Articolo G.A. Partial Differential Equations and Boundary Value Problems with Maple V
Подождите немного. Документ загружается.
328 Chapter 5
x = 0 and y = 1 are at a fixed temperature of zero, the side y = 0 is insulated, and the
side x = 1 has a given temperature distribution.
u(0,y)= 0 and u(1,y)= 1 −y
2
u
y
(x, 0) = 0 and u(x, 1) = 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.5. We seek the electrostatic potential in a charge-free rectangular domain
D ={(x,y) |0 <x<1, 0 <y<1}. The sides x = 0,y= 0, and y = 1 are held at a
fixed potential of zero, and the side x = 1 has a given potential distribution. The
boundary conditions are
u(0,y)= 0 and u(1,y)= y(1 −y)
u(x, 0) = 0 and u(x, 1) = 0
Develop the graphics for the three-dimensional equipotential surfaces showing the
equipotential lines.
5.6. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The sides
x = 0 and x = 1 are at the fixed temperature of zero, the side y = 0 is insulated, and
the side y = 1 has a given temperature distribution. The boundary conditions are
u(0,y)= 0 and u(1,y)= 0
u
y
(x, 0) = 0 and u(x, 1) = x(1 −x)
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.7. We seek the electrostatic potential in a charge-free rectangular domain
D ={(x,y) |0 <x<1, 0 <y<1}. The sides x = 1,y= 0, and y = 1 are held at a
fixed potential of zero, and the side x = 0 has a given potential distribution. The
boundary conditions are
u(0,y)= y(1 −y) and u(1,y)= 0
u(x, 0) = 0 and u(x, 1) = 0
Develop the graphics for the three-dimensional equipotential surfaces showing the
equipotential lines.
5.8. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The sides
The Laplace Par tial Differential Equation 329
x = 0 and y = 1 are held at the fixed temperature of zero, the side x = 1 is losing heat
by convection into a surrounding temperature of zero, and the side y = 0 has a given
temperature distribution. The boundary conditions are
u(0,y)= 0 and u(1,y)+u
x
(1,y)= 0
u(x, 0) = x
1 −
2x
3
and u(x, 1) = 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.9. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The sides
y = 0 and y = 1 are at a fixed temperature of zero, the side x = 1 is losing heat by
convection into a surrounding temperature of zero, and the side x = 0 has a given
temperature distribution. The boundary conditions are
u(0,y)= y(1 −y) and u(1,y)+u
x
(1,y)= 0
u(x, 0) = 0 and u(x, 1) = 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.10. We seek the electrostatic potential in a charge-free rectangular domain
D ={(x,y) |0 <x<1, 0 <y<1}. The sides x = 0,x= 1, and y = 0 are held at a fixed
potential of zero, and the side y = 1 has a given potential distribution. The boundary
conditions are
u(0,y)= 0 and u(1,y)= 0
u(x, 0) = 0 and u(x, 1) = x(1 −x)
Develop the graphics for the three-dimensional equipotential surfaces showing the
equipotential lines.
5.11. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The side
y = 0 is at a fixed temperature of zero, the side y = 1 is insulated, the side x = 0is
losing heat by convection into a surrounding medium of zero, and the side x = 1 has a
given temperature distribution. The boundary conditions are
u(0,y)
−u
x
(0,y)= 0 and u(1,y)= y
1 −
y
2
u(x, 0) = 0 and u
y
(x, 1) = 0
330 Chapter 5
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.12. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The sides
x = 0 and y = 0 are at a fixed temperature of zero, the side y = 1 is losing heat by
convection into a surrounding temperature of zero, and the side x = 1 has a given
temperature distribution. The boundary conditions are
u(0,y)= 0 and u(1,y)= y
1 −
2y
3
u(x, 0) = 0 and u(x, 1) +u
y
(x, 1) = 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.13. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The side
y = 1 is insulated, the side x = 0 is losing heat by convection into a surrounding
temperature of zero, the side x = 1 is at a fixed temperature of zero, and the side y = 0
has a given temperature distribution. The boundary conditions are
u
x
(0,y)−u(0,y)= 0 and u(1,y)= 0
u(x, 0) =−
2x
2
3
+
x
3
+
1
3
and u
y
(x, 1) = 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.14. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The sides
x = 0 and x = 1 are at a fixed temperature of zero, the side y = 1 is losing heat by
convection into a surrounding temperature of zero, and the side y = 0 has a given
temperature distribution. The boundary conditions are
u(0,y)= 0 and u(1,y)= 0
u(x, 0) = x(1 −x) and u(x, 1) +u
y
(x, 1) = 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.15. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The side
x = 0 is at a fixed temperature of zero, the side y = 0 is losing heat by convection into a
The Laplace Par tial Differential Equation 331
surrounding temperature of zero, the side x = 1 is insulated, and the side y = 1 has a
given temperature distribution. The boundary conditions are
u(0,y)= 0 and u
x
(1,y)= 0
u
y
(x, 0) −u(x, 0) = 0 and u(x, 1) = x
1 −
x
2
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.16. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The side
x = 1 is insulated, the side y = 0 is losing heat by convection into a surrounding
temperature of zero, the side y = 1 is at a fixed temperature of zero, and the side x = 0
has a given temperature distribution. The boundary conditions are
u(0,y)=−
2y
2
3
+
y
3
+
1
3
and u
x
(1,y)= 0
u
y
(x, 0) −u(x, 0) = 0 and u(x, 1) = 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
Superposition of Solutions
In all the preceding exercises, homogeneous boundary conditions occurred with respect to
either the x or y coordinate. We now consider problems whereby we do not have a set of
homogeneous boundary conditions. To treat these problems, we must write the solution as a
superposition of two solutions; that is, we set u(x, y) = u1(x, y) +u2(x, y), and we partition
the two subsequent problems in a manner that gives homogeneous boundary conditions.
For example, consider the problem of finding the steady-state temperature in a two-dimensional
domain D ={(x,y) |0 <x<1, 0 <y<1} that satisfies the boundary conditions
u(0,y)= g(y) and u(1,y)= 0
and
u(x, 0) = 0 and u(x, 1) = f(x)
Note that we lack a set of homogeneous boundary conditions with respect to both the x and y
coordinates. We write the solution as the superposition of two solutions:
u(x, y) = u1(x, y) +u2(x, y)
332 Chapter 5
We partition the boundary conditions in the following manner so as to give a set of
homogeneous boundary conditions for each problem
u1(0,y)= g(y) and u1(1,y)= 0 and u1(x, 0) = 0 and u1(x, 1) = 0
and
u2(0,y)= 0 and u2(1,y)= 0 and u2(x, 0) = 0 and u2(x, 1) = f(x)
The solution for u1(x,y), for a specific g(y), has been evaluated in Exercise 5.7, and the
solution for u2(x,y), for a specific f(x), has been evaluated in Exercise 5.10. Thus, the sum of
these two solutions gives the answer to our original problem.
The following exercises require the solution to be partitioned as a superposition of two
solutions u1(x,y) and u2(x,y). The problems are such that they can be resolved into two
problems, each of which has already been solved in Exercises 5.1 through 5.16. Seek those
preceding solutions, and write the general solution as a superposition.
5.17. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The side
x
= 1 is insulated, the side y = 0 is at a fixed temperature of zero, and the sides x = 0
and y = 1 have given temperature distributions. The boundary conditions are
u(0,y)= y(1 −y) and u
x
(1,y)= 0
u(x, 0) = 0 and u(x, 1) = x
1 −
x
2
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.18. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The side
y = 0 is insulated, the side x = 0 is at a fixed temperature of zero, and the sides x = 1
and y = 1 have given temperature distributions. The boundary conditions are
u(0,y)= 0 and u(1,y)= 1 −y
2
u
y
(x, 0) = 0 and u(x, 1) = x(1 −x)
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.19. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The side
y = 1 is at a fixed temperature of zero, the side x = 1 is losing heat by convection into a
The Laplace Par tial Differential Equation 333
surrounding temperature of zero, and the sides x = 0 and y = 0 have given temperature
distributions. The boundary conditions are
u(0,y)= y(1 −y) and u(1,y)+u
x
(1,y)= 0
u(x, 0) = x
1 −
2x
3
and u(x, 1) = 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.20. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x, y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The side
x = 0 is at a fixed temperature of zero, the side y = 1 is losing heat by convection into a
surrounding temperature of zero, and the sides x = 1 and y = 0 have given temperature
distributions. The boundary conditions are
u(0,y)= 0 and u(1,y)= y
1 −
2y
3
u(x, 0) = x(1 −x) and u(x, 1) +u
y
(x, 1) = 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.21 We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The side
y = 1 is insulated, the side x = 0 is losing heat by convection into a surrounding
temperature of zero, and the sides x = 1 and y =0 have given temperature distributions.
The boundary conditions are
u(0,y)−u
x
(0,y)= 0 and u(1,y)= y
1 −
y
2
u(x, 0) =−
2x
2
3
+
x
3
+
1
3
and u
y
(x, 1) = 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.22. We seek the steady-state temperature distribution in a thin plate over the rectangular
domain D ={(x,y) |0 <x<1, 0 <y<1} whose lateral surface is insulated. The side
x = 1 is insulated, the side y = 0 is losing heat by convection into a surrounding
temperature of zero, and the sides x = 0 and y =1 have given temperature distributions.
334 Chapter 5
The boundary conditions are
u(0,y)=−
2y
2
3
+
y
3
+
1
3
and u
x
(1,y)= 0
u(x, 0) −u
y
(x, 0) = 0 and u(x, 1) = x
1 −
x
2
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
Exercises in Cylindrical Coordinates
5.23. We seek the steady-state temperature distribution in a thin plate over the cylindrical
domain D ={(r, θ) |0 <r<1, 0 <θ<π/3} whose lateral surface is insulated. The
temperature at the origin is finite. The side θ = 0 is at a fixed temperature of zero, the
side θ = π/3 is insulated, and the side r = 1 has a given temperature distribution. The
boundary conditions are
|
u(0,θ)
|
< ∞,u(1,θ)= θ
1 −
3θ
2π
u(r, 0) = 0 and u
θ
r,
π
3
= 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.24. We seek the steady-state temperature distribution in a thin plate over the cylindrical
domain D ={(r, θ) |0 <r<1, 0 <θ<π/3} whose lateral surface is insulated. The
temperature at the origin is finite. The sides θ = 0 and θ = π/3 are at a fixed
temperature of zero, and the side r = 1 has a given temperature distribution. The
boundary conditions are
|
u(0,θ)
|
< ∞ and u(1,θ)= θ
1 −
3θ
π
u(r, 0) = 0 and u
r,
π
3
= 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.25 We seek the steady-state temperature distribution in a thin plate over the cylindrical
domain D ={(r, θ) |0 <r<1, 0 <θ<π/3} whose lateral surface is insulated. The
temperature at the origin is finite. The side θ = 0 is insulated, the side θ = π/3isata
fixed temperature of zero, and the side r = 1 has a given temperature distribution. The
The Laplace Par tial Differential Equation 335
boundary conditions are
|
u(0,θ)
|
< ∞ and u(1,θ)= 1 −
9θ
2
π
2
u
θ
(r, 0) = 0 and u
r,
π
3
= 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.26. We seek the steady-state temperature distribution in a thin plate over the cylindrical
domain D ={(r,θ) |1 <r<2, 0 <θ<π/2} whose lateral surface is insulated. The sides
r = 1,r = 2, and θ = 0 are at a fixed temperature of zero, and the side θ = π/2 has a
given temperature distribution. The boundary conditions are
u(1,θ)= 0 and u(2,θ)= 0
u(r, 0) = 0 and u
r,
π
2
= (r −1)(2 −r)
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.27. We seek the steady-state temperature distribution in a thin plate over the cylindrical
domain D ={(r,θ) |1 <r<2, 0 <θ<π/2} whose lateral surface is insulated. The sides
r = 1 and θ = 0 are at a fixed temperature of zero, the side r = 2 is insulated, and the
side θ = π/2 has a given temperature distribution. The boundary conditions are
u(1,θ)= 0 and u
r
(2,θ)= 0
u(r, 0) = 0 and u
r,
π
2
=−r
2
+4r −3
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.28 We seek the steady-state temperature distribution in a thin plate over the cylindrical
domain D ={(r,θ) |1 <r<2, 0 <θ<π/2} whose lateral surface is insulated. The sides
r = 1,r = 2, and θ = π/2 are at a fixed temperature of zero, and the side θ =0 has a
given temperature distribution. The boundary conditions are
u(1,θ)= 0 and u(2,θ)= 0
u(r, 0) = (r −1)(2 −r) and u
r,
π
2
= 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
336 Chapter 5
5.29. We seek the steady-state temperature distribution in a thin plate over the cylindrical
domain D ={(r,θ) |1 <r<2, 0 <θ<π/2} whose lateral surface is insulated. The sides
r = 1 and θ = π/2 are at a fixed temperature of zero, the side r = 2 is insulated, and the
side θ = 0 has a given temperature distribution. The boundary conditions are
u(1,θ)= 0 and u
r
(2,θ)= 0
u(r, 0) =−r
2
+4r −3 and u
r,
π
2
= 0
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.30. We seek the steady-state temperature distribution in a thin plate over the cylindrical
domain D ={(r,θ) |1 <r<2, 0 <θ<π/2} whose lateral surface is insulated. The sides
r = 2 and θ = 0 are at a fixed temperature of zero, the side r = 1 is insulated, and the
side θ = π/2 has a given temperature distribution. The boundary conditions are
u
r
(1,θ)= 0 and u(2,θ)= 0
u(r, 0) = 0 and u
r,
π
2
=−r
2
+2r
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
Superposition of Solutions in Cylindrical Coordinates
5.31. We seek the steady-state temperature distribution in a thin plate over the cylindrical
domain D ={(r,θ) |1 <r<2, 0 <θ<π/2} whose lateral surface is insulated. The sides
r = 1 and r = 2 are at a fixed temperature of zero, and the sides θ = 0 and θ =π/2 have
given temperature distributions. The boundary conditions are
u(1,θ)= 0 and u(2,θ)= 0
u(r, 0) = (r −1)(2 −r) and u
r,
π
2
= 1
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.
5.32. We seek the steady-state temperature distribution in a thin plate over the cylindrical
domain D ={(r,θ) |1 <r<2, 0 <θ<π/2} whose lateral surface is insulated. The side
r =1 is at a fixed temperature of zero, the side r = 2 is insulated, and the sides θ = 0 and
The Laplace Par tial Differential Equation 337
θ = π/2 have given temperature distributions. The boundary conditions are
u(1,θ)= 0 and u
r
(2,θ)= 0
u(r, 0) =−r
2
+4r −3 and u
r,
π
2
= 1
Develop the graphics for the three-dimensional temperature surface showing the
isotherms.