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Boundary-Value Problems
in Rectangular Coordinates
EXERCISES 13.
Separable Partial Differential Equations
- Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields X′Y = XY ′. Separating variables and using the separation constant −λ, where λ = 0, we obtain X′ X
= Y^
′ Y = −λ.
When λ = 0 X′^ + λX = 0 and Y ′^ + λY = 0 so that X = c 1 e−λx^ and Y = c 2 e−λy^. A particular product solution of the partial differential equation is u = XY = c 3 e−λ(x+y), λ = 0. When λ = 0 the differential equations become X′^ = 0 and Y ′^ = 0, so in this case X = c 4 , Y = c 5 , and u = XY = c 6.
- Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields X′Y + 3XY ′^ = 0. Separating variables and using the separation constant −λ we obtain X′ − 3 X
Y ′
Y
= −λ.
When λ = 0 X′^ − 3 λX = 0 and Y ′^ + λY = 0 so that X = c 1 e^3 λx^ and Y = c 2 e−λy^. A particular product solution of the partial differential equation is u = XY = c 3 eλ(3x−y). When λ = 0 the differential equations become X′^ = 0 and Y ′^ = 0, so in this case X = c 4 , Y = c 5 , and u = XY = c 6.
- Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields X′Y + XY ′^ = XY. Separating variables and using the separation constant −λ we obtain X′ X
Y − Y ′
Y
= −λ.
13.1 Separable Partial Differential Equations
Then X′^ + λX = 0 and Y ′^ − (1 + λ)Y = 0 so that X = c 1 e−λx^ and Y = c 2 e(1+λ)y^. A particular product solution of the partial differential equation is u = XY = c 3 ey+λ(y−x).
- Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields X′Y = XY ′^ + XY. Separating variables and using the separation constant −λ we obtain X′ X
= Y^ +^ Y^
′ Y = −λ.
Then X′^ + λX = 0 and y′^ + (1 + λ)Y = 0 so that X = c 1 e−λx^ and Y = c 2 e−(1+λ)y^ = 0. A particular product solution of the partial differential equation is u = XY = c 3 e−y−λ(x+y).
- Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields xX′Y = yXY ′. Separating vari- ables and using the separation constant −λ we obtain xX′ X
yY ′ Y = −λ. When λ = 0 xX′^ + λX = 0 and yY ′^ + λY = 0 so that X = c 1 x−λ^ and Y = c 2 y−λ. A particular product solution of the partial differential equation is u = XY = c 3 (xy)−λ. When λ = 0 the differential equations become X′^ = 0 and Y ′^ = 0, so in this case X = c 4 , Y = c 5 , and u = XY = c 6.
- Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields yX′Y + xXY ′^ = 0. Separating variables and using the separation constant −λ we obtain X′ xX
Y ′
yY = −λ.
When λ = 0 X′^ + λxX = 0 and Y ′^ − λyY = 0 so that X = c 1 eλx (^2) / 2 and Y = c 2 e−λy (^2) / 2 . A particular product solution of the partial differential equation is u = XY = c 3 eλ(x (^2) −y (^2) )/ 2 .
13.1 Separable Partial Differential Equations
I. If λ = 0 then X′′^ = 0 and X(x) = c 2 x + c 3 , so
u = XT = A 1 x + A 2. II. If λ = −α^2 < 0, then X′′^ − α^2 X = 0, and X(x) = c 4 cosh αx + c 5 sinh αx, so u = XT = (A 3 cosh αx + A 4 sinh αx)ekα (^2) t .
III. If λ = α^2 > 0, then X′′^ + α^2 X = 0, and X(x) = c 6 cos αx + c 7 sin αx, so u = XT = (A 5 cos αx + A 6 sin αx)e−kα (^2) t .
- Substituting u(x, t) = X(x)T (t) into the partial differential equation yields a^2 X′′T = XT ′′. Separating variables and using the separation constant −λ we obtain X′′ X
= T^
′′ a^2 T = −λ.
Then X′′^ + λX = 0 and T ′′^ + a^2 λT = 0. We consider three cases: I. If λ = 0 then X′′^ = 0 and X(x) = c 1 x + c 2. Also, T ′′^ = 0 and T (t) = c 3 t + c 4 , so
u = XT = (c 1 x + c 2 )(c 3 t + c 4 ). II. If λ = −α^2 < 0, then X′′^ − α^2 X = 0, and X(x) = c 5 cosh αx + c 6 sinh αx. Also, T ′′^ − α^2 a^2 T = 0 and T (t) = c 7 cosh αat + c 8 sinh αat, so u = XT = (c 5 cosh αx + c 6 sinh αx)(c 7 cosh αat + c 8 sinh αat).
III. If λ = α^2 > 0, then X′′^ + α^2 X = 0, and X(x) = c 9 cos αx + c 10 sin αx. Also, T ′′^ + α^2 a^2 T = 0 and T (t) = c 11 cos αat + c 12 sin αat, so u = XT = (c 9 cos αx + c 10 sin αx)(c 11 cos αat + c 12 sin αat).
- Substituting u(x, t) = X(x)T (t) into the partial differential equation yields a^2 X′′T = XT ′′^ +2kXT ′. Separating variables and using the separation constant −λ we obtain X′′ X
= T^
′′ (^) + 2kT ′ a^2 T = −λ.
Then X′′^ + λX = 0 and T ′′^ + 2kT ′^ + a^2 λT = 0. We consider three cases: I. If λ = 0 then X′′^ = 0 and X(x) = c 1 x + c 2. Also, T ′′^ + 2kT ′^ = 0 and T (t) = c 3 + c 4 e−^2 kt, so
u = XT = (c 1 x + c 2 )(c 3 + c 4 e−^2 kt). II. If λ = −α^2 < 0, then X′′^ − α^2 X = 0, and X(x) = c 5 cosh αx + c 6 sinh αx. The auxiliary equation of T ′′^ + 2kT ′^ − α^2 a^2 T = 0 is m^2 + 2km − α^2 a^2 = 0. Solving for m we obtain m = −k ±
k^2 + α^2 a^2 , so T (t) = c 7 e(−k+
√k (^2) +α (^2) a (^2) )t
√k (^2) +α (^2) a (^2) )t
. Then
u = XT = (c 5 cosh αx + c 6 sinh αx)
c 7 e(−k+
√k (^2) +α (^2) a (^2) )t
√k (^2) +α (^2) a (^2) )t) .
III. If λ = α^2 > 0, then X′′^ + α^2 X = 0, and X(x) = c 9 cos αx + c 10 sin αx. The auxiliary equation
13.1 Separable Partial Differential Equations
of T ′′^ + 2kT ′^ + α^2 a^2 T = 0 is m^2 + 2km + α^2 a^2 = 0. Solving for m we obtain m = −k ±
k^2 − α^2 a^2. We consider three possibilities for the discriminant k^2 − α^2 a^2 : (i) If k^2 − α^2 a^2 = 0 then T (t) = c 11 e−kt^ + c 12 te−kt^ and
u = XT = (c 9 cos αx + c 10 sin αx)(c 11 e−kt^ + c 12 te−kt). From k^2 − α^2 a^2 = 0 we have α = k/a so the solution can be written
u = XT = (c 9 cos kx/a + c 10 sin kx/a)(c 11 e−kt^ + c 12 te−kt).
(ii) If k^2 − α^2 a^2 < 0 then T (t) = e−kt^
c 13 cos
α^2 a^2 − k^2 t + c 14 sin
α^2 a^2 − k^2 t
and
u = XT = (c 9 cos αx + c 10 sin αx)e−kt^
c 13 cos
α^2 a^2 − k^2 t + c 14 sin
α^2 a^2 − k^2 t
(iii) If k^2 − α^2 a^2 > 0 then T (t) = c 15 e(−k+
√k (^2) −α (^2) a (^2) )t
√k (^2) −α (^2) a (^2) )t and
u = XT = (c 9 cos αx + c 10 sin αx)
c 15 e(−k+
√k (^2) −α (^2) a (^2) )t
√k (^2) −α (^2) a (^2) )t) .
- Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields X′′Y + XY ′′^ = 0. Separating variables and using the separation constant −λ we obtain
− X
′′ X
= Y^
′′ Y = −λ.
Then X′′^ − λX = 0 and Y ′′^ + λY = 0. We consider three cases: I. If λ = 0 then X′′^ = 0 and X(x) = c 1 x + c 2. Also, Y ′′^ = 0 and Y (y) = c 3 y + c 4 so
u = XY = (c 1 x + c 2 )(c 3 x + c 4 ). II. If λ = −α^2 < 0 then X′′^ + α^2 X = 0 and X(x) = c 5 cos αx + c 6 sin αx. Also, Y ′′^ − α^2 Y = 0 and Y (y) = c 7 cosh αx + c 8 sinh αx so u = XY = (c 5 cos αx + c 6 sin αx)(c 7 cosh αx + c 8 sinh αx).
III. If λ = α^2 > 0 then X′′^ − α^2 X = 0 and X(x) = c 9 cosh αx + c 10 sinh αx. Also, Y ′′^ + α^2 Y = 0 and Y (y) = c 11 cos αy + c 12 sin αy so
u = XY = (c 9 cosh αx + c 10 sinh αx)(c 11 cos αy + c 12 sin αy).
- Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields x^2 X′′Y + XY ′′^ = 0. Separating variables and using the separation constant −λ we obtain
− x
2 X′′
X
= Y^
′′ Y = −λ.
Then x^2 X′′^ − λX = 0 and Y ′′^ + λY = 0. We consider three cases: I. If λ = 0 then x^2 X′′^ = 0 and X(x) = c 1 x + c 2. Also, Y ′′^ = 0 and Y (y) = c 3 y + c 4 so
u = XY = (c 1 x + c 2 )(c 3 y + c 4 ).
13.1 Separable Partial Differential Equations
Y ′′^ − (1 + α^2 )Y = 0 and Y (y) = c 15 cosh
1 + α^2 y + c 16 sinh
1 + α^2 y so
u = XY = (c 13 cos αx + c 14 sin αx)
c 15 cosh
1 + α^2 y + c 16 sinh
1 + α^2 y
- Substituting u(x, t) = X(x)T (t) into the partial differential equation yields a^2 X′′T − g = XT ′′, which is not separable.
- Identifying A = B = C = 1, we compute B^2 − 4 AC = − 3 < 0. The equation is elliptic.
- Identifying A = 3, B = 5, and C = 1, we compute B^2 − 4 AC = 13 > 0. The equation is hyperbolic.
- Identifying A = 1, B = 6, and C = 9, we compute B^2 − 4 AC = 0. The equation is parabolic.
- Identifying A = 1, B = −1, and C = −3, we compute B^2 − 4 AC = 13 > 0. The equation is hyperbolic.
- Identifying A = 1, B = −9, and C = 0, we compute B^2 − 4 AC = 81 > 0. The equation is hyperbolic.
- Identifying A = 0, B = 1, and C = 0, we compute B^2 − 4 AC = 1 > 0. The equation is hyperbolic.
- Identifying A = 1, B = 2, and C = 1, we compute B^2 − 4 AC = 0. The equation is parabolic.
- Identifying A = 1, B = 0, and C = 1, we compute B^2 − 4 AC = − 4 < 0. The equation is elliptic.
- Identifying A = a^2 , B = 0, and C = −1, we compute B^2 − 4 AC = 4a^2 > 0. The equation is hyperbolic.
- Identifying A = k > 0, B = 0, and C = 0, we compute B^2 − 4 AC = − 4 k < 0. The equation is elliptic.
- Substituting u(r, t) = R(r)T (t) into the partial differential equation yields
k
R′′T +
r
R′T
= RT ′.
Separating variables and using the separation constant −λ we obtain rR′′^ + R′ rR =^
T ′
kT =^ −λ. Then rR′′^ + R′^ + λrR = 0 and T ′^ + λkT = 0. Letting λ = α^2 and writing the first equation as r^2 R′′^ + rR′^ = α^2 r^2 R = 0 we see that it is a parametric Bessel equation of order 0. As discussed in Chapter 5 of the text, it has solution R(r) = c 1 J 0 (αr) + c 2 Y 0 (αr). Since a solution of T ′^ + α^2 kT is T (t) = e−kα (^2) t , we see that a solution of the partial differential equation is u = RT = e−kα (^2) t [c 1 J 0 (αr) + c 2 Y 0 (αr)].
- Substituting u(r, θ) = R(r)Θ(θ) into the partial differential equation yields
R′′Θ +
r R
′Θ +^1
r^2 RΘ
Separating variables and using the separation constant −λ we obtain r^2 R′′^ + rR′ R
= −λ.
Then r^2 R′′^ + rR′^ + λR = 0 and Θ′′^ − λΘ = 0. Letting λ = −α^2 we have the Cauchy-Euler equation r^2 R′′+rR′−α^2 R = 0 whose solution is R(r) = c 3 rα+c 4 r−α. Since the solution of Θ′′^ + α^2 Θ = 0 is Θ(θ) = c 1 cos αθ + c 2 sin αθ we see that a solution of the partial differential equation is u = RΘ = (c 1 cos αθ + c 2 sin αθ)(c 3 rα^ + c 4 r−α).
13.1 Separable Partial Differential Equations
- For u = A 1 + B 1 x we compute ∂^2 u/∂x^2 = 0 = ∂u/∂y. Then ∂^2 u/∂x^2 = 4 ∂u/∂y.
For u = A 2 eα^2 y^ cos 2αx + B 2 eα^2 y^ sin 2αx we compute ∂u ∂x = 2αA 2 eα (^2) y sinh 2αx + 2αB 2 eα (^2) y cosh 2αx ∂^2 u ∂x^2 = 4α^2 A 2 eα (^2) y cosh 2αx + 4α^2 B 2 eα (^2) y sinh 2αx and ∂u ∂y = α^2 A 2 eα (^2) y cosh 2αx + α^2 B 2 eα (^2) y sinh 2αx.
Then ∂^2 u/∂x^2 = 4 ∂u/∂y. For u = A 3 e−α^2 y^ cosh 2αx + B 3 e−α^2 y^ sinh 2αx we compute ∂u ∂x = − 2 αA 3 e−α (^2) y sin 2αx + 2αB 3 e−α (^2) y cos 2αx ∂^2 u ∂x^2 = − 4 α^2 A 3 e−α (^2) y cos 2αx − 4 α^2 B 3 e−α (^2) y sin 2αx and ∂u ∂y = −α^2 A 3 e−α (^2) y cos 2αx − α^2 B 3 e−α (^2) y sin 2αx.
Then ∂^2 u/∂x^2 = 4 ∂u/∂y.
- We identify A = xy + 1, B = x + 2y, and C = 1. Then B^2 − 4 AC = x^2 + 4y^2 − 4. The equation x^2 + 4y^2 = 4 defines an ellipse. The partial differential equation is hyperbolic outside the ellipse, parabolic on the ellipse, and elliptic inside the ellipse.
- Assuming u(x, y) = X(x)Y (y) and substituting into ∂^2 u/∂x^2 − u = 0 we get X′′Y − XY = 0 or Y (X′′^ − X) = 0. This implies X(x) = c 1 ex^ or X(x) = c 2 e−x. For these choices of X, Y can be any function of y. Two solutions of the partial differential equation are then
u 1 (x, y) = A(y)ex^ and u 2 (x, y) = B(y)e−x.
Since the partial differential equation is linear and homogeneous the superposition principle indicates that another solution is u(x, y) = u 1 (x, y) + u 2 (x, y) = A(y)ex^ + B(y)e−x.
- Assuming u(x, y) = X(x)Y (y) and substituting into ∂^2 u/∂x∂y + ∂u/∂x = 0 we get X′Y ′^ + X′Y = 0 or X′(Y ′^ + Y ) = 0. This implies Y (y) = c 1 e−y^. For this choice of Y , X can be any function of x. A solution of the partial differential equation is then u(x, y) = A(x)e−y^. In addition, noting that the partial differential equation can be written ∂ ∂x
[
∂u ∂y
]
any function u 2 (x, y) = B(y) will satisfy the partial differential equation since, in this case, ∂u 2 /∂y + u 2 = B′(y) + B(y) and the x-partial of B′(y) + B(y) is 0. Thus, using the superposition principle, a solution of the partial differential equation is
u(x, y) = u 1 (x, y) + u 2 (x, y) = A(x)e−y^ + B(y).
13.3 Heat Equation
- a^2 ∂^2 u ∂x^2
∂^2 u ∂t^2 , 0 < x < L, t > 0 u(0, t) = 0, u(L, t) = 0, t > 0
u(x, 0) = 0, ∂u ∂t
t=
= sin πx L , 0 < x < L
- a^2 ∂^2 u ∂x^2 − 2 β ∂u ∂t
∂^2 u ∂t^2 , 0 < x < L, t > 0 u(0, t) = 0, u(L, t) = sin πt, t > 0
u(x, 0) = f (x), ∂u ∂t
t=
= 0, 0 < x < L
- a^2 ∂^2 u ∂x^2
- Ax = ∂^2 u ∂t^2 , 0 < x < L, t > 0 , A a constant u(0, t) = 0, u(L, t) = 0, t > 0
u(x, 0) = 0, ∂u ∂t
t=
= 0, 0 < x < L
∂^2 u ∂x^2
∂^2 u ∂y^2 = 0, 0 < x < 4 , 0 < y < 2
∂u ∂x
x=
= 0, u(4, y) = f (y), 0 < y < 2
∂u ∂y
y=
= 0, u(x, 2) = 0, 0 < x < 4
∂^2 u ∂x^2
∂^2 u ∂y^2 = 0, 0 < x < π, y > 0
u(0, y) = e−y^ , u(π, y) =
100 , 0 < y ≤ 1 0 , y > 1 u(x, 0) = f (x), 0 < x < π
EXERCISES 13.
Heat Equation
- Using u = XT and −λ as a separation constant we obtain X′′^ + λX = 0, X(0) = 0, X(L) = 0, and T ′^ + kλT = 0.
This leads to X = c 1 sin nπ L x and T = c 2 e−kn (^2) π (^2) t/L 2
13.3 Heat Equation
for n = 1, 2, 3,... so that u =
∑^ ∞
n=
Ane−kn (^2) π (^2) t/L 2 sin nπ L x.
Imposing u(x, 0) =
∑^ ∞
n=
An sin nπ L x gives An =^2 L
∫ L/ 2
0
sin nπ L x dx = 2 nπ
1 − cos nπ 2
for n = 1, 2, 3,... so that u(x, t) =
π
∑^ ∞
n=
1 − cos nπ 2 n e−kn (^2) π (^2) t/L 2 sin nπ L x.
- Using u = XT and −λ as a separation constant we obtain X′′^ + λX = 0, X(0) = 0, X(L) = 0, and T ′^ + kλT = 0. This leads to X = c 1 sin nπ L x and T = c 2 e−kn (^2) π (^2) t/L 2
for n = 1, 2, 3,... so that u =
∑^ ∞
n=
Ane−kn
(^2) π (^2) t/L 2 sin nπ L x. Imposing u(x, 0) =
∑^ ∞
n=
An sin nπ L x gives An =
L
∫ L
0
x(L − x) sin nπ L x dx =
4 L^2
n^3 π^3 [1 − (−1)n]
for n = 1, 2, 3,... so that u(x, t) =^4 L
2 π^3
∑^ ∞
n=
1 − (−1)n n^3 e−kn (^2) π (^2) t/L 2 sin nπ L x.
- Using u = XT and −λ as a separation constant we obtain X′′^ + λX = 0, X′(0) = 0, X′(L) = 0, and T ′^ + kλT = 0. This leads to X = c 1 cos nπ L x and T = c 2 e−kn (^2) π (^2) t/L 2
(^0 20 40 ) (^80 ) x
0 50 100 150 200 t
0
10
20
30
40
u
(^0 20 40 )
1 2 3 4 5 6 t
20
40
60
80
100
u
x= 3 pê 4 x= 5 pê 6 x= 11 pê 12 x=p
13.3 Heat Equation
where
An =
[∫ 50
0
- 8 x sin nπ 100 x dx +
50
0 .8(100 − x) sin nπ 100 x dx
]
n^2 π^2 sin nπ 2
Thus, u(x, t) =^320 π^2
∑^ ∞
n=
n^2
sin nπ 2
e−kn (^2) π (^2) t/ 1002 sin nπ 100 x.
(b) Since An = 0 for n even, the first five nonzero terms correspond to n = 1, 3, 5, 7, 9. In this case sin(nπ/2) = sin(2p − 1)/2 = (−1)p+1^ for p = 1, 2, 3, 4, 5, and
u(x, t) =^320 π^2
∑^ ∞
p=
(−1)p+ (2p − 1)^2 e(−^1 .6352(2p−1) (^2) π (^2) / 1002 )t sin (2p^ −^ 1)π 100 x.
13.4 Wave Equation
EXERCISES 13.
Wave Equation
- Using u = XT and −λ as a separation constant we obtain
X′′^ + λX = 0, X(0) = 0, X(L) = 0, and T ′′^ + λa^2 T = 0, T ′(0) = 0.
Solving the differential equations we get
X = c 1 sin nπ L x + c 2 cos nπ L x and T = c 3 cos nπa L t + c 4 sin nπa L t
for n = 1, 2, 3,.. .. The boundary and initial conditions give
u =
∑^ ∞
n=
An cos nπa L t sin nπ L x.
Imposing
u(x, 0) =
x(L − x) =
∑^ ∞
n=
An sin nπ L x
gives An =
L^2
n^3 π^3 [1 − (−1)n] for n = 1, 2, 3,... so that
u(x, t) =
L^2
π^3
∑^ ∞
n=
1 − (−1)n n^3 cos nπa L t sin nπ L x.
- Using u = XT and −λ as a separation constant we obtain
X′′^ + λX = 0, X(0) = 0, X(L) = 0, and T ′′^ + λa^2 T = 0, T (0) = 0.
Solving the differential equations we get
X = c 1 sin nπ L x + c 2 cos nπ L x and T = c 3 cos nπa L t + c 4 sin nπa L t
13.4 Wave Equation
and u(x, t) =
π^2
cos πa L t sin π L x −
cos 5 πa L t sin 5 π L x +
cos 7 πa L t sin 7 π L x − · · ·
- Using u = XT and −λ as a separation constant we obtain
X′′^ + λX = 0, X(0) = 0, X(π) = 0, and T ′′^ + λa^2 T = 0, T ′(0) = 0. Solving the differential equations we get X = c 1 sin nx + c 2 cos nx and T = c 3 cos nat + c 4 sin nat for n = 1, 2, 3,.. .. The boundary and initial conditions give
u =
∑^ ∞
n=
An cos nt sin nx.
Imposing u(x, 0) =
x(π^2 − x^2 ) =
∑^ ∞
n=
An sin nx and ut(x, 0) = 0
gives An =
n^3 (−1)
n+
for n = 1, 2, 3,... so that u(x, t) = 2
∑^ ∞
n=
(−1)n+ n^3 cos^ nat^ sin^ nx.
- Using u = XT and −λ as a separation constant we obtain
X′′^ + λX = 0, X(0) = 0, X(π) = 0, and T ′′^ + λa^2 T = 0, T ′(0) = 0. Solving the differential equations we get X = c 1 sin nx + c 2 cos nx and T = c 3 cos nat + c 4 sin nat for n = 1, 2, 3,.. .. The boundary and initial conditions give
u =
∑^ ∞
n=
An cos nt sin nx.
Imposing ut(x, 0) = sin x =
∑^ ∞
n=
Bnna sin nx
13.4 Wave Equation
gives B 1 = 1 a^2 , and Bn = 0 for n = 2, 3, 4,... so that u(x, t) =
a sin at sin x.
- Using u = XT and −λ as a separation constant we obtain X′′^ + λX = 0, X(0) = 0, X(1) = 0, and T ′′^ + λa^2 T = 0, T ′(0) = 0. Solving the differential equations we get X = c 1 sin nπx + c 2 cos nπx and T = c 3 cos nπat + c 4 sin nπat for n = 1, 2, 3,.. .. The boundary and initial conditions give
u =
∑^ ∞
n=
An cos nt sin nx.
Imposing u(x, 0) = 0.01 sin 3πx =
∑^ ∞
n=
An sin nπx
gives A 3 = 0.01, and An = 0 for n = 1, 2, 4, 5, 6,... so that u(x, t) = 0.01 sin 3πx cos 3πat.
- Using u = XT and −λ as a separation constant we obtain X′′^ + λX = 0, X(0) = 0, X(L) = 0, and T ′′^ + λa^2 T = 0, T ′(0) = 0. Solving the differential equations we get X = c 1 sin nπ L x + c 2 cos nπ L x and T = c 3 cos nπa L t + c 4 sin nπa L t
for n = 1, 2, 3,.. .. The boundary and initial conditions give
u =
∑^ ∞
n=
An cos nπa L t^ sin^
nπ L x. Imposing u(x, 0) =
∑^ ∞
n=
An sin nπ L x
13.4 Wave Equation
Solving the differential equations we get
X = c 1 sin nx + c 2 cos nx and T = e−βt^
c 3 cos
n^2 − β^2 t + c 4 sin
n^2 − β^2 t
The boundary conditions on X imply c 2 = 0 so
X = c 1 sin nx and T = e−βt^
c 3 cos
n^2 − β^2 t + c 4 sin
n^2 − β^2 t
and u =
∑^ ∞
n=
e−βt^
An cos
n^2 − β^2 t + Bn sin
n^2 − β^2 t
sin nx.
Imposing
u(x, 0) = f (x) =
∑^ ∞
n=
An sin nx
and ut(x, 0) = 0 =
∑^ ∞
n=
Bn
n^2 − β^2 − βAn
sin nx
gives
u(x, t) = e−βt
∑^ ∞
n=
An
cos
n^2 − β^2 t + √ β n^2 − β^2
sin
n^2 − β^2 t
sin nx,
where An =
π
∫ (^) π
0
f (x) sin nx dx.
- Using u = XT and −λ = as a separation constant leads to X′′^ + α^2 X = 0, X(0) = 0, X(π) = 0 and T ′′^ + (1 + α^2 )T = 0, T ′(0) = 0. Then X = c 2 sin nx and T = c 3 cos
n^2 + 1 t for n = 1, 2, 3,... so that
u =
∑^ ∞
n 1
Bn cos
n^2 + 1 t sin nx.
Imposing u(x, 0) =
∑^ ∞
n=
Bn sin nx gives
Bn =^2 π
∫ (^) π/ 2
0
x sin nx dx +^2 π
∫ (^) π
π/ 2
(π − x) sin nx dx = 4 πn^2 sin nπ 2
0 , n even, 4 πn^2 (−1)(n+3)/^2 , n = 2k − 1, k = 1, 2, 3,.. ..
Thus with n = 2k − 1,
u(x, t) =^4 π
∑^ ∞
n=
sin nπ 2 n^2 cos
n^2 + 1 t sin nx =^4 π
∑^ ∞
k=
(−1)k+ (2k − 1)^2 cos
(2k − 1)^2 + 1 t sin(2k − 1)x.
- From (8) in the text we have
u(x, t) =
∑^ ∞
n=
An cos nπa L t + Bn sin nπa L t
sin nπ L x.
13.4 Wave Equation
Since ut(x, 0) = g(x) = 0 we have Bn = 0 and
u(x, t) =
∑^ ∞
n=
An cos nπa L t sin nπ L x
∑^ ∞
n=
An
[
sin
( (^) nπ L x + nπa L t
( (^) nπ L x − nπa L t
)]
∑^ ∞
n=
An
[
sin nπ L (x + at) + sin nπ L (x − at)
]
From u(x, 0) = f (x) =
∑^ ∞
n=
An sin nπ L x
we identify
f (x + at) =
∑^ ∞
n=
An sin nπ L (x^ +^ at) and f (x − at) =
∑^ ∞
n=
An sin nπ L (x^ −^ at), so that u(x, t) =
[f (x + at) + f (x − at)].
- (a) We note that ξx = ηx = 1, ξt = a, and ηt = −a. Then
∂u ∂x = ∂u ∂ξ
∂ξ ∂x
∂η ∂x = uξ + uη and ∂^2 u ∂x^2
∂x (uξ + uη ) = ∂uξ ∂ξ
∂ξ ∂x
∂uξ ∂η
∂η ∂x
∂uη ∂ξ
∂ξ ∂x
∂uη ∂η
∂η ∂x = uξξ + 2uξη + uηη. Similarly ∂^2 u ∂t^2 = a^2 (uξξ − 2 uξη + uηη ). Thus a^2 ∂^2 u ∂x^2
∂^2 u ∂t^2 becomes ∂^2 u ∂ξ∂η
(b) Integrating ∂^2 u ∂ξ∂η
∂η uξ = 0
we obtain (^) ∫ ∂ ∂η uξ dη =
0 dη
uξ = f (ξ). Integrating this result with respect to ξ we obtain ∫ ∂u ∂ξ dξ =
f (ξ) dξ
u = F (ξ) + G(η).
