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Solving partial differential equations in Maple and Mathematica

Nasser M. Abbasi

August 14, 2018   Compiled on August 14, 2018 at 9:13pm

Contents

1 First order PDE
 1.1 Linear PDE, the transport equation
 1.2 Linear PDE
 1.3 Linear PDE, initial value problem
 1.4 Initial-boundary value problem
 1.5 Linear PDE, the transport equation with initial conditions
 1.6 General solution for a quasilinear first-order PDE
 1.7 quasilinear first-order PDE, scalar conservation law
 1.8 quasilinear first-order PDE, scalar conservation law with initial value
 1.9 nonlinear first-order PDE, the Clairaut equation
 1.10 nonlinear first-order PDE, the Clairaut equation with initial value
 1.11 Another example of nonlinear Clairaut equation
 1.12 Recover a function from its gradient vector
 1.13 General solution of a first order nonlinear PDE
 1.14 Nonlinear first order PDE
2 Heat PDE
 2.1 Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.
 2.2 Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.
 2.3 Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.
 2.4 Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.
 2.5 Heat PDE on bar, homogeneous Neumann boundary conditions, No source.
 2.6 Heat PDE on bar, homogeneous Dirichlet boundary conditions with heat sink
 2.7 Heat PDE on bar, homogeneous Neumann boundary conditions, No source
 2.8 Heat PDE on bar, homogeneous Neumann boundary conditions, No source
 2.9 Heat PDE on bar, homogeneous Neumann boundary conditions, No source
 2.10 Heat PDE on bar, homogeneous Neumann boundary conditions, No source
 2.11 Heat PDE on bar, homogeneous Neumann on left and Dirichlet on right, No source
 2.12 Heat PDE on bar, semi-infinite domain, No source
 2.13 Heat PDE on bar, periodic boundary conditions, No source
 2.14 Heat PDE on bar, semi-infinite domain, zero initial condition, No source
 2.15 Heat PDE on bar, semi-infinite domain, non-zero initial condition, No source
 2.16 Heat PDE on bar, heat absorption radiation in bounded domain, No source
 2.17 Heat PDE infinite domain
 2.18 Heat PDE on bar, with domain from -1 to +1, no source
 2.19 Heat PDE on bar, Dirichlet nonhomogeneous BC, no source term
 2.20 Heat PDE on bar, nonhomogeneous Dirichlet BC, with constant source term
 2.21 Heat PDE on bar, homogeneous Dirichlet BC, non zero initial conditions, with extra term
 2.22 Heat PDE on bar with initial conditions sum of sine terms, homogeneous Dirichlet BC, no source
 2.23 Heat PDE on bar, homogeneous Dirichlet BC, initial condition is piecewise function, no source
 2.24 Heat PDE on bar, inhomogeneous Dirichlet BC, initial condition is piecewise function, no source
 2.25 Heat PDE on bar, inhomogeneous Dirichlet BC which depends on time. Zero initial condition, no source
 2.26 Heat PDE on bar, homogeneous Neumann boundary conditions, non zero initial conditions, with source as sin function that depends on space only.
 2.27 Heat PDE on bar, homogeneous Neumann boundary conditions, nonzero initial conditions, with source that depends on time only
 2.28 Heat PDE on bar, homogeneous Neumann boundary conditions, nonzero initial conditions, with source that depends on time and space
 2.29 Heat PDE on bar, non-homogeneous, time dependent, Neumann boundary conditions, with source that depends on time and space
 2.30 Heat PDE on bar, non-homogeneous, not time dependent Neumann boundary conditions, No source term
 2.31 Heat PDE on bar, homogeneous Neumann boundary conditions, Source term that depends on both time and space
 2.32 Heat PDE on bar, homogeneous Neumann boundary conditions, Source term that depends on both time and space
 2.33 Heat PDE on bar, Dirichlet boundary conditions that depends on time with source that depends on space only
 2.34 Heat PDE on bar, homogeneous Dirichlet boundary conditions, with source that depends on time and space
 2.35 Heat/Diffusion PDE in 2D, inside rectangle with initial and boundary conditions
 2.36 Heat/Diffusion PDE in 2D, inside rectangle with initial and boundary conditions with heat loss
 2.37 Heat PDE inside disk, with no θ  dependency. initial and boundary conditions given
 2.38 Heat PDE on whole line with no intial nor boundary conditions specified
 2.39 Heat PDE in 1D on the whole real line with initial position specified
 2.40 Heat PDE in 1D on the whole real line, with linear adevction
 2.41 Heat PDE in 1D on the whole real line with initial position as UnitBox
 2.42 Heat PDE on half the line with non-zero initial conditions and Dirichlet boundary conditions
 2.43 Heat PDE on half the line with zero initial conditions and time dependent boundary conditions
 2.44 Initial value problem for the heat PDE with a Neumann condition on the half-line
3 Laplace PDE
 3.1 Laplace PDE inside quarter-circle
 3.2 Laplace PDE inside semi-circle
 3.3 Laplace PDE inside rectangle
 3.4 Laplace PDE inside rectangle
 3.5 Laplace PDE inside rectangle
 3.6 Laplace PDE inside rectangle
 3.7 Laplace PDE inside rectangle
 3.8 Laplace PDE inside rectangle, top/bottom edges non-zero
 3.9 Laplace PDE inside circular annulus, Neumann boundary conditions using unspecified functions
 3.10 Laplace PDE inside circular annulus, Dirichlet boundary conditions using specified functions
 3.11 Laplace PDE example 18 from Maple help page
 3.12 Laplace PDE on rectangle with one edge at infinity
 3.13 Laplace PDE inside a disk, periodic boundary conditions
 3.14 Dirichlet problem for the Laplace equation in upper half plan
 3.15 Dirichlet problem for the Laplace equation in right half-plane:
 3.16 Dirichlet problem for the Laplace equation in the first quadrant
 3.17 Neumann problem for the Laplace equation in the upper half-plane
 3.18 Dirichlet problem for the Laplace equation in a rectangle
 3.19 Laplace PDE outside a disk, periodic boundary conditions
 3.20 Laplace equation in spherical coordinates
4 Poisson PDE
 4.1 Dirichlet problem for the Poisson equation in a rectangle
5 Helmholtz PDE
 5.1 Dirichlet problem for the Helmholtz equation in a rectangle
6 Wave PDE
 6.1 General solution for a second-order hyperbolic PDE on real line
 6.2 Hyperbolic PDE with non-rational coefficients
 6.3 Inhomogeneous hyperbolic PDE with constant coefficients
 6.4 system of 2 inhomogeneous linear hyperbolic system with constant coefficients
 6.5 Wave PDE on string (finite domain) with zero initial position and velocity, and with source term
 6.6 Wave PDE on string, one end fixed, another free, both initial conduitions non zero, and source that depends on time and space
 6.7 Wave PDE on string (finite domain), fixed ends, no initial conduitions give and no source
 6.8 Wave PDE on string (finite domain), one fixed end, one free end, initial position not zero, initial velocity zero, no source
 6.9 Wave PDE on string (finite domain), both ends fixed end, initial conditions zero, with source as generic function that depends on time and space
 6.10 Wave PDE on string (finite domain), both ends fixed, initial conditions both not zero, No source
 6.11 Wave PDE on string (finite domain), both ends fixed end, initial conditions both not zero, and with constant source
 6.12 Wave PDE on string (finite domain), both ends fixed end, with source
 6.13 Wave PDE on semi-infinite domain, with one end having a moving boundary condition
 6.14 Telegraphy PDE, a wave PDE on string, both ends fixed with damping
 6.15 Wave PDE, on string, both ends fixed. Initial velocity zero. Dispersion term present
 6.16 Wave PDE on string with fixed ends, non-zero initial position
 6.17 Wave PDE homogeneous in square, given initial position but with zero initial velocity
 6.18 Wave PDE homogeneous in square with damping. Given zero initial position but with non-zero initial velocity
 6.19 Wave PDE inside rectangle. All 4 edges are fixed and given non-zero initial position with zero initial velocity
 6.20 Wave PDE inside disk. fixed edge of disk, no θ  dependency, with initial position and velocity given
 6.21 Wave PDE inside disk. fixed edge of disk, with θ  dependency, zero initial velocity
 6.22 Wave PDE on infinite domain with initial conditions specified, no source
 6.23 Wave PDE on infinite domain with initial conditions specified, with source term
 6.24 Wave PDE initial value with a Dirichlet condition on the half-line
 6.25 Wave PDE Initial value problem with a Neumann condition on the half-line
 6.26 non-linear wave PDE (Solitons)
7 Schrodinger PDE
 7.1 Schrodinger PDE with zero potential
 7.2 Schrodinger PDE with initial and boundary conditions
 7.3 Initial value problem with Dirichlet boundary conditions
 7.4 Solve a Schrodinger equation with potential over the whole real line
8 Beam PDE
 8.1 Beam PDE with zero initial velocity
9 Burger’s PDE
 9.1 viscous fluid flow with no initial conditions
 9.2 viscous fluid flow with initial conditions
 9.3 viscous fluid flow with initial conditions as UnitBox
10 Black Scholes PDE
 10.1 classic Black Scholes model from finance
 10.2 Boundary value problem for the Black Scholes equation
11 Korteweg-deVries PDE
 11.1 Korteweg-deVries (waves on shallow water surfaces) with no initial conditions
12 Tricomi PDE
 12.1 Boundary value problem for the Tricomi equation
13 Cauchy Riemann PDE’s
 13.1 Cauchy Riemann PDE with Prescribe the values of u  and v  on the x  axis
 13.2 Cauchy Riemann PDE With extra term on right side
14 Hamilton-Jacobi PDE
 14.1 Hamilton-Jacobi type PDE
15 Other second order PDE’s
 15.1 A second order PDE

This report gives the result of running a number of partial differential equations in Mathematica and Maple.

The following systems were used at this time.

  1. Mathematica 11.3 (64 bit).
  2. Maple 2018.1 (64 bit) with Physics version MapleCloud 84.

No time limit was used.

All possible options, assumptions and HINTS were tried to obtain a solution. The command DSolve was used in Mathematica and the command pdsolve in Maple.

It is possible I missed some option, assumption or HINT, which could help make the CAS able to solve a given PDE now marked as unsolved. Will correct such a case if found. I have verified some, but not all, solutions returned by Maple or Mathematica.

Number of problems is [ 122 ]. Mathematica solved 82 or 67.21%. Maple solved 102 or 83.61%.






Table 1: Breakdown of results for each PDE





#

PDE

description

Mathematica

Maple






1

First order PDE

Linear PDE, the transport equation

Solved

Solved






2

First order PDE

Linear PDE

Solved

Solved






3

First order PDE

Linear PDE, initial value problem

Solved

Solved






4

First order PDE

Initial-boundary value problem

Solved

Did not solve






5

First order PDE

Linear PDE, the transport equation with initial conditions

Solved

Solved






6

First order PDE

General solution for a quasilinear first-order PDE

Solved

Solved






7

First order PDE

quasilinear first-order PDE, scalar conservation law

Solved, solution in implicit form

Solved






8

First order PDE

quasilinear first-order PDE, scalar conservation law with initial value

Solved

Solved






9

First order PDE

nonlinear first-order PDE, the Clairaut equation

Solved

Solved






10

First order PDE

nonlinear first-order PDE, the Clairaut equation with initial value

Solved

Solved






11

First order PDE

Another example of nonlinear Clairaut equation

Solved

Solved






12

First order PDE

Recover a function from its gradient vector

Solved

Solved






13

First order PDE

General solution of a first order nonlinear PDE

Did not solve

Solved






14

First order PDE

Nonlinear first order PDE

Solved

Solved






15

Heat PDE

Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.

Solved

Solved






16

Heat PDE

Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.

Solved

Solved






17

Heat PDE

Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.

Solved

Solved






18

Heat PDE

Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.

Solved

Solved






19

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, No source.

Solved

Solved






20

Heat PDE

Heat PDE on bar, homogeneous Dirichlet boundary conditions with heat sink

Did not solve

Solved






21

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, No source

Solved

Solved






22

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, No source

Solved

Solved






23

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, No source

Solved

Solved






24

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, No source

Solved

Solved






25

Heat PDE

Heat PDE on bar, homogeneous Neumann on left and Dirichlet on right, No source

Solved

Solved






26

Heat PDE

Heat PDE on bar, semi-infinite domain, No source

Solved

Solved






27

Heat PDE

Heat PDE on bar, periodic boundary conditions, No source

Did not solve

Solved






28

Heat PDE

Heat PDE on bar, semi-infinite domain, zero initial condition, No source

Solved

Solved






29

Heat PDE

Heat PDE on bar, semi-infinite domain, non-zero initial condition, No source

Solved

Solved






30

Heat PDE

Heat PDE on bar, heat absorption radiation in bounded domain, No source

Did not solve

Solved






31

Heat PDE

Heat PDE infinite domain

Solved

Solved






32

Heat PDE

Heat PDE on bar, with domain from -1 to +1, no source

Did not solve

Solved






33

Heat PDE

Heat PDE on bar, Dirichlet nonhomogeneous BC, no source term

Solved

Solved






34

Heat PDE

Heat PDE on bar, nonhomogeneous Dirichlet BC, with constant source term

Did not solve

Solved






35

Heat PDE

Heat PDE on bar, homogeneous Dirichlet BC, non zero initial conditions, with extra term

Did not solve

Solved






36

Heat PDE

Heat PDE on bar with initial conditions sum of sine terms, homogeneous Dirichlet BC, no source

Solved

Solved






37

Heat PDE

Heat PDE on bar, homogeneous Dirichlet BC, initial condition is piecewise function, no source

Solved

Solved






38

Heat PDE

Heat PDE on bar, inhomogeneous Dirichlet BC, initial condition is piecewise function, no source

Solved

Solved






39

Heat PDE

Heat PDE on bar, inhomogeneous Dirichlet BC which depends on time. Zero initial condition, no source

Solved

Solved






40

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, non zero initial conditions, with source as sin function that depends on space only.

Did not solve

Solved






41

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, nonzero initial conditions, with source that depends on time only

Did not solve

Solved






42

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, nonzero initial conditions, with source that depends on time and space

Did not solve

Solved






43

Heat PDE

Heat PDE on bar, non-homogeneous, time dependent, Neumann boundary conditions, with source that depends on time and space

Did not solve

Solved






44

Heat PDE

Heat PDE on bar, non-homogeneous, not time dependent Neumann boundary conditions, No source term

Did not solve

Solved






45

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, Source term that depends on both time and space

Solved

Solved






46

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, Source term that depends on both time and space

Solved

Solved






47

Heat PDE

Heat PDE on bar, Dirichlet boundary conditions that depends on time with source that depends on space only

Did not solve

Solved






48

Heat PDE

Heat PDE on bar, homogeneous Dirichlet boundary conditions, with source that depends on time and space

Did not solve

Solved






49

Heat PDE

Heat/Diffusion PDE in 2D, inside rectangle with initial and boundary conditions

Did not solve

Solved






50

Heat PDE

Heat/Diffusion PDE in 2D, inside rectangle with initial and boundary conditions with heat loss

Did not solve

Solved






51

Heat PDE

Heat PDE inside disk, with no θ  dependency. initial and boundary conditions given

Solved

Did not solve






52

Heat PDE

Heat PDE on whole line with no intial nor boundary conditions specified

Did not solve

Solved, returning a solution that is not the most general one






53

Heat PDE

Heat PDE in 1D on the whole real line with initial position specified

Solved

Solved






54

Heat PDE

Heat PDE in 1D on the whole real line, with linear adevction

Solved

Solved






55

Heat PDE

Heat PDE in 1D on the whole real line with initial position as UnitBox

Solved

Solved






56

Heat PDE

Heat PDE on half the line with non-zero initial conditions and Dirichlet boundary conditions

Solved

Solved, but has unresolved inverse Laplace transforms






57

Heat PDE

Heat PDE on half the line with zero initial conditions and time dependent boundary conditions

Solved

Solved, but has unresolved inverse Laplace transforms






58

Heat PDE

Initial value problem for the heat PDE with a Neumann condition on the half-line

Solved

Did not solve






59

Laplace PDE

Laplace PDE inside quarter-circle

Did not solve

Did not solve






60

Laplace PDE

Laplace PDE inside semi-circle

Did not solve

Solved






61

Laplace PDE

Laplace PDE inside rectangle

Solved

Solved






62

Laplace PDE

Laplace PDE inside rectangle

Solved

Solved






63

Laplace PDE

Laplace PDE inside rectangle

Did not solve

Solved






64

Laplace PDE

Laplace PDE inside rectangle

Did not solve

Solved






65

Laplace PDE

Laplace PDE inside rectangle

Did not solve

Solved






66

Laplace PDE

Laplace PDE inside rectangle, top/bottom edges non-zero

Solved

Solved






67

Laplace PDE

Laplace PDE inside circular annulus, Neumann boundary conditions using unspecified functions

Did not solve

Did not solve






68

Laplace PDE

Laplace PDE inside circular annulus, Dirichlet boundary conditions using specified functions

Solved

Did not solve






69

Laplace PDE

Laplace PDE example 18 from Maple help page

Solved

Solved






70

Laplace PDE

Laplace PDE on rectangle with one edge at infinity

Did not solve

Solved






71

Laplace PDE

Laplace PDE inside a disk, periodic boundary conditions

Solved

Solved






72

Laplace PDE

Dirichlet problem for the Laplace equation in upper half plan

Solved

Did not solve






73

Laplace PDE

Dirichlet problem for the Laplace equation in right half-plane:

Solved

Did not solve






74

Laplace PDE

Dirichlet problem for the Laplace equation in the first quadrant

Solved

Did not solve






75

Laplace PDE

Neumann problem for the Laplace equation in the upper half-plane

Solved

Did not solve






76

Laplace PDE

Dirichlet problem for the Laplace equation in a rectangle

Solved

Solved






77

Laplace PDE

Laplace PDE outside a disk, periodic boundary conditions

Did not solve

Solved






78

Laplace PDE

Laplace equation in spherical coordinates

Did not solve

Solved, but not verified






79

Poisson PDE

Dirichlet problem for the Poisson equation in a rectangle

Solved

Solved






80

Helmholtz PDE

Dirichlet problem for the Helmholtz equation in a rectangle

Solved

Solved






81

Wave PDE

General solution for a second-order hyperbolic PDE on real line

Solved

Solved






82

Wave PDE

Hyperbolic PDE with non-rational coefficients

Solved

Did not solve. Tried all HINTS






83

Wave PDE

Inhomogeneous hyperbolic PDE with constant coefficients

Solved

Solved






84

Wave PDE

system of 2 inhomogeneous linear hyperbolic system with constant coefficients

Solved

Did not solve






85

Wave PDE

Wave PDE on string (finite domain) with zero initial position and velocity, and with source term

Solved

Solved






86

Wave PDE

Wave PDE on string, one end fixed, another free, both initial conduitions non zero, and source that depends on time and space

Did not solve

Solved






87

Wave PDE

Wave PDE on string (finite domain), fixed ends, no initial conduitions give and no source

Did not solve

Solved






88

Wave PDE

Wave PDE on string (finite domain), one fixed end, one free end, initial position not zero, initial velocity zero, no source

Did not solve

Solved






89

Wave PDE

Wave PDE on string (finite domain), both ends fixed end, initial conditions zero, with source as generic function that depends on time and space

Did not solve

Solved






90

Wave PDE

Wave PDE on string (finite domain), both ends fixed, initial conditions both not zero, No source

Solved

Solved






91

Wave PDE

Wave PDE on string (finite domain), both ends fixed end, initial conditions both not zero, and with constant source

Did not solve

Solved






92

Wave PDE

Wave PDE on string (finite domain), both ends fixed end, with source

Did not solve

Solved






93

Wave PDE

Wave PDE on semi-infinite domain, with one end having a moving boundary condition

Solved

Solved






94

Wave PDE

Telegraphy PDE, a wave PDE on string, both ends fixed with damping

Did not solve

Solved, But n = 1  should not be included.






95

Wave PDE

Wave PDE, on string, both ends fixed. Initial velocity zero. Dispersion term present

Did not solve due to adding dispersion term

Solved






96

Wave PDE

Wave PDE on string with fixed ends, non-zero initial position

Solved but sum should not include n = 2

Solved






97

Wave PDE

Wave PDE homogeneous in square, given initial position but with zero initial velocity

Did not solve

Solved






98

Wave PDE

Wave PDE homogeneous in square with damping. Given zero initial position but with non-zero initial velocity

Did not solve

Solved






99

Wave PDE

Wave PDE inside rectangle. All 4 edges are fixed and given non-zero initial position with zero initial velocity

Solved

Solved






100

Wave PDE

Wave PDE inside disk. fixed edge of disk, no θ  dependency, with initial position and velocity given

Solved

Did not solve






101

Wave PDE

Wave PDE inside disk. fixed edge of disk, with θ  dependency, zero initial velocity

Did not solve

Did not solve






102

Wave PDE

Wave PDE on infinite domain with initial conditions specified, no source

Solved

Solved






103

Wave PDE

Wave PDE on infinite domain with initial conditions specified, with source term

Solved

Solved






104

Wave PDE

Wave PDE initial value with a Dirichlet condition on the half-line

Solved

Solved






105

Wave PDE

Wave PDE Initial value problem with a Neumann condition on the half-line

Solved

Did not solve






106

Wave PDE

non-linear wave PDE (Solitons)

Solved. build a special solution.

Solved. Returning a solution that is not the most general one






107

Schrodinger PDE

Schrodinger PDE with zero potential

Solved

Solved






108

Schrodinger PDE

Schrodinger PDE with initial and boundary conditions

Solved

Did not solve






109

Schrodinger PDE

Initial value problem with Dirichlet boundary conditions

Solved

Solved






110

Schrodinger PDE

Solve a Schrodinger equation with potential over the whole real line

Solved

Did not solve. Maple does not support ∞ in boundary conditions






111

Beam PDE

Beam PDE with zero initial velocity

Solved

Solved






112

Burger’s PDE

viscous fluid flow with no initial conditions

Solved

Solved






113

Burger’s PDE

viscous fluid flow with initial conditions

Solved

Solved, but has unresolved integrals






114

Burger’s PDE

viscous fluid flow with initial conditions as UnitBox

Solved

Solved, but has unresolved integrals






115

Black Scholes PDE

classic Black Scholes model from finance

Solved

Did not solve






116

Black Scholes PDE

Boundary value problem for the Black Scholes equation

Solved

Did not solve






117

Korteweg-deVries PDE

Korteweg-deVries (waves on shallow water surfaces) with no initial conditions

Solved

Solved






118

Tricomi PDE

Boundary value problem for the Tricomi equation

Solved

Solved






119

Cauchy Riemann PDE’s

Cauchy Riemann PDE with Prescribe the values of u  and v  on the x  axis

Solved

Did not solve






120

Cauchy Riemann PDE’s

Cauchy Riemann PDE With extra term on right side

Did not Solve

Solved






121

Hamilton-Jacobi PDE

Hamilton-Jacobi type PDE

Did not Solve

Solved






122

Other second order PDE’s

A second order PDE

Did not Solve

Solved











1 First order PDE

_________________________________________________________________________________

1.1 Linear PDE, the transport equation

problem number 1

Taken from Mathematica Symbolic PDE document

Solve for u(x,t)

∂u   ∂u
---+ ---= 0
∂t   ∂x

Mathematica
{{u(x,t) → c1(t− x)}}

Result Solved

Maple
u (x,t) =-F1 (− x+ t)

Result Solved

_________________________________________________________________________________

1.2 Linear PDE

problem number 2

Taken from Mathematica help pages

Solve for u(x,y)

 ∂u-   ∂u-
3∂x + 5∂y = 0

Mathematica
{ {          (    (         )     ) }}
   u(x,y) → 1  6c1 1 (3y − 5x) + x2
            6      3

Result Solved

Maple
u(x,y) = 1∕6x2 +-F1 (− 5∕3x+ y)

Result Solved

_________________________________________________________________________________

1.3 Linear PDE, initial value problem

problem number 3

Taken from Mathematica help pages

Solve for u(x,y)

  ∂u-   ∂u-
x ∂x + y ∂y = − 4xyu(x,y)

with initial value            2
u(x,0) = e−x

Mathematica
{ {         −x2−y2}}
   u(x,y) → e

Result Solved

Maple
         −x2−y2
u(x,y) = e

Result Solved

_________________________________________________________________________________

1.4 Initial-boundary value problem

problem number 4

Taken from Mathematica help pages

Solve for u(x,t)

∂u-  ∂u-
∂t + ∂x = 0

with initial value u(x,0) = sinx  and boundary value u (0,t) = 0

Mathematica
{{u(x,t) → (θ(t− x)− 1)sin(t − x)}}

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

1.5 Linear PDE, the transport equation with initial conditions

problem number 5

Taken from Mathematica help pages

Solve for u(x,t)

∂u-  ∂u-
∂t + c∂x = 0

With initial conditions            2
u (x,0) = e−x

Mathematica
{{          −(x−ct)2}}
   u(x,t) → e

Result Solved

Maple
         −(tc−x)2
u(x,t) = e

Result Solved

_________________________________________________________________________________

1.6 General solution for a quasilinear first-order PDE

problem number 6

Taken from Mathematica help pages

Solve for u(x,y)

 ∂u-   ∂u-   2
2∂x +5 ∂y = u (x,y)+ 1

Mathematica
{{            (  (    (         )    ) )}}
   u(x,y) → tan 1  2c1 1 (2y − 5x) + x
                2      2

Result Solved

Maple
u(x,y) = tan(x∕2 +1∕2-F 1(− 5∕2x+ y))

Result Solved

_________________________________________________________________________________

1.7 quasilinear first-order PDE, scalar conservation law

problem number 7

Taken from Mathematica Symbolic PDE document

Solve for u(x,y)

∂u-        ∂u-
∂x + u(x,y)∂y = 0

Mathematica
    [          (          )       ]
Solve u(x,y) = c1 x− ---y--  ,u(x,y)
                    u(x,y)

Result Solved, solution in implicit form

Maple
− y+ xu (x,y)+ -F1 (u (x,y)) = 0

Result Solved

_________________________________________________________________________________

1.8 quasilinear first-order PDE, scalar conservation law with initial value

problem number 8

Taken from Mathematica Symbolic PDE document

Solve for u(x,y)

∂u-        ∂u-
∂x + u(x,y)∂y = 0

With u(x,0) = x1+1

Mathematica
{{              }}
   u(x,y) → y-+-1
           x + 1

Result Solved

Maple
        y-+-1
u(x,y) = x + 1

Result Solved

_________________________________________________________________________________

1.9 nonlinear first-order PDE, the Clairaut equation

problem number 9

Taken from Mathematica Symbolic PDE document

Solve for u(x,y)

 ∂u    ∂u   1 (( ∂u)2   (∂u )2)
x---+ y---+ -    ---  +  ---    = 0
 ∂x    ∂y   2    ∂x      ∂y

Mathematica
{{                   1 ( 2   2)} }
   u(x,y) → c1x+ c2y+ 2  c1 + c2

Result Solved

Maple
              2     ∘ --------     (    ∘ -------)           2     ∘ --------     (    ∘ -------)
u(x,y) = − 1∕2x − 1∕2x x2 + 2-c1−-c1 ln  x+   x2 + 2-c1 +-C1− 1∕2y− 1∕2y y2 − 2 c1+ c1ln y + y2 − 2 c1 + C2

Result Solved

_________________________________________________________________________________

1.10 nonlinear first-order PDE, the Clairaut equation with initial value

problem number 10

Taken from Mathematica Symbolic PDE document

Solve for u(x,y)

 ∂u    ∂u   1 (( ∂u)2   (∂u )2)
x---+ y---+ -    ---  +  ---    = 0
 ∂x    ∂y   2    ∂x      ∂y

With         1     2
u(x,0) = 2(1− x )

Mathematica
{{          1(   2       )}}
   u(x,y) → 2 − x − 2y+ 1

Result Solved

Maple
− 1∕2(x − y+ 1)(x− y − 1) = − 1∕2(1+ x + y)(x+ y− 1)

Result Solved

_________________________________________________________________________________

1.11 Another example of nonlinear Clairaut equation

problem number 11

Taken from Mathematica DSolve help pages

Solve for u(x,y)

         ∂u    ∂u      (∂u   ∂u )
u(x,y) = x∂x-+ y∂y-+ sin ∂x-+ ∂y-

Mathematica
{{u(x,y) → c1x + c2y + sin(c1 + c2)}}

Result Solved

Maple
u(x,y) = x-c1 + y-c2 + sin(-c1 + c2)

Result Solved

_________________________________________________________________________________

1.12 Recover a function from its gradient vector

problem number 12

Taken from Mathematica DSolve help pages

Solve for f(x,y)

pict

Mathematica
{{f(x,y) → c1 + x sin(xy)+ e−y}}

Result Solved

Maple
{                           }
 f (x,y) = x sin(yx)+ e−y + C1

Result Solved

_________________________________________________________________________________

1.13 General solution of a first order nonlinear PDE

problem number 13

Taken from Maple pdsolve help pages

Solve for f(x,y)

pict

Mathematica
      [   (0,1)        (1,0)       g(x)f(x,y)2            ]
DSolve xf    (x,y)− f   (x,y) =   h(y)   ,f(x,y),{x,y}

Result Did not solve

Maple
        (∫ x                                        )−1
f (x,y) =    --------g( a)-------d-a+ -F 1(1∕2x2 + y)
             h(− 1∕2-a2 + 1∕2x2 + y)

Result Solved

_________________________________________________________________________________

1.14 Nonlinear first order PDE

problem number 14

Taken from Maple pdsolve help pages, probem 5

Solve for f(x,y,z)

pict

Mathematica
((             (                (                )                    (                )     ) ))
{{                                   c1y(z)+cc21(z(z))+x−1                          c1y(z)+cc21(z(z))+x−1        }}
   f(x,y,z) → 1(c1 (z)2ProductLog( − e------------) 2 + 2c1(z)2ProductLog( − e-----------) − 4z)
((            4                         c1(z)                                  c1(z)             ))

Result Solved

Maple
            e−xz-C52-+-ex-C32-+-C3y--C5+-z-C4-C5-+--C1-C5
f (x,y,z) = −                   C52e −x

Result Solved

2 Heat PDE

_________________________________________________________________________________

2.1 Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.

problem number 15

This is problem 2.3.3, part (a) from Richard Haberman applied partial differential equations, 5th edition.

Consider the heat equation

       2
∂u-= k∂-u2-
∂t    ∂x

Subject to boundary conditions u(0,t) = 0  and u(L,t) = 0  with the temperature initially u(x,0) = 6 sin(3πx)
              L

Mathematica
{{                    (    )}}
   u(x,t) → 6e− 81πL22ktsin 9πx-
                        L

Result Solved

Maple
            ( πx-) −81kLπ22t
u(x,t) = 6sin 9 L  e

Result Solved

_________________________________________________________________________________

2.2 Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.

problem number 16

This is problem 2.3.3, part (b) from Richard Haberman applied partial differential equations, 5th edition.

Consider the heat equation

       2
∂u-= k∂-u-
∂t    ∂x2

Subject to boundary conditions u(0,t) = 0  and u(L,t) = 0  with the temperature initially u(x,0) = 3 sin πx − sin 3πx-
             L      L

Mathematica
{{                       (             (    )    )} }
            − 9πL2k2t  (πx-)    8π2Lk2t        2πx-
   u(x,t) → e     sin  L    3e    − 2cos   L   − 1

Result Solved

Maple
            (   )           (    )
u (x,t) = 3 sin πx- e− kLπ22t− sin 3πx- e−9kπL22t
              L                L

Result Solved

_________________________________________________________________________________

2.3 Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.

problem number 17

This is problem 2.3.3, part (c) from Richard Haberman applied partial differential equations, 5th edition.

Consider the heat equation

∂u    ∂2u
---= k--2-
∂t    ∂x

Subject to boundary conditions u(0,t) = 0  and u(L,t) = 0  with the temperature initially u(x,0) = 2 cos 3πx
             L

Mathematica
({({          ∞             − kn2π2t    (nπx))} )}
   u(x,t) → ∑  4(1-+(−-1)n)e---L2--nsin---L--
((         n=1          (n2 − 9)π         ) )

Result Solved

Maple
           (
        ∞∑  {0                          n = 3
u(x,t) =    (  n((−1)n+1)-  (nπx) − π2kn22t-
        n=1 4  π(n2−9) sin  L  e   L    otherwise

Result Solved

_________________________________________________________________________________

2.4 Heat PDE on bar, homogeneous Dirichlet boundary conditions, No source.

problem number 18

This is problem 2.3.3, part (d) from Richard Haberman applied partial differential equations, 5th edition.

Consider the heat equation

       2
∂u-= k∂-u-
∂t    ∂x2

Subject to boundary conditions u(0,t) = 0  and u(L,t) = 0  with the temperature initially          { 1    0 < x ≤ L
u (x,0) =   2    L-< x ≤ 2L
                2

Mathematica
({({         ∑∞   − kn2π22t (   (nπ)   )  2 (nπ)   (nπx))} )}
   u(x,t) →    4e---L---4-cos--2--+-3-sin---4--sin---L--
((         n=1                  nπ                  ) )

Result Solved

Maple
         ∞                       1+n   (    )
u(x,t) = ∑  2cos(1∕2nπ-)+-2+-4(− 1)---sin  nπx- e− π2Lkn22t
        n=1            nπ                 L

Result Solved

_________________________________________________________________________________

2.5 Heat PDE on bar, homogeneous Neumann boundary conditions, No source.

problem number 19

This is problem 2.3.7, from Richard Haberman applied partial differential equations, 5th edition.

Consider the heat equation

∂u    ∂2u
---= k--2-
∂t    ∂x

Subject to boundary conditions ∂∂ux(0,t) = 0  ∂∂ux(L,t) = 0  with the temperature initially u (x,0) = f(x )

Mathematica
( (          ∞      22    (   )∫     (   )                 ) )
||{ ||{         2∑  e− knLπ2-tcos nπLx L0 cos nπLx f(x)dx   ∫L      ||} ||}
   u (x,t) → -n=1--------------------------------+  0-f(x)dx-
||( ||(                          L                        L    ||) ||)

Result Solved

Maple
         1 (∑∞ (  1 ∫ L       ( nπx)      (n πx)   π2kn2t)    ∫ L       )
u (x,t) = --      2--   f (x) cos  ---- dxcos ----  e−  L2   L +    f (x)dx
         L  n=1   L  0           L          L                 0

Result Solved

_________________________________________________________________________________

2.6 Heat PDE on bar, homogeneous Dirichlet boundary conditions with heat sink

problem number 20

This is problem 2.3.8, from Richard Haberman applied partial differential equations, 5th edition.

Consider the heat equation

∂u-= k ∂2u− αu
 ∂t    ∂x2

This corresponds to a one-dimentional rod either with heat loss through the lateral sides with outside temperature zero degrees (α > 0  ) or with insulated sides with a heat sink propertional to the temperature.

Suppose the boundary conditions are u(0,t) = 0,u(L,t) = 0  , solve with the temperature initially u (x,0) = f(x )  if α > 0

Mathematica
      [{                                                               }            ]
DSolve  u(0,1)(x,t) = ku(2,0)(x,t)− au(x,t),{u(0,t) = 0,u(L,t) = 0},u(x,0) = f(x) ,u(x,t),{x,t}

Result Did not solve

Maple
           (                                            )
        ∞∑     1 ∫ L       (nπx )     ( nπx)  − t(π2kn2+L2a)
u(x,t) =    2 L-   f (x)sin -L-- dx sin  -L-- e     L2
        n=1      0

Result Solved

_________________________________________________________________________________

2.7 Heat PDE on bar, homogeneous Neumann boundary conditions, No source

problem number 21

This is problem 2.4.1 part(a) from Richard Haberman applied partial differential equations, 5th edition.

Consider the heat equation

∂u    ∂2u
---= k--2-
∂t    ∂x

The boundary conditions are ∂∂ux(0,t) = 0  ∂∂ux(L,t) = 0  with the temperature initially          { 0    x < L-
u (x,0) =   1    x > 2L-
                    2

Mathematica
(| (|           ∞∑    − kn2Lπ22t   (nπx)  (nπ)    )| )|
||{ ||{         2   −  e-----L-cosnπL--sin--2--   ||} ||}
    u(x,t) → -n=1----------------------- + 1
|||( |||(                      L                2|||) |||)

Result Solved

Maple
     ∑∞    sin(1∕2nπ)    (nπx)   π2kn2t
1∕2+    − 2----------cos  ---- e−  L2
     n=1      nπ          L

Result Solved

_________________________________________________________________________________

2.8 Heat PDE on bar, homogeneous Neumann boundary conditions, No source

problem number 22

This is problem 2.4.1 part(b) from Richard Haberman applied partial differential equations, 5th edition.

Solve the heat equation

       2
∂u-= k∂-u2-
∂t    ∂x

The boundary conditions are ∂∂ux(0,t) = 0  ∂∂ux(L,t) = 0  with the temperature initially                 (3πx-)
u(x,0) = 6 + 4cos L

Mathematica
{ {          − 9π2kt   (3πx )   } }
   u(x,t) → 4e  L2 cos -L--  + 6

Result Solved

Maple
                (    )   kπ2t
u (x,t) = 6 + 4cos 3πx e−9-L2
                   L

Result Solved

_________________________________________________________________________________

2.9 Heat PDE on bar, homogeneous Neumann boundary conditions, No source

problem number 23

This is problem 2.4.1 part(c) from Richard Haberman applied partial differential equations, 5th edition.

Solve the heat equation

∂u    ∂2u
---= k--2-
∂t    ∂x

The boundary conditions are ∂∂ux(0,t) = 0  ∂∂ux(L,t) = 0  with the temperature initially u(x,0) = − 2sin πLx

Mathematica
( (                       kn2π2t            ) )
||| |||         2 ∞∑ 2(1+-(−1)n)e−--L2--Lcos(nπLx)    ||| |||
{ {         -n=1---------(n2−1)π---------   4} }
|| || u(x,t) →              L             −  π|| ||
|( |(                                        |) |)

Result Solved

Maple
          ( ∞  ({                                   )
u(x,t) = 1-( ∑   0                  2 2  n ≤ 1 π− 4)
        π   n=1( 4(−π(n1)2n−+11) cos(nπLx)e− kπLn2-t 1 < n

Result Solved

_________________________________________________________________________________

2.10 Heat PDE on bar, homogeneous Neumann boundary conditions, No source

problem number 24

This is problem 2.4.1 part(d) from Richard Haberman applied partial differential equations, 5th edition.

Solve the heat equation

∂u    ∂2u
∂t-= k∂x2-

The boundary conditions are ∂∂ux(0,t) = 0  and ∂u∂x-(L,t) = 0  with the temperature initially u(x,0) = − 3cos 8πLx

Mathematica
{{              64π2kt   (    ) }}
   u(x,t) → − 3e− -L2--cos  8πx-
                          L

Result Solved

Maple
      ( πx)  −64kπL22t
− 3 cos 8L  e

Result Solved

_________________________________________________________________________________

2.11 Heat PDE on bar, homogeneous Neumann on left and Dirichlet on right, No source

problem number 25

This is problem 2.4.2 from Richard Haberman applied partial differential equations, 5th edition.

Solve the heat equation

       2
∂u-= k∂-u-
∂t    ∂x2

The boundary conditions are ∂u(0,t) = 0
∂x  u(L,t) = 0  with the temperature initially u(x,0) = f(x)

Mathematica
((           ∞∑  − k(2n+1)2π2t   ((2n+1)πx)∫ L   ((2n+1)πx)       ))
||{||{         2   e    4L2   cos ---2L---  0 cos ---2L--  f(x)dx ||}||}
   u(x,t) → -n=0---------------------------------------------
||(||(                                 L                        ||)||)

Result Solved

Maple
           (   ∫          (             )      (             )              )
        ∞∑     1- L            (1-+2n-)πx            (1-+-2n)πx-  −1∕4kπ2(1+L22n)2t
u(x,t) =     2 L 0  f (x)cos 1∕2    L      dxcos  1∕2     L      e
        n=0

Result Solved

_________________________________________________________________________________

2.12 Heat PDE on bar, semi-infinite domain, No source

problem number 26

This is problem at page 76 from David J Logan text book.

Solve the heat equation

∂u   ∂2u
---= ---2
∂t   ∂x

The boundary conditions are u(0,t) = f (t)  and initial conditions u(x,0) = 0

Mathematica
(| (|          ∫ t   − 4(xt−2z)  )| )|
|{ |{         x 0 f(z()te−z)3∕2--dz|} |}
| |u (x,t) → ------2√π------| |
|( |(                        |) |)

Result Solved

Maple
               ∫ t
u (x,t) = 1∕2 x√--   --f (-U1)-e− 4tx−24 U1d-U1
             π  0 (t− U 1)3∕2

Result Solved

_________________________________________________________________________________

2.13 Heat PDE on bar, periodic boundary conditions, No source

problem number 27

Solve the heat equation

∂u-= k∂2u-
∂t    ∂x2

For − L < x < L  and t > 0  . The boundary conditions are

pict

And initial conditions u(x,0) = f(x)

Mathematica
      [{ (0,1)        (2,0)    {                 (1,0)         (1,0)    }             }                                           ]
DSolve  u   (x,t) = ku  (x,t), u(− L,t) = u(L,t),u  (− L,t) = u  (L,t) ,u(x,0) = f(x) ,u(x,t),{x,t},Assumptions → {− L ≤ x ≤ L,t > 0}

Result Did not solve

Maple
             (  ∞  (  ( ∫                             ∫                            )        )     ∫         )
u(x,t) = 1∕21 2 ∑   -1    L f (x)sin (nπx) dxsin(n-πx) + L f (x) cos( nπx) dxcos(n-πx)  e− kπ2Ln22t L +  L f (x )dx
           L    n=1  L    −L          L          L       −L          L          L                   −L

Result Solved

_________________________________________________________________________________

2.14 Heat PDE on bar, semi-infinite domain, zero initial condition, No source

problem number 28

Solve the heat equation

∂u    ∂2u
∂t-= k∂x2-

For x > 0  and t > 0  . The boundary conditions is u(0,t) = 1  and And initial condition u(x,0) = 0

Mathematica
{{            (      )}}
   u(x,t) → erfc  -√x--
                2  kt

Result Solved

Maple
               (      x  )
u (x,t) = 1 − Erf 1∕2√-√---
                      t k

Result Solved

_________________________________________________________________________________

2.15 Heat PDE on bar, semi-infinite domain, non-zero initial condition, No source

problem number 29

Solve the heat equation

∂u    ∂2u
∂t-= k∂x2-

For x > 0  and t > 0  . The boundary conditions is u(0,t) = μ  and And initial condition u(x,0) = λ

Mathematica
{{            (   x )        (  x  )} }
   u(x,t) → μerf  -√--- + λerfc -√---
                2  kt         2  kt

Result Solved

Maple
                    (     x   )
u (x,t) = (− λ+ μ)Erf 1∕2 √t√k-  +λ

Result Solved

_________________________________________________________________________________

2.16 Heat PDE on bar, heat absorption radiation in bounded domain, No source

problem number 30

Solve the heat equation

∂u-   ∂2u-
∂t = k∂x2

For 0 < x < L  and t > 0  . The boundary conditions are

pict

And initial condition u(x,0) = f(x)

Mathematica
DSolve[{u(0,1)(x,t) = ku(2,0)(x,t),{u (1,0)(0,t)+ u(0,t) = 0,u(1,0)(L,t)+ u(L,t) = 0},u(x,0) = f(x)},u(x,t),{x,t},Assumptions → {t ≥ 0,k &#

Result Did not solve

Maple
        ∑∞ (              ∫ L     (       (    )     (    ) )   (       (    )      (    ) )     22 )
u(x,t) =     2---2-12----2-   f (x)  − πn cos nπx + sin  nπx- L  dx − πn cos  nπx- + sin  nπx- L  e− kπLn2-t
        n=1   L(π n + L  ) 0                L          L                   L          L

Result Solved

_________________________________________________________________________________

2.17 Heat PDE infinite domain

problem number 31

Solve the heat equation

∂u-   ∂2u-
∂t = k∂x2 + m

For − ∞ < x < ∞ and t > 0  . The boundary conditions are

Initial condition is u(x,0) = sin(x)

Mathematica
{{u (x,t) → e−ktsin(x)+ mt}}

Result Solved

Maple
u (x,t) = sin(x)e−tk + mt

Result Solved

_________________________________________________________________________________

2.18 Heat PDE on bar, with domain from -1 to +1, no source

problem number 32

Solve the heat equation

∂u-= ∂2u-
∂t   ∂x2

For − 1 < x < 1  and t > 0  . The boundary conditions are zero at both ends. Initial condition is u(x,0) = f(x)

Mathematica
      [{                                                       }            ]
DSolve  u (0,1)(x,t) = u (2,0)(x,t),{u(− 1,t) = 0,u(1,t) = 0},u(x,0) = f(x) ,u(x,t),{x,t}

Result Did not solve

Maple
        ∑∞ (( ∫ 1                           2    2   ∫ 1                                           2 2)      2(  2    ))
u(x,t) =         f (x )sin(nπx)dx sin(nπx )e1∕4π (2n−1)t +    f (x)cos(1∕2πx (2n− 1))dxcos(1∕2πx(2n − 1))eπ nt e−1∕4π t8n −4n+1
        n=1    −1                                     −1

Result Solved

_________________________________________________________________________________

2.19 Heat PDE on bar, Dirichlet nonhomogeneous BC, no source term

problem number 33

Taken from Maple PDE help pages

Solve the heat equation

       2
∂u-= ∂-u-
∂t   ∂x2

For 0 < x < 1  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = 0

Mathematica
(| (|            ∞∑  (20−50(−1)n)e−n2π2tsin(nπx)-         )| )|
|{ |{           2n=1          n                    |} |}
| | u(x,t) → − ------------π-------------+ 30x+ 20| |
|( |(                                              |) |)

Result Solved

Maple
                  ∞          n                22
u(x,t) = 20+ 30x + ∑ (100(− 1)-−-40)sin-(n-πx)ekπn-t
                 n=1              nπ

Result Solved

_________________________________________________________________________________

2.20 Heat PDE on bar, nonhomogeneous Dirichlet BC, with constant source term

problem number 34

This is problem 8.2.1 par(d) from Richard Haberman applied partial differential equations 5th edition.

Solve the heat equation

∂u    ∂2u
∂t-= k∂x2-+ k

For 0 < x < 1  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = f(x)

Mathematica
      [{ (0,1)        (2,0)                                            }          ]
DSolve  u   (x,t) = ku  (x,t)+ k,u(x,0) = f(x),{u(0,t) = A0,u(L0,t) = B0} ,u(x,t),x,t

Result Did not solve

Maple
             (     (                                                                                )                                 )
           1    ∞∑      1 ∫ L (              2   (     2    )            )   (nπx )     ( nπx)  kπ2n2t      2    (  2     )
u(x,t) = 1∕2L 2     − L2-   2 − f (x)L + 1∕2L x+  − 1∕2x + A L − x(A − B) sin  -L-- dx sin  -L-- e L2   L + L x + − x + 2A  L− 2x (A− B )
                n=1        0

Result Solved

_________________________________________________________________________________

2.21 Heat PDE on bar, homogeneous Dirichlet BC, non zero initial conditions, with extra term

problem number 35

Solve the heat equation

               2
∂u-+ u(x,t) = k ∂-u
∂t            ∂x2

For 0 < x < L  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = f(x)

Mathematica
DSolve[{u(0,1)(x,t)+ u(x,t) = u(2,0)(x,t),u(x,0) = f(x),{u(0,t) = 0,u(L,t) = 0}},u(x,t),x,t]

Result Did not solve

Maple
         ∞  (   ∫ L        (    )     (    )   t(π2n2+L2))
u (x,t) = ∑   2-1    f (x)sin nπx- dxsin n-πx  e− ---L2---
         n=1   L  0           L           L

Result Solved

_________________________________________________________________________________

2.22 Heat PDE on bar with initial conditions sum of sine terms, homogeneous Dirichlet BC, no source

problem number 36

added Feb 10, 2018.

Solve the heat equation

∂u-+ u(x,t) = 100 ∂2u-
∂t              ∂x2

For 0 < x < L  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = sin(2πx)− sin(5πx )

Mathematica
u(x,t) → e−400π2tsin(2πx)− e−2500π2tsin(5πx)

Result Solved

Maple
                     2                 2
u (x,t) = sin(2πx)e−400πt − sin (5πx)e−2500π t

Result Solved

_________________________________________________________________________________

2.23 Heat PDE on bar, homogeneous Dirichlet BC, initial condition is piecewise function, no source

problem number 37

added Feb 10, 2018.

Solve the heat equation

∂u   ∂2u
---= ---2
∂t   ∂x

For 0 < x < L  and t > 0  . The boundary conditions are

pict

Initial condition is

        {
u (x,0) =     x     0 ≤ x < 20
           40− x  20 ≥ x ≤ 40

Mathematica
{ {                                            } }
            ∞∑  640e− n126π200tcos(nπ) sin3(nπ)sin(nπx)
    u(x,t) →    --------------4n2π2----4------40--
            n=1

Result Solved

Maple
         ∞∑     sin(1∕2nπ)sin(1∕40nπx)  π2n2t
u (x,t) =    160--------π2n2---------e− 1600
         n=1

Result Solved

_________________________________________________________________________________

2.24 Heat PDE on bar, inhomogeneous Dirichlet BC, initial condition is piecewise function, no source

problem number 38

Added July 2, 2018, taken from Maple 2018.1 improvement to PDE document.

Solve the heat equation

∂u   ∂2u
---= ---2
∂t   ∂x

For 0 < x < 1  and t > 0  . The boundary conditions are

pict

Initial condition is

        {
u (x,0) =   1   x = 1
           0  otherwise

Mathematica
({ ({           ∞      (           )    2   2          )} )}
    u(x,t) →  ∑   − 2--− 1+-(−-1)K-[1]e−-πtK[1]-sin(πxK-[1])
( (         K[1]=1                 πK [1]              ) )

Result Solved

Maple
                           2 2
            ∑∞  sin-(nπx)e−π-n-t
u (x,t) = x +   2      nπ
            n=1

Result Solved

_________________________________________________________________________________

2.25 Heat PDE on bar, inhomogeneous Dirichlet BC which depends on time. Zero initial condition, no source

problem number 39

added March 8, 2018. Exam problem

Solve the heat equation

∂u   ∂2u
∂t-= ∂x2-

For 0 < x < π  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = 0  .

Mathematica
((               (        )               ))
{{         ∑∞     2− 2e−n2t sin(nx)        }}
   u(x,t) →    −  ------------------− tx-+ t
((         n=1          n3π          π    ))

Result Solved

Maple
             (  ∞             2                              )
u (x,t) = 1∕61  6∑  2sin-(nx)e−n-tπ− 6(− π + x)(1∕6x2 − 1∕3πx + t)
           π    n=1     πn3

Result Solved

_________________________________________________________________________________

2.26 Heat PDE on bar, homogeneous Neumann boundary conditions, non zero initial conditions, with source as sin function that depends on space only.

problem number 40

added March 18, 2018.

This is problem 8.2.1, part(f) from Richard Haberman applied partial differential equations 5th edition.

Solve the heat equation

∂u    ∂2u      (2πx )
---= k--2-+ sin  ----
∂t    ∂x         L

For 0 < x < L  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = f(x)  .

Mathematica
      [{                          (2πx )              {                         }}                                          ]
DSolve  u(0,1)(x,t) = ku(2,0)(x,t)+ sin -L--  ,u (x,0) = f (x), u(1,0)(0,t) = 0,u(1,0)(L,t) = 0 ,u(x,t),x,t,Assumptions → {L > 0,k > 0,t > 0}

Result Did not solve

Maple
                (     (                                                                                         )                                                                                     )
             1     ∞∑          1  ∫ L 2   (  πx)   ( nπx)                                (nπx )      (nπx ) π2kn2t   2          2     3   (  πx)     2    ∫ L         2         2    2   (  πx)
u(x,t) = 1∕4π2kL 4     − 1∕2π2kL-   L  sin 2 L--cos  -L-- − 2π(− 2πk C2 + 2πkf (x )+ Lx)cos-L-- dx cos -L-- e  L2    π kL +4-C2π  kL+ L  sin 2 L-- − 2L xπ−     − 4f (x)π k + 4 C2 π k+ L sin 2 L- − 2
                   n=1            0                                                                                                                      0

Result Solved

_________________________________________________________________________________

2.27 Heat PDE on bar, homogeneous Neumann boundary conditions, nonzero initial conditions, with source that depends on time only

problem number 41

Added July 2, 2018. Taken from Maple 2018.1 document, originally exercise 6.25 from Pinchover and Rubinstein.

Solve the heat equation

        2
∂u-= k ∂-u+ cos(wt)
 ∂t    ∂x2

For 0 < x < L  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = x  .

Mathematica
      [{                                        {                         }}                                          ]
DSolve  u(0,1)(x,t) = ku(2,0)(x,t)+ cos(tw),u(x,0) = x, u(1,0)(0,t) = 0,u(1,0)(L,t) = 0 ,u(x,t),x,t,Assumptions → {L > 0,t > 0,k > 0}

Result Did not solve

Maple
        ({      ∑ ∞    L((−1)n−-1)-  (nπx)  − kπL2n22t
u(x,t) =  L∕2+ ( n=1 2  n2π2   cos  L   e      + t2 2            )  w = 0
        ( 1∕21- Lw + 2∑ ∞n=12L((−12)n2−1)cos(nπx)e− kπLn2-tw + 2sin(wt)   otherwise
             w                 n π        L

Result Solved

_________________________________________________________________________________

2.28 Heat PDE on bar, homogeneous Neumann boundary conditions, nonzero initial conditions, with source that depends on time and space

problem number 42

added March 18, 2018.

Solve the heat equation

       2    (      (    ) )
∂u-= k∂-u2-+  e−ctsin  2πx-
∂t    ∂x              L

For 0 < x < L  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = f(x)  .

Mathematica
      [{                              (    )              {                         }}                                          ]
DSolve  u(0,1)(x,t) = ku(2,0)(x,t)+ e−ctsin 2πx- ,u(x,0) = f(x), u(1,0)(0,t) = 0,u(1,0)(L,t) = 0 ,u(x,t),x,t,Assumptions → {L > 0,k > 0,t > 0}
                                        L

Result Did not solve

Maple
          (∑∞  (   ∫ L        (    )     (    )     22 )    ∫ t     (  − cτL2−π2k(t−τ)   (   )    ∞∑        n1    −π2k(t−τ)n12−cτL2    (     ) )      ∫ L       )
u(x,t) = 1      21-   f (τ)cos nπτ- dτ cos nπx- e− kπLn2-t L+    1∕31- 8e-----L2---- cos πx- + 3    4-(−(1)2-−-1)e------L2----- cos n1πx- π  dτL +    f (τ)dτ
        L  n=1   L  0           L           L                0    π                     L     n1=3 π  n1 − 4                      L            0

Result Solved

_________________________________________________________________________________

2.29 Heat PDE on bar, non-homogeneous, time dependent, Neumann boundary conditions, with source that depends on time and space

problem number 43

added July 2, 2018.

Pinchover and Rubinstein’s exercise 6.17. Taken from Maple document for new improvements in Maple 2018.1

Solve the heat equation

            2
∂-u(x,t)− -∂2u (x,t) = 1 + xcos(t)
∂t        ∂x

For 0 < x < 1  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = 1+ cos(2πx)  .

Mathematica
      [{                                                   {                                } }          ]
DSolve  u(2,0)(x,t)+ xcos(t)+ 1 = u(1,0)(x,t),cos(2πx )+ 1 = u(x,0), sin(t) = u(1,0)(0,t),sin(t) = u(1,0)(1,t) ,u(x,t),x,t

Result Did not solve

Maple
u (x,t) = 1 + cos(2πx )e− 4π2t + t+ x sin(t)

Result Solved

_________________________________________________________________________________

2.30 Heat PDE on bar, non-homogeneous, not time dependent Neumann boundary conditions, No source term

problem number 44

added July 2, 2018.

Second example from Maple document for new improvements in Maple 2018.1

Solve the heat equation

∂u-    ∂2u-
 ∂t = 13∂x2

For 0 < x < 1  and t > 0  . The boundary conditions are

pict

Initial condition is         1 2
u(x,0) = 2x + x  .

Mathematica
      [{                                                                }          ]
         (1,0)         (2,0)             x2    {  (1,0)         (1,0)       }
DSolve  u    (x,t) = 13u   (x,t),u(x,0) = 2 + x, u    (0,t) = 0,u  (1,t) = 1  ,u(x,t),x,t

Result Did not solve

Maple
             ∑∞  ((− 1)n −-1)cos(nπx)e−13π2n2t         2
u(x,t) = 1∕2+    2           π2n2            +13t+ 1∕2x
             n=1

Result Solved

_________________________________________________________________________________

2.31 Heat PDE on bar, homogeneous Neumann boundary conditions, Source term that depends on both time and space

problem number 45

added July 2, 2018.

4th example from Maple document for new improvements in Maple 2018.1, originally taken from Pinchover and Rubinstein’s exercise 6.23 .

Solve the heat equation on bar

∂u   ∂2u
∂t-= ∂x2 + g(x,t)

Where g(x,t) = e3tcos(17πx)  for 0 < x < 1  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = f(x)  where f(x) = 3 cos(42πx)  .

Mathematica
{ {        e3tcos(17πx )  e−289π2tcos(17πx )         2         }}
   u(x,t) →---------2- − -----------2----+ 3e−1764π tcos(42πx)
             3+ 289π         3+ 289π

Result Solved

Maple
        (    2   )          −1764π2t          ( 3t   −289π2t)
u(x,t) = -867π-+-9-cos(42πx)e-------+-cos(17πx)--e-−-e--------
                             289π2 + 3

Result Solved

_________________________________________________________________________________

2.32 Heat PDE on bar, homogeneous Neumann boundary conditions, Source term that depends on both time and space

problem number 46

added July 2, 2018.

Taken from Maple document for new improvements in Maple 2018.1, originally taken from Pinchover and Rubinstein’s exercise 6.21

Solve the heat equation on bar

∂u   ∂2u
∂t-= ∂x2 + g(x,t)

Where g(x,t) = tcos(2001x)  for 0 < x < π  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = f(x)  where f(x) = π cos(2x)  .

Mathematica
{ {                      e−4004001tcos(2001x)  tcos(2001x)     cos(2001x)  } }
   u(x,t) → πe −4tcos(2x)+------------------+ ---------- − --------------
                           16032024008001       4004001    16032024008001

Result Solved

Maple
        (4004001t+ e−4004001t − 1)cos(2001x)
u(x,t) =----------16032024008001----------+ π cos(2x)e−4t

Result Solved

_________________________________________________________________________________

2.33 Heat PDE on bar, Dirichlet boundary conditions that depends on time with source that depends on space only

problem number 47

added March 28, 2018. A problem from my PDE animation page.

Solve the heat equation

∂u-   ∂2u-
∂t = k∂x2 + x

For 0 < x < π  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = 60− 20x  .

Mathematica
      [{                                       {                                }}                                     ]
         (0,1)       (2,0)                                1               1-
DSolve  u    (x,t) = u   (x,t)+ x,u(x,0) = 60− 2x, u(0,t) = 5 tsin(t),u(π,t) = 10tcos(t)  ,u(x,t),x,t,Assumptions → {t > 0,x > 0}

Result Did not solve

Maple
              (                             (     (                      )          2                2     (                                                             )          ))
            1                                 ∞∑    (− 1)nπ + 30(− 1)n+1 + 30 sin(nx)e−n t ∫ t ∑∞    en1 (− t+τ) − (− 1)n1sin (τ)τ + 10π(− 1)1+n1 +(− 1)n1cos(τ)&
u(x,t) = 1∕10π-(cos(t)tx + 2(π− x)tsin(t) +10π (   4-----------------nπ-----------------+        1∕5 -------------------------------------πn1---------------------------------------dτ))
                                              n=1                                         0 n1=1

Result Solved

_________________________________________________________________________________

2.34 Heat PDE on bar, homogeneous Dirichlet boundary conditions, with source that depends on time and space

problem number 48

Taken from Maple PDE help pages

Solve the heat equation for u(x,t)

∂u-   ∂2u-
∂t = k∂x2 + f(x,t)

For 0 < x < 1  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,0) = g(x)

Mathematica
      [{ (0,1)                (2,0)                                     }                                               ]
DSolve  u   (x,t) = f(x,t)+ ku  (x,t),u(x,0) = g(x),{u(0,t) = 0,u(1,t) = 0} ,u(x,t),x,t,Assumptions → {k > 0,t > 0,x > 0,x < 1}

Result Did not solve

Maple
        ∑∞ (    ∫ l      (     )     (     )    2 2 )  ∫ t∑∞ (   ∫ l         (    )     (    )    22    )
u(x,t) =     21   g (x )sin  πmx- dx sin  πmx- e− kπlm2-t +         21   f (x,τ)sin nπx dx sin  nπx- ekπnl(−2t+τ) dτ
        m=1   l  0          l           l               0 n=1   l 0             l          l

Result Solved

_________________________________________________________________________________

2.35 Heat/Diffusion PDE in 2D, inside rectangle with initial and boundary conditions

problem number 49

Taken from Maple help pages on PDE

Solve the heat equation for u(x,y,t)

∂u       ( ∂2u  ∂2u )
∂t-= 1∕10  ∂x2 + ∂y2

For 0 < x < 1  and 0 < y < 1  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,y,0) = x(1− x)(1− y)y  .

Mathematica
      [{                                                                                                                    }                ]
         (0,0,1)         1-( (0,2,0)         (2,0,0)      )
DSolve  u     (x,y,t) = 10 u     (x,y,t)+ u     (x,y,t) ,u(x,y,0) = (1 − x)x(1 − y)y,{u(0,y,t) = 0,u (1,y,t) = 0,u(x,0,t) = 0,u(x,1,t) = 0} ,u(x,y,t),{x,y,t}

Result Did not solve

Maple
           (                                        )        (                                          )
        ∑∞    1 ∫ l      ( πmx )     ( πmx ) − kπ2m2t   ∫ t∑∞    1∫ l         (nπx )     ( nπx)  kπ2n2(−t+τ)
u(x,t) =     2l   g (x )sin  -l-- dx sin  -l-- e   l2   +         2l   f (x,τ)sin--l- dx sin  -l-- e   l2     dτ
        m=1      0                                      0 n=1     0

Result Solved

_________________________________________________________________________________

2.36 Heat/Diffusion PDE in 2D, inside rectangle with initial and boundary conditions with heat loss

problem number 50

Taken from Maple help pages on PDE

Solve the heat equation for u(x,y,t)

         (  2     2 )
∂u-= 1∕10  ∂-u-+ ∂-u- − 1u(x,y,t);
∂t         ∂x2   ∂y2    5

For 0 < x < 1  and 0 < y < 1  and t > 0  . The boundary conditions are

pict

Initial condition is u(x,y,0) = (1− x2)(1 − 12y)y  .

Mathematica
      [{                                                                                                                                          }                ]
         (0,0,1)         1-( (0,2,0)         (2,0,0)      )  1                  (    2)(    y)   { (1,0,0)                                 (0,1,0)         }
DSolve  u     (x,y,t) = 10 u     (x,y,t)+ u     (x,y,t) − 5u (x,y,t),u(x,y,0) = 1− x    1− 2  y, u    (0,y,t) = 0,u(1,y,t) = 0,u(x,0,t) = 0,u  (x,1,t) = 0  ,u(x,y,t),{x,y,t}

Result Did not solve

Maple
           ∞ ( ∞                                                   ( ( 2  2        ) 2))
          ∑    ∑     (− 1)ncos(1∕2(1+-2n)πx-)sin(1∕2(1+-2m-)πy)e−1∕10t-2+-n-+m-+n+m+1-∕2-π-
u(x,y,t) = m=0  n=0512                       (1 +2n )3π6 (1+ 2m)3

Result Solved

_________________________________________________________________________________

2.37 Heat PDE inside disk, with no θ  dependency. initial and boundary conditions given

problem number 51

Taken from Mathematica DSolve help pages

Solve the heat equation in polar coordinates for u(r,t)

∂u   ∂2u   1 ∂u
---= --2-+ - ---
∂t   ∂r    r ∂r

For 0 < r < 1  and t > 0  . The boundary conditions are

pict

Initial condition is u(r,0) = 1− r  .

Mathematica
((                            (                             ) ))
{{         ∑∞ 2e−t(j0,n)2J0(rj0,n)  J1(j0,n)-− 11F2(3;1, 5;− 1(j0,n)2) }}
   u(r,t) →    -------------------j0,n2---3----22--2---4--------
((         n=1               J0(j0,n) + J1(j0,n)                ))

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

2.38 Heat PDE on whole line with no intial nor boundary conditions specified

problem number 52

Solve the heat equation for u(x,t)

∂u   ∂2u
∂t-= ∂x2-

Mathematica
      [ (0,1)       (2,0)               ]
DSolve u   (x,t) = u  (x,t),u(x,t),{x,t}

Result Did not solve

Maple
                   √ ---  -C3e-c1t-C2
u (x,t) = C3e c1t C1e c1x +---√-c1x--
                             e

Result Solved, returning a solution that is not the most general one

_________________________________________________________________________________

2.39 Heat PDE in 1D on the whole real line with initial position specified

problem number 53

From Mathematica DSolve help pages. Solve the heat equation for u(x,t)  on real line with t > 0

∂u   ∂2u
---= ---2
∂t   ∂x

With initial condition

u (x,0) = e− x2

Mathematica
{{           −-x2- }}
   u(x,t) → √e-4t+1-
             4t+ 1

Result Solved

Maple
           1      x2
u (x,t) = √-----e−4t+1
          4t+ 1

Result Solved

_________________________________________________________________________________

2.40 Heat PDE in 1D on the whole real line, with linear adevction

problem number 54

From Mathematica DSolve help pages. Solve the heat equation for u(x,t)  on real line with t > 0

∂u     ∂2u      ∂u
∂t-= 12∂x2-+ sin t∂x-

With initial condition

u(x,0) = x

Mathematica
{{u(x,t) → − cos(t)+ x+ 1}}

Result Solved

Maple
u(x,t) = − cos(t)+ x+ 1

Result Solved

_________________________________________________________________________________

2.41 Heat PDE in 1D on the whole real line with initial position as UnitBox

problem number 55

From Mathematica DSolve help pages. Solve the heat equation for u(x,t)  on real line with t > 0

       2
∂u-= ∂-u2
∂t   ∂x

With initial condition

u(x,0) = UnitBox[x]

Where UnitBox is equal to 1 if |x| ≤ 12  and zero otherwise.

Mathematica
{{           (   (      )     (      ) )}}
   u(x,t) → 1  erf 1-−√ 2x  +erf  2x+√-1-
           2       4  t          4 t

Result Solved

Maple
               (   2x + 1)         (    2x− 1)
u(x,t) = 1∕2Erf 1∕4--√t--  − 1∕2Erf   1∕4-√t---

Result Solved

_________________________________________________________________________________

2.42 Heat PDE on half the line with non-zero initial conditions and Dirichlet boundary conditions

problem number 56

From Mathematica DSolve help pages.

Solve the heat equation for u(x,t)  on half the line x > 0  and t > 0

∂u-  ∂2u-
∂t = ∂x2

With initial condition

u (x,0) = cosx

And boundary conditions

u(0,t) = 1

Mathematica
({ ({               − x2( (2t−ix) (2t+ix))      (   )        )} )}
   u(x,t) →  {   ie-4t-F--2√t√--−F--2√t--+ erfc -x√-   x > 0
( (                        πIndeterminate      2 t   True  ) )

Result Solved

Maple
                                            (  √-     )           ( √ -    )
                  ( √sx          )            e-sx-                e--sx                ( −√sx         )         −t
u(x,t) = − invlaplace e -F1 (s),s,t − invlaplace s+ 1,s,t +invlaplace   s ,s,t +invlaplace e    -F1(s),s,t+cos (x)e

Result Solved, but has unresolved inverse Laplace transforms

_________________________________________________________________________________

2.43 Heat PDE on half the line with zero initial conditions and time dependent boundary conditions

problem number 57

Solve the heat equation for u(x,t)  on half the line x > 0  and t > 0

∂u    ∂2u
---= k--2-
∂t    ∂x

With initial condition

u(x,0) = 0

And boundary conditions

pict

The last condition above means it is bounded at infinity.

Mathematica
(|(|          (      2)    (  x )   2x√kte− x42kt )|)|
{{          2kt-+x---erfc-2√kt--−----√π----}}
|(|(u (x,t) →              2k               |)|)

Result Solved

Maple
           -1-(   √-√ -√-- −1∕4x2   ( (        2)    (    --x--)            (        √√sx-   )             ( − √s√x         )            2)  )
u(x,t) = 1∕2πk − 2 π  t kxe    kt − 2  kt+ 1∕2x   Erf  1∕2√k-√t  − invlaplace -F1 (s)e  k ,s,t k + invlaplace e   k-F1 (s),s,t k − kt− 1∕2x  π

Result Solved, but has unresolved inverse Laplace transforms

_________________________________________________________________________________

2.44 Initial value problem for the heat PDE with a Neumann condition on the half-line

problem number 58

From Mathematica DSolve help pages.

Solve the heat equation for u(x,t)  on half the line x > 0  and t > 0

∂u   ∂2u
---= ---2
∂t   ∂x

With initial condition

u(x,0) = UnitTriagle[x- 3]

And boundary conditions

∂u(0,t) = 0
∂x

Mathematica
((                (                                                                                                (      2       2       2       2       2       2)  )         ) )
||||||                  erf(|x−√4|)(x− 4)2           (   )            (   )           (   )   2(x−3)2erf(|x−√3|)   (x−2)2erf(|x−√2|)  2 e− (x−44)t-−2e− (x−3
{{              12 |( ---2|4t−x|----+ (x+ 2)erf x2+√2t − 2(x +3)erf x2+√3t  + (x + 4)erf x2+√4t  − -----|3−x|2-t-+  ----|2−x|2t--+ -------------------------√π-------------------------|)   x > 0 } }
|||| u(x,t) → {                                                                                                                                                                   || ||
|(|(                                                                                    Indeterminate                                                                        True  |) |)

Result Solved

Maple
sol = ()

Result Did not solve

3 Laplace PDE

_________________________________________________________________________________

3.1 Laplace PDE inside quarter-circle

problem number 59

This is problem 2.5.5 part (c) from Richard Haberman applied partial differential equations, 5th edition

Solve Laplace equation

∂2u-  1∂u-  -1∂2u-
∂r2 + r∂r + r2∂ θ2 = 0

Inside quarter circle of radius 1 with         π
0 ≤ θ ≤ 2  and 0 ≤ r ≤ 1  , with following boundary conditions

pict

Mathematica
      [{ u(0,2)(r,θ)u(1,0)(r,θ)   (2,0)        { (1,0)            (  π)             } }                          {                  π}]
DSolve          r3        + u   (r,θ) = 0, u  (1,θ) = f(θ),u r,2 = 0,u(r,0) = 0   ,u(r,θ),{r,θ},Assumptions →  0 ≤ r ≤ 1∧ 0 ≤ θ ≤ 2

Result Did not solve

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

3.2 Laplace PDE inside semi-circle

problem number 60

Solve Laplace equation

∂2u-  1∂u-  -1∂2u-
∂r2 + r∂r + r2∂ θ2 = 0

Inside semi-circle of radius 1 with 0 ≤ θ ≤ π  and 0 ≤ r ≤ 1  , with following boundary conditions

pict

Mathematica
      [{  (0,2)     (1,0)                                                             }                                               ]
DSolve   u---(r,θ)u---(r,θ)+ u(2,0)(r,θ) = 0,{u(r,0) = 0,u(r,π) = 0,u(0,θ) = 0,u(1,θ) = f (θ)} ,u(r,θ),{r,θ},Assumptions → {0 ≤ r ≤ 1 &
                r3

Result Did not solve

Maple
        ∞∑  ( ∫ π              n       )
u(r,θ) =     2-0-sin(nθ)f (θ)dθr-sin(nθ)
        n=1               π

Result Solved

_________________________________________________________________________________

3.3 Laplace PDE inside rectangle

problem number 61

This is problem 2.5.1 part (a) from Richard Haberman applied partial differential equations, 5th edition

Solve Laplace equation

∂2u   ∂2u
∂x2-+ ∂y2-= 0

inside a rectangle 0 ≤ x ≤ L, 0 ≤ y ≤ H  , with following boundary conditions

pict

Mathematica
( (                (    )    (   )(∫ L   (   )      )    (   )             ) )
{ {         ∑∞ 2 cos  nLπx- csch  HnLπ-  0 cos nπLx f(x)dx  sinh nπLy    y∫0Lf(x)dx} }
( (u (x,y) →   -----------------------L----------------------- + ---HL-----) )
            n=1

Result Solved

Maple
            (     (                                           (    (     ) )  )                )
        -1--   ∞∑          ( nπx) ∫ L       ( nπx)       (πyn-)       nπH--  −1      ∫ L
u(x,y) = HL  4     1∕21 cos   L    0 f (x )cos   L   dxsinh  L     sinh   L         H +  0 f (x)dxy
               n=1

Result Solved

_________________________________________________________________________________

3.4 Laplace PDE inside rectangle

problem number 62

This is problem 2.5.1 part (b) from Richard Haberman applied partial differential equations, 5th edition

Solve Laplace equation

 2     2
∂-u2-+ ∂-u2-= 0
∂x    ∂y

inside a rectangle 0 ≤ x ≤ L, 0 ≤ y ≤ H  , with following boundary conditions

pict

Mathematica
((                    (       )          (                )         ))
{{          ∞∑    2cosh  nπ(L−x)-csch(Lnπ)  ∫H g(y)sin(nπy)dy  sin (nπy)}}
   u(x,y) →    − ---------H----------H-----0---------H----------H---
((          n=1                          nπ                         ))

Result Solved

Maple
           (               ∫                   (    (           )      (           )                         )(     (     )       (     )    )  )
        ∑∞      1--  (n-πy)  H    (n-πy)              nπ(2L-−-x)         nπ(2L-−-x)       ( nπx)       (nπx-)         nπL-          nπL-      −1
u(x,y) =     − 2 nπ sin H   0  sin  H   g (y)dy  cosh      H       +sinh      H       +cosh   H   + sinh   H      cosh  2 H    + sinh  2 H    − 1
        n=1

Result Solved

_________________________________________________________________________________

3.5 Laplace PDE inside rectangle

problem number 63

This is problem 2.5.1 part (c) from Richard Haberman applied partial differential equations, 5th edition

Solve Laplace equation

∂2u   ∂2u
∂x2-+ ∂y2-= 0

inside a rectangle 0 ≤ x ≤ L, 0 ≤ y ≤ H  , with following boundary conditions

pict

Mathematica
      [{                        {                                              }}                                                ]
DSolve  u(0,2)(x,y)+ u(2,0)(x,y) = 0, u(1,0)(0,y) = 0,u (L,y) = g(y),u(x,0) = 0,u(x,H) = 0 ,u(x,y),{x,y},Assumptions → {0 ≤ x ≤ L ∧0 ≤ y ≤ H }

Result Did not solve

Maple
           (                                 (    (    )       (    ) )           (    (      )      (     )    )   )
        ∑∞    1-   (nπy-)∫ H    (nπy-)              nπL-         nπL-      ( nπx)         nπL-          nπL-      −1
u(x,y) =     4H sin  H    0  sin  H   g (y)dy  cosh   H   + sinh   H      cosh   H    cosh  2 H   + sinh 2  H   + 1
        n=1

Result Solved

_________________________________________________________________________________

3.6 Laplace PDE inside rectangle

problem number 64

This is problem 2.5.1 part (d) from Richard Haberman applied partial differential equations, 5th edition

Solve Laplace equation

 2     2
∂-u2-+ ∂-u2-= 0
∂x    ∂y

inside a rectangle 0 ≤ x ≤ L, 0 ≤ y ≤ H  , with following boundary conditions

pict

Mathematica
      [{                        {                                           }}                                                 ]
DSolve  u(0,2)(x,y)+ u(2,0)(x,y) = 0, u(0,y) = 0,u(L,y) = 0,u(0,1)(x,0) = 0,u(x,H ) = 0 ,u(x,y),{x,y},Assumptions → {0 ≤ x ≤ L∧ 0 ≤ y ≤ H }

Result Did not solve

Maple
           (                                                                                                                                                                                                                )
        ∑∞      1   (    π(2yn + H + y))∫ H    (   π (2yn+ H + y))       (      (    π(2L − x)(1+ 2n))      (    π(2L − x)(1+ 2n))       (   π(1 +2n )x)      (    π�
u(x,y) =     − 2 H-sin 1∕2------H-------     sin 1∕2------H-------  g(y)dy  − cosh 1∕2-------H--------  − sinh  1∕2-------H--------  + cosh  1∕2----H-----  + sinh  1∕2----H-----    cosh  ----H-----  + sinh  ----H-----  − 1
        n=0                              0

Result Solved

_________________________________________________________________________________

3.7 Laplace PDE inside rectangle

problem number 65

This is problem 2.5.1 part (e) from Richard Haberman applied partial differential equations, 5th edition

Solve Laplace equation

∂2u   ∂2u
--2-+ --2-= 0
∂x    ∂y

inside a rectangle 0 ≤ x ≤ L, 0 ≤ y ≤ H  , with following boundary conditions

pict

Mathematica
      [{ (0,2)       (2,0)        {                             (0,1)                     }}                                                ]
DSolve  u   (x,y)+ u    (x,y) = 0, u(0,y) = 0,u(L,y) = 0,u(x,0)− u (x,0) = 0,u(x,H ) = f(x) ,u(x,y),{x,y},Assumptions → {0 ≤ x ≤ L ∧0 ≤ y ≤ H}

Result Did not solve

Maple
           (                                                                                                                                                                       )
        ∑∞    1 (    ( nπH )       (nπH ) ) ∫ L   (nπx )          (nπx )(      (πyn )         (πyn ))(      ( n πH )         (  nπH )          ( nπH  )        ( n πH )        &
u(x,y) =     4L- cosh  -L--- + sinh  --L--      sin -L-- f (x)dxsin  -L--  π cosh  -L-- n + Lsinh  -L--    πsinh  2--L-- n + πcosh  2-L--- n+ L sinh  2--L-- + L cosh  2--L--  +n π− L
        n=1                                  0

Result Solved

_________________________________________________________________________________

3.8 Laplace PDE inside rectangle, top/bottom edges non-zero

problem number 66

Taken from Mathematica DSolve help pages.

Solve Laplace equation

 2     2
∂-u-+ ∂-u-= 0
∂x2   ∂y2

inside a rectangle 0 ≤ x ≤ 1,0 ≤ y ≤ 2  , with following boundary conditions

pict

Mathematica
{ {         ∞∑  8csch(2nπ)sin (nπ)sin(nπx )(sinh (nπ (2 − y)) +sinh(nπy))}}
   u(x,y) →    --------------2--------2-2-----------------------
            n=1                      n π

Result Solved

Maple
                               (                                )
         ∞∑   sin (1∕2n π)sin(nπx )eπn(3y− 2) − eπn(3y− 4) + eπyn − eπn(y−2) e−2πn(y− 2)
u (x,y) =   8------------------------π2n2-(e4nπ −-1)-----------------------
         n=1

Result Solved

_________________________________________________________________________________

3.9 Laplace PDE inside circular annulus, Neumann boundary conditions using unspecified functions

problem number 67

This is problem 2.5.8 part (b) from Richard Haberman applied partial differential equations, 5th edition

Solve Laplace equation

 2             2
∂-u-+ 1∂u-+ -1∂-u-= 0
∂r2   r∂r   r2∂ θ2

Inside circular annulus a < r < b  subject to the following boundary conditions

pict

Mathematica
      [{  (0,2)        (1,0)                    {                        }}                                   ]
DSolve   u---(r,θ)-+ u---(r,θ)-+ u(2,0)(r,θ) = 0, u(1,0)(a,θ) = 0,u(b,θ) = g(θ) ,u(r,θ),{r,θ},Assumptions → a < r ≤ b
            r2          r

Result Did not solve

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

3.10 Laplace PDE inside circular annulus, Dirichlet boundary conditions using specified functions

problem number 68

Solve Laplace equation

∂2u   1∂u    1∂2u
∂r2-+ r∂r-+ r2∂-θ2-= 0

Inside circular annulus 1 < r < 2  subject to the following boundary conditions

pict

Mathematica
{ {              2(r2−1)sin(θ)             }}
   u(r,θ) → {    ----3r----   1 ≤ r ≤ 2
                Indeterminate    True

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

3.11 Laplace PDE example 18 from Maple help page

problem number 69

Solve Laplace equation

∂2u   ∂2u
∂x2-+ ∂y2-= 0

With boundary conditions

pict

Mathematica
{ {                                             }}
   u(x,y) → (sinh(x)−-cosh(x))(x-cos(y)−-y-sin(y))-+x-
                           x2 + y2

Result Solved

Maple
        sin(− y +ix)+-F2 (y − ix )(y − ix) + (− y+ ix)-F2(y+ ix)
u(x,y) =----------------------−-y+-ix----------------------

Result Solved

_________________________________________________________________________________

3.12 Laplace PDE on rectangle with one edge at infinity

problem number 70

Solve Laplace equation

∂2u-  ∂2u-
∂x2 + ∂y2 = 0

With boundary conditions

pict

Mathematica
DSolve[{u(0,2)(x,y)+ u(2,0)(x,y) = 0,{u(x,0) = 0,u(x,a) = 0,u(0,y) = sin(y),u (∞, y) = 0}} ,u(x,y),{x,y},Assumptions → a > 0]

Result Did not solve

Maple
         ∞    { a(−(−1)nsin(a)+nπ+a)                  (    )
u(x,y) = ∑ 2 1-      2nπ+2a       a = nπ   e− nπaxsin πyn
        n=1  a  − π(−1π)n2n2si−na(a2)na     otherwise           a

Result Solved

_________________________________________________________________________________

3.13 Laplace PDE inside a disk, periodic boundary conditions

problem number 71

Solve Laplace equation in polar coordinates inside a disk

Solve for u(r,θ)

pict

Boundary conditions

pict

Mathematica
                         (               )      (                )
({({         ∑∞   ( cos(nθ) ∫π cos(nθ)f (θ)dθ a− n    ∫π f(θ)sin(nθ)dθ sin (nθ)a−n )   ∫π       )} )}
   u(r,θ) →   rn( ---------−π-----------------+ --−-π---------------------- ) + -−πf(θ)dθ
((         n=1                 π                             π                     2π    ) )

Result Solved

Maple
             (     (∫ π                      ∫ π                           )              )
           1-  ∑∞   -−π-f (θ)sin-(nθ)dθ-sin(nθ)+-−π-f (θ)cos(nθ)dθcos(nθ)(a-)−n    ∫ π
u(r,θ) = 1∕2π 2                             π                         r      π+  −π f (θ)dθ
               n=1

Result Solved

_________________________________________________________________________________

3.14 Dirichlet problem for the Laplace equation in upper half plan

problem number 72

Taken from Mathematica DSolve help pages

Solve for u(x,y)

pict

Boundary conditions u(x,0) = 1  for − 12 ≤ x ≤ 12  and x = 0  otherwise. This is called UnitBox in Mathematica.

Mathematica
(|(|                      −1 (12−x)     − 1(x+12-)            1        1          )| )|
|{|{                {  tan     y   + tan     y     y > 0 ∨x > 2 ∨ x < − 2        |} |}
|| u(x,y) →  {   ----------------0----------------------True----------  y ≥ 0 | |
|(|(                                   Indeterπminate                       True  |) |)

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

3.15 Dirichlet problem for the Laplace equation in right half-plane:

problem number 73

Taken from Mathematica DSolve help pages

Solve for u(x,y)

pict

Boundary conditions u(0,y) = sinc(y)  .

Mathematica
{{              x+-(xcos(y)−y-sin(y))(sinh(x)−-cosh(x))        }}
   u(x,y) → {               x2+y2             x ≥ 0
                        Indeterminate          True

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

3.16 Dirichlet problem for the Laplace equation in the first quadrant

problem number 74

Taken from Mathematica DSolve help pages

Solve for u(x,y)

pict

Boundary conditions u(0,y) = sinc(y)  .

Mathematica
((               (                                                                                                                                                                                                                   (  (   )  (        )             )    ()  ( ( -       -        -   ( -  )  ( -       -  )   ) (( -    (
|{|{               2( 3(y(3π(x+1)(x4−4x3+2(y2+12)x2−4(y2+10)x+y4−16y2+100)+x((6tan−1(3)−log(10))x4+2((6tan−1(3)−log(10))y2+10(6tan−1(3)+log&
   u(x,y) →  {                                                                                     x2−2x+y2−6y+10 x2+2x+y2−6y+10 x2−2x+y2+6y+10 x2+2x+y2+6y+10                                                                                                                                                                                x4− 8x3+2y2+15x2−8y2+7x
|(|(                                                                                                                                                                                                                                                   3π                                                                                                                                                                                                                                      x
                                                                                                                                                                                                                                                Indeterminate                                                                                                                                                                                                                                     Tru

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

3.17 Neumann problem for the Laplace equation in the upper half-plane

problem number 75

Taken from Mathematica DSolve help pages

Solve for u(x,y)

pict

Boundary conditions u(x,0) = UnitBox[x]
y  where UnitBox[x]  is 1  for − 1≤ x ≤ 1
  2      2  and 0  otherwise. This is called UnitBox in Mathematica.

Mathematica
((                      (   )       (    )                                                                                 ) )
{{               −4ytan−1 x−y 12 +4ytan−1 x+y12 −2xlog(4x2−4x+4y2+1)+log(4x2−4x+4y2+1)+2xlog(4x2+4x+4y2+1)+log(4x2+4x+4y2+1)−2log(4)− 4        } }
((u (x,y) →  {   ------------------------------------------------4π------------------------------------------------- y ≥ 0  ) )
                                                            Indeterminate                                              True

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

3.18 Dirichlet problem for the Laplace equation in a rectangle

problem number 76

Taken from Mathematica DSolve help pages

Solve for u(x,y)

pict

Boundary conditions u(x,0) = x2(1− x),u(x,2) = 0,u (0,y) = 0,u(1,y) = 0  .

Mathematica
{{                                                        } }
           ∑∞    4(1+-2(− 1)n)csch(2nπ)sin(n-πx)sinh(nπ(2−-y))
   u(x,y) →    −                    n3π3
           n=1

Result Solved

Maple
               (            )
         ∞∑      (− 1)1+n − 1∕2 (− e− πn(y−4) + eπyn)sin(nπx )
u (x,y) =    − 8----------------------------------------
         n=1                n3π3(e4nπ − 1)

Result Solved

_________________________________________________________________________________

3.19 Laplace PDE outside a disk, periodic boundary conditions

problem number 77

Solve Laplace equation in polar coordinates outside a disk

Solve for u(r,θ)

pict

Boundary conditions

pict

Mathematica
      [{ u(0,2)(r,θ)-  u(1,0)(r,θ)-   (2,0)         {                             (0,1)         (0,1)    }}                                       ]
DSolve      r2    +     r    + u   (r,θ) = 0, u(a,θ) = f(θ),u(r,− π) = u(r,π),u  (r,− π) = u   (r,π)   ,u(r,θ),{r,θ},Assumptions → {a > 0,r > a}

Result Did not solve

Maple
             (     (∫ π                      ∫ π                          )    ∫         )
           1-  ∑∞   -−π-f (θ)sin-(nθ)dθ-sin(nθ)+-−π-f (θ)cos(nθ)dθcos(nθ)(a-)n      π
u(r,θ) = 1∕2π 2                             π                         r     π+  −π f (θ)dθ
               n=1

Result Solved

_________________________________________________________________________________

3.20 Laplace equation in spherical coordinates

problem number 78

Taken from Maple pdsolve help pages

Solve for u(r,θ,ϕ)

pict

Mathematica
      ⌊                                              (                                                 )                                          ⌋
        f(0,2,0)(r,θ,ϕ)-+f (1,0,0)(r,θ,ϕ)                 csc(θ) sin(θ)f(1,0,0)(r,θ,ϕ)+ cos(θ)f(0,r1,0)(r,θ,ϕ)+ csc(θ)f(0,0,2)(r,θ,ϕ
DSolve⌈-----r------------------ + f(2,0,0)(r,θ,ϕ )+ -------------------------------------------------------- = 0,f (r,θ,ϕ),{r,θ,ϕ},Assumptions → 0 ≤ θ ≤ π ⌉
                   r                                                       r

Result Did not solve

Maple
          √ -      √ ---      √---(      (√ --- )        (√ ---))(       √-----          √-----)(            √ ---    √ -------        √ ---     √-------                                        √ ---    √ -------        √ ---    √ -------                         )
            2(− 1)1∕2 c2 (sin (θ)) c2 C5 sin   -c2ϕ  + C6 cos   c2ϕ   -C1r1∕2 1+4 c1 +-C2r−1∕2 1+4 c1 cos(θ)2F1(1∕2 c2 + 1∕4  1+ 4-c1 + 3∕4,1∕2 -c2 − 1∕4 1 +4 c1 + 3∕4;3∕2;1∕2 cos (2&#x
F (r,θ,ϕ) =-----------------------------------------------------------------------------------------------------------------------------√r-----------------------------------------------------------------------------------------------------------------------------

Result Solved, but not verified

4 Poisson PDE

_________________________________________________________________________________

4.1 Dirichlet problem for the Poisson equation in a rectangle

problem number 79

Taken from Mathematica DSolve help pages.

Solve for u(x,y)

pict

Boundary conditions

pict

Mathematica
{{u(x,y) → x3 + 11x − y3 + 1}}

Result Solved

Maple
         3   3
u(x,y) = x − y + 11x+ 1

Result Solved

5 Helmholtz PDE

_________________________________________________________________________________

5.1 Dirichlet problem for the Helmholtz equation in a rectangle

problem number 80

Taken from Mathematica DSolve help pages.

Solve for u(x,y)

pict

Boundary conditions

pict

Mathematica
{{          1∞∑  128(cos(nπ)+ cos(3nπ))csch (1√n2π2-−-80)sin3(nπ-)sin(n-πx-)sinh(1-√n2π2-− 80(2 − y))} }
   u(x,y) → -   --------8---------8--------2----------2-2---8-------4-------4-----------------
            2n=1                                    n π

Result Solved

Maple
        ∑∞  (8 sin(3∕4nπ)+ 8sin(1∕4nπ)− 16sin(1∕2nπ))sin(1∕4nπx)sin(1∕4√−-π2n2 +-80(y − 2))
u(x,y) =    ----------------------------(---√----------)-------------------------------
        n=1                          sin 1∕2 − π2n2 + 80 π2n2

Result Solved

6 Wave PDE

_________________________________________________________________________________

6.1 General solution for a second-order hyperbolic PDE on real line

problem number 81

From Mathematica DSolve help pages (slightly modified)

Solve for u(x,t)  with t > 0  on real line

 2     2       2
∂-u+ -∂-u-= c2∂-u-
∂t2  ∂t∂x     ∂x2

Mathematica
{{           (    (√4c2 +-1 − 1)x)    (    (− √4c2-+-1− 1)x) }}
   u(x,t) → c1  t− -------2------  + c2 t − --------2-------
                        2c                        2c

Result Solved

Maple
           (    2tc2 + x√4c2 +-1 +x )     (    2tc2 − x√4c2-+-1+ x)
u(x,t) =-F1  1∕2---------2--------  + F 2  1∕2 --------2---------
                        c                             c

Result Solved

_________________________________________________________________________________

6.2 Hyperbolic PDE with non-rational coefficients

problem number 82

From Mathematica DSolve help pages

Solve for u(x,y)

 2           2          2
∂-u2-− 2sin x-∂-u-− cos2x∂-u2-− cos x∂u-= 0
∂x         ∂x∂y        ∂y        ∂y

Mathematica
{{u(x,y) → c(x − cos(x)+ y)+ c (− x− cos(x)+ y)}}
            1                2

Result Solved

Maple
sol = ()

Result Did not solve. Tried all HINTS

_________________________________________________________________________________

6.3 Inhomogeneous hyperbolic PDE with constant coefficients

problem number 83

From Mathematica DSolve help pages

Solve for u(x,t)

3∂2u-− ∂2u-+ ∂2u--= 1
 ∂x2   ∂t2   ∂x∂t

Mathematica
{{           (    1(    √--)  )    (    1(    √--)  )   x2}}
   u(x,t) → c1 t−  6 1 +  13 x  + c2  t− 6  1−  13  x  + 6-

Result Solved

Maple
           (    (    √ --)     )     (   (    √--    )        √ -)      √ --(   (     √ -)     )(    (    √--   )         √--)
u(x,t) =-F2 1∕6  − 1 + 13 x + t +-F1  1∕2 1∕13 13 + 1 x − 3∕13t  13 +1∕13  13  1∕6 − 1+   13  x+ t   1∕2  1∕13 13 + 1 x− 3∕13t 13

Result Solved

_________________________________________________________________________________

6.4 system of 2 inhomogeneous linear hyperbolic system with constant coefficients

problem number 84

From Mathematica DSolve help pages

Solve for u(x,t),v(x,t)

pict

With initial conditions

pict

Mathematica
{{                                                                                    }}
                        1                    1
   u(x,t) → sinh(t)cos(x)+ 2 cosh(2t)cos(2x)+ t+  2,v(x,t) → cosh(t)sin(x)(2 sinh(t) cos(x)+ 1)− t

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

6.5 Wave PDE on string (finite domain) with zero initial position and velocity, and with source term

problem number 85

This is problem at page 115, David J Logan textbook, applied PDE textbook.

Falling cable lying on a table that is suddenly removed.

∂2u     ∂2u
--2-= c2--2-− g
 ∂t     ∂x

With boundary condition

pict

And initial conditions

pict

Mathematica
{{           (                     )            } }
           1   (   x)2  (   x-)   2      (    x)
   u(x,t) → 2g   t− c   θ t−  c − t   − c1δ t− c

Result Solved

Maple
u(x,t) = 1∕2-g(Heaviside (t− x) (tc− x)2 − c2t2)
           c2               c

Result Solved

_________________________________________________________________________________

6.6 Wave PDE on string, one end fixed, another free, both initial conduitions non zero, and source that depends on time and space

problem number 86

Added July 2, 2018. Taken from Maple 2018.1 improvement to PDE document.

Solve

   2             2       (               )
− ∂-u2-+ u(x,t) = ∂-u2 + 2e−t x− 1x2 + 1t− 1
  ∂t            ∂x            2     2

With boundary condition

pict

And initial conditions

pict

Mathematica
      [{                                                   {                                                 }                          }          ]
                 (0,2)       (2,0)        −t( t  x2       )            2               (  1)    x2− x   3x2− 3x   {           (1,0)       }
DSolve   u(x,t)− u   (x,t) = u  (x,t)+ 2e   2 − -2 + x− 1  , u(x,0) = x − 2x,u(x,1) = u x,2 + -2-e-- − -4√e--2- , u(0,t) = 0,u   (1,t) = 0  ,u(x,t),x,t

Result Did not solve

Maple
             −t
u(x,t) = − 1∕2e x(− 2 + x)(t− 2)

Result Solved

_________________________________________________________________________________

6.7 Wave PDE on string (finite domain), fixed ends, no initial conduitions give and no source

problem number 87

This is problem at page 28, David J Logan textbook, applied PDE textbook.

∂2u     ∂2u
∂t2-= c2∂x2-

With boundary condition

pict

Mathematica
      [{                                          }                                 ]
DSolve  u(0,2)(x,t) = c2u(2,0)(x,t),{u(0,t) = 0,u(L,t) = 0} ,u(x,t),{x,t},Assumptions → {L > 0}

Result Did not solve

Maple
         ∞∑    ( nπx) (         ( cnπt)            ( cnπt ))
u (x,t) =    sin  ----  -C1 (n) sin  ----  + C5 (n)cos  ----
        n=1      L                L                  L

Result Solved

_________________________________________________________________________________

6.8 Wave PDE on string (finite domain), one fixed end, one free end, initial position not zero, initial velocity zero, no source

problem number 88

This is problem at page 130, David J Logan textbook, applied PDE textbook.

∂2u    2∂2u
∂t2-= c ∂x2-

With boundary conditions

pict

With initial conditions

pict

Mathematica
      [{ (0,2)       2 (2,0)     {           (1,0)        } {  (0,1)                    } }                                    ]
DSolve  u   (x,t) = c u  (x,t), u(0,t) = 0,u  (L,t) = 0 , u   (x,0) = 0,u(x,0) = f(x ) ,u(x,t),{x,t},Assumptions → {0 ≤ x ≤ L}

Result Did not solve

Maple
           ( (                                                                  )
        ∑∞   {0                                                            L ≤ 0
u(x,t) =   ( (  1-  (   (1+2n)πx)   (   πc(1+2n)t)∫L    (   (1+2n)πx)              )
        n=0   2 L sin 1∕2   L    cos 1∕2   L      0 sin  1∕2   L    f (x)dx  0 < L

Result Solved

_________________________________________________________________________________

6.9 Wave PDE on string (finite domain), both ends fixed end, initial conditions zero, with source as generic function that depends on time and space

problem number 89

This is problem at page 149, David J Logan textbook, applied PDE textbook.

∂2u-   2∂2u-
∂t2 = c ∂x2 + p(x,t)

With boundary conditions

pict

With initial conditions

pict

Mathematica
      [{                                                  {                       }}            ]
DSolve  u(0,2)(x,t) = c2u(2,0)(x,t) + p(x,t),{u(0,t) = 0,u(π,t) = 0}, u(x,0) = 0,u(0,1)(x,0) = 0 ,u(x,t),{x,t}

Result Did not solve

Maple
               ( ∫                                     )
        ∫ t ∞∑    -0πsin(nx)p(x,τ)dxsin(nx)sin-(c(t-− τ)-n)
u (x,t) = 0      2                 nπc                   dτ
            n=1

Result Solved

_________________________________________________________________________________

6.10 Wave PDE on string (finite domain), both ends fixed, initial conditions both not zero, No source

problem number 90

Added July 2, 2018.

Taken from Maple 2018.1 improvements to PDE’s document.

Solve

∂2v   ∂2v
∂t2 = ∂x2-

For t > 0  and 0 < x < 1  . With boundary conditions

pict

With initial conditions

pict

Where          2  x+1    1− x
f(x) = − e-x−-ee2−−x1+e--  and            2  x+1   1−x
g(x) = 1 + ex−e-e2−−1x+e--

Mathematica
{{         ∑∞ ( 2(− 1)ncos(nπt)  (− 2(− 1+ (− 1)n)π2n2 − 4(− 1)n + 2)sin(nπt))     }}
   v(x,t) →      ---3-3--------+ ---------------4-4----22---------------  sin(nπx)
           n=1    π n  +πn                    π n + π n

Result Solved

Maple
        ∑∞    sin (nπx)((π2(− 1)nn2 − π2n2 + 2(− 1)n − 1) sin(πtn)− (− 1)ncos(πtn)πn)
v (x,t) =   − 2---------------------------2-2--2-2------------------------------
        n=1                             π n (π n + 1)

Result Solved

_________________________________________________________________________________

6.11 Wave PDE on string (finite domain), both ends fixed end, initial conditions both not zero, and with constant source

problem number 91

Added July 2, 2018.

Third example, from Maple 2018.1 improvements to PDE’s document.

Solve

∂2u-   2∂2u-
 ∂t2 = c ∂x2 + 1

For t > 0  and 0 < x < L  . With boundary conditions

pict

With initial conditions

pict

Mathematica
DSolve[{u(0,2)(x,t) = c2u(2,0)(x,t) + 1,{u(0,t) = 0,u(L, t) = 0},{u (x,0) = f(x ),u(0,1)(x,0) = g(x)}} ,u(x,t),{x,t},Assumptions → L > 0]

Result Did not solve

Maple
             (  ∑∞ (         (    ) (   ∫ L   (    )          (     )     ∫ L   (    ) (                 )     (     )  ))            )
u(x,t) = 1∕2 12 2    --12--sin  nπx-  2L     sin  nπx- g(x)dxsin  n-πct c− π     sin  nπx- − 2f (x)c2 + Lx − x2 dx cos nπct n   c2 + Lx − x2
           c    n=1  nπc L      L        0       L               L         0       L                              L

Result Solved

_________________________________________________________________________________

6.12 Wave PDE on string (finite domain), both ends fixed end, with source

problem number 92

This is problem at page 213, David J Logan textbook, applied PDE textbook.

 2      2
∂-u2 = c2∂-u2 + Ax
∂t     ∂x

With boundary conditions

pict

With initial conditions

pict

Mathematica
DSolve[{u(0,2)(x,t) = c2u(2,0)(x,t) + Ax,{u(0,t) = 0,u(L,t) = 0},{u(x,0) = 0,u(0,1)(x,0) = 0}} ,u(x,t),{x,t}]

Result Did not solve

Maple
             (                  (     (   )        (                                      (     (   ) )            )            )
                             ∞∑  { 2csgn3-L3−21-sin(nπx-) 1∕2π3sin (nπct)csgn(L −1)-C1(n)n3c2 + cos cπcsgn-L−1-tn- (− 1)1+n L3A    L = 0
u(x,t) = 1∕6 12(AL2x − Ax3 + 6      n π c    ( L )   (    )   L                                 L                             c2)
           c                 n=1( 2(−n1)3nπL3c32Asin nπLx cos nπLct                                                            otherwise

Result Solved

_________________________________________________________________________________

6.13 Wave PDE on semi-infinite domain, with one end having a moving boundary condition

problem number 93

Solve for u(x,t)  with t > 0  and x > 0

∂2u     ∂2u
∂t2-= c2∂x2-

With boundary conditions

pict

With initial conditions

pict

Mathematica
( (                                  ) )
{ {                 ( 0  )    x > ct } }
( (u (x,t) →  {     g t− xc     x ≤ ct ) )
                 Indeterminate   True

Result Solved

Maple
                 (   x)  ( tc − x)
u(x,t) = Heaviside  t− c- g  --c--

Result Solved

_________________________________________________________________________________

6.14 Telegraphy PDE, a wave PDE on string, both ends fixed with damping

problem number 94

Solve

∂2u    ∂u    2∂2u
∂t2-+ 2∂t-= c ∂x2-

With boundary conditions

pict

With initial conditions

pict

Mathematica
      [{                                                     {                         }}          ]
DSolve  2u(0,1)(x,t)+ u(0,2)(x,t) = u(2,0)(x,t),{u(0,t) = 0,u (π,t) = 0}, u(0,1)(x,0) = 0,u(x,0) = f(x) ,u(x,t),x,t

Result Did not solve

Maple
           (                 (                  (√-----  )    (   √-----)              )       )
         ∞   ∫π sin (nx) f (x)dx (− 1+ √ − n2-+-1)e− −n2+1+1 t + e −1+ −n2+1 t(√−-n2 +-1+ 1) sin(nx)
        ∑  || -0--------------------------------------------------------------------------------||
u(x,t) =   (                                     √ − n2 + 1π                                   )
        n=1

Result Solved, But n = 1  should not be included.

_________________________________________________________________________________

6.15 Wave PDE, on string, both ends fixed. Initial velocity zero. Dispersion term present

problem number 95

Solve

 1∂2u    2         2∂2u
a2-∂t2-+ γ u(x,t) = c ∂x2-

Dispersion term γ2u(x,t)  causes the shape of the original wave to distort with time.

With 0 < x < π  and t > 0  and with boundary conditions

pict

With initial conditions

pict

Mathematica
      [{  (0,2)                                              {                            }}            ]
DSolve   u---(x,t)-+ γ2u(x,t) = u(2,0)(x,t),{u (0,t) = 0,u(π,t) = 0}, u(0,1)(x,0) = 0,u(x,0) = sin2(x) ,u(x,t),{x,t}
            a2

Result Did not solve due to adding dispersion term

Maple
             (                                (  ∘ -2---2-)     n            )
            1(     ( ∘ -2----)         ∑∞   cos--a--γ-+-n-t-((−-1)-−-1)sin-(nx-) )
u (x,t) = 1∕3 π 8cos a  γ + 1t sin(x)+ 3   4            nπ(n2 − 4)           π
                                       n=3

Result Solved

_________________________________________________________________________________

6.16 Wave PDE on string with fixed ends, non-zero initial position

problem number 96

Added March 9, 2018.

Solve

∂2u    ∂2u
-∂t2-= 4∂x2-

With boundary conditions

pict

With initial conditions

pict

Mathematica
{{                                      }}
           ∑∞ 4(cos(nπ) − 1)cos(2nt)sin(nx)
   u(x,t) →    --------(n3 −-4n)π--------
           n=1

Result Solved but sum should not include n = 2

Maple
            1(                 ∑∞   ((− 1)n − 1)sin(nx)cos(2tn) )
u (x,t) = 1∕3 -- 8sin (x )cos(2t)+ 3   4 -----------2------------π
            π                  n=3         nπ(n  − 4)

Result Solved

_________________________________________________________________________________

6.17 Wave PDE homogeneous in square, given initial position but with zero initial velocity

problem number 97

Taken from Maple PDE help pages. This wave PDE inside square with free to move on left edge and right edge, and top and bottom edges are fixed. It has zero initial velocity, but given a non-zero initial position. Where 0 < x < π  and 0 < y < π  and t > 0  .

Solve

 2     (  2     2 )
∂-u2 = 1  ∂-u2-+ ∂-u2-
∂t    4  ∂x    ∂y

With boundary conditions

pict

With initial conditions

pict

Mathematica
      [{                (                          ) {                                                      } {                                   }}                ]
DSolve  u(0,0,2)(x,y,t) = 1 u(0,2,0)(x,y,t)+ u(2,0,0)(x,y,t) , u(1,0,0)(0,y,t) = 0,u(1,0,0)(π,y,t) = 0,u(x,0,t) = 0,u(x,π,t) = 0 , u(0,0,1)(x,y,0) = 0,u(x,y,0�
                       4

Result Did not solve

Maple
          ∞         n                       ∞ (  ∞   (     n+m       m       n   )                  (  √ --2---2) )
u(x,y,t) = ∑  − 2 ((− 1)-−-1)sin(ny)cos(1∕2tn)+ ∑ ( ∑  8 -− (−-1)--+-(−-1)--+-(− 1)-−-1-cos-(mx-)sin-(ny-)cos-1∕2--m--+-n-t-)
          n=1              n3              n=1  m=1                             π2m2n3

Result Solved

_________________________________________________________________________________

6.18 Wave PDE homogeneous in square with damping. Given zero initial position but with non-zero initial velocity

problem number 98

Taken from Maple PDE help pages. This wave PDE inside square with damping present.

Membrane is free to move on the right edge and also on top edge. But fixed at left edge and bottom edge.

It has zero initial position, but given a non-zero initial velocity. Where 0 < x < 1  and 0 < y < 1  and t > 0  .

Solve

∂2u   1 (∂2u   ∂2u)    1 ∂u
∂t2-= 4  ∂x2-+ ∂y2- − 10 ∂t-

With boundary conditions

pict

With initial conditions

pict

Mathematica
      [{                                                                                                                                                                   }                ]
         (0,0,2)         1( (0,2,0)         (2,0,0)      )  -1  (0,0,1)       {            (1,0,0)                      (0,1,0)         }  {             (0,0,1)        (    x)  (   y-) }
DSolve  u     (x,y,t) = 4 u     (x,y,t)+ u     (x,y,t) − 10u     (x,y,t), u(0,y,t) = 0,u    (1,y,t) = 0,u(x,0,t) = 0,u    (x,1,t) = 0 , u(x,y,0) = 0,u   (x,y,0) = 1−  2 x  1− 2  y   ,u(x,y,t),{x

Result Did not solve

Maple
             (                                                   (    ∘ ---------------------------------------))
          ∑∞   ∑∞     sin (1∕2 (1 + 2n)πx)sin (1∕2(1+ 2m)πy )e−t∕20sin  1∕20t − 1+ (100m2 +100n2 + 100m + 100n + 50)π2
u(x,y,t) =    (    5120----------------∘----------2-------2-------------------2-6--------3-------3---------------)
          m=0  n=0                     − 1+ (100m  + 100n + 100m + 100n+ 50)π π (1+ 2m ) (1+ 2n)

Result Solved

_________________________________________________________________________________

6.19 Wave PDE inside rectangle. All 4 edges are fixed and given non-zero initial position with zero initial velocity

problem number 99

Taken from Mathematica helps pages on DSolve

Solve for u(x,y,t)  with 0 < x < 1  and 0 < y < 2  and t > 0  .

Solve

∂2u-  ∂2u-  ∂2u-
∂t2 = ∂x2 + ∂y2

With boundary conditions

pict

With initial conditions

pict

Mathematica
( (                                             ( ∘-------  )           (   ) ))
||{ ||{          ∑∞ ∑∞  32(− 1+ (− 1)m )(− 1+ (− 1)n)cos  m42+ n2πt  sin(nπx)sin mπ2y- ||}||}
   u(x,y,t) →       ----------------------------------------------------------
||( ||(          n=1m=1                          5m3n3π6                          ||)||)

Result Solved

Maple
                    (                                   )
          ∑∞ ( ∑∞    − 32 (− 1)n+m +32 (− 1)m + 32 (− 1)n − 32 sin(nπx)sin (1∕2m πy)cos(1∕2π√m2-+4n2t))
u(x,y,t) =    (    − ---------------------------------------------------------------------------)
          m=1  n=1                                    5n3π6m3

Result Solved

_________________________________________________________________________________

6.20 Wave PDE inside disk. fixed edge of disk, no θ  dependency, with initial position and velocity given

problem number 100

Taken from Mathematica helps pages on DSolve

Solve for u(r,t)  with 0 < r < 1  and t > 0  .

∂2u     ( ∂2u   1∂u)
∂t2-= c2  ∂r2 + r∂r-

With boundary conditions

pict

With initial conditions

pict

Mathematica
((                     (                                             (       ) )) )
{{         ∑∞ 2J0(rj0,n) 9√c2J1 (j0,n)cos(ctj0,n)+ 1F2(3;1, 5;− 1 (j0,n)2) sin √c2tj0,n } }
   u(r,t) →    --------------------√----------------2---2--4---------------------
((         n=1                   9  c2(J0 (j0,n)2 + J1(j0,n)2)j0,n                   ) )

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

6.21 Wave PDE inside disk. fixed edge of disk, with θ  dependency, zero initial velocity

problem number 101

Solve for u(r,θ,t)  with 0 < r < a  and t > 0  and − π < θ < π

        (                  )
∂2u-   2  ∂2u- 1 ∂u-  1-∂2u-
∂t2 = c   ∂r2 + r∂r + r2∂θ2

With boundary conditions

pict

With initial conditions

pict

Mathematica
      [{ (0,0,2)         2( u(0,2,0)(r,θ,t)   u(1,0,0)(r,θ,t)   (2,0,0)      ) {                 (0,0,1)         } {                                (0,1,0)           (0,1,0)     } }                                                            &
DSolve  u     (r,θ,t) = c  ----r2-----+  -----r-----+ u     (r,θ,t) , u(r,θ,0) = f(r,θ),u    (r,θ,0) = 0 , u(a,θ,t) = 0,u(r,− π,t) = u(r,π,t),u   (r,− π,t) = u   (r,π,t)

Result Did not solve

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

6.22 Wave PDE on infinite domain with initial conditions specified, no source

problem number 102

Taken from Mathematica DSolve help pages.

Solve initial value wave PDE on infinite domain

∂2u   ∂2u
∂t2-= ∂x2-

With initial conditions

pict

Mathematica
{{                                }}
           1 ( −(x− t)2   −(t+x)2)
   u(x,t) → 2  e      + e       + t

Result Solved

Maple
u(x,t) = 1∕2e− (−x+t)2 + t+ 1∕2e− (t+x)2

Result Solved

_________________________________________________________________________________

6.23 Wave PDE on infinite domain with initial conditions specified, with source term

problem number 103

Taken from Mathematica DSolve help pages.

Solve initial value wave PDE on infinite domain

∂2u   ∂2u
∂t2-= ∂x2-+ m

With initial conditions

pict

Mathematica
{{                                                                            } }
           1 (  − |x−6t|             − |t+x6|                              )   mt2-
   u(x,t) → 2  − e    cos(3(x − t))− e     cos(3(t+ x))− sin (t− x) +sin(t+ x) +  2

Result Solved

Maple
                           ((                          )                                                        )
u(x,t) = 1∕2e−1∕6|−x+t|−1∕6|t+x| mt2 + sin(t+ x)− sin(− x + t) e1∕6|t+x|+1∕6|−x+t| − e1∕6|t+x|cos(3t− 3x)− cos(3t+ 3x)e1∕6|−x+t|

Result Solved

_________________________________________________________________________________

6.24 Wave PDE initial value with a Dirichlet condition on the half-line

problem number 104

Taken from Mathematica DSolve help pages.

Solve for u(x,t)  initial value wave PDE on infinite domain with t > 0  and x > 0  .

∂2u     ∂2u
--2-= c2--2-
∂t      ∂x

With initial conditions

pict

And boundary conditions u(0,t) = 0

Mathematica
((                ((         (√ --   )          √--       )   (         (√--    )      √ --          ))                ))
||||||||              1     {  sin2   c2t− x   π < x −  c2t < 2π   +   {   sin2   c2t+ x   π <   c2t+ x < 2π       x > √c2t ≥ 0 ||||||||
||{||{              2 ((         ( 0     )        True        )   (         ( 0     )        True        ))                ||}||}
   u(x,t) → {            sin2 √c2t+ x   π < √c2t +x < 2π            sin2  √c2t− x   π < √c2t− x < 2π              √--
||||||||              12     {                                     −   {                                         0 ≤ x ≤ c2t  ||||||||
||(||(                             0              True                        0              True                          ||)||)
                                                      Indeterminate                                           True

Result Solved

Maple
        (| (|0                tc+ x ≤ π   (| 0                 tc − x ≤ π
        |||| {              2              {                2
        |||| |(1∕2 (sin(tc + x))   tc+ x < 2π + |( − 1∕2(sin (tc− x)) tc − x < 2π x < tc
        {  0                2π ≤ tc+ x     0                 2π ≤ tc− x
u(x,t) = || (|0               tc+ x ≤ π   (| 0               − tc + x ≤ π
        |||| {              2              {              2
        |||( |(1∕2 (sin(tc + x))   tc+ x < 2π + |( 1∕2(sin(tc− x)) − tc + x < 2π tc < x
           0                2π ≤ tc+ x     0               2π ≤ − tc+ x

Result Solved

_________________________________________________________________________________

6.25 Wave PDE Initial value problem with a Neumann condition on the half-line

problem number 105

Taken from Mathematica DSolve help pages.

Solve initial value wave PDE on infinite domain

∂2u-   2∂2u-
∂t2 = c ∂x2

With initial conditions

pict

And boundary conditions ∂u(0,t) = 1
∂x

Mathematica
(|  (    (√--    )      (√ --    ))  2√c2t− 20e−x∕10sinh(∘c102t)                               √ --
{ 12 sin3(∘--c2t)+ x  −( sin∘3)  c2t− x  + ---------2√c2--------                           x >   c2t ≥ 0
|( 10e 110−-c2t−x-+10e√ 110-x−-c2t+2√c2t−-20  √-2(    √x-)  1 (  3(√ -2    )     3(√-2    ))         √ -2
               2 c2              −  c  t −  c2 + 2  sin    c t− x + sin    c t+ x    0 ≤ x ≤  c t

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

6.26 non-linear wave PDE (Solitons)

problem number 106

This was first solved analytically by (Krvskal, Zabrsky 1965).

Solve

                 3
∂u-+ 6u(x,t)∂u-+ ∂-u3-= 0
∂t         ∂x   ∂x

Mathematica
{ {          12c3tanh2 (c t+ c x+ c )− 8c3+ c }}
   u(x,t) → − ---1------2----1----3-----1---2
                           6c1

Result Solved. build a special solution.

Maple
              2                       2     8-C23 −-C3-
u(x,t) = − 2 C2 (tanh ( C2x +-C3t+-C1)) + 1∕6    C2

Result Solved. Returning a solution that is not the most general one

7 Schrodinger PDE

_________________________________________________________________________________

7.1 Schrodinger PDE with zero potential

problem number 107

From page 30, David J Logan textbook, applied PDE textbook.

Solve

  ∂f     h2 ∂2f
Ih---= − -----2-
  ∂t     2m ∂x

With boundary conditions

pict

Mathematica
{ {         ∞∑    ihn2π2t     (n πx)} }
    f(x,t) →    e− 2L2m cnsin ----
            n=1               L

Result Solved

Maple
        ∞∑           (nπx ) −-i∕2hπ2n2t
f (x,t) =   -C1(n)sin ---- e   mL2
        n=1           L

Result Solved

_________________________________________________________________________________

7.2 Schrodinger PDE with initial and boundary conditions

problem number 108

Solve for f(x,y,t)

 ∂f     ℏ2 (∂2f   ∂2f )
I∂t-= − 2m- ∂x2-+ ∂y2-

With boundary conditions

pict

And initial conditions           √-
f(x,y,0) = 2 (sin(2πx )sin(πy)+ sin(πx)sin(2πy ))

Mathematica
{{          √ - − 5iπ2h2mBar2t                              }}
  f(x,y,t) →  2e        (sin(2πx)sin(πy) + sin(πx)sin(2πy))

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

7.3 Initial value problem with Dirichlet boundary conditions

problem number 109

Taken from Mathematica DSolve help pages

Solve for f(x,t)

 ∂f      ∂2f
I-∂t = − 2∂x2

With boundary conditions

pict

And initial conditions f(x,2) = f(x)  where f(x) = − 350+ 155x− 22x2 +x3

Mathematica
{{         ∑∞             n  −225in2π2(t−2)   (1        )} }
   f(x,t) →    100(7+-8(− 1)-)e----------sin--5nπ(x−-5)-
           n=1                  n3π3

Result Solved

Maple
        ∑∞               n            − 225iπ2n2(−2+t)
f (x,t) =    (800-+-700(− 1)-)sin-(13∕53nπx-)e-----------
        n=1                  n π

Result Solved

_________________________________________________________________________________

7.4 Solve a Schrodinger equation with potential over the whole real line

problem number 110

Taken from Mathematica DSolve help pages

Solve for f(x,t)

         2
I∂f-= − ∂-f-+ 2x2f(x,t)
 ∂t    ∂x2

With boundary conditions

pict

Mathematica
{{                                            } }
           ∑∞  − x√2−2i√2(n+ 1)t         (  √4- )
   f(x,t) →    e  2        2 cnHermiteH  n,  2x
           n=0

Result Solved

Maple
sol = ()

Result Did not solve. Maple does not support ∞ in boundary conditions

8 Beam PDE

_________________________________________________________________________________

8.1 Beam PDE with zero initial velocity

problem number 111

Added January 20, 2018.

Solve

∂2u   ∂4u
--2-+ --4-= 0
∂t    ∂x

With boundary conditions

pict

And initial conditions

pict

Mathematica
{{          4    2}}
  u (x,t) → x − 12t

Result Solved

Maple
u(x,t) = x4 − 12t2

Result Solved

9 Burger’s PDE

_________________________________________________________________________________

9.1 viscous fluid flow with no initial conditions

problem number 112

From Mathematica symbolic PDE document.

Solve for u(x,t)

                 2
∂u+ u(x,t)∂u-= μ∂-u-
∂t        ∂x    ∂x2

Mathematica
{{           2c2μ tanh (ct+ c x+ c )+ c }}
   u(x,t) → − -1-------2----1----3----2-
                        c1

Result Solved

Maple
u (x,t) = − 2μ C2 tanh(-C2x + C3t +-C1)−-C3-
                                       -C2

Result Solved

_________________________________________________________________________________

9.2 viscous fluid flow with initial conditions

problem number 113

From Mathematica symbolic PDE document.

Solve for u(x,t)

∂u        ∂u    ∂2u
--+ u(x,t)---= μ--2-
∂t        ∂x    ∂x

With initial conditions

        {
u (x,0) =   1    x < 0
           0    x ≥ 0

Mathematica
(|(|                              )|)|
|||{|||{                              |||}|||}
   u(x,t) → ---------1----------
||||||          e− t−42μx(erf(2x√μt)+1)    ||||||
|(|(          ---erf(-t−√x)+1-----+1 |)|)
                  2 μt

Result Solved

Maple
                    (        )   ∫  ∫                ∂
                        --x√--     t  ∞       u(ζ,τ)∂ζ√u-(ζ,τ) 1∕4 (μ(x−−ζt)+2τ)
u(x,t) = 1∕2− 1∕2Erf  1∕2√ μ- t +  0   −∞ − 1∕2 √ π√ μ- t− τ e        dζdτ

Result Solved, but has unresolved integrals

_________________________________________________________________________________

9.3 viscous fluid flow with initial conditions as UnitBox

problem number 114

From Mathematica DSolve help pages.

Solve for u(x,t)

∂u        ∂u    ∂2u
--+ u(x,t)---= μ--2-
∂t        ∂x    ∂x

With initial conditions

        {
u (x,0) =   1     |x| ≤ 12
           0    otherwise

Mathematica
((                                  (   (      )     (       ))                     ) )
{{                              et+4μ1 erf 2t−√2x−1- − erf 2t−2√x+1-                      } }
   u(x,t) → -x---(----)----t+1---(------)--4-μt+t1---(------4)μt-x+1--(-----)---x----x+1-
((          e2μerf 14−√2μxt  + e4μ erf 2t−4√2xμ−t1 − e4μ erf 2t−4√2x+μ1t + e-2μ-erf  2x4+√1μt − e2μ − e-2μ-) )

Result Solved

Maple
               (         )        (          )
                   2x + 1              2x − 1   ∫ t ∫ ∞     u(ζ,τ) ∂∂ζu(ζ,τ) 1∕4-(x−ζ)2
u(x,t) = 1∕2Erf 1∕4√-μ√t- − 1∕2Erf  1∕4√-μ√t- +  0  −∞ − 1∕2-√-π√μ√t-−-τ--e  μ(− t+τ)dζdτ

Result Solved, but has unresolved integrals

10 Black Scholes PDE

_________________________________________________________________________________

10.1 classic Black Scholes model from finance

problem number 115

From Mathematica symbolic PDE document.

Solve for V(S,t)  where V  is the price of the option as a function of stock price S  and time t  . r  is the risk-free interest rate, and σ  is the volatility of the stock.

∂V   1     ∂2V          ∂V
---+ -σ2S2 --2-= rV − rS---
∂t   2     ∂S           ∂S

With boundary condition V(S,T ) = max{S − k,0}

Reference https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_equation

Mathematica
{{               (        (          (     2)               )         (          (     2)               )) }}
   V(S,t) → 1e−rT  SerTerfc  2log(k-)+-2r√+-σ√-(t−-T)−-2log(S)  − kerterfc  2log(k)+-2r√−-σ√-(t−-T)−-2log(S)
            2                         2  2σ  T − t                                2  2σ  T − t

Result Solved

Maple
sol = ()

Result Did not solve

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10.2 Boundary value problem for the Black Scholes equation

problem number 116

From Mathematica DSolve help pages.

Solve for V (t,s)

∂v   1  22∂2v          ∂v
∂t-+ 2σ s ∂s2 + (r− q)s∂s − rv(t,s) = 0

With boundary condition v(T,s) = ψ(s)

Reference https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_equation

Mathematica
( (               ∫     (   )    (  (         (      σ2)     )2)      ) )
|||| ||||         er(t−T) ∞−∞ ψ eK [1] exp  − -−K-[1]+(T−t)2−σ2(q+Tr−−t)2-+log(s)-  dK [1]|||| ||||
{ {         -------------------------∘-------------------------------} }
|| || v(t,s) →                       √2π  σ2(T − t)                      || ||
||( ||(                                                                  ||) ||)

Result Solved

Maple
sol = ()

Result Did not solve

11 Korteweg-deVries PDE

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11.1 Korteweg-deVries (waves on shallow water surfaces) with no initial conditions

problem number 117

From Mathematica symbolic PDE document.

Solve for u(x,t)

∂3u   ∂u         ∂u
∂x3-+ ∂t-− 6u(x,t)∂x-= 0

Reference https://en.wikipedia.org/wiki/Korteweg%E2%80%93de_Vries_equation

Mathematica
{{            3    2                 3    } }
   u(x,t) → 12c1tanh-(c2t+-c1x-+c3)−-8c1 +-c2
                         6c1

Result Solved

Maple
                                          8-C23 − C3
u(x,t) = 2-C22(tanh( C2x +-C3t + C1 ))2 − 1∕6---------
                                              -C2

Result Solved

12 Tricomi PDE

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12.1 Boundary value problem for the Tricomi equation

problem number 118

From Mathematica DSolve helps pages.

Solve for u(x,y)

∂2u+ y ∂2u= 0
∂x2    ∂y2

With boundary conditions

pict

Mathematica
{{            (    2)}}
  u(x,y) → − y y − x

Result Solved

Maple
u(x,y) = y (x2 − y)

Result Solved

13 Cauchy Riemann PDE’s

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13.1 Cauchy Riemann PDE with Prescribe the values of u  and v  on the x  axis

problem number 119

From Mathematica DSolve helps pages.

Solve for u(x,y),v(x,y

pict

With boundary conditions

pict

Mathematica
{{u (x,y) → x3 − 3xy2,v(x,y) → 3x2y − y3}}

Result Solved

Maple
sol = ()

Result Did not solve

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13.2 Cauchy Riemann PDE With extra term on right side

problem number 120

Solve for u(x,y),v(x,y

pict

Mathematica
      [{                                            }                     ]
DSolve  u(1,0)(x,y) = v(0,1)(x,y),u(0,1)(x,y) = y − v(1,0)(x,y) ,{u(x,y),v(x,y)},{x,y}

Result Did not Solve

Maple
{u (x,y) = F 1(y− ix)+ -F2(y +ix),v(x,y) = i-F1(y − ix)− i-F2(y + ix)+ 1∕2y2 + C1 }

Result Solved

14 Hamilton-Jacobi PDE

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14.1 Hamilton-Jacobi type PDE

problem number 121

Taken from Maple pdsolve help pages, which is taken from Landau, L.D. and Lifshitz, E.M. Translated by Sykes, J.B. and Bell, J.S. Mechanics. Oxford: Pergamon Press, 1969

Solve for S (t,ξ,η,ϕ )

pict

Mathematica
      [                                          (      )                  (     )                                               ]
         (1,0,0,0)           --s(0,0,0,1)(t,ζ,η,ϕ)2---  -− η2 −-1-s(0,0,1,0)(t,ζ,η,ϕ)2-ζ2-−-1-s(0,1,0,0)(t,ζ,η,ϕ)2   a(ζ)-+b(ζ)-
DSolve − s     (t,ζ,η,ϕ) = 2(ζ2 − 1)(− η2 − 1)m σ2 +   2m σ2(ζ2 − η2)     +      2m σ2(ζ2 − η2)     +  ζ2 − η2 ,s(t,ζ,η,ϕ),{t,ζ,η,ϕ

Result Did not Solve

Maple
                                          ∘ ------------------------------------------------------------------------      ∘------------------------------------------------------------------------
                                        ∫   − 2η4m σ2-c1 + 2b(η)η2m σ2 +2η2-c1σ2m − 2 c3η2σ2m − 2b(η)σ2m + 2 c3σ2m −-c42  ∫   − 2m σ2ξ4-c1 − 2a(ξ)m σ2ξ2 + 2ξ2 c1σ2m − 2m &
S(t,ξ,η,ϕ) = c4ϕ+ c1t+-C1+ -C2+ C3+ -C4−   ----------------------------------η2 −-1----------------------------------dη−   ----------------------------------ξ2-−-1----------------------------------dξ

Result Solved

15 Other second order PDE’s

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15.1 A second order PDE

problem number 122

Taken from Maple pdsolve help pages, problem 4.

Solve for S (x,y)

pict

Mathematica
      [ (0,1)     (1,0)             (1,1)                    ]
DSolve s   (x,y)s   (x,y) + s(x,y)s   (x,y) = 1,s(x,y),{x,y}

Result Did not Solve

Maple
         √ ---------∘ ------2------
S (x,y) = --2 c1x+-C1---C2-c1-+-y-c1
                     c1

Result Solved