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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
Identify the eigenvalue problem and solve it.
The eigenvalue problem is:𝑋′′ + 𝜆𝑋 = 0 𝑋 0 = 0, 𝑋 𝑎 = 0
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
Identify the eigenvalue problem and solve it.
The eigenvalue problem is:𝑋′′ + 𝜆𝑋 = 0 𝑋 0 = 0, 𝑋 𝑎 = 0
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
And we have solutions of the form 𝑋𝑛 𝑥 = 𝐵𝑛 sin𝑛𝜋𝑥
𝑎
with eigenvalues 𝜆𝑛 =𝑛2𝜋2
𝑎2, 𝑛 = 1, 2, 3…
And we have solutions of the form 𝑋𝑛 𝑥 = 𝐵𝑛 sin𝑛𝜋𝑥
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
∞
𝐾𝑛 sin𝑛𝜋𝑥
𝑎sinh
𝑛𝜋𝑦
𝑎
𝑧 𝑥, 𝑏 =
𝑛=1
∞
𝐾𝑛 sin𝑛𝜋𝑥
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
Case 1: 𝜆 = −𝑝2
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
𝑦 = 𝑏: 𝑌 𝑏 = 𝜃0 sin𝜋𝑥
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
𝑦 = 𝑏: 𝑌 𝑏 = 𝜃0 sin𝜋𝑥
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
with eigenvalues 𝜆𝑛 =2𝑛−1 2𝜋2
4𝑎2, 𝑛 = 1, 2, 3…
𝑦 = 0: 𝑌 0 = 0
𝑦 = 𝑏: 𝑌 𝑏 = 𝜃0 sin𝜋𝑥
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
with eigenvalues 𝜆𝑛 =2𝑛−1 2𝜋2
4𝑎2, 𝑛 = 1, 2, 3…
𝑦 = 0: 𝑌 0 = 0
𝑦 = 𝑏: 𝑌 𝑏 = 𝜃0 sin𝜋𝑥
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… .
Take linear combination of these solutions.
𝜃 𝑥, 𝑦 =
𝑛=1
∞
𝐾𝑛 sin2𝑛 − 1 𝜋𝑥
2𝑎sinh
2𝑛 − 1 𝜋𝑦
2𝑎
𝜃 𝑥, 𝑦 =
𝑛=1
∞
𝐾𝑛 sin2𝑛 − 1 𝜋𝑥
2𝑎sinh
2𝑛 − 1 𝜋𝑦
2𝑎
𝑌 𝑏 = 𝜃0 sin𝜋𝑥
2𝑎
𝜃 𝑥, 𝑏 =
𝑛=1
∞
𝐾𝑛 sin2𝑛 − 1 𝜋𝑥
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
Case 3: 𝜆 = −𝑝2
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
𝑌𝑛 𝑦 = 𝐵𝑛 sin 𝑛𝑦 with eigenvalues 𝜆𝑛 = 𝑛2, 𝑛 = 1, 2, 3
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… .
Take linear combination of these solutions.
𝑢 𝑥, 𝑦 =
𝑛=1
∞
𝐾𝑛 sinh𝑛𝑥 sin 𝑛𝑦
𝑌𝑛 𝑦 = 𝐵𝑛 sin 𝑛𝑦 with eigenvalues 𝜆𝑛 = 𝑛2, 𝑛 = 1, 2, 3
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
𝑌𝑛 𝑦 = 𝐵𝑛 sin 𝑛𝑦 with eigenvalues 𝜆𝑛 = 𝑛2, 𝑛 = 1, 2, 3
𝑥 = 0: 𝑋 0 = 𝑦(𝜋2 − 𝑦2)𝑥 = 2: 𝑋 2 = 0𝑦 = 0: 𝑌 0 = 0𝑦 = 𝜋: 𝑌 𝜋 = 0
𝑌𝑛 𝑦 = 𝐵𝑛 sin 𝑛𝑦 with eigenvalues 𝜆𝑛 = 𝑛2, 𝑛 = 1, 2, 3
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
𝑌𝑛 𝑦 = 𝐵𝑛 sin 𝑛𝑦 with eigenvalues 𝜆𝑛 = 𝑛2, 𝑛 = 1, 2, 3
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
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