laplace’s and poisson’s equation - open university · boundary conditions suppose that...

Laplace’s and Poisson’s equation

Laplace:

𝛻2𝑢 = 0

In 2D – rectangular coordinates:

𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2= 0

Poisson:𝛻2𝑢 = 𝑓

Applications

• Irrotational flow of an inviscid, incompressible fluid:

𝛻2𝜙 = 0where 𝜙 is the velocity potential function 𝐮 = 𝛁𝜙

• Heat flow, 3D diffusion

𝜕𝑢

𝜕𝑡− 𝑘𝛻2𝑢 = 0 or in steady state: 𝛻2𝑢 = 0

• Membrane motion, no external forces: 𝛻2𝑢 = 0

Boundary conditions

Suppose that Laplace’s equation, ∇2u = 0, holds throughout a region B with a bounding surface S.

(i) If u is a specified function on the bounding surface S, the problem is a Dirichlet problem.

(ii) If ∂u/∂n is a specified function on the bounding surface S, the problem is a Neumann problem.

(iii)If u is a specified function on part of the bounding surface S, and ∂u/∂n is a specified function on the remaining part of the surface,the problem is a mixed (Dirichlet–Neumann) problem.

Procedure

• Write down problem in mathematical form.

• Look for solutions of form u(x, y)= X(x) Y(y).

• Substitute in partial differential equation and separate to obtain two ordinary differential equations.

• Substitute in homogeneous boundary conditions and find resulting conditions on X(x) and Y(y).

• Identify the eigenvalue problem and solve it.

• Solve the other ordinary differential equation with the corresponding boundary condition.

• Take product of the solutions in steps above to find solutions of partial differential equation and homogeneous boundary conditions

• Take linear combination of these solutions.

• Substitute in inhomogeneous boundary condition. Use methods of Unit 10 to find the coefficients.

• Write down solution to the problem

Identify the eigenvalue problem and solve it.

The eigenvalue problem is:𝑋′′ + 𝜆𝑋 = 0 𝑋 0 = 0, 𝑋 𝑎 = 0

Case 1: 𝜆 = −𝑝2

𝑋′′ − 𝑝2𝑋 = 0𝑋 = 𝐴 cosh 𝑝𝑥 + 𝐵 sinh 𝑝𝑥

If 𝑋 0 = 0, then 𝐴 cosh0 = 0, ie 𝐴 = 0If 𝑋 𝑎 = 0, then 𝐵 sinh 𝑝𝑎 = 0, ie 𝐵 = 0So no solution of this form



Case 2: 𝜆 = 0𝑋′′ = 0

𝑋 = 𝐴𝑥 + 𝐵If 𝑋 0 = 0, then 𝐵 = 0If 𝑋 𝑎 = 0, then 𝐴𝑎 = 0, ie 𝐴 = 0So no solution of this form



Case 3: 𝜆 = 𝑝2

𝑋′′ + 𝑝2𝑋 = 0𝑋 = 𝐴 cos 𝑝𝑥 + 𝐵 sin 𝑝𝑥

If 𝑋 0 = 0, then 𝐴 cos 0 = 0, ie 𝐴 = 0

If 𝑋 𝑎 = 0, then 𝐵 sin 𝑝𝑎 = 0, ie 𝑝 =𝑛𝜋

𝑎

And we have solutions of the form 𝑋𝑛 𝑥 = 𝐵𝑛 sin𝑛𝜋𝑥

𝑎

with

eigenvalues 𝜆𝑛 =𝑛2𝜋2

𝑎2, 𝑛 = 1, 2, 3…

Solve the other ordinary differential equation with the corresponding boundary condition

We now need to solve:𝑌′′ − 𝜆𝑌 = 0

𝑌𝑛 = 𝐶𝑛 cosh𝑛𝜋𝑦

𝑎+ 𝐷𝑛 sinh

𝑛𝜋𝑦

𝑎

If 𝑌𝑛 0 = 0, then 𝐶𝑛 cosh0 = 0, ie 𝐶𝑛 = 0So we are left with

𝑌𝑛 = 𝐷𝑛 sinh𝑛𝜋𝑦

𝑎


𝑎

with eigenvalues 𝜆𝑛 =𝑛2𝜋2

𝑎2, 𝑛 = 1, 2, 3…


𝑎

with eigenvalues 𝜆𝑛 =𝑛2𝜋2

𝑎2, 𝑛 = 1, 2, 3…

𝑌𝑛 = 𝐷𝑛 sinh𝑛𝜋𝑦

𝑎

Take product of the solutions in steps above to find solutions of partial differential equation and homogeneous boundary conditions

𝑧𝑛 𝑥, 𝑦 = 𝑋𝑛 𝑥 𝑌𝑛 = 𝐾𝑛 sin𝑛𝜋𝑥

𝑎sinh

𝑛𝜋𝑦

𝑎𝑛 = 1, 2, 3… .

Take linear combination of these solutions.

𝑧 𝑥, 𝑦 =

𝑛=1

∞

𝐾𝑛 sin𝑛𝜋𝑥

𝑎sinh

𝑛𝜋𝑦

𝑎

𝑧 𝑥, 𝑦 =

𝑛=1

∞


𝑎sinh

𝑛𝜋𝑦

𝑎

𝑧 𝑥, 𝑏 =

𝑛=1

∞


𝑎sinh

𝑛𝜋𝑏

𝑎= 𝑧0 sin

𝜋𝑥

𝑎

So we must have 𝐾1 =𝑧0

sinh𝜋𝑏

𝑎

and 𝐾𝑛 = 0, 𝑛 ≠ 1 so our particular solution is:

𝑧 𝑥, 𝑦 =𝑧0

sinh𝜋𝑏𝑎

sin𝜋𝑥

𝑎sinh

𝜋𝑦

𝑎

𝜕2𝜃

𝜕𝑥2+𝜕2𝜃

𝜕𝑦2= 0 (0 < 𝑥 < 𝑎, 0 < 𝑦 < 𝑏)

𝜃 0, 𝑦 = 0 0 < 𝑦 < 𝑏𝜕𝜃

𝜕𝑥𝑎, 𝑦 = 0 0 < 𝑦 < 𝑏

𝜃 𝑥, 0 = 0 0 < 𝑥 < 𝑎

𝜃 𝑥, 𝑏 = 𝜃0 sin𝜋𝑥

2𝑎(0 < 𝑥 < 𝑎)

𝜕2𝜃

𝜕𝑥2+𝜕2𝜃

𝜕𝑦2= 0 (0 < 𝑥 < 𝑎, 0 < 𝑦 < 𝑏)

𝑥 = 0: 𝑋 0 = 0𝑥 = 𝑎: 𝑋′ 𝑎 = 0𝑦 = 0: 𝑌 0 = 0

𝑦 = 𝑏: 𝑌 𝑏 = 𝜃0 sin𝜋𝑥

2𝑎

𝑋′′𝑌 + 𝑌′′𝑋 = 0𝑋′′

𝑋= −

𝑌′′

𝑌= −𝜆

𝑋′′ + 𝜆𝑋 = 0𝑌′′ − 𝜆𝑌 = 0


𝑋′′ − 𝑝2𝑋 = 0𝑋 = 𝐴 cosh𝑝𝑥 + 𝐵 sinh𝑝𝑥

If 𝑋 0 = 0, then 𝐴 cosh0 = 0, ie 𝐴 = 0If 𝑋′ 𝑎 = 0, then 𝐵𝑝 cosh 𝑝𝑎 = 0, ie 𝐵 = 0So no solution of this form

𝜕2𝜃

𝜕𝑥2+𝜕2𝜃

𝜕𝑦2= 0 (0 < 𝑥 < 𝑎, 0 < 𝑦 < 𝑏)

𝑥 = 0: 𝑋 0 = 0𝑥 = 𝑎: 𝑋′ 𝑎 = 0𝑦 = 0: 𝑌 0 = 0


2𝑎

𝑋′′𝑌 + 𝑌′′𝑋 = 0𝑋′′

𝑋= −

𝑌′′

𝑌= −𝜆

𝑋′′ + 𝜆𝑋 = 0𝑌′′ − 𝜆𝑌 = 0Case 2: 𝜆 = 0

𝑋′′ = 0𝑋 = 𝐴𝑥 + 𝐵

If 𝑋 0 = 0, then 𝐵 = 0If 𝑋′ 𝑎 = 0, then 𝐴 = 0, ie 𝐴 = 0So no solution of this form

𝜕2𝜃

𝜕𝑥2+𝜕2𝜃

𝜕𝑦2= 0 (0 < 𝑥 < 𝑎, 0 < 𝑦 < 𝑏)

𝑥 = 0: 𝑋 0 = 0𝑥 = 𝑎: 𝑋′ 𝑎 = 0𝑦 = 0: 𝑌 0 = 0


2𝑎

𝑋′′ + 𝑝2𝑋 = 0𝑋 = 𝐴 cos 𝑝𝑥 + 𝐵 sin 𝑝𝑥

If 𝑋 0 = 0, then 𝐴 cos 0 = 0, ie 𝐴 = 0

If 𝑋′ 𝑎 = 0, then 𝐵 cos 𝑝𝑎 = 0, ie 𝑝 =2𝑛−1 𝜋

2𝑎𝑛 = 1,2,3… .

And we have solutions of the form 𝑋𝑛 𝑥 = 𝐵𝑛 sin2𝑛−1 𝜋𝑥

2𝑎

with eigenvalues 𝜆𝑛 =2𝑛−1 2𝜋2

4𝑎2, 𝑛 = 1, 2, 3…

𝑋𝑛 𝑥 = 𝐵𝑛 sin2𝑛−1 𝜋𝑥

2𝑎


4𝑎2, 𝑛 = 1, 2, 3…

𝑦 = 0: 𝑌 0 = 0


2𝑎

𝑌′′ − 𝜆𝑌 = 0

𝑌𝑛 = 𝐶𝑛 cosh2𝑛 − 1 𝜋𝑦

2𝑎+ 𝐷𝑛 sinh

2𝑛 − 1 𝜋𝑦

2𝑎If 𝑌𝑛 0 = 0, then 𝐶n cosh0 = 0, ie 𝐶𝑛 = 0So we are left with

𝑌𝑛 = 𝐷𝑛 sinh2𝑛 − 1 𝜋𝑦

2𝑎

𝑋𝑛 𝑥 = 𝐵𝑛 sin2𝑛−1 𝜋𝑥

2𝑎


4𝑎2, 𝑛 = 1, 2, 3…

𝑦 = 0: 𝑌 0 = 0


2𝑎

𝑌𝑛 = 𝐷𝑛 sinh2𝑛 − 1 𝜋𝑦

2𝑎

𝜃𝑛 𝑥, 𝑦 = 𝑋𝑛 𝑥 𝑌𝑛 = 𝐾𝑛 sin2𝑛 − 1 𝜋𝑥

2𝑎sinh

2𝑛 − 1 𝜋𝑦

2𝑎𝑛 = 1, 2, 3… .


𝜃 𝑥, 𝑦 =

𝑛=1

∞

𝐾𝑛 sin2𝑛 − 1 𝜋𝑥

2𝑎sinh

2𝑛 − 1 𝜋𝑦

2𝑎

𝜃 𝑥, 𝑦 =

𝑛=1

∞


2𝑎sinh

2𝑛 − 1 𝜋𝑦

2𝑎

𝑌 𝑏 = 𝜃0 sin𝜋𝑥

2𝑎

𝜃 𝑥, 𝑏 =

𝑛=1

∞


2𝑎sinh

2𝑛 − 1 𝜋𝑏

2𝑎= 𝜃0 sin

𝜋𝑥

2𝑎

So we must have 𝐾1 =𝜃0

sinh𝜋𝑏

2𝑎

and 𝐾𝑛 = 0, 𝑛 ≠ 1 so our particular solution is:

𝜃 𝑥, 𝑦 =𝜃0

sinh𝜋𝑏2𝑎

sin𝜋𝑥

2𝑎sinh

𝜋𝑦

2𝑎

Superposition

• The Dirichlet problem

𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2= 0 (0 < 𝑥 < 𝐿1, 0 < 𝑦 < 𝐿2)

𝑢 0, 𝑦 = 𝑓1 𝑦 0 < 𝑦 < 𝐿2𝑢 𝐿1, 𝑦 = 𝑓2 𝑦 0 < 𝑦 < 𝐿2𝑢 𝑥, 0 = 𝑓3 𝑥 0 < 𝑥 < 𝐿1𝑢 𝑥, 𝐿2 = 𝑓4 𝑥 (0 < 𝑥 < 𝐿1),

Solution 𝑢 = 𝑢1 + 𝑢2 + 𝑢3 + 𝑢4, where 𝑢1, 𝑢2, 𝑢3, 𝑢4 are each solutions of Laplace’s equation

Superposition

𝑢1, 𝑢2, 𝑢3, 𝑢4 are each solutions of Laplace’s equation such that

• 𝑢1 = 0 on each boundary except 𝑥 = 0 where:𝑢1 0, 𝑦 = 𝑓1 𝑦 0 < 𝑦 < 𝐿2

• 𝑢2 = 0 on each boundary except 𝑥 = 𝐿1 where:𝑢2 𝐿1, 𝑦 = 𝑓2 𝑦 0 < 𝑦 < 𝐿2

• 𝑢3 = 0 on each boundary except y= 0 where:𝑢3 𝑥, 0 = 𝑓3 𝑥 0 < 𝑥 < 𝐿1

• 𝑢4 = 0 on each boundary except 𝑦 = 𝐿2 where:𝑢4 𝑥, 𝐿2 = 𝑓4 𝑥 (0 < 𝑥 < 𝐿1),

Consider the function 𝑢 𝑥, 𝑦 that satisfies:𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2= 0 (0 < 𝑥 < 2, 0 < 𝑦 < 𝜋)

𝑢 𝑥, 0 = 𝑢 𝑥, 𝜋 = 0 0 < 𝑥 < 2𝑢 0, 𝑦 = 0 (0 < 𝑦 < 𝜋)

𝑢 2, 𝑦 = sin 3𝑦 (0 < 𝑦 < 𝜋)

𝑋′′𝑌 + 𝑌′′𝑋 = 0𝑋′′

𝑋= −

𝑌′′

𝑌= −𝜆

𝑋′′ + 𝜆𝑋 = 0𝑌′′ − 𝜆𝑌 = 0

𝑥 = 0: 𝑋 0 = 0𝑥 = 2: 𝑋 2 = sin 3𝑦𝑦 = 0: 𝑌 0 = 0𝑦 = 𝜋: 𝑌 𝜋 = 0

𝑋′′𝑌 + 𝑌′′𝑋 = 0𝑋′′

𝑋= −

𝑌′′

𝑌= −𝜆

𝑋′′ + 𝜆𝑋 = 0𝑌′′ − 𝜆𝑌 = 0

𝑥 = 0: 𝑋 0 = 0𝑥 = 2: 𝑋 2 = sin 3𝑦𝑦 = 0: 𝑌 0 = 0𝑦 = 𝜋: 𝑌 𝜋 = 0

The eigenvalue problem is:𝑌′′ − 𝜆𝑌 = 0 𝑌 0 = 0, 𝑌 𝜋 = 0Case 1: 𝜆 = 𝑝2

𝑋′′ − 𝑝2𝑋 = 0𝑋 = 𝐴 cosh𝑝𝑥 + 𝐵 sinh𝑝𝑥

If 𝑌 0 = 0, then 𝐴 cosh0 = 0, ie 𝐴 = 0If 𝑌 0 = 0, then 𝐵 sinh 𝑝𝜋 = 0, ie 𝐵 = 0So no solution of this form

Case 2: 𝜆 = 0𝑌′′ = 0

𝑌 = 𝐴𝑦 + 𝐵If 𝑌 0 = 0, then 𝐵 = 0If 𝑌 0 = 0, then 𝐴𝜋 = 0, ie 𝐴 = 0So no solution of this form

𝑋′′𝑌 + 𝑌′′𝑋 = 0𝑋′′

𝑋= −

𝑌′′

𝑌= −𝜆

𝑋′′ + 𝜆𝑋 = 0𝑌′′ − 𝜆𝑌 = 0

𝑥 = 0: 𝑋 0 = 0𝑥 = 2: 𝑋 2 = sin 3𝑦𝑦 = 0: 𝑌 0 = 0𝑦 = 𝜋: 𝑌 𝜋 = 0


𝑌′′ + 𝑝2𝑌 = 0𝑌 = 𝐴 cos𝑝𝑦 + 𝐵 sin 𝑝𝑦

If 𝑌 0 = 0, then 𝐴 cos 0 = 0, ie 𝐴 = 0If 𝑌 𝜋 = 0, then 𝐵 sin 𝑝𝜋 = 0, ie 𝑝 = 𝑛, 𝑛 = 1,2,3. .And we have solutions of the form 𝑌𝑛 𝑦 = 𝐵𝑛 sin 𝑛𝑦 with eigenvalues 𝜆𝑛 = 𝑛2,

𝑛 = 1, 2, 3

𝑌𝑛 𝑦 = 𝐵𝑛 sin 𝑛𝑦 with eigenvalues 𝜆𝑛 = 𝑛2, 𝑛 = 1, 2, 3

𝑥 = 0: 𝑋 0 = 0𝑥 = 2: 𝑋 2 = sin 3𝑦

We now need to solve:𝑋′′ − 𝑛2𝑋 = 0

Which has the general solution:𝑋𝑛 = 𝐶𝑛 cosh𝑛𝑥 + 𝐷𝑛 sinh 𝑛𝑥

If 𝑋𝑛 0 = 0, then 𝐶𝑛 cosh0 = 0, ie 𝐶𝑛 = 0So we are left with

𝑋𝑛 = 𝐷𝑛 sinh 𝑛𝑥


𝑥 = 0: 𝑋 0 = 0𝑥 = 2: 𝑋 2 = sin 3𝑦

𝑋𝑛 = 𝐷𝑛 sinh 𝑛𝑥

So now we have solutions to the PDE with form:

𝑢𝑛 𝑥, 𝑦 = 𝑋𝑛 𝑥 𝑌𝑛(𝑦) = 𝐾𝑛 sinh𝑛𝑥 sin 𝑛𝑦 𝑛 = 1, 2, 3… .


𝑢 𝑥, 𝑦 =

𝑛=1

∞

𝐾𝑛 sinh𝑛𝑥 sin 𝑛𝑦


𝑥 = 0: 𝑋 0 = 0𝑥 = 2: 𝑋 2 = sin 3𝑦

𝑢 𝑥, 𝑦 =

𝑛=1

∞

𝐾𝑛 sinh 𝑛𝑥 sin 𝑛𝑦

Using:𝑋 2 = sin 3𝑦

𝑢 2, 𝑦 =

𝑛=1

∞

𝐾𝑛 sinh2𝑛 sin 𝑛𝑦 = sin 3𝑦

So we must have 𝐾3 =1

sinh 6and 𝐾𝑛 = 0, 𝑛 ≠ 3 so our particular solution is:

𝑢 𝑥, 𝑦 =1

sinh6sinh 3𝑥 sin 3𝑦

Consider the function 𝑢 𝑥, 𝑦 that satisfies:𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2= 0 (0 < 𝑥 < 2, 0 < 𝑦 < 𝜋)

𝑢 𝑥, 0 = 𝑢 𝑥, 𝜋 = 0 0 < 𝑥 < 2𝑢 0, 𝑦 = 𝑦(𝜋2 − 𝑦2) (0 < 𝑦 < 𝜋)

𝑢 2, 𝑦 = 0 (0 < 𝑦 < 𝜋)

𝑋′′𝑌 + 𝑌′′𝑋 = 0𝑋′′

𝑋= −

𝑌′′

𝑌= −𝜆

𝑋′′ + 𝜆𝑋 = 0𝑌′′ − 𝜆𝑌 = 0

𝑥 = 0: 𝑋 0 =𝑦(𝜋2 − 𝑦2)𝑥 = 2: 𝑋 2 = 0𝑦 = 0: 𝑌 0 = 0𝑦 = 𝜋: 𝑌 𝜋 = 0


𝑥 = 0: 𝑋 0 = 𝑦(𝜋2 − 𝑦2)𝑥 = 2: 𝑋 2 = 0𝑦 = 0: 𝑌 0 = 0𝑦 = 𝜋: 𝑌 𝜋 = 0


𝑋′′ − 𝑛2𝑋 = 0

𝑋𝑛 = 𝐶𝑛 cosh𝑛(2 − 𝑥) + 𝐷𝑛 sinh 𝑛(2 − 𝑥)

And then when 𝑋𝑛 2 = 0, then 𝐶𝑛 = 0So we are left with

𝑋𝑛 = 𝐷𝑛 sinh𝑛(2 − 𝑥)

𝑥 = 0: 𝑋 0 = 𝑦(𝜋2 − 𝑦2)𝑥 = 2: 𝑋 2 = 0𝑦 = 0: 𝑌 0 = 0𝑦 = 𝜋: 𝑌 𝜋 = 0


𝑋𝑛 = 𝐷𝑛 sinh𝑛(2 − 𝑥)

𝑢𝑛 𝑥, 𝑦 = 𝑋𝑛 𝑥 𝑌𝑛(𝑦) = 𝐾𝑛 sinh 𝑛(2 − 𝑥) sin 𝑛𝑦 𝑛 = 1, 2, 3… .Take linear combination of these solutions.

𝑢 𝑥, 𝑦 =

𝑛=1

∞

𝐾𝑛 sinh𝑛(2 − 𝑥) sin 𝑛𝑦 (𝑛 = 1, 2, 3, … . . )

𝑢 𝑥, 𝑦 =

𝑛=1

∞

𝐾𝑛 sinh 𝑛(2 − 𝑥) sin 𝑛𝑦 𝑛 = 1, 2, 3, … . .

𝑋 0 = 𝑦(𝜋2 − 𝑦2)

𝑢 0, 𝑦 =

𝑛=1

∞

𝐾𝑛 sinh2𝑛 sin 𝑛𝑦 = 𝑦(𝜋2 − 𝑦2)

𝐾𝑛 sinh2𝑛 =2

𝜋න0

𝜋

𝑦 𝜋2 − 𝑦2 sin 𝑛𝑦 𝑑𝑦 = −2

𝜋×6𝜋

𝑛3−1 𝑛

𝑢 𝑥, 𝑦 =

𝑛=1

∞

−12

𝑛3−1 𝑛

sinh 𝑛 2 − 𝑥 sin 𝑛𝑦

sinh2𝑛(𝑛 = 1, 2, 3,… . . )

න0

𝜋

𝑦 𝜋2 − 𝑦2 sin 𝑛𝑦 𝑑𝑦 = −6𝜋

𝑛3−1 𝑛 𝑛 = 1,2,3… .

By applying an appropriate result and using your answers to parts (a) and (b), find the solution to the problem when:𝑓 𝑦 = 𝑥 𝜋2 − 𝑦2 and 𝑔 𝑦 = sin 3𝑦.

𝑢1 𝑥, 𝑦 =1

sinh6sinh3𝑥 sin 3𝑦

𝑢2 𝑥, 𝑦 =

𝑛=1

∞

−12

𝑛3−1 𝑛

sinh𝑛 2 − 𝑥 sin 𝑛𝑦

sinh 2𝑛(𝑛 = 1, 2, 3, … . . )

Using the principle of superposition, the solution is then the sum of the results from a and b:

𝑢 𝑥, 𝑦 =1

sinh6sinh3𝑥 sin 3𝑦 −

𝑛=1

∞12

𝑛3−1 𝑛

sinh 𝑛 2 − 𝑥 sin 𝑛𝑦

sinh2𝑛

(𝑛 = 1, 2, 3, … . . )

Assuming that the series found for 𝑢 0, 𝑦 in b converges pointwise for 0 ≤ 𝑦 ≤ 𝜋 show that:

1 −1

33+

1

53−

1

73+. . =

𝜋3

32

Choose a value of 𝑦 =𝜋

2:

𝜋

2𝜋2 −

𝜋2

4= −

𝑛=1

∞12

𝑛3−1 𝑛 sin

𝑛𝜋

2

3𝜋3

8= −12

1

13−1 sin

𝜋

2+

1

33−1 3 sin

3𝜋

2+

1

53−1 5 sin

5𝜋

2+

1

73−1 7 sin

7𝜋

2+ ⋯

ie

1 −1

33+

1

53−

1

73+. . =

𝜋3

32

laplace’s and poisson’s equation - open university · boundary conditions suppose that...

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