal eigenfunctions, obtained from
Example 2.5.1, are
> lambda[n]:=(n*Pi/a)ˆ2;
λ
n
:= n
2
π
2
(8.92)
> X[n](x):=sqrt(2/a)*sin(n*Pi/a*x);X[m](x):=subs(n=m,X[n](x)):
X
n
(x) :=
√
2 sin(nπx) (8.93)
for n = 1, 2, 3,....
Statement of orthonormality with the respective weight function w(x) = 1
> w(x):=1:Int(X[n](x)*X[m](x)*w(x),x=0..a)=delta(n,m);
1
0
2 sin(nπx) sin(mπx)dx = δ(n, m) (8.94)
Time-dependent equation
> diff(T[n](t),t,t)+cˆ2*lambda[n]*T[n](t)=Q[n](t);
d
2
dt
2
T
n
(t) +
1
4
n
2
π
2
T
n
(t) = Q
n
(t) (8.95)
> q(x,t):=h(x,t)−diff(s(x,t),t,t);
q(x, t) := −sin(t)x +sin(t) (8.96)
> Q[n](t):=Int(q(x,t)*X[n](x),x=0..a);Q[n](t):=expand(value(%)):
Q
n
(t) :=
1
0
(
−sin(t)x +sin(t)
)
√
2 sin(nπx) dx (8.97)
> Q[n](t):=simplify(factor(subs({sin(n*Pi)=0,cos(n*Pi)=(−1)ˆn},Q[n](t))));
Q[n](tau):=subs(t=tau,%):
Q
n
(t) :=
√
2 sin(t)
nπ
(8.98)
Basis vectors
> T1(t):=cos(c*n*Pi*t/a);
T 1(t) := cos
1
2
nπt
(8.99)
526 Chapter 8
> T2(t):=sin(c*n*Pi*t/a);
T 2(t) := sin
1
2
nπt
(8.100)
Second-order Green’s function
> G2(t,tau):=sin(c*n*Pi/a*(t−tau))/(c*n*Pi/a);
G2(t, τ) :=
2 sin
1
2
nπ
(
t −τ
)
nπ
(8.101)
Time-dependent solution
> T[n](t):=C(n)*T1(t)+D(n)*T2(t)+Int(G2(t,tau)*Q[n](tau),tau=0..t);
T
n
(t) := C(n) cos
1
2
nπt
+D(n) sin
1
2
nπt
+
t
0
2 sin
1
2
nπ
(
t −τ
)
√
2 sin(τ)
n
2
π
2
dτ (8.102)
> T[n](t):=value(%);v[n](x,t):=(T[n](t)*X[n](x)):
T
n
(t) := C(n) cos
1
2
nπt
+D(n) sin
1
2
nπt
+
4
√
2
−2 sin
1
2
nπt
+sin(t)nπ
n
2
π
2
n
2
π
2
−4
(8.103)
Variable portion of solution
> v(x,t):=Sum(v[n](x,t),n=1..infinity);
v(x, t) :=
∞
n=1
C(n) cos
1
2
nπt
+D(n) sin
1
2
nπt
+
4
√
2
−2 sin
1
2
nπt
+sin(t)nπ
n
2
π
2
n
2
π
2
−4
⎞
⎠
√
2 sin(nπx) (8.104)
The Fourier coefficients C(n) and D(n) are to be determined from the initial conditions on
v(x, t). We now substitute the initial conditions v(x, 0) = f(x) −s(x, 0) and
v
t
(x, 0) = g(x) −s
t
(x, 0) into the integrals for C(n) and D(n) for the special case where
> f(x):=0;
f(x) := 0 (8.105)
> g(x):=0;
g(x) := 0 (8.106)
Nonhomogeneous Partial Differential Equations 527
The integrals for C(n) and D(n) are
> C(n):=Int((f(x)−s(x,0))*X[n](x),x=0..a);C(n):=expand(value(%)):
C(n) :=
1
0
0dx (8.107)
> C(n):=simplify(subs({sin(n*Pi)=0,cos(n*Pi)=(−1)ˆn},C(n)));
C(n) := 0 (8.108)
> D(n):=Int((g(x)−s[t](x,0))/(c*n*Pi/a)*X[n](x),x=0..a);D(n):=expand(value(%)):
D(n) :=
1
0
2(−1 +x)
√
2 sin(nπx)
nπ
dx (8.109)
> D(n):=simplify(subs({sin(n*Pi)=0,cos(n*Pi)=(−1)ˆn},D(n)));
D(n) := −
2
√
2
n
2
π
2
(8.110)
for n = 1, 2, 3,....
Final solution (linear plus variable portion)
> u(x,t):=simplify(Sum(eval(v[n](x,t)),n=1..infinity))+s(x,t);
u(x, t) :=
∞
n=1
4
−sin
1
2
nπt
nπ +2 sin(t)
sin(nπx)
nπ
n
2
π
2
−4
−sin(t)x +sin(t) (8.111)
First few terms of sum
> u(x,t):=simplify(sum(eval(v[n](x,t)),n=1..3)+s(x,t)):
ANIMATION
> animate(u(x,t),x=0..a,t=0..5,thickness=3);
The preceding animation command displays the spatial-time-dependent solution of u(x, t) for
the given boundary conditions and initial conditions. The animation sequence here and in
Figure 8.6 shows snapshots of the animation at times t = 0, 1, 2, 3, 4, 5. Note how the solution
satisfies the given boundary and initial conditions.
ANIMATION SEQUENCE
> u(x,0):=subs(t=0,u(x,t)):u(x,1):=subs(t=1,u(x,t)):
> u(x,2):=subs(t=2,u(x,t)):u(x,3):=subs(t=3,u(x,t)):