Homework 1
Variational Formulation of Boundary Value Problems
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The Galerkin Finite Element Method (GFEM) is used to solve differential equations. Below are
the steps for the GFEM.
i.
ii.
iii.
iv.
v.
Obtain the variation form or a weak form.
Finite element mesh and basis functions.
Galerkin’s method.
Solve system of algebraic equations.
Quality of the approximate solution.
Below are the steps to get the variational or weak form.
i.
ii.
iii.
iv.
Introduce a “test” function 𝑣(𝑥). 𝑣(𝑥) is somewhat similar to 𝑢(𝑥). The detailed
characteristics of 𝑣(𝑥) is to be determined.
Multiply the left hand side (LHS) and right hand side (RHS) of ordinary differential
equations by 𝑣(𝑥). That is −𝑢". 𝑣 = 1 . 𝑣
Integrate both sides between 𝑥 = 0 and 𝑥 = 1.
Perform integration by parts on LHS only.
The GFEM is used to convert a continuous operator problem such as differential equations to a
discrete problem. The GFEM breaks down a complex problems into a simple algebraic problems
that is easy to solve. The differential equations and boundary values problems are converted
into a matrix of the form𝐴𝑋 = 𝐵.
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𝑑2 𝑢
− 𝑑𝑥 2 = −𝑢"
−𝑢" = 𝑓(𝑥)
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∫0 −𝑢" . 𝑣 𝑑𝑥 = ∫0 𝑓(𝑥) . 𝑣 𝑑𝑥
1 𝑑 𝑑𝑢
( )
𝑑𝑥 𝑑𝑥
∫0
1
. 𝑣 𝑑𝑥 = ∫0 𝑓(𝑥) . 𝑣 𝑑𝑥, where 𝑢′ =
𝑑𝑢
𝑑𝑥
Integrating by parts the left hand side only provides:
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[−𝑢′ . 𝑣]10 − ∫0 − 𝑢′ . 𝑑𝑣 𝑑𝑥 = ∫0 𝑓(𝑥) . 𝑣 𝑑𝑥
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[−𝑢′ . 𝑣] 𝑎𝑡 𝑥 = 1 − [𝑢′ . 𝑣]𝑎𝑡 𝑥 = 0 + ∫0 𝑢′ . 𝑣 ′ 𝑑𝑥 = ∫0 𝑓(𝑥) . 𝑣 𝑑𝑥
[−𝑢′(1) . 𝑣(1)] = 0. This is provided.
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−[ℎ(𝑥) . 𝑣(0)] + ∫0 𝑢′ . 𝑣 ′ 𝑑𝑥 = ∫0 𝑓(𝑥) . 𝑣 𝑑𝑥
This is a variational form of the given problem.
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−∇2 𝑢 = 𝑓(𝑥, 𝑦)
𝜕2 𝑢
−∇2 𝑢 = − 𝜕𝑥 2 −
𝜕2 𝑢
𝜕2 𝑦
Test function 𝑣(𝑥, 𝑦)
∇𝑢 =
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
. 𝑗⃗
𝜕𝑣
𝜕𝑦
. 𝑗⃗
. 𝑖⃗ +
𝜕𝑢 𝜕𝑢
∇𝑢 = (𝜕𝑥 , 𝜕𝑦)
∇𝑣 =
𝜕𝑣
𝜕𝑥
. 𝑖⃗ +
𝜕𝑣 𝜕𝑣
∇𝑣 = (𝜕𝑥 , 𝜕𝑦)
∇𝑢 . ∇𝑣 =
𝑏
𝜕𝑢
𝜕𝑥
.
𝜕𝑢
𝜕𝑦
+
𝜕𝑣
𝜕𝑥
𝑑
.
𝜕𝑣
𝜕𝑦
𝑏
𝑑
∫𝑎 ∫𝑐 ∇𝑢 . ∇𝑣 𝑑𝑥 𝑑𝑦 + [ ] = ∫𝑎 ∫𝑐 𝑓(𝑥, 𝑦) . 𝑣(𝑥, 𝑦) 𝑑𝑥𝑑𝑦
The second term on the left hand side of the equation above is 0, i.e., [ ] = 0.
𝑏
𝑑
𝑏
𝑑
Variational Form is ∫𝑎 ∫𝑐 ∇𝑢 . ∇𝑣 𝑑𝑥 𝑑𝑦 = ∫𝑎 ∫𝑐 𝑓(𝑥, 𝑦) . 𝑣(𝑥, 𝑦) 𝑑𝑥𝑑𝑦