Articolo G.A. Partial Differential Equations and Boundary Value Problems with Maple V
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528 Chapter 8
> u(x,4):=subs(t=4,u(x,t)):u(x,5):=subs(t=5,u(x,t)):
> plot({u(x,0),u(x,1),u(x,2),u(x,3),u(x,4),u(x,5)},x=0..a,thickness=10);
x
1
0
0.5
0.2 0.4 0.6 0.8 1
20.5
21
21.5
Figure 8.6
EXAMPLE 8.7.2: (Applied force on a magnetic wire) We seek the wave amplitude u(x, t) for
transverse wave motion along a taut magnetic wire over the interval I ={x |0 <x<1}. Both
ends of the wire are held fixed. A nonuniform external applied magnetic force term h(x, t) acts
on the wire. There is no damping in the system. The wire has an initial displacement
distribution u(x, 0) = f(x) and initial speed distribution u
t
(x, 0) = g(x) given as follows. The
wave speed is c = 1/5.
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 homogeneous type 1 at the left and homogeneous type 1 at the
right:
u(0,t)= 0 and u(1,t)= 0
The initial displacement distribution is
u(x, 0) = x(1 −x)
and the initial speed distribution is
u
t
(x, 0) = x
The external applied force is
h(x, t) = x sin(t)
Nonhomogeneous Partial Differential Equations 529
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)
where
q(x, t) = h(x, t) −
∂
2
∂t
2
s(x, t)
Assignment of system parameters
> restart: with(plots):a:=1:c:=1/5:h(x,t):=x*sin(t):
> A(t):=0: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) := 0 (8.112)
> 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) := 0 (8.113)
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) := 0 (8.114)
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
530 Chapter 8
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.
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
(8.115)
> 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) (8.116)
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) (8.117)
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
25
n
2
π
2
T
n
(t) = Q
n
(t) (8.118)
> q(x,t):=h(x,t)−diff(s(x,t),t,t);
q(x, t) := x sin(t) (8.119)
> Q[n](t):=Int(q(x,t)*X[n](x),x=0..a);Q[n](t):=expand(value(%)):
Q
n
(t) :=
1
0
x sin(t)
√
2 sin(nπx) dx (8.120)
> 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) := −
sin(t)
√
2(−1)
n
nπ
(8.121)
Basis vectors
> T1(t):=cos(c*n*Pi*t/a);
T 1(t) := cos
1
5
nπt
(8.122)
Nonhomogeneous Partial Differential Equations 531
> T2(t):=sin(c*n*Pi*t/a);
T 2(t) := sin
1
5
nπt
(8.123)
Second-order Green’s function
> G2(t,tau):=sin(c*n*Pi/a*(t−tau))/(c*n*Pi/a);
G2(t, τ) :=
5 sin
1
5
nπ
(
t −τ
)
nπ
(8.124)
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
5
nπt
+D(n) sin
1
5
nπt
+
t
0
⎛
⎝
−
5 sin
1
5
nπ
(
t −τ
)
sin(τ)
√
2 (−1)
n
n
2
π
2
⎞
⎠
dτ
(8.125)
> T[n](t):=value(%);
T
n
(t) := C(n) cos
1
5
nπt
+D(n) sin
1
5
nπt
+
25(−1)
1+n
√
2
−5 sin
1
5
nπt
+sin(t)nπ
n
2
π
2
n
2
π
2
−25
(8.126)
Generalized series terms
> v[n](x,t):=(T[n](t)*X[n](x)):
Variable portion of solution
> v(x,t):=Sum(v[n](x,t),n=1..infinity);
v(x, t) :=
∞
n=1
C(n) cos
1
5
nπt
+D(n) sin
1
5
nπt
(8.127)
+
25(−1)
1+n
√
2
−5 sin
1
5
nπt
+sin(t)nπ
n
2
π
2
n
2
π
2
−25
⎞
⎠
√
2 sin(nπx)
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
532 Chapter 8
> f(x):=x*(1−x);
f(x) := x(1 −x) (8.128)
> g(x):=x;
g(x) := x (8.129)
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
x(1 −x)
√
2 sin(nπx) dx (8.130)
> C(n):=simplify(subs({sin(n*Pi)=0,cos(n*Pi)=(−1)ˆn},C(n)));
C(n) := −
2
√
2
−1 +(−1)
n
n
3
π
3
(8.131)
> 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
5x
√
2 sin(nπx)
nπ
dx (8.132)
> D(n):=simplify(subs({sin(n*Pi)=0,cos(n*Pi)=(−1)ˆn},D(n)));
D(n) :=
5(−1)
1+n
√
2
n
2
π
2
(8.133)
for n = 1, 2, 3,....
Final solution (linear plus variable portion)
> u(x,t):=Sum(eval(v[n](x,t)),n=1..infinity)+s(x,t);
u(x, t) :=
∞
n=1
⎛
⎝
−
2
√
2
−1 +(−1)
n
cos
1
5
nπt
n
3
π
3
+
5(−1)
1+n
√
2 sin
1
5
nπt
n
2
π
2
(8.134)
+
25(−1)
1+n
√
2
−5 sin
1
5
nπt
+sin(t)nπ
n
2
π
2
n
2
π
2
−25
⎞
⎠
√
2 sin(nπx)
First few terms of sum
> u(x,t):=sum(eval(v[n](x,t)),n=1..3)+s(x,t):
Nonhomogeneous Partial Differential Equations 533
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 8.7 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)):
> u(x,4):=subs(t=4,u(x,t)):u(x,5):=subs(t=5,u(x,t)):
> plot({u(x,0),u(x,1),u(x,2),u(x,3),u(x,4),u(x,5)},x=0..a,thickness=10);
0
0.5
1
1.5
2
0.2 0.4 0.6 1
x
0.8
Figure 8.7
EXAMPLE 8.7.3: (Longitudinal wave motion) We seek the wave amplitude u(x, t) for
longitudinal wave motion in a rigid bar over the interval I ={x |0 <x<1}. The left end of the
bar is held fixed, and the right end experiences an oscillatory compression (Young’s modulus
E = 1). There is no external applied force acting on the system. The bar has an initial
displacement distribution f(x) and initial speed distribution g(x) given as follows. The wave
speed is c = 1/5.
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 homogeneous type 1 at the left and nonhomogeneous type 2 at the
right:
u(0,t)= 0 and u
x
(1,t)= sin(t)
534 Chapter 8
The initial displacement distribution is
u(x, 0) = x
1 −
x
2
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)
where
q(x, t) = h(x, t) −
∂
2
∂t
2
s(x, t)
Assignment of system parameters
> restart: with(plots):a:=1:c:=1/5:h(x,t):=0:
> A(t):=0:B(t):=sin(t):kappa[1]:=1:kappa[2]:=0:kappa[3]:=0:kappa[4]:=1:
> 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) := 0 (8.135)
> 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) (8.136)
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 (8.137)
By the method of separation of variables, the variable solution is
v(x, t) =
∞
n=0
X
n
(x)T
n
(t)
Nonhomogeneous Partial Differential Equations 535
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
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 2 at the right.
The allowed eigenvalues and corresponding orthonormal eigenfunctions, obtained from
Example 2.5.2, are
> lambda[n]:=((2*n−1)*Pi/(2*a))ˆ2;
λ
n
:=
1
4
(
2 n −1
)
2
π
2
(8.138)
> X[n](x):=sqrt(2/a)*sin((2*n−1)*Pi/(2*a)*x);X[m](x):=subs(n=m,X[n](x)):
X
n
(x) :=
√
2 sin
1
2
(
2n −1
)
πx
(8.139)
forn=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
1
2
(2 n −1)πx
sin
1
2
(2 m −1)πx
dx = δ(n, m) (8.140)
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
100
(2 n −1)
2
π
2
T
n
(t) = Q
n
(t) (8.141)
> q(x,t):=h(x,t)−diff(s(x,t),t,t);
q(x, t) := sin(t)x (8.142)
536 Chapter 8
> 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
√
2 sin
1
2
(
2n −1
)
πx
dx (8.143)
> 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) :=
4(−1)
1+n
sin(t)
√
2
π
2
(2n −1)
2
(8.144)
Basis vectors
> T1(t):=cos(c*(2*n−1)*Pi*t/(2*a));
T 1(t) := cos
1
10
(2n −1)πt
(8.145)
> T2(t):=sin(c*(2*n−1)*Pi*t/(2*a));
T 2(t) := sin
1
10
(2n −1)πt
(8.146)
Second-order Green’s function
> G2(t,tau):=sin(c*(2*n−1)*Pi/(2*a)*(t−tau))/(c*(2*n−1)*Pi/(2*a));
G2(t, τ) :=
10 sin
1
10
(2n −1)π(t −τ)
(2n −1)π
(8.147)
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):=value(%):
v[n](x,t):=(T[n](t)*X[n](x)):
T
n
(t) := C(n) cos
1
10
(2n −1)πt
+D(n) sin
1
10
(2n −1)πt
(8.148)
+
t
0
40 sin
1
10
(2n −1)π(t −τ)
(−1)
1+n
sin(τ)
√
2
(2n −1)
3
π
3
dτ
Variable portion of solution
Nonhomogeneous Partial Differential Equations 537
> v(x,t):=Sum(v[n](x,t),n=1..infinity);
v(x, t) :=
∞
n=1
C(n) cos
1
10
(2n −1)πt
+D(n) sin
1
10
(2n −1)πt
+
400(−1)
1+n
2 sin(t)πn −sin(t) −10 sin
1
5
πnt
cos
1
10
πt
√
2
16n
4
π
2
−32n
3
π
2
+24π
2
n
2
−8π
2
n +π
2
−400n
2
+400 n −100
(2n −1)π
3
√
2 sin
1
2
(2n −1)πx
(8.149)
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):=x*(1−x/2);
f(x) := x
1 −
1
2
x
(8.150)
> g(x):=0;
g(x) := 0 (8.151)
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
x
1 −
1
2
x
√
2 sin
1
2
(2n −1)πx
dx (8.152)
> C(n):=simplify(subs({sin(n*Pi)=0,cos(n*Pi)=(−1)ˆn},C(n)));
C(n) :=
8
√
2
π
3
8n
3
−12n
2
+6n −1
(8.153)
> D(n):=Int((g(x)−s[t](x,0))/(c*(2*n−1)*Pi/(2*a))*X[n](x),x=0..a);D(n):=expand(value(%)):
D(n) :=
1
0
⎛
⎝
−
10x
√
2 sin
1
2
(2n −1)πx
(2n −1)π
⎞
⎠
dx (8.154)