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Solving partial differential equations in Maple and Mathematica

Nasser M. Abbasi

April 10, 2018   Compiled on April 10, 2018 at 12:32am

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, No source, nonhomogeneous BC
 2.20 Heat PDE on bar, with source, nonhomogeneous BC
 2.21 Heat PDE on bar with extra term
 2.22 Heat PDE on bar with initial conditions sum of sine terms, no source
 2.23 Heat PDE on bar, with initial conditions as piecewise function, no source
 2.24 Heat PDE on bar, with time dependent B.C, no source
 2.25 Heat PDE on bar, homogeneous Neumann boundary conditions, with source as sin function
 2.26 Heat PDE on bar, homogeneous Neumann boundary conditions, with source that depends on time and space
 2.27 Heat PDE on bar, Dirichlet boundary conditions that depends on time with source that depends on space only
 2.28 Heat PDE on bar, homogeneous Dirichlet boundary conditions, with source that depends on time and space
 2.29 Heat/Diffusion PDE in 2D, inside rectangle with initial and boundary conditions
 2.30 Heat/Diffusion PDE in 2D, inside rectangle with initial and boundary conditions with heat loss
 2.31 Heat PDE inside disk, with no θ  dependency. initial and boundary conditions given
 2.32 Heat PDE on whole line with no intial nor boundary conditions specified
 2.33 Heat PDE in 1D on the whole real line with initial position specified
 2.34 Heat PDE in 1D on the whole real line with extra term
 2.35 Heat PDE in 1D on the whole real line with initial position as UnitBox
 2.36 Heat PDE on half the line with non-zero initial conditions and Dirichlet boundary conditions
 2.37 Heat PDE on half the line with zero initial conditions and time dependent boundary conditions
 2.38 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 source
 6.6 Wave PDE on string (finite domain), fixed ends
 6.7 Wave PDE on string (finite domain), one fixed end, one free end
 6.8 Wave PDE on string (finite domain), both ends fixed end, with source
 6.9 Wave PDE on string (finite domain), both ends fixed end, with source
 6.10 Wave PDE on semi-infinite domain, with one end having a moving boundary condition
 6.11 Telegraphy PDE, a wave PDE on string, both ends fixed with damping
 6.12 Wave PDE, on string, both ends fixed. Initial velocity zero. Dispersion term present
 6.13 Wave PDE on string with fixed ends, non-zero initial position
 6.14 Wave PDE homogeneous in square, given initial position but with zero initial velocity
 6.15 Wave PDE homogeneous in square with damping. Given zero initial position but with non-zero initial velocity
 6.16 Wave PDE inside rectangle. All 4 edges are fixed and given non-zero initial position with zero initial velocity
 6.17 Wave PDE inside disk. fixed edge of disk, no θ  dependency, with initial position and velocity given
 6.18 Wave PDE inside disk. fixed edge of disk, with θ  dependency, zero initial velocity
 6.19 Wave PDE on infinite domain with initial conditions specified, no source
 6.20 Wave PDE on infinite domain with initial conditions specified, with source term
 6.21 Wave PDE initial value with a Dirichlet condition on the half-line
 6.22 Wave PDE Initial value problem with a Neumann condition on the half-line
 6.23 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). Windows 7.
  2. Maple 2018.0 (64 bit). Windows 7 with ”Physics Updates.maple” version March 28, 2018.

The PC used is an Intel i7-3930k running at 3.20 GHz with 16 GB memory. 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 [ 113 ]. Mathematica solved 78 or 69.03%. Maple solved 89 or 78.76%.






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, No source, nonhomogeneous BC

Solved

Solved






34

Heat PDE

Heat PDE on bar, with source, nonhomogeneous BC

Did not solve

Solved






35

Heat PDE

Heat PDE on bar with extra term

Did not solve

Solved






36

Heat PDE

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

Solved

Solved






37

Heat PDE

Heat PDE on bar, with initial conditions as piecewise function, no source

Solved

Solved






38

Heat PDE

Heat PDE on bar, with time dependent B.C, no source

Solved

Did not solve






39

Heat PDE

Heat PDE on bar, homogeneous Neumann boundary conditions, with source as sin function

Did not solve

Solved






40

Heat PDE

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

Did not solve

Solved






41

Heat PDE

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

Did not solve

Did not solve






42

Heat PDE

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

Did not solve

Solved






43

Heat PDE

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

Did not solve

Solved






44

Heat PDE

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

Did not solve

Solved






45

Heat PDE

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

Solved

Did not solve






46

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






47

Heat PDE

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

Solved

Solved






48

Heat PDE

Heat PDE in 1D on the whole real line with extra term

Solved

Solved






49

Heat PDE

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

Solved

Solved






50

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






51

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






52

Heat PDE

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

Solved

Did not solve






53

Laplace PDE

Laplace PDE inside quarter-circle

Did not solve

Did not solve






54

Laplace PDE

Laplace PDE inside semi-circle

Did not solve

Solved






55

Laplace PDE

Laplace PDE inside rectangle

Solved

Solved






56

Laplace PDE

Laplace PDE inside rectangle

Solved

Solved






57

Laplace PDE

Laplace PDE inside rectangle

Did not solve

Solved






58

Laplace PDE

Laplace PDE inside rectangle

Did not solve

Solved






59

Laplace PDE

Laplace PDE inside rectangle

Did not solve

Solved






60

Laplace PDE

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

Solved

Solved






61

Laplace PDE

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

Did not solve

Did not solve






62

Laplace PDE

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

Solved

Did not solve






63

Laplace PDE

Laplace PDE example 18 from Maple help page

Solved

Solved






64

Laplace PDE

Laplace PDE on rectangle with one edge at infinity

Did not solve

Solved






65

Laplace PDE

Laplace PDE inside a disk, periodic boundary conditions

Solved

Solved






66

Laplace PDE

Dirichlet problem for the Laplace equation in upper half plan

Solved

Did not solve






67

Laplace PDE

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

Solved

Did not solve






68

Laplace PDE

Dirichlet problem for the Laplace equation in the first quadrant

Solved

Did not solve






69

Laplace PDE

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

Solved

Did not solve






70

Laplace PDE

Dirichlet problem for the Laplace equation in a rectangle

Solved

Solved






71

Laplace PDE

Laplace PDE outside a disk, periodic boundary conditions

Did not solve

Solved






72

Laplace PDE

Laplace equation in spherical coordinates

Did not solve

Solved, but not verified






73

Poisson PDE

Dirichlet problem for the Poisson equation in a rectangle

Solved

Solved






74

Helmholtz PDE

Dirichlet problem for the Helmholtz equation in a rectangle

Solved

Did not solve






75

Wave PDE

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

Solved

Solved






76

Wave PDE

Hyperbolic PDE with non-rational coefficients

Solved

Did not solve. Tried all HINTS






77

Wave PDE

Inhomogeneous hyperbolic PDE with constant coefficients

Solved

Solved






78

Wave PDE

system of 2 inhomogeneous linear hyperbolic system with constant coefficients

Solved

Did not solve






79

Wave PDE

Wave PDE on string (finite domain) with source

Solved

Solved






80

Wave PDE

Wave PDE on string (finite domain), fixed ends

Did not solve

Solved






81

Wave PDE

Wave PDE on string (finite domain), one fixed end, one free end

Did not solve

Solved






82

Wave PDE

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

Did not solve

Solved






83

Wave PDE

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

Did not solve

Solved






84

Wave PDE

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

Solved

Solved






85

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.






86

Wave PDE

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

Did not solve due to adding dispersion term

Solved






87

Wave PDE

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

Solved but sum should not include n = 2

Solved, but sum should not include n = 2






88

Wave PDE

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

Did not solve

Solved






89

Wave PDE

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

Did not solve

Solved






90

Wave PDE

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

Solved

Solved






91

Wave PDE

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

Solved

Did not solve






92

Wave PDE

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

Did not solve

Did not solve






93

Wave PDE

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

Solved

Solved






94

Wave PDE

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

Solved

Solved






95

Wave PDE

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

Solved

Solved






96

Wave PDE

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

Solved

Did not solve






97

Wave PDE

non-linear wave PDE (Solitons)

Solved. build a special solution.

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






98

Schrodinger PDE

Schrodinger PDE with zero potential

Solved

Solved






99

Schrodinger PDE

Schrodinger PDE with initial and boundary conditions

Solved

Did not solve






100

Schrodinger PDE

Initial value problem with Dirichlet boundary conditions

Solved

Solved






101

Schrodinger PDE

Solve a Schrodinger equation with potential over the whole real line

Solved

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






102

Beam PDE

Beam PDE with zero initial velocity

Solved

Solved






103

Burger’s PDE

viscous fluid flow with no initial conditions

Solved

Solved






104

Burger’s PDE

viscous fluid flow with initial conditions

Solved

Solved, but has unresolved integrals






105

Burger’s PDE

viscous fluid flow with initial conditions as UnitBox

Solved

Solved, but has unresolved integrals






106

Black Scholes PDE

classic Black Scholes model from finance

Solved

Did not solve






107

Black Scholes PDE

Boundary value problem for the Black-Scholes equation

Solved

Did not solve






108

Korteweg-deVries PDE

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

Solved

Solved






109

Tricomi PDE

Boundary value problem for the Tricomi equation

Solved

Did not solve






110

Cauchy-Riemann PDE’s

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

Solved

Did not solve






111

Cauchy-Riemann PDE’s

Cauchy-Riemann PDE With extra term on right side

Did not Solve

Solved






112

Hamilton-Jacobi PDE

Hamilton-Jacobi type PDE

Did not Solve

Solved






113

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 u(x,0) = e−x2

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

Result Solved

Maple
           2  2
u(x,y) = e−x −y

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∂x-+5 ∂y-= u2(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)

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

Mathematica
{{                   1 (     )} }
   u(x,y) → c1x+ c2y+ 2  c21 + c22

Result Solved

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

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

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

Mathematica
{{          1(           )}}
   u(x,y) → 2 − x2 − 2y+ 1

Result Solved

Maple
− 1∕2(x − y+ 1)(x− y − 1) = − 1∕2(x +y + 1)(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) → c + x sin(xy)+ e−y}}
           1

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

Result Did not solve

Maple
        (∫                                          )
           x --------g( a)---------   -   (    2   ) −1
f (x,y) =    h(− 1∕2-a2 + 1∕2x2 + y)d a+ F 1 1∕2x + 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
({({             (                (    -y-+c2(z)+x−1)                    (    -y-+c2(z)+x−1)     ) )})}
   f(x,y,z) → 1(c1 (z)2ProductLog( − e-c1(z)-c1(z)---) 2 + 2c1(z)2ProductLog( − ec1(z)-c1(z)---) − 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

∂u    ∂2u
∂t-= k∂x2-

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

Mathematica
{{            81π2kt   (    )}}
   u(x,t) → 6e−-L2--sin  9πx-
                        L

Result Solved

Maple
            ( πx ) −81kπ2t
u(x,t) = 6sin 9-L- e   L2

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

∂u-= k∂2u-
∂t    ∂x2

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

Mathematica
{{          − 9π2k2t  (πx-)(   8π2k2t      ( 2πx)    )} }
   u(x,t) → e  L  sin  L    3e L  − 2cos   L   − 1

Result Solved

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

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

       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) = 2 cos 3πx
             L

Mathematica
((                                        ) )
{{         ∑∞ 4(1 +(− 1)n)e− kn2Lπ22tnsin (nπx)} }
   u(x,t) →    ------------2------------L--
((         n=1          (n − 9)π          ) )

Result Solved

Maple
           (
        ∞∑  {0                          n = 3
u(x,t) =       n((−1)n+1)   (   ) − π2kn2t-
        n=1(4 -π(n2−9)-sin nπLx e   L2   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

∂u-= k∂2u-
∂t    ∂x2

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

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-
∂t = k∂x2

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

Mathematica
( (                                                        ) )
|| ||          ∞∑   − knL2π22t  (nπx)∫L    (nπx)         ∫       || ||
{ {         2n=1e------cos--L---0-cos--L--f(x)dx   0Lf(x)dx} }
|| ||u (x,t) →                  L                  +     L    || ||
( (                                                        ) )

Result Solved

Maple
           (∑∞ (    ∫ L       (    )      (    )    2 2 )    ∫ L       )
u (x,t) = 1-      21-   f (x) cos  nπx- dxcos n-πx  e− πkLn2-t 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     ∂2u
---= k --2 − αu
 ∂t    ∂x

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
      [{ (0,1)        (2,0)                                              }            ]
DSolve  u   (x,t) = ku  (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
        ∞∑  (    ∫ L       (    )     (    )   t(π2kn2+L2a))
u(x,t) =    2 1-   f (x)sin nπx- dx sin  nπx- e−----L2----
        n=1   L  0          L           L

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

       2
∂u-= k∂-u-
∂t    ∂x2

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

Mathematica
( (                  kn2π2t   (  )  (  )    ) )
||| |||         2 ∞∑ −  e−--L2-L-cos-nπLx-sin-nπ2--   ||| |||
{ {         -n=1------------nπ---------   1} }
|| || u(x,t) →              L              + 2|| ||
|( |(                                        |) |)

Result Solved

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

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

∂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) = 6 + 4cos(3πxL-)

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

Result Solved

Maple
                (    )     2
u (x,t) = 6 + 4cos 3πx e−9kπL2t
                   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
∂t-= k∂x2-

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

Mathematica
(| (|           ∞∑        n − kn2πL22t  (nπx)    )| )|
||{ ||{         2   2(1+-(−1)-)e(n2−1)π-Lcos-L--    ||} ||}
    u(x,t) → -n=1-----------------------−  4-
|||( |||(                      L                π|||) |||)

Result Solved

Maple
          ( ∞  ({ 0                       n ≤ 1     )
u(x,t) = 1-( ∑       n     (   )    2 2        π− 4)
        π   n=1( 4(−π(n1)2−+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
---= k--2-
∂t    ∂x

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

Mathematica
{{                2    (    ) }}
   u(x,t) → − 3e− 64πL2kt-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

∂u    ∂2u
∂t-= k∂x2-

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

Mathematica
((           ∞∑    k(2n+1)2π2t   (       )∫     (       )       ))
||{||{         2   e−   4L2   cos (2n+12L)πx  0Lcos (2n+21L)πx  f(x)dx ||}||}
   u(x,t) → -n=0---------------------------------------------
||(||(                                 L                        ||)||)

Result Solved

Maple
           (                                                                )
        ∞∑     1∫ L        (   (1 +2n )πx)      (    (1 + 2n)πx)  −1∕4kπ2(1+2n)2t
u(x,t) =     2 L-   f (x)cos 1∕2----L----  dxcos  1∕2 ----L----- e       L2
        n=0     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

       2
∂u-= ∂-u-
∂t   ∂x2

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

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

Result Solved

Maple
            x  ∫ t  f (-U1)  − -x2---
u (x,t) = 1∕2 √π-   --------3∕2e  4t−4 U1d-U1
                0 (t− U 1)

Result Solved

_________________________________________________________________________________

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

problem number 27

Solve the heat equation

∂u    ∂2u
∂t-= k∂x2-

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

pict

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

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

Result Did not solve

Maple
           1 (  ∞∑  ( 1( ∫ L        (nπx)      (n πx)  ∫ L        ( nπx)      (n πx))   kπ2n2t)     ∫ L       )
u(x,t) = 1∕2- 2     --      f (x)sin ---- dxsin ----  +    f (x) cos  ---- dxcos ----   e−  L2   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-= k∂2u-
∂t    ∂x2

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

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

Result Solved

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

       2
∂u-= k∂-u2-
∂t    ∂x

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  2√kt- + λerfc 2√kt-

Result Solved

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

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
---= k--2-
∂t    ∂x

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 > 0,x ≥ 0,x ≤

Result Did not solve

Maple
        ∑∞ (       1      ∫ L     (       (nπx )     (nπx ) )   (       ( nπx)      (nπx ) )   kπ2n2t)
u(x,t) =     2---2-2----2-   f (x)  − πn cos---- + sin  ---- L  dx − πn cos  ---- + sin  ---- L  e− L2
        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-= k∂2u-+ m
∂t    ∂x2

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
               −kt
u (x,t) = sin(x)e   + 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
      [{  (0,1)        (2,0)                                       }            ]
DSolve  u    (x,t) = u   (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                        1∕4π2(2n−1)2t  ∫ 1                                          π2n2t)  −1∕4π2t(8n2−4n+1))
u(x,t) =         f (x )sin(nπx)dx sin(nπx )e          +     f (x)cos(1∕2xπ (2n− 1))dxcos(1∕2xπ(2n − 1))e    e
        n=1    −1                                     −1

Result Solved

_________________________________________________________________________________

2.19 Heat PDE on bar, No source, nonhomogeneous BC

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)          ) )
||{ ||{           2   ----------n-----------         ||} ||}
    u(x,t) → − -n=1----------------------+ 30x+ 20
||( ||(                       π                      ||) ||)

Result Solved

Maple
                  ∞∑  (100(− 1)n −-40)sin-(n-πx)ekπ2n2t
u(x,t) = 20+ 30x +                nπ
                 n=1

Result Solved

_________________________________________________________________________________

2.20 Heat PDE on bar, with source, nonhomogeneous BC

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
      [{                                                             }          ]
DSolve  u(0,1)(x,t) = ku(2,0)(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 with extra term

problem number 35

Solve the heat equation

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

For 0 < x < L  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)+ u(x,t) = u   (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, no source

problem number 36

added Feb 10, 2018.

Solve the heat equation

                 2
∂u-+ u(x,t) = 100 ∂-u-
∂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
u (x,t) = sin(2πx)e−400π2t − sin (5πx)e−2500π2t

Result Solved

_________________________________________________________________________________

2.23 Heat PDE on bar, with initial conditions as piecewise function, no source

problem number 37

added Feb 10, 2018.

Solve the heat equation

∂u-  ∂2u-
∂t = ∂x2

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
{ {         ∞∑      − n126π200t  (nπ)   3(nπ)   (nπx)} }
    u(x,t) →    640e-----cos--42-si2n---4-sin--40--
            n=1              n π

Result Solved

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

Result Solved

_________________________________________________________________________________

2.24 Heat PDE on bar, with time dependent B.C, no source

problem number 38

added March 8, 2018. Exam problem

Solve the heat equation

       2
∂u-= ∂-u-
∂t   ∂x2

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

pict

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

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

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

2.25 Heat PDE on bar, homogeneous Neumann boundary conditions, with source as sin function

problem number 39

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 )
∂t-= k∂x2-+ sin  -L--

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

pict

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

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

Result Did not solve

Maple
          (    (               (     )                   )         (     (                    (     )      (      )          )                   )                   )
        1- ∑∞    1-∫ L          nπ-τ1        (nπx-) kπ2n22t     ∫ t 1- ∑∞      1-∫ L   (  πx)     n1πx-        n1πx-   kπ2n12(2t−τ1)     ∫ L   ( πx-)           ∫ L
u(x,t) = L      2L  0 f (τ1)cos   L   dτ 1cos  L   e  L    L + 0  L       − 2L  0  sin 2 L  cos   L    dx cos    L   e    L      L −  0  sin 2 L  dx  dτ1L +  0 f (τ1)dτ1
           n=1                                                       n1=1

Result Solved

_________________________________________________________________________________

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

problem number 40

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

Result Did not solve

Maple
          ( ∞  (   ∫           (     )                    )     ∫    ( ∞  (                   ∫               (     )      (     ) )         ∫             )        ∫          )
        1- ∑     1-  L          nπ-τ1        (nπx-) − kπ2Ln22t       t1-  ∑     1-−kπ2(t−τ1)Ln212−cτ1L2  L    ( πx)     n1πx-         n1πx-       −cτ1  L   ( πx-)            L
u(x,t) = L n=1  2L  0 f (τ1)cos   L   dτ 1cos  L   e        L +  0 L  n1=1  2Le                0  sin  2L   cos   L    dxcos   L     L + e     0 sin 2 L  dx  dτ1L +  0 f (τ1)dτ1

Result Solved

_________________________________________________________________________________

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

problem number 41

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

Solve the heat equation

∂u    ∂2u
---= k--2-+ x
∂t    ∂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
sol = ()

Result Did not solve

_________________________________________________________________________________

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

problem number 42

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
      [{                                                             }                                               ]
DSolve  u(0,1)(x,t) = f(x,t)+ ku(2,0)(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
           (                                        )        (                                            )
        ∑∞    1 ∫ l      ( πmx-)     ( πmx-) − kπ2m22t   ∫ t∑∞    1∫ l          (nπx-)     ( nπx)  kπ2n2(−2t+τ1)
u(x,t) =     2l  0 g (x )sin  l   dx sin   l   e   l    +  0      2l 0 f (x,τ1)sin   l  dx sin   l   e    l      dτ1
        m=1                                               n=1

Result Solved

_________________________________________________________________________________

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

problem number 43

Taken from Maple help pages on PDE

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

         (  2     2 )
∂u-= 1∕10  ∂-u2 + ∂-u2
∂t         ∂x   ∂y

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-) − kπ2lm22t     t∑     1  l          (nπx-)     ( nπx)  kπ2n2(−l2t+τ1)
u(x,t) = m=1 2l  0 g (x )sin  l   dx sin   l   e        +  0 n=1  2l 0 f (x,τ1)sin   l  dx sin   l   e           dτ1

Result Solved

_________________________________________________________________________________

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

problem number 44

Taken from Maple help pages on PDE

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

∂u       ( ∂2u   ∂2u)   1
---= 1∕10  --2-+ --2- − -u(x,y,t);
∂t         ∂x    ∂y     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
      [{               1 (                          )  1                  (     )(    y)   {                                                     }}                ]
DSolve  u(0,0,1)(x,y,t) = -- u(0,2,0)(x,y,t)+ u(2,0,0)(x,y,t) − -u (x,y,t),u(x,y,0) = 1− x2   1− -- y, u(1,0,0)(0,y,t) = 0,u(1,y,t) = 0,u(x,0,t) = 0,u(0,1,0)(x,1,t) 
                       10                              5                              2

Result Did not solve

Maple
          ∑∞ ( ∞∑     (− 1)ncos(1∕2(1+ 2n)πx )sin(1∕2(1+ 2m )πy)e−1∕10t(2+(n2+m2+n+m+1 ∕2)π2))
u(x,y,t) =         512------------------------------3--6-------3------------------------
          m=0  n=0                          (1 +2n )π  (1+ 2m)

Result Solved

_________________________________________________________________________________

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

problem number 45

Taken from Mathematica DSolve help pages

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

∂u-= ∂2u-+ 1 ∂u-
∂t   ∂r2   r ∂r

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

pict

Initial condition is u(r,0) = 1− r  .

Mathematica
((                                                    (                                        ({  } {   }                   ))) )
{{         ∑∞ 2e−tBesselJZero(0,n)2BesselJ(0,rBesselJZero(0,n))  BesselBJe(1s,BseelsJZseelroJZ(e0r,no()0,n))− 13HypergeometricPFQ    32 , 1, 52 ,− 14BesselJZero(0,n)2 } }
((u (r,t) →    -------------------------------------------------------2-------------------------2-------------------------------) )
           n=1                               BesselJ(0,BesselJZero(0,n)) + BesselJ(1,BesselJZero(0,n))

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

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

problem number 46

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
             c t   √ c-x  -C3e-c1t-C2
u (x,t) = C3e  1- C1e  1 + ---e√-c1x--

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

_________________________________________________________________________________

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

problem number 47

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

∂u-= ∂2u-
∂t   ∂x2

With initial condition

         − x2
u (x,0) = e

Mathematica
{{                }}
            e−4x2t+1
   u(x,t) → √------
             4t+ 1

Result Solved

Maple
u (x,t) = √-1---e−1x+24t
          1+ 4t

Result Solved

_________________________________________________________________________________

2.34 Heat PDE in 1D on the whole real line with extra term

problem number 48

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

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

_________________________________________________________________________________

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

problem number 49

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

∂u-  ∂2u-
∂t = ∂x2

With initial condition

u(x,0) = UnitBox[x]

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

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

Result Solved

Maple
                 (         )         (         )
u (x,t) = − 1∕2Erf 1∕42x√− 1 + 1∕2Erf  1∕4 2x√+-1-
                        t                    t

Result Solved

_________________________________________________________________________________

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

problem number 50

From Mathematica DSolve help pages.

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

       2
∂u-= ∂-u2
∂t   ∂x

With initial condition

u (x,0) = cosx

And boundary conditions

u(0,t) = 1

Mathematica
( (                                                                ) )
{ {              ie− x42t(DawsonF(2t−√ix)−DawsonF(2t+√ix))     (   )        } }
   u (x,t) →  {   -------------2t√π----------2-t--+ erfc  2x√t   x > 0
( (                             Indeterminate                 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

_________________________________________________________________________________

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

problem number 51

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∕4xk2t      (        2)        --x--                      √√sxk-                   − √√sxk-                      2
u(x,t) = 1∕2πk − 2 π  t kxe      − 2π   kt+ 1∕2x  Erf  1∕2√k-√t  − invlaplace  F 1(s) e  ,s,t k + invlaplace e    -F1 (s),s,t k − kt− 1∕2x

Result Solved, but has unresolved inverse Laplace transforms

_________________________________________________________________________________

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

problem number 52

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) = UnitTriagle[x- 3]

And boundary conditions

∂u
--(0,t) = 0
∂x

Mathematica
(|(|                (  (    )                                                                (    )         (    )   ( − (x−4)2  − (x−3)2 − (x−2)2- − (x+2)2 − (x+3)2- − (x+4)2)√-)        )| )|
||{||{              1 | erf|x2−√4|t-(x−-4)2-          (x+√2)            (x+√3)           (x+√4)   2(x−3)2erf|x2−√3|t-   (x−2)2erf|x−2√2t|-- 2 e   4t −2e  4t +e   4t +e   4t  −2e  4t +e   4
   u(x,t) → {   2 (     |4−x|    + (x+ 2)erf 2 t − 2(x +3)erf 2 t  + (x + 4)erf 2 t  −      |3−x|    +      |2−x|    +                           π                         )   x > 0
|||(|||(                                                                                                                                                                              |||) |||)
                                                                                      Indeterminate                                                                        True

Result Solved

Maple
sol = ()

Result Did not solve

3 Laplace PDE

_________________________________________________________________________________

3.1 Laplace PDE inside quarter-circle

problem number 53

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 0 ≤ θ ≤ π2  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 54

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
      [{                                                                           }                                               ]
DSolve   u(0,2)(r,θ)u(1,0)(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 → &
                r3

Result Did not solve

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

Result Solved

_________________________________________________________________________________

3.3 Laplace PDE inside rectangle

problem number 55

This is problem 2.5.1 part (a) 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
( (                               (                 )                      ) )
{ {         ∑∞ 2 cos(n-πx-)csch (Hnπ) ∫ Lcos(nπx)f(x)dx  sinh(nπy)    ∫ L      } }
    u(x,y) →   ------L---------L----0------L---------------L-- + y-0-f(x)dx
( (         n=1                       L                             HL     ) )

Result Solved

Maple
            (     (            ∫                           (     (    ) )  )     ∫          )
        -1--   ∞∑        ( πyn)   L       ( nπx)      (n-πx)       nπH--  −1        L
u(x,y) = HL  2     1 sinh   L    0 f (x)cos   L   dxcos  L     sinh    L        H +  0  f (x)dxy
               n=1

Result Solved

_________________________________________________________________________________

3.4 Laplace PDE inside rectangle

problem number 56

This is problem 2.5.1 part (b) 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
((                    (       )    (   ) (∫        (   )  )    (   )))
{{          ∞∑    2cosh  nπ(LH−x)-csch LnHπ    H0 g(y)sin nπHy dy  sin  nHπy }}
   u(x,y) →    − ---------------------------------------------------
((          n=1                          nπ                         ))

Result Solved

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

Result Solved

_________________________________________________________________________________

3.5 Laplace PDE inside rectangle

problem number 57

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

Solve Laplace equation

 2     2
∂-u-+ ∂-u-= 0
∂x2   ∂y2

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 ∫ H    (nπy )         ( nπy) (    ( πxn )      ( πxn )   ) (    ( πn(L − x))      ( πn(L − x))) (    (  nπL )      (  nπL )   ) −1�
u(x,y) =     2--    sin ---- g (y)dy sin  ----  cosh  2---- + sinh  2---- + 1   cosh  ---------  +sinh  ---------    cosh  2---- + sinh  2---- + 1
        n=1   H  0       H               H           H            H                  H                 H                 H             H

Result Solved

_________________________________________________________________________________

3.6 Laplace PDE inside rectangle

problem number 58

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

Solve Laplace equation

∂2u-+ ∂2u-= 0
∂x2   ∂y2

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 ∫ H    (   π (2yn+ H + y))          (   π (2yn+ H + y)) (    ( (1 + 2n)π(L − x))      ( (1+ 2n)π (L − x))    )(     (   (1+ 2n)πx )      (    &
u(x,y) =     2--    sin 1∕2--------------  g(y)dysin 1∕2--------------   cosh  ---------------  + sinh  ---------------  − 1   cosh  1∕2----------  + sinh  1∕2----------    cosh  ----------  + sinh  ----------  − 1
        n=0   H  0               H                            H                      H                       H                            H                     H                  H                  H

Result Solved

_________________________________________________________________________________

3.7 Laplace PDE inside rectangle

problem number 59

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

Solve Laplace equation

∂2u-+ ∂2u-= 0
∂x2   ∂y2

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(x,0)− u(0,1)(x,0) = 0,u(x,H ) = f(x)}},u(x,y),{x,y},Assumptions → {0 ≤ x ≤ L

Result Did not solve

Maple
        ∑∞ (    (     (     )         (     )          (     )       (     )         )   (    )(    (          )       (          )) ∫ L   (    )        (     (      )          (     )          (      )        (      )         )
u(x,y) =     21- πcosh  2πyn- n+ π sinh 2 πyn- n+ L cosh  2πyn- + L sinh 2 πyn- + nπ− L  sin  nπx-  cosh  πn-(H--− y) + sinh   πn(H-−-y)-      sin  nπx- f (x)dx πsinh  2nπH-- n+ π cosh  2nπH-- n + Lsin
        n=1   L           L               L               L              L                 L              L                 L         0       L                    L                 L                L               L

Result Solved

_________________________________________________________________________________

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

problem number 60

Taken from Mathematica DSolve help pages.

Solve Laplace equation

∂2u-  ∂2u-
∂x2 + ∂y2 = 0

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

pict

Mathematica
{ {         ∞              (  )                                 }}
            ∑  8csch(2nπ)sin--nπ2-sin(nπx-)(sinh-(nπ-(2-−-y))-+sinh(nπy))
   u(x,y) → n=1                      n2π2

Result Solved

Maple
         ∞∑                     (πn(3y− 2)   πn(3y− 4)   πyn   πn(y−2)) −2πn(y− 2)
u (x,y) =   8sin-(1∕2n-π)sin(nπx-)e--------− e------+-e---−-e-------e--------
         n=0                         π2n2 (e4nπ − 1)

Result Solved

_________________________________________________________________________________

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

problem number 61

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

Solve Laplace equation

 2             2
∂-u2-+ 1∂u-+ -12 ∂-u2-= 0
∂r    r∂r   r ∂ θ

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

pict

Mathematica
      [{ u(0,2)(r,θ)   u(1,0)(r,θ)                {                        }}                                   ]
DSolve   ---r2----+ ----r----+ 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

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 62

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 63

Solve Laplace equation

∂2u   ∂2u
∂x2-+ ∂y2-= 0

With boundary conditions

pict

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

Result Solved

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

Result Solved

_________________________________________________________________________________

3.12 Laplace PDE on rectangle with one edge at infinity

problem number 64

Solve Laplace equation

∂2u   ∂2u
--2-+ --2-= 0
∂x    ∂y

With boundary conditions

pict

Mathematica
      [{ (0,2)       (2,0)                                                     }                                ]
DSolve  u   (x,y)+ u    (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
         ∞∑   (− 1)1+nπ-sin(a)n-− nπxa   (πyn)
u (x,y) =    2    π2n2 − a2  e     sin   a
         n=0

Result Solved

_________________________________________________________________________________

3.13 Laplace PDE inside a disk, periodic boundary conditions

problem number 65

Solve Laplace equation in polar coordinates inside a disk

Solve for u(r,θ)

pict

Boundary conditions

pict

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

Result Solved

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

Result Solved

_________________________________________________________________________________

3.14 Dirichlet problem for the Laplace equation in upper half plan

problem number 66

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 (1−x)     − 1(x+1 )            1        1          ) )
||{||{                {  tan    2y-- + tan    -y2-   y > 0 ∨x > 2 ∨ x < − 2        ||} ||}
   u(x,y) →  {   ----------------0----------------------True----------
||(||(                                         π                            y ≥ 0 ||) ||)
                                     Indeterminate                       True

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

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

problem number 67

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 68

Taken from Mathematica DSolve help pages

Solve for u(x,y)

pict

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

Mathematica
(|(|               ( 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&#
{{               2( -------------------------------------------------------------------------------(x2−2x+y2−6y+10)(x2+2x+y2−6y+10)(x2−2x+y2+6y+10)(x2+2x+y2+6y+10)-----------------------------------------------------------------------------x-− -----------------------------------------------------------------------------------------
|(|(u (x,y) →  {   --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
                                                                                                                                                                                                                                                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 69

Taken from Mathematica DSolve help pages

Solve for u(x,y)

pict

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

Mathematica
((                                                                         (                       (      )    )                                                                               ) )
||||                     (         (    )          (   )                  − i4itanh−1(2x)x+ 2πx + ilog  14 − x2 − 2i                                                 )  − 12 ≤ x ≤ 0∧ y ≤ 0         || ||
{{                {   1 − 4y tan−1 x−-12 + 4ytan−1  x+12- − 2xlog(4x2 − 4x+ 4y2 +1) + log(4x2 − 4x + 4y2 + 1)+ 2x log(4x2 + 4x + 4y2 + 1)+ log(4x2 + 4x+ 4y2 +1) − 2log(4)− 4    True                 } }
|||| u(x,y) →  {   -----2------------y---------------y-----------------------------------------------2π-----------------------------------------------------------------------------------  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 70

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) →    −  -------------------3-3--------------------
           n=1                     n π

Result Solved

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

Result Solved

_________________________________________________________________________________

3.19 Laplace PDE outside a disk, periodic boundary conditions

problem number 71

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 72

Taken from Maple pdsolve help pages

Solve for u(r,θ,ϕ)

pict

Mathematica
      ⌊                                              (                                                 )                                          ⌋
        f(0,2,0)(r,θ,ϕ)-   (1,0,0)                       csc(θ) sin(θ)f(1,0,0)(r,θ,ϕ)+ cos(θ)f(0,1,0)(r,θ,ϕ)+ csc(θ)f(0,0,2)(r,θ,ϕ)
DSolve⌈-----r-----+f-----(r,θ,ϕ) + f(2,0,0)(r,θ,ϕ )+ -------------------------------r----------------r------- = 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
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 73

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
u(x,y) = x3 − y3 + 11x+ 1

Result Solved

5 Helmholtz PDE

_________________________________________________________________________________

5.1 Dirichlet problem for the Helmholtz equation in a rectangle

problem number 74

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) → 2   --------8---------8--------2--------n2π2----8-------4-------4-----------------
             n=1

Result Solved

Maple
sol = ()

Result Did not solve

6 Wave PDE

_________________________________________________________________________________

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

problem number 75

From Mathematica DSolve help pages (slightly modified)

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

∂2u- -∂2u-   2∂2u-
∂t2 + ∂t∂x = c ∂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 76

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) → c1(x − cos(x)+ y)+ c2(− x− cos(x)+ y)}}

Result Solved

Maple
sol = ()

Result Did not solve. Tried all HINTS

_________________________________________________________________________________

6.3 Inhomogeneous hyperbolic PDE with constant coefficients

problem number 77

From Mathematica DSolve help pages

Solve for u(x,t)

  2     2      2
3∂-u-− ∂-u-+ ∂-u--= 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 78

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 source

problem number 79

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 (finite domain), fixed ends

problem number 80

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

 2       2
∂-u2-= c2∂-u2-
∂t      ∂x

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  -L--  -C1 (n) sin  -L--  + C5 (n)cos  -L--
        n=1

Result Solved

_________________________________________________________________________________

6.7 Wave PDE on string (finite domain), one fixed end, one free end

problem number 81

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

∂2u     ∂2u
∂t2-= c2∂x2-

With boundary conditions

pict

With initial conditions

pict

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

Result Did not solve

Maple
           (                                                                      )
        ∑∞    1 ∫ L   (   (1+ 2n) πx)          (    (1 + 2n)πx)    (   cπ (1 + 2n)t)
u(x,t) =     2L-   sin 1∕2----L-----  f (x)dxsin 1∕2 ----L----- cos 1∕2-----L-----
        n=0      0

Result Solved

_________________________________________________________________________________

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

problem number 82

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

∂2u     ∂2u
--2-= c2--2-+ p(x,t)
∂t      ∂x

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∑∞ (  ∫π sin (nx )p(x,τ1)dxsin (nx )sin(c(t− τ1)n))
u(x,t) =        2-0-------------------------------------  dτ1
         0 n=1                     nπc

Result Solved

_________________________________________________________________________________

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

problem number 83

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
        ∫ t∑∞ (      ∫ L   (    )       (    )   (           ))
u(x,t) =        2-A--   sin nπx- xdx sin  nπx- sin  cπ-(t−-τ1)n-   dτ1
         0 n=1   nπc  0      L            L            L

Result Solved

_________________________________________________________________________________

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

problem number 84

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

 2       2
∂-u-= c2∂-u-
∂t2     ∂x2

With boundary conditions

pict

With initial conditions

pict

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

Result Solved

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

Result Solved

_________________________________________________________________________________

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

problem number 85

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              ((     √ ------) − (1+√ −n2+1)t   (−1+√ −n2+1)t(    √-------)))
        ∑∞ | sin (nx ) 0 sin(nx)f (x )dx  − 1+   − n2 + 1 e           +e             1 +  − n2 + 1 |
u(x,t) =   |( ------------------------------------√---2-----------------------------------------|)
        n=1                                        − n + 1π

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

_________________________________________________________________________________

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

problem number 86

Solve

    2                2
-1∂-u-+ γ2u(x,t) = c2∂-u-
a2 ∂t2               ∂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
                 n              ( ∘ -------)
        ∑∞   ((− 1) − 1)sin(nx)cos a  γ2 + n2t
u (x,t) =   4 -----------nπ(n2-− 4)-----------
        n=1

Result Solved

_________________________________________________________________________________

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

problem number 87

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)n − 1)sin(nx)cos(2tn)
u (x,t) =   4 -------nπ(n2-− 4)-------
        n=1

Result Solved, but sum should not include n = 2

_________________________________________________________________________________

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

problem number 88

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

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

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

_________________________________________________________________________________

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

problem number 89

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

_________________________________________________________________________________

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

problem number 90

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

_________________________________________________________________________________

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

problem number 91

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
((                                      (                                                                                                     (                  )) ))
{{         ∑∞ 2BesselJ(0,rBesselJZero(0,n)) 9√c2BesselJ(1,BesselJZero(0,n))cos(ctBesselJZero(0,n))+ HypergeometricPFQ  ({ 3},{1, 5} ,− 1BesselJZero(0,n )2)sin √c2tBesselJZero
   u(r,t) →    ----------------------------------------√--------------------------------------------------------2-----2----4-----------------------------------------
((         n=1                                       9  c2 (BesselJ(0,BesselJZero(0,n))2 + BesselJ(1,BesselJZero(0,n))2)BesselJZero(0,n)                                      ))

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

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

problem number 92

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

_________________________________________________________________________________

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

problem number 93

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

_________________________________________________________________________________

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

problem number 94

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|−x+t|+1∕6|t+x| − cos(3t− 3x)e1∕6|t+x| − e1∕6|−x+t|cos(3t+ 3x)

Result Solved

_________________________________________________________________________________

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

problem number 95

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

Result Solved

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

Result Solved

_________________________________________________________________________________

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

problem number 96

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
(                                    √--           (∘--)
|{ 1(  3 (√-2    )     3(√ -2    ))  2-c2t−-20e−x√∕10sinh--c102t-                               √ 2-
  2 s1in(∘-2c t)+ x 1−( sin∘2)  c√-t− x  + --(      2)c2   (   (  --    )      ( --    ))  x >   ct ≥-0
|( 10e10−-c-t−x-+10e√10-x−--ct+2-c2t−-20-− √c2 t − √x- + 1  sin3 √ c2t− x + sin3  √c2t+ x    0 ≤ x ≤ √ c2t
               2 c2                         c2   2

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

6.23 non-linear wave PDE (Solitons)

problem number 97

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

Solve

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

Mathematica
{ {             3    2                 3    }}
   u(x,t) → − 12c1tanh--(c2t+-c1x+-c3)−-8c1 +-c2
                           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 98

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

Solve

          2  2
Ih∂f-= − h--∂-f-
  ∂t     2m ∂x2

With boundary conditions

pict

Mathematica
{ {         ∞∑      2 2     (    )} }
    f(x,t) →    e− ihn2L2πmtcnsin n-πx
            n=1               L

Result Solved

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

Result Solved

_________________________________________________________________________________

7.2 Schrodinger PDE with initial and boundary conditions

problem number 99

Solve for f(x,y,t)

         2 (  2    2  )
I∂f-= − ℏ-- ∂-f-+ ∂-f-
 ∂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
{{           √- − 5iπ2hmBar2t(                             5iπ2hB2mar2t)} }
   f(x,y,t) →  2e          sin(πx)sin(3πy )+ sin(2πx)sin(πy)e

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

7.3 Initial value problem with Dirichlet boundary conditions

problem number 100

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  −-2in2π2(t−2)   (1        )} }
   f(x,t) →    100(7+-8(− 1)-)e-25-------sin--5nπ(x−-5)-
           n=1                  n3π3

Result Solved

Maple
        ∑∞               n            − 2iπ2n2(−2+t)
f (x,t) =    (800-+-700(− 1)-)sin-(1∕5nπx-)e-25--------
        n=1                  n3π3

Result Solved

_________________________________________________________________________________

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

problem number 101

Taken from Mathematica DSolve help pages

Solve for f(x,t)

         2
I∂f-= − ∂-f2 + 2x2f(x,t)
 ∂t    ∂x

With boundary conditions

pict

Mathematica
{{                                            } }
           ∑∞  − x√2−2i√2(n+ 12)t         (  √4- )
   f(x,t) →    e  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 102

Added January 20, 2018.

Solve

∂2u-  ∂4u-
∂t2 + ∂x4 = 0

With boundary conditions

pict

And initial conditions

pict

Mathematica
{{                }}
  u (x,t) → x4 − 12t2

Result Solved

Maple
         4    2
u(x,t) = x − 12t

Result Solved

9 Burger’s PDE

_________________________________________________________________________________

9.1 viscous fluid flow with no initial conditions

problem number 103

From Mathematica symbolic PDE document.

Solve for u(x,t)

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

Mathematica
{{            2                       }}
   u(x,t) → − 2c1μ-tanh-(c2t+-c1x+-c3)+-c2-
                        c1

Result Solved

Maple
            -       -      -     -     -C3-
u (x,t) = − 2μC2 tanh( C2x + C3t + C1)− -C2

Result Solved

_________________________________________________________________________________

9.2 viscous fluid flow with initial conditions

problem number 104

From Mathematica symbolic PDE document.

Solve for u(x,t)

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

With initial conditions

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

Mathematica
((                              ))
||||||                              ||||||
|{|{                   1          |}|}
|| u(x,t) → −-t−42μx(--(-x-)--)----||
|||(|||(          e----(erf-2)√μt-+1--+1 |||)|||)
               erf2t−√xμt+1

Result Solved

Maple
                    (      x  )   ∫ t∫ ∞       u(ζ1,τ1) ∂-u (ζ1,τ1)    (x−ζ1)2-
u(x,t) = 1∕2− 1∕2Erf  1∕2 √-√--- +         − 1∕2---√-√--∂√ζ1--------e1∕4μ(−t+τ1)dζ1dτ1
                          μ  t     0  −∞           π  μ  t− τ1

Result Solved, but has unresolved integrals

_________________________________________________________________________________

9.3 viscous fluid flow with initial conditions as UnitBox

problem number 105

From Mathematica DSolve help pages.

Solve for u(x,t)

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

With initial conditions

        {  1     |x| ≤ 1
u (x,0) =   0    otherw2ise

Mathematica
((                               t+4μ1(   (2t−-2x−1)     ( 2t−2x+1))                     ) )
{{ u(x,t) → -----(----)---------e(----erf)--4√μt---−(-erf--4√)μt------(-----)-----------} }
((          e x2μerf 1−√2x + et+41μ erf 2t−√2x−1 − et+41μ erf 2t−√2x+1 + ex+2μ1erf  2x+√1- − e x2μ − ex+2μ1) )
                  4 μt            4 μt              4 μt            4 μt

Result Solved

Maple
               (         )        (          )
                   2x + 1              2x − 1   ∫ t ∫ ∞     u(ζ1,τ 1) ∂∂ζ1u (ζ1,τ1) 1∕4 (x−ζ1)2
u(x,t) = 1∕2Erf 1∕4√-μ√t- − 1∕2Erf  1∕4√-μ√t- +        − 1∕2---√π-√μ√t-−-τ1---e   μ(−t+τ1)dζ1dτ1
                                                 0  −∞

Result Solved, but has unresolved integrals

10 Black Scholes PDE

_________________________________________________________________________________

10.1 classic Black Scholes model from finance

problem number 106

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
{{          1    (        ( 2log(k )+ (2r + σ2)(t− T)− 2log(S))         ( 2log(k)+ (2r − σ2)(t− T)− 2log(S))) }}
   V(S,t) → -e−rT  SerTerfc  -----------√---√----------------  − kerterfc  -----------√---√----------------
            2                         2  2σ  T − t                                2  2σ  T − t

Result Solved

Maple
sol = ()

Result Did not solve

_________________________________________________________________________________

10.2 Boundary value problem for the Black-Scholes equation

problem number 107

From Mathematica DSolve help pages.

Solve for V (t,s)

            2
∂v-+ 1σ2s2∂-v2 + (r− q)s∂v− rv(t,s) = 0
∂t   2    ∂s           ∂s

With boundary condition v(T,s) = ψ(s)

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

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

Result Solved

Maple
sol = ()

Result Did not solve

11 Korteweg-deVries PDE

_________________________________________________________________________________

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

problem number 108

From Mathematica symbolic PDE document.

Solve for u(x,t)

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

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

Mathematica
{{         12c31tanh2(c2t+-c1x-+c3)−-8c31 +-c2} }
   u(x,t) →               6c1

Result Solved

Maple
                                              3
u(x,t) = 2-C22(tanh( C2x +-C3t + C1 ))2 − 1∕68-C2-−-C3
                                              -C2

Result Solved

12 Tricomi PDE

_________________________________________________________________________________

12.1 Boundary value problem for the Tricomi equation

problem number 109

From Mathematica DSolve helps pages.

Solve for u(x,y)

∂2u-   ∂2u-
∂x2 + y ∂y2 = 0

With boundary conditions

pict

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

Result Solved

Maple
sol = ()

Result Did not solve

13 Cauchy-Riemann PDE’s

_________________________________________________________________________________

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

problem number 110

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

_________________________________________________________________________________

13.2 Cauchy-Riemann PDE With extra term on right side

problem number 111

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

_________________________________________________________________________________

14.1 Hamilton-Jacobi type PDE

problem number 112

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

Result Did not Solve

Maple
                                        ∫ ∘ ------------------------------------------------------------------------   ∫  ∘------------------------------------------------------------------------
                                          --−-2η4m-σ2-c1 +-2b(η)η2m-σ2-+2η2-c1σ2m-−-2η2 c3σ2m-−-2b(η)σ2m-+-2 c3σ2m-−-c42      − 2m σ2ξ4-c1 − 2a(ξ)m σ2ξ2 + 2ξ2 c1σ2m − 2m σ2
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

_________________________________________________________________________________

15.1 A second order PDE

problem number 113

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-c-x+--C1∘ -C2-c-2 +-cy-
S (x,y) = ----1-------------1-----1-
                     c1

Result Solved