TWO-DIMENSIONAL PARTIAL ELLIPTIC
DIFFERENTIAL EQUATIONS IN MAPLE

Alexei V. Tikhonenko

General Physics Department,
State Technical University of Nuclear Power Engineering,
Obninsk, Russia

futureprint@obninsk.com 

This work introduces functional programming method in MAPLE for boundary
problem solving of two-dimensional partial elliptic differential equations in polar coordinates. 

1. Introduction 

1.1. Two-dimensional Laplace's equation in polar coordinates 

>

 

Solution of boundary-value problem for Laplace's equation 

Δu = 0 

for circle and ring can be solved by the variable separation method. 

For this purpose we will use "VectorCalculus" package 

 

0 < r < ∞, 0 < φ < 2π, 

>

 

and polar coordinate system: 

>

 

(1.1.1)

 

So programming of Laplace's equation in these coordinates gives: 

>

 

(1.1.2)

 

1.2. Variable separation and general solution 

Laplace's equation admits the variable separation and with MAPLE aids we will have separated equations: 

>

 

(1.2.1)

 

After constant replacement 

>

 

(1.2.2)

 

we will solve separated equations. 

Angular equation: 

>

 

(1.2.3)

 

>

 

(1.2.4)

 

Solution for angular function Φ(φ) will satisfy to periodicity condition with integer n. 

Radial equation: 

a)  n ≠ 0: 

>

 

(1.2.5)

 

>

 

(1.2.6)

 

b)  n= 0: 

>

 

(1.2.7)

 

>

 

(1.2.8)

 

So solutions of angular and radial equations will have forms (with changed constants): 

>

 

(1.2.9)

 

>

 

(1.2.10)

 

>

 

(1.2.11)

 

Thus general solution of two-dimensional Laplace's equation in polar coordinates is: 

>

 

(1.2.12)

 

2. The first boundary-value problem for Laplace's equation in circle 

2.1. The first boundary-value problem statement (inside of circle) 

 

Consider the first boundary-value problem for Laplace's equation inside of circle (internal Dirichlet problem for circle). 

This problem implies that function u satisfies the Laplace's equation: 

Δu = 0 

and boundary condition on the circle boundary S: 

u(S) = f, 

where f is given function and S is circumference of radius R₀. 

General solution of two-dimensional Laplace's equation in polar coordinates is (see point 1.2): 

>

 

(2.1.1)

 

and boundary conditions are: 

1) boundedness of solution at origin of coordinates: 

u < ∞; 

2) value of function u on the circle boundary S is: 

u(R₀,φ) = f(φ). 

2.2. Accounting of boundary conditions (inside of circle) 

1) Boundedness of solution at origin of coordinates implies: 

B2₀ = 0, B2_n = 0, 

whence it follows (with redefinition of A1_n, A2_n and B1_n): 

>

 

 

 

	(2.2.1)

 

>

 

 

 

	(2.2.2)

 

So we have: 

>

 

(2.2.3)

 

>

 

(2.2.4)

 

2) The second boundary condition implies: 

>

 

(2.2.5)

 

>

 

(2.2.6)

 

>

 

(2.2.7)

 

Using these relations we will find: 

>

 

(2.2.8)

 

>

 

(2.2.9)

 

2.3. Calculation of coefficients (inside of circle) 

Using of formulas (6.1) - (6.6) (see Appendix 1) - orthonormality conditions - gives: 

>

 

(2.3.1)

 

>

 

(2.3.2)

 

>

 

 

 

	(2.3.3)

 

As a result of calculations we find: 

>

 

 

 

	(2.3.4)

 

or 

>

 

 

 

	(2.3.5)

 

And solution of the first boundary-value problem for Laplace's equation inside of circle is: 

>

 

(2.3.6)

 

2.4. Poisson integral (inside of circle) 

This solution can be represented by the Poisson integral. For this purpose we will perform the transformation: 

>

 

(2.4.1)

 

and use variables 

r = ρ⋅R₀, 

α = φ - ψ. 

and formulas: 

>

 

(2.4.2)

 

>

 

(2.4.3)

 

After additional transformations we have: 

>

 

(2.4.4)

 

Using denotation (kernel of  Poisson integral): 

>

 

(2.4.5)

 

we receive integral form of the solution (Poisson integral): 

>

 

(2.4.6)

 

2.5. The first boundary-value problem statement (outside of circle) 

>

 

Consider the first boundary-value problem for Laplace's equation outside of circle (external Dirichlet problem for circle). 

This problem implies that function u satisfies the Laplace's equation: 

Δu = 0 

and boundary condition on the circle boundary S: 

u(S) = f, 

where f is given function and S is circumference of radius R₀. 

General solution of two-dimensional Laplace's equation in polar coordinates is (see point 1.2): 

>

 

(2.5.1)

 

and boundary conditions are: 

1) boundedness of solution at infinity: 

u < ∞; 

2) value of function u on the circle boundary S is: 

u(R0,φ) = f(φ). 

2.6. Accounting of boundary conditions (outside of circle) 

1) Boundedness of solution at infinity implies: 

B1₀ = 0, 

B1_n = 0, 

whence it follows (with redefinition of A1_n, A2_n and B2_n): 

>

 

 

 

	(2.6.1)

 

>

 

 

 

	(2.6.2)

 

So we have: 

>

 

(2.6.3)

 

>

 

(2.6.4)

 

2) The second boundary condition implies: 

>

 

(2.6.5)

 

>

 

(2.6.6)

 

>

 

(2.6.7)

 

Using these relations we will find: 

>

 

(2.6.8)

 

>

 

(2.6.9)

 

2.7. Calculation of coefficients (outside of circle) 

Using of formulas (6.1) - (6.6) (see Appendix 1) - orthonormality conditions - gives: 

>

 

(2.7.1)

 

>

 

(2.7.2)

 

>

 

 

 

	(2.7.3)

 

As a result of calculations we find: 

>

 

 

 

	(2.7.4)

 

or 

>

 

 

 

	(2.7.5)

 

And solution of the first boundary-value problem for Laplace's equation outside of circle is: 

>

 

(2.7.6)

 

2.8. Poisson integral (outside of circle) 

This solution can be represented by the Poisson integral. For this purpose we will perform the transformation: 

>

 

(2.8.1)

 

and use variables 

R₀ = ρ⋅r, 

α = φ - ψ. 

and formulas: 

>

 

(2.8.2)

 

>

 

(2.8.3)

 

After additional transformations we have: 

>

 

(2.8.4)

 

Using denotation (kernel of  Poisson integral): 

>

 

(2.8.5)

 

we receive integral form of the solution (Poisson integral): 

>

 

(2.8.6)

 

2.9. General representation of the solution in expanded form for circle 

>

 

The solution inside of circle is: 

>

 

(2.9.1)

 

where 

>

 

 

 

	(2.9.2)

 

Now we redefine coefficients: 

>

 

 

 

	(2.9.3)

 

and get the formula: 

>

 

(2.9.4)

 

The solution outside of circle is: 

>

 

(2.9.5)

 

where 

>

 

 

 

	(2.9.6)

 

Now we redefine coefficients: 

>

 

 

 

	(2.9.7)

 

and get the formula: 

>

 

(2.9.8)

 

Thus we have the solution in expanded form for circle: 

>

 

(2.9.9)

 

2.10. Poisson integral for circle 

>

 

Poisson integral for circle inside of circle is: 

>

 

(2.10.1)

 

>

 

(2.10.2)

 

Poisson integral for circle outside of circle is: 

>

 

(2.10.3)

 

>

 

(2.10.4)

 

So the Poisson integral for circle is: 

>

 

(2.10.5)

 

>

 

(2.10.6)

 

3. The second boundary-value problem for Laplace's equation in circle 

3.1. The second boundary-value problem statement (inside of circle) 

>

 

Consider the first boundary-value problem for Laplace's equation inside of circle (internal Neumann problem for circle). 

This problem implies that function u satisfies the Laplace's equation: 

Δu = 0 

and boundary condition on the circle boundary S: 

= g, 

where g is given function and S is circumference of radius R₀. 

General solution of two-dimensional Laplace's equation in polar coordinates is (see point 1.2): 

>

 

(3.1.1)

 

and boundary conditions are: 

1) boundedness of solution at origin of coordinates: 

u < ∞; 

2) value of derivative of function u on the circle boundary S is: 

= g(φ). 

3.2. Accounting of boundary conditions (inside of circle) 

1) Boundedness of solution at origin of coordinates implies: 

B2₀ = 0, 

B2_n = 0, 

whence it follows (with redefinition of A1_n, A2_n and B1_n): 

>

 

 

 

	(3.2.1)

 

>

 

 

 

	(3.2.2)

 

So we have: 

>

 

(3.2.3)

 

and 

>

 

(3.2.4)

 

2) The second boundary condition implies: 

>

 

(3.2.5)

 

>

 

(3.2.6)

 

>

 

(3.2.7)

 

>

 

(3.2.8)

 

Using these relations we will find: 

>

 

(3.2.9)

 

>

 

(3.2.10)

 

3.3. Calculation of coefficients (inside of circle) 

Using of formulas (6.1) - (6.6) (see Appendix 1) - orthonormality conditions - gives: 

>

 

(3.3.1)

 

>

 

(3.3.2)

 

>

 

 

	(3.3.3)

 

As a result of calculations we find: 

>

 

 

	(3.3.4)

 

or 

>

 

 

	(3.3.5)

 

And solution of the second boundary-value problem for Laplace's equation inside of circle is: 

>

 

(3.3.6)

 

3.4. Poisson integral (inside of circle) 

This solution can be represented by the Poisson integral. For this purpose we will perform the transformation: 

>

 

(3.4.1)

 

and use variables 

r = ρ⋅R₀, 

α = φ - ψ. 

and formulas: 

>

 

(3.4.2)

 

Now we have 

>

 

(3.4.3)

 

Consider kernel of  Poisson integral: 

>

 

(3.4.4)

 

and relation: 

>

 

(3.4.5)

 

Integration gives: 

>

 

(3.4.6)

 

and 

>

 

(3.4.7)

 

So we receive integral form of the solution (Poisson integral): 

>

 

(3.4.8)

 

>

 

(3.4.9)

 

3.5. The second boundary-value problem statement (outside of circle) 

>

 

Consider the second boundary-value problem for Laplace's equation inside of circle (external Neumann problem for circle). 

This problem implies that function u satisfies the Laplace's equation: 

Δu = 0 

and boundary condition on the circle boundary S: 

= g, 

where g is given function and S is circumference of radius R₀. 

General solution of two-dimensional Laplace's equation in polar coordinates is (see point 1.2): 

>

 

(3.5.1)

 

and boundary conditions are: 

1) boundedness of solution at infinity: 

u < ∞; 

2) value of derivative of function u on the circle boundary S is: 

= g(φ). 

3.6. Accounting of boundary conditions (inside of circle) 

1) Boundedness of solution at origin of coordinates implies:: 

B2₀ = 0, 

B1_n = 0, 

whence it follows (with redefinition of A1_n, A2_n and B1_n): 

>

 

 

 

	(3.6.1)

 

>

 

 

 

	(3.6.2)

 

So we have: 

>

 

(3.6.3)

 

and 

>

 

(3.6.4)

 

2) The second boundary condition implies: 

>

 

(3.6.5)

 

>

 

(3.6.6)

 

>

 

(3.6.7)

 

>

 

(3.6.8)

 

Using these relations we will find: 

>

 

(3.6.9)

 

>

 

(3.6.10)

 

3.7. Calculation of coefficients (inside of circle) 

Using of formulas (6.1) - (6.6) (see Appendix 1) - orthonormality conditions - gives: 

>

 

(3.7.1)

 

>

 

(3.7.2)

 

>

 

 

	(3.7.3)

 

As a result of calculations we find: 

>

 

 

	(3.7.4)

 

or 

>

 

 

	(3.7.5)

 

And solution of the second boundary-value problem for Laplace's equation outside of circle is: 

>

 

(3.7.6)

 

3.8. Poisson integral (outside of circle) 

This solution can be represented by the Poisson integral. For this purpose we will perform the transformation: 

>

 

(3.8.1)

 

and use variables 

R₀ = ρ⋅r, 

α = φ - ψ. 

and formulas: 

>

 

(3.8.2)

 

Now we have 

>

 

(3.8.3)

 

Consider kernel of  Poisson integral: 

>

 

(3.8.4)

 

and relation: 

>

 

(3.8.5)

 

Integration gives: 

>

 

(3.8.6)

 

and 

>

 

(3.8.7)

 

So we receive integral form of the solution (Poisson integral): 

>

 

(3.8.8)

 

>

 

(3.8.9)

 

3.9. General representation of the solution in expanded form for circle 

>

 

The solution inside of circle is: 

>

 

(3.9.1)

 

where B1 is arbitrary constant and 

>

 

 

	(3.9.2)

 

Now we redefine coefficients: 

>

 

 

	(3.9.3)

 

and get the formula: 

>

 

(3.9.4)

 

The solution outside of circle is: 

>

 

(3.9.5)

 

where 

>

 

 

	(3.9.6)

 

Now we redefine coefficients: 

>

 

 

	(3.9.7)

 

and get the formula: 

>

 

(3.9.8)

 

Thus we have the solution in expanded form for circle: 

>

 

(3.9.9)

 

3.10. General representation of the solution in integral form 

>

 

Solution for circle inside of circle is: 

>

 

(3.10.1)

 

>

 

(3.10.2)

 

Solution for circle outside of circle is: 

>

 

(3.10.3)

 

>

 

(3.10.4)

 

So the the integral form of the solution for circle is: 

>

 

(3.10.5)

 

>

 

(3.10.6)

 

4. The third boundary-value problem for Laplace's equation in circle 

4.1. The third boundary-value problem statement (inside of circle) 

>

 

Consider the third boundary-value problem for Laplace's equation inside of circle. 

This problem implies that function u satisfies the Laplace's equation: 

Δu = 0 

and boundary condition on the circle boundary S: 

= h, 

where h is given function and S is circumference of radius R₀. 

General solution of two-dimensional Laplace's equation in polar coordinates is (see point 1.2): 

>

 

(4.1.1)

 

and boundary conditions are: 

1) boundedness of solution at origin of coordinates: 

u < ∞; 

2) value of derivative of function u on the circle boundary S is: 

= h(φ). 

4.2. Accounting of boundary conditions (inside of circle) 

1) Boundedness of solution at origin of coordinates implies: 

B2₀ = 0, 

B2_n = 0, 

whence it follows (with redefinition of  A1_n, A2_n and B1_n): 

>

 

 

 

	(4.2.1)

 

>

 

 

 

	(4.2.2)

 

So we have: 

>

 

(4.2.3)

 

and 

>

 

(4.2.4)

 

2) The second boundary condition implies: 

>

 

(4.2.5)

 

>

 

(4.2.6)

 

>

 

(4.2.7)

 

>

 

(4.2.8)

 

Using these relations we will find: 

>

 

(4.2.9)

 

>

 

(4.2.10)

 

4.3. Calculation of coefficients (inside of circle) 

Using of formulas (6.1) - (6.6) (see Appendix 1) - orthonormality conditions - gives: 

>

 

(4.3.1)

 

>

 

(4.3.2)

 

>

 

(4.3.3)

 

>

 

(4.3.4)

 

>

 

(4.3.5)

 

As a result of calculations we find: 

>

 

 

 

	(4.3.6)

 

or 

>

 

 

 

	(4.3.7)

 

And solution of the third boundary-value problem for Laplace's equation inside of circle is: 

>

 

(4.3.8)

 

or 

>

 

(4.3.9)

 

4.4. The third boundary-value problem statement (outside of circle) 

>

 

Consider the third boundary-value problem for Laplace's equation inside of circle. 

This problem implies that function u satisfies the Laplace's equation: 

Δu = 0 

and boundary condition on the circle boundary S: 

= h, 

where h is given function and S is circumference of radius R₀. 

General solution of two-dimensional Laplace's equation in polar coordinates is (see point 1.2): 

>

 

(4.4.1)

 

and boundary conditions are: 

1) boundedness of solution: 

u < ∞; 

2) value of derivative of function u on the circle boundary S is: 

= h(φ). 

4.5. Accounting of boundary conditions (outside of circle) 

1) Boundedness of solution at origin of coordinates implies: 

B2₀ = 0, 

B1_n = 0, 

whence it follows (with redefinition of  A1_n, A2_n and B1_n): 

>

 

 

 

	(4.5.1)

 

>

 

 

 

	(4.5.2)

 

So we have: 

>

 

(4.5.3)

 

and 

>

 

(4.5.4)

 

2) The second boundary condition implies: 

>

 

(4.5.5)

 

>

 

(4.5.6)

 

>

 

(4.5.7)

 

>

 

(4.5.8)

 

Using these relations we will find: 

>

 

(4.5.9)

 

>

 

(4.5.10)

 

4.6. Calculation of coefficients (outside of circle) 

Using of formulas (6.1) - (6.6) (see Appendix 1) - orthonormality conditions - gives: 

>

 

(4.6.1)

 

>

 

(4.6.2)

 

>

 

(4.6.3)

 

>

 

(4.6.4)

 

>

 

(4.6.5)

 

As a result of calculations we find: 

>

 

 

 

	(4.6.6)

 

or 

>

 

 

 

	(4.6.7)

 

And solution of the third boundary-value problem for Laplace's equation outside of circle is: 

>

 

(4.6.8)

 

or 

>

 

(4.6.9)

 

4.7. General representation of the solution in expanded form 

>

 

The solution inside of circle is: 

>

 

(4.7.1)

 

where 

>

 

 

 

	(4.7.2)

 

Now we redefine coefficients: 

>

 

 

 

	(4.7.3)

 

and get the formula: 

>

 

(4.7.4)

 

The solution outside of circle is: 

>

 

(4.7.5)

 

where 

>

 

 

 

	(4.7.6)

 

Now we redefine coefficients: 

>

 

 

 

	(4.7.7)

 

and get the formula: 

>

 

(4.7.8)

 

Thus we have the solution in expanded form for circle: 

>

 

(4.7.9)

 

or 

>

 

 

	(4.7.10)

 

>

 

(4.7.11)

 

5. The first boundary-value problem for Laplace's equation in ring 

5.1. The second boundary-value problem statement in ring 

>

 

Consider the first boundary-value problem for Laplace's equation inside of circle (internal Dirichlet problem for circle). 

This problem implies that function u satisfies the Laplace's equation: 

Δu = 0 

and boundary conditions on the circle boundaries S₁ and S₂: 

u(S₁) = f₁, u(S₂) = f₂ 

where f₁ and f₂ are given functions and S₁ and S₂ are circumferences of radiuses R₁ and R₂ correspondingly. 

General solution of two-dimensional Laplace's equation in polar coordinates is (see point 1.2): 

>

 

and boundary conditions are: 

1) value of function u on the internal boundary S₁ is: 

u(R₁,φ) = f₁(φ); 

2) value of function u on the external boundary S₂ is: 

u(R₁,φ) = f₁(φ). 

Now we redefine coefficients: 

>

 

and find 

>

 

(5.1.1)

 

5.2. Accounting of boundary conditions 

Boundary conditions are represented as 

>

 

(5.2.1)

 

>

 

(5.2.2)

 

>

 

(5.2.3)

 

so we have 

>

 

(5.2.4)

 

>

 

(5.2.5)

 

>

 

(5.2.6)

 

>

 

(5.2.7)

 

5.3. Calculation of coefficients 

Using of formulas (6.1) - (6.6) (see Appendix 1) - orthonormality conditions - gives: 

?) k = 0: 

>

 

(5.3.1)

 

>

 

(5.3.2)

 

?) k ≠ 0: 

>

 

(5.3.3)

 

>

 

(5.3.4)

 

>

 

(5.3.5)

 

>

 

(5.3.6)

 

and 

>

 

(5.3.7)

 

>

 

(5.3.8)

 

>

 

(5.3.9)

 

>

 

(5.3.10)

 

As a result of calculations we find: 

>

 

(5.3.11)

 

>

 

(5.3.12)

 

>

 

(5.3.13)

 

or 

>

 

(5.3.14)

 

>

 

(5.3.15)

 

>

 

(5.3.16)

 

>

 

(5.3.17)

 

>

 

(5.3.18)

 

>

 

(5.3.19)

 

And solution of the first boundary-value problem is: 

>

 

(5.3.20)

 

5.4. Integral representation of the solution 

This solution can be represented in integral form. We will use relations: 

>

 

(5.4.1)

 

>

 

(5.4.2)

 

and get formulas 

>

 

>

 

(5.4.3)

 

Using denotation: 

>

 

(5.4.4)

 

>

 

(5.4.5)

 

we get solution in integral form: 

>

 

(5.4.6)

 

Conclusions 

This work demonstrates universal possibilities of MAPLE in solving the partial elliptic differential equations. 

Appendix 1 

Orthonormality conditions for sin(k ) and cos(k ). 

>

 

(7.1)

 

>

 

(7.2)

 

>

 

(7.3)

 

>

 

(7.4)

 

>

 

(7.5)

 

>

 

(7.6)

 

References 

1. A.N. Tikhonov, A.A. Samarski. Equations of mathematical physics. Moskow. MSU, 1999. 

2. B.N. Budak, A.N. Tikhonov, A.A. Samarski. Problems on mathematical physics. Moskow. Fizmatlit, 2004. 

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