Introduction to Computational Linear Algebra

212 Introduction to Computational Linear Algebra

Let then for each interior node i, 2 ≤ i ≤ n − 1:

ψ

i

(x) =



































(x − x

i+1

)(x − x

i+

1

2

)

h

2

i

2

x ∈ [x

i

, x

i+1

]

(x − x

i−1

)(x − x

i−

1

2

)

h

2

i−1

2

x ∈ [x

i−1

, x

i

]

0 otherwise

and at the boundary nodes:

ψ

1

(x) =











(x − x

2

)(x − x

3

2

)

h

2

1

2

x ∈ [x

1

, x

2

]

0 otherwise

ψ

n

(x) =











(x − x

n−1

)(x − x

n−

1

2

)

h

2

n−1

2

x ∈ [x

n−1

, x

n

]

0 otherwise

Note that ψ

i

(x

i

) = 1 and ψ

i

(x

j

) = 0, ∀j 6= i.

A Lagrangian basis of S

2

(Π) consists of:

{ψ

i

(x) |i = 1 : n} ∪ {η

i

(x) |i = 1 : m}

where:

η

i

(x) =











(x − x

i

)(x

i+1

− x)

h

2

i

4

x

i

≤ x ≤ x

i+1

0 otherwise

Obviously, one has:

∀ϕ ∈ S

2

(Π), ϕ(x) =

n

X

i=1

ϕ

i

ψ

i

(x) +

m

X

i=1

ϕ

i+

1

2

η(x).

Implementation

The finite element solution ...