Problem 4
edit
Consider the problem of solving a nonlinear system of ODE
y
′
=
f
(
t
,
y
)
{\displaystyle y'=f(t,y)\!\,}
by an implicit method. The
n
{\displaystyle n\!\,}
-th step consists of solving for the unknown
y
{\displaystyle y\!\,}
a non-linear algebraic system of the form
y
=
α
h
f
(
t
n
,
y
)
+
g
n
−
1
(
1
)
{\displaystyle y=\alpha hf(t_{n},y)+g_{n-1}\quad (1)\!\,}
where
g
n
−
1
∈
R
n
{\displaystyle g_{n-1}\in \mathbb {R} ^{n}\!\,}
is known,
α
>
0
{\displaystyle \alpha >0\!\,}
and
h
{\displaystyle h\!\,}
is the stepsize. Let
f
∈
C
1
{\displaystyle f\in C^{1}\!\,}
Problem 4a
edit
Solution 4a
edit
Fixed point iteration
edit
Equation
(
1
)
{\displaystyle (1)\!\,}
is conveniently in fixed point iteration form.
y
n
j
+
1
=
α
h
f
(
t
n
,
y
n
j
)
+
g
n
−
1
⏟
ϕ
(
y
n
j
)
{\displaystyle y_{n}^{j+1}=\underbrace {\alpha hf(t_{n},y_{n}^{j})+g_{n-1}} _{\phi (y_{n}^{j})}\!\,}
Notice that the right hand side is only a function of
y
n
j
{\displaystyle y_{n}^{j}\!\,}
since
α
,
h
,
t
n
,
g
n
−
1
{\displaystyle \alpha ,h,t_{n},g_{n-1}\!\,}
are fixed when solving for the fixed point
y
n
∗
{\displaystyle y_{n}^{*}\!\,}
where
ϕ
(
y
n
∗
)
=
y
n
∗
{\displaystyle \phi (y_{n}^{*})=y_{n}^{*}\!\,}
Also note that
j
{\displaystyle j\!\,}
is the fixed point iteration index.
Conditions for local convergence
edit
The fixed point iteration will converge when the norm of the Jacobian of
ϕ
{\displaystyle \phi \!\,}
is less than 1 i.e.
‖
D
(
ϕ
)
‖
<
1
{\displaystyle \|D(\phi )\|<1\!\,}
Since
‖
D
(
ϕ
)
‖
=
‖
α
h
D
(
f
)
‖
{\displaystyle \|D(\phi )\|=\|\alpha hD(f)\|\!\,}
, we equivalently have the condition
‖
D
(
f
)
‖
<
1
α
h
{\displaystyle \|D(f)\|<{\frac {1}{\alpha h}}\!\,}
Problem 4b
edit
Solution 4b
edit
Newton iteration
edit
The Newton iteration solves
ψ
(
y
)
=
0
{\displaystyle \psi (y)=0\!\,}
and the iteration is given by
y
i
+
1
=
y
i
−
D
(
ψ
(
y
i
)
)
−
1
ψ
(
y
i
)
{\displaystyle y_{i+1}=y_{i}-D(\psi (y_{i}))^{-1}\psi (y_{i})\!\,}
Let
ψ
(
y
)
=
y
−
ϕ
(
y
)
{\displaystyle \psi (y)=y-\phi (y)\!\,}
Conditions for local convergence
edit
If
D
(
ψ
(
y
)
)
−
1
{\displaystyle D(\psi (y))^{-1}\!\,}
exists, i.e.
D
(
ψ
(
y
)
)
{\displaystyle D(\psi (y))\!\,}
is invertible or equivalently non-singular, then local convergence is guaranteed.
Note that
D
(
ψ
(
y
)
)
=
I
−
D
(
ϕ
(
y
)
)
{\displaystyle D(\psi (y))=I-D(\phi (y))\!\,}
Conditions for quadratic convergence
edit
If
D
(
ψ
(
y
)
)
{\displaystyle D(\psi (y))\!\,}
is Lipschitz, then we have quadratic convergence and
ψ
(
y
)
{\displaystyle \psi (y)\!\,}
is twice continuously differentiable in a neighborhood of the root
Problem 5
edit
This problem is about choosing between a specific single-step and a specific multi-step methods for solving ODE:
y
′
=
f
(
t
,
y
)
{\displaystyle y'=f(t,y)\!\,}
Problem 5a
edit
Write the trapezoid method, define its local truncation error and estimate it.
Solution 5a
edit
Trapezoid method (Implicit, Adams-Moulton)
edit
y
n
+
1
=
y
n
+
1
2
h
(
f
(
t
n
+
1
,
y
n
+
1
)
+
f
(
t
n
,
y
n
)
)
{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{2}}h{\big (}f(t_{n+1},y_{n+1})+f(t_{n},y_{n}){\big )}\!\,}
Define Local Truncation Error
edit
The local truncation error is given as follows:
error
=
(
y
(
t
n
+
1
)
−
y
(
t
n
)
)
−
1
2
h
(
f
(
t
n
+
1
,
y
(
t
n
+
1
)
)
+
f
(
t
n
,
y
n
)
)
{\displaystyle {\mbox{error}}=(y(t_{n+1})-y(t_{n}))-{\tfrac {1}{2}}h{\big (}f(t_{n+1},y(t_{n+1}))+f(t_{n},y_{n}){\big )}\!\,}
Find Local Truncation Error Using Taylor Expansion
edit
Note that
y
i
≈
y
(
t
i
)
{\displaystyle y_{i}\approx y(t_{i})\!\,}
. The uniform step size is
h
{\displaystyle h\!\,}
. Hence,
t
i
+
1
=
t
i
+
h
{\displaystyle t_{i+1}=t_{i}+h\!\,}
Therefore, the given equation may be written as
y
(
t
i
+
h
)
−
y
(
t
i
)
≈
1
2
h
(
y
′
(
t
i
+
h
)
+
y
′
(
t
n
)
)
{\displaystyle y(t_{i}+h)-y(t_{i})\approx {\frac {1}{2}}h(y'(t_{i}+h)+y'(t_{n}))\!\,}
Expand Left Hand Side
edit
Expanding about
t
n
{\displaystyle t_{n}\!\,}
, we get
Order
y
(
t
n
+
h
)
−
y
(
t
n
)
Σ
0
y
(
t
n
)
−
y
(
t
n
)
0
1
y
′
(
t
n
)
h
0
y
′
(
t
n
)
h
2
1
2
!
y
″
(
t
n
)
h
2
0
1
2
y
″
(
t
n
)
h
2
3
1
3
!
y
‴
(
t
n
)
h
3
0
1
6
y
‴
(
t
n
)
h
3
4
O
(
h
4
)
0
O
(
h
4
)
{\displaystyle {\begin{array}{|c|c|c|c|}{\mbox{Order}}&y(t_{n}+h)&-y(t_{n})&\Sigma \\\hline &&&\\0&y(t_{n})&-y(t_{n})&0\\&&&\\1&y'(t_{n})h&0&y'(t_{n})h\\&&&\\2&{\frac {1}{2!}}y''(t_{n})h^{2}&0&{\frac {1}{2}}y''(t_{n})h^{2}\\&&&\\3&{\frac {1}{3!}}y'''(t_{n})h^{3}&0&{\frac {1}{6}}y'''(t_{n})h^{3}\\&&&\\4&{\mathcal {O}}(h^{4})&0&{\mathcal {O}}(h^{4})\\&&&\\\hline \end{array}}}
Expand Right Hand side
edit
Also expanding about
t
n
{\displaystyle t_{n}\!\,}
gives
Order
h
1
2
h
y
′
(
t
n
+
h
)
1
2
h
y
′
(
t
n
)
Σ
0
0
0
0
1
1
2
h
⋅
y
′
(
t
n
)
1
2
h
y
′
(
t
n
)
y
′
(
t
n
)
h
2
1
2
h
⋅
y
″
(
t
n
)
h
0
1
2
y
″
(
t
n
)
h
2
3
1
2
h
⋅
y
‴
2
(
t
n
)
h
2
0
y
‴
4
(
t
n
)
h
3
4
O
(
h
4
)
O
(
h
4
)
O
(
h
4
)
{\displaystyle {\begin{array}{|c|c|c|c|}{\mbox{Order }}h&{\frac {1}{2}}hy'(t_{n}+h)&{\frac {1}{2}}hy'(t_{n})&\Sigma \\\hline &&&\\0&0&0&0\\&&&\\1&{\frac {1}{2}}h\cdot y'(t_{n})&{\frac {1}{2}}hy'(t_{n})&y'(t_{n})h\\&&&\\2&{\frac {1}{2}}h\cdot y''(t_{n})h&0&{\frac {1}{2}}y''(t_{n})h^{2}\\&&&\\3&{\frac {1}{2}}h\cdot {\frac {y'''}{2}}(t_{n})h^{2}&0&{\frac {y'''}{4}}(t_{n})h^{3}\\&&&\\4&{\mathcal {O}}(h^{4})&{\mathcal {O}}(h^{4})&{\mathcal {O}}(h^{4})\\\hline \end{array}}}
Calculate local truncation error
edit
Since the order 3 terms of
h
{\displaystyle h\!\,}
do not agree (
y
‴
(
t
n
)
6
h
3
≠
y
‴
(
t
n
)
4
h
3
{\displaystyle {\frac {y'''(t_{n})}{6}}h^{3}\neq {\frac {y'''(t_{n})}{4}}h^{3}\!\,}
), the error is of order
h
2
{\displaystyle h^{2}\!\,}
.
Problem 5b
edit
Show that the truncation error for the following multistep method is of the same order as in (a):
y
n
+
1
=
2
y
n
−
y
n
−
1
−
h
f
(
t
n
−
1
,
y
n
−
1
)
+
h
f
(
t
n
,
y
n
)
{\displaystyle y_{n+1}=2y_{n}-y_{n-1}-hf(t_{n-1},y_{n-1})+hf(t_{n},y_{n})\!\,}
Solution 5b
edit
We need to show that
y
(
t
n
+
1
)
−
2
y
(
t
n
)
+
y
(
t
n
−
1
)
+
h
y
′
(
t
n
−
1
)
−
h
y
′
(
t
n
)
=
O
(
h
3
)
{\displaystyle y(t_{n+1})-2y(t_{n})+y(t_{n-1})+hy'(t_{n-1})-hy'(t_{n})={\mathcal {O}}(h^{3})\!\,}
Again, note that
t
n
+
1
=
t
n
+
h
t
n
−
1
=
t
n
−
h
{\displaystyle {\begin{aligned}t_{n+1}&=t_{n}+h\\t_{n-1}&=t_{n}-h\end{aligned}}\!\,}
Order of
h
y
(
t
n
+
h
)
−
2
y
(
t
n
)
y
(
t
n
−
h
)
h
y
′
(
t
n
−
h
)
−
h
y
′
(
t
n
)
Σ
0
y
(
t
n
)
−
2
y
(
t
n
)
y
(
t
n
)
0
0
0
1
y
′
(
t
n
)
h
0
−
y
′
(
t
n
)
h
y
′
(
t
n
)
h
−
y
′
(
t
n
)
0
2
1
2
y
″
(
t
n
)
h
2
0
1
2
y
″
(
t
n
)
h
2
−
y
″
(
t
n
)
h
2
0
0
3
1
6
y
‴
(
t
n
)
h
3
0
−
1
6
y
‴
(
t
n
)
h
3
1
2
y
‴
(
t
n
)
h
3
0
1
2
y
‴
(
t
n
)
h
3
{\displaystyle {\begin{array}{|c|c|c|c|c|c|c|}{\mbox{Order of }}h&y(t_{n}+h)&-2y(t_{n})&y(t_{n}-h)&hy'(t_{n}-h)&-hy'(t_{n})&\Sigma \\\hline 0&y(t_{n})&-2y(t_{n})&y(t_{n})&0&0&0\\&&&&&&\\1&y'(t_{n})h&0&-y'(t_{n})h&y'(t_{n})h&-y'(t_{n})&0\\&&&&&&\\2&{\frac {1}{2}}y''(t_{n})h^{2}&0&{\frac {1}{2}}y''(t_{n})h^{2}&-y''(t_{n})h^{2}&0&0\\&&&&&&\\3&{\frac {1}{6}}y'''(t_{n})h^{3}&0&-{\frac {1}{6}}y'''(t_{n})h^{3}&{\frac {1}{2}}y'''(t_{n})h^{3}&0&{\frac {1}{2}}y'''(t_{n})h^{3}\\&&&&&&\\\hline \end{array}}}
So this method is also consistency order 2.
Problem 5c
edit
What could be said about the global convergence rate for these two methods? Justify your conclusions for both methods.
Solution 5c
edit
The trapezoid is stable because its satisfies the root condition. (The root of the characteristic equation is 1 and has a simple root)
The second method is not stable because the characteristic equation has a double root of 1.
Both the trapezoid method and second method are consistent with order
h
2
{\displaystyle h^{2}\!\,}
Note that convergence occurs if and only the method is both stable and consistent.
Therefore, the trapezoid method converges in general but the second method does not.
mkmkmkmlmklml
Problem 6
edit
Consider the boundary value problem
L
(
u
)
=
−
u
″
+
b
u
=
f
,
x
∈
I
≡
[
0
,
1
]
,
u
(
0
)
=
u
(
1
)
=
0
(
2
)
{\displaystyle L(u)=-u''+bu=f,\;\;x\in I\equiv [0,1],\;\;u(0)=u(1)=0\qquad (2)\!\,}
where
b
≥
0
{\displaystyle b\geq 0\!\,}
is constant. Let
{
0
=
x
0
<
x
1
<
⋯
<
x
n
=
1
}
{\displaystyle \{0=x_{0}<x_{1}<\dots <x_{n}=1\}\!\,}
be a uniform meshsize
h
{\displaystyle h\!\,}
.
Let
V
h
=
{
v
∈
C
[
0
,
1
]
:
v
|
[
x
i
−
1
,
x
i
]
is linear for each i,
v
(
0
)
=
v
(
1
)
=
0
}
{\displaystyle V_{h}=\{v\in C[0,1]:\left.v\right|_{[x_{i-1},x_{i}]}{\mbox{ is linear for each i, }}v(0)=v(1)=0\}\!\,}
be the corresponding finite element space, and let
u
h
=
R
h
(
u
)
{\displaystyle u_{h}=R_{h}(u)\!\,}
be the corresponding finite element solution of (2). Note that
R
h
{\displaystyle R_{h}\!\,}
is a projection operator, the Ritz projector, onto the finite dimensional space
V
h
{\displaystyle V_{h}\!\,}
with respect to the element scalar product
a
(
⋅
,
⋅
)
{\displaystyle a(\cdot ,\cdot )\!\,}
induced by problem (2).
Problem 6a
edit
Let
|
⋅
|
1
{\displaystyle |\cdot |_{1}\!\,}
be the
H
1
{\displaystyle H^{1}\!\,}
-seminorm, namely
|
v
|
1
2
=
∫
I
|
v
′
|
2
{\displaystyle |v|_{1}^{2}=\int _{I}|v'|^{2}\!\,}
for all
v
∈
H
0
1
(
I
)
{\displaystyle v\in H_{0}^{1}(I)\!\,}
. Find the constant
Λ
{\displaystyle \Lambda \!\,}
in terms of the parameter
b
{\displaystyle b\!\,}
such that
|
u
h
|
1
≤
Λ
|
u
|
1
{\displaystyle |u_{h}|_{1}\leq \Lambda |u|_{1}\!\,}
Hint: recall the Poincare inequality
‖
v
‖
0
≤
1
2
|
v
|
1
{\displaystyle \|v\|_{0}\leq {\frac {1}{2}}|v|_{1}\!\,}
for all
v
∈
H
0
1
(
I
)
{\displaystyle v\in H_{0}^{1}(I)\!\,}
where
‖
⋅
‖
0
{\displaystyle \|\cdot \|_{0}\!\,}
denotes the
L
2
{\displaystyle L^{2}\!\,}
-norm
Solution 6a
edit
Weak Form
edit
Integrating by parts gives, for all
v
∈
V
{\displaystyle v\in V\!\,}
∫
u
′
v
′
+
b
∫
u
v
=
∫
f
v
{\displaystyle \int u'v'+b\int uv=\int fv\!\,}
Specifically,
∫
u
′
u
h
′
+
b
∫
u
u
h
=
∫
f
u
h
{\displaystyle \int u'u_{h}'+b\int uu_{h}=\int fu_{h}\!\,}
Discretized Form (Finite Element Formulation)
edit
Similarly, the finite element formulation is find
u
h
∈
V
h
{\displaystyle u_{h}\in V_{h}\!\,}
such that for all
v
h
∈
V
h
{\displaystyle v_{h}\in V_{h}\!\,}
∫
u
h
′
v
h
′
+
b
∫
u
h
v
h
=
∫
f
v
h
{\displaystyle \int u_{h}'v_{h}'+b\int u_{h}v_{h}=\int fv_{h}\!\,}
Specifically,
∫
u
h
′
u
h
′
+
b
∫
u
h
u
h
=
∫
f
u
h
{\displaystyle \int u_{h}'u_{h}'+b\int u_{h}u_{h}=\int fu_{h}\!\,}
Equate Both Sides and Apply Inequalities
edit
∫
0
1
u
h
′
u
h
′
+
b
∫
0
1
u
h
u
h
=
∫
0
1
u
′
u
h
′
+
b
∫
0
1
u
u
h
≤
|
u
|
1
|
u
h
|
1
+
b
‖
u
‖
0
‖
u
h
‖
0
Cauchy-Schwartz
≤
|
u
|
1
|
u
h
|
1
+
b
1
2
|
u
|
1
1
2
|
u
h
|
1
Poincare
=
|
u
h
|
1
(
1
+
b
4
)
|
u
|
1
{\displaystyle {\begin{aligned}\int _{0}^{1}u_{h}'u_{h}'+b\int _{0}^{1}u_{h}u_{h}&=\int _{0}^{1}u'u_{h}'+b\int _{0}^{1}uu_{h}\\&\leq |u|_{1}|u_{h}|_{1}+b\|u\|_{0}\|u_{h}\|_{0}{\mbox{ Cauchy-Schwartz}}\\&\leq |u|_{1}|u_{h}|_{1}+b{\frac {1}{2}}|u|_{1}{\frac {1}{2}}|u_{h}|_{1}{\mbox{ Poincare}}\\&=|u_{h}|_{1}(1+{\frac {b}{4}})|u|_{1}\end{aligned}}}
Hence we have,
|
u
h
|
1
2
≤
|
u
h
|
1
2
+
b
‖
u
h
‖
0
2
≤
|
u
h
|
1
(
1
+
b
4
)
|
u
|
1
|
u
h
|
1
2
≤
|
u
h
|
1
(
1
+
b
4
)
|
u
|
1
|
u
h
|
1
≤
(
1
+
b
4
)
|
u
|
1
{\displaystyle {\begin{aligned}|u_{h}|_{1}^{2}\leq |u_{h}|_{1}^{2}+b\|u_{h}\|_{0}^{2}&\leq |u_{h}|_{1}(1+{\frac {b}{4}})|u|_{1}\\|u_{h}|_{1}^{2}&\leq |u_{h}|_{1}(1+{\frac {b}{4}})|u|_{1}\\|u_{h}|_{1}&\leq (1+{\frac {b}{4}})|u|_{1}\end{aligned}}}
Problem 6b
edit
If
I
h
u
∈
V
h
{\displaystyle I_{h}u\in V_{h}\!\,}
is the Lagrange interpolant of
u
{\displaystyle u\!\,}
, then prove
R
h
(
I
h
u
)
=
I
h
u
{\displaystyle R_{h}(I_{h}u)=I_{h}u\!\,}
. Deduce
|
u
h
−
I
h
u
|
1
≤
Λ
|
u
−
I
h
u
|
1
{\displaystyle |u_{h}-I_{h}u|_{1}\leq \Lambda |u-I_{h}u|_{1}\!\,}
Solution 6b
edit
Prove equality
edit
We have for all
v
∈
H
0
1
(
0
,
1
)
{\displaystyle v\in H_{0}^{1}(0,1)\!\,}
a
(
u
,
v
)
=
∫
u
′
v
′
d
x
+
b
∫
u
v
d
x
=
∫
f
v
{\displaystyle a(u,v)=\int u'v'dx+b\int uvdx=\int fv\!\,}
Specifically, for all
v
h
∈
V
h
{\displaystyle v_{h}\in V_{h}\!\,}
a
(
u
,
v
h
)
=
∫
u
′
v
h
′
d
x
+
b
∫
u
v
h
d
x
=
∫
f
v
h
(
1
)
{\displaystyle a(u,v_{h})=\int u'v_{h}'dx+b\int uv_{h}dx=\int fv_{h}\quad \quad (1)\!\,}
The discrete form of the energy scalar product is for all
u
h
,
v
h
∈
V
h
{\displaystyle u_{h},v_{h}\in V_{h}\!\,}
a
(
u
h
,
v
h
)
=
∫
u
h
′
v
h
′
d
x
+
b
∫
u
h
v
h
d
x
=
∫
f
v
h
(
2
)
{\displaystyle a(u_{h},v_{h})=\int u_{h}'v_{h}'dx+b\int u_{h}v_{h}dx=\int fv_{h}\quad \quad (2)\!\,}
Subtracting equation (2) from equation (1), we have
a
(
u
−
u
h
,
v
h
)
=
0
{\displaystyle a(u-u_{h},v_{h})=0\!\,}
Let
R
h
(
I
h
u
)
=
u
h
¯
∈
V
h
{\displaystyle R_{h}(I_{h}u)={\overline {u_{h}}}\in V_{h}\!\,}
. Note that by hypothesis
I
h
u
∈
V
h
{\displaystyle I_{h}u\in V_{h}\!\,}
. Then,
a
(
I
h
u
−
u
h
¯
,
I
h
u
−
u
h
¯
)
=
0
{\displaystyle a(I_{h}u-{\overline {u_{h}}},I_{h}u-{\overline {u_{h}}})=0\!\,}
By ellipticity,
0
=
a
(
I
h
u
−
u
h
¯
,
I
h
u
−
u
h
¯
)
≥
α
‖
I
h
u
−
u
h
¯
‖
2
{\displaystyle 0=a(I_{h}u-{\overline {u_{h}}},I_{h}u-{\overline {u_{h}}})\geq \alpha \|I_{h}u-{\overline {u_{h}}}\|^{2}\!\,}
which implies
I
h
u
−
u
h
¯
=
0
{\displaystyle I_{h}u-{\overline {u_{h}}}=0\!\,}
i.e.
I
h
u
=
u
h
¯
{\displaystyle I_{h}u={\overline {u_{h}}}\!\,}
Deduce inequality
edit
Hence we have
R
h
(
u
−
I
h
u
)
=
u
h
−
I
h
u
{\displaystyle R_{h}(u-I_{h}u)=u_{h}-I_{h}u\!\,}
Arguing as we did in part (a), we have
|
u
h
−
I
h
u
|
1
≤
Λ
|
u
−
I
h
u
|
1
{\displaystyle |u_{h}-I_{h}u|_{1}\leq \Lambda |u-I_{h}u|_{1}\!\,}
Problem 6c
edit
Use (b) to derive the error estimate
|
u
−
u
h
|
1
≤
(
1
+
Λ
)
|
u
−
I
h
u
|
1
{\displaystyle |u-u_{h}|_{1}\leq (1+\Lambda )|u-I_{h}u|_{1}\!\,}
and bound the right hand side by suitable power of
h
{\displaystyle h\!\,}
. Make explicit the required regularity of
u
{\displaystyle u\!\,}
Solution 6c
edit
Show inequality
edit
|
u
−
u
h
|
1
=
|
u
−
I
h
u
+
I
h
u
−
u
h
|
1
≤
|
u
−
I
h
u
|
1
+
|
I
h
u
−
u
h
|
1
≤
|
u
−
I
h
u
|
1
+
Λ
|
u
−
I
h
u
|
1
=
(
1
+
Λ
)
|
u
−
I
h
u
|
1
{\displaystyle {\begin{aligned}|u-u_{h}|_{1}&=|u-I_{h}u+I_{h}u-u_{h}|_{1}\\&\leq |u-I_{h}u|_{1}+|I_{h}u-u_{h}|_{1}\\&\leq |u-I_{h}u|_{1}+\Lambda |u-I_{h}u|_{1}\\&=(1+\Lambda )|u-I_{h}u|_{1}\end{aligned}}}
Bound Right Hand Side
edit
For
x
∈
[
x
i
−
1
,
x
i
]
{\displaystyle x\in [x_{i-1},x_{i}]\!\,}
, Newton's polynomial interpolation error gives for some
ξ
i
∈
[
x
i
−
1
,
x
i
]
{\displaystyle \xi _{i}\in [x_{i-1},x_{i}]\!\,}
|
u
−
I
h
u
|
1
2
=
|
u
(
2
)
(
ξ
i
)
2
(
x
−
x
i
)
(
x
−
x
i
−
1
)
|
1
2
d
x
=
∫
0
1
(
[
u
(
2
)
(
ξ
i
)
2
(
x
−
x
i
)
(
x
−
x
i
−
1
)
]
′
)
2
=
(
u
(
2
)
(
ξ
i
)
2
)
2
∫
0
1
[
(
x
−
x
i
−
1
)
⏟
≤
h
(
x
−
x
i
)
⏟
≤
h
]
d
x
≤
(
u
(
2
)
(
ξ
i
)
h
)
2
{\displaystyle {\begin{aligned}|u-I_{h}u|_{1}^{2}&=\left|{\frac {u^{(2)}(\xi _{i})}{2}}(x-x_{i})(x-x_{i-1})\right|_{1}^{2}dx\\&=\int _{0}^{1}\left(\left[{\frac {u^{(2)}(\xi _{i})}{2}}(x-x_{i})(x-x_{i-1})\right]'\right)^{2}\\&=\left({\frac {u^{(2)}(\xi _{i})}{2}}\right)^{2}\int _{0}^{1}[\underbrace {(x-x_{i-1})} _{\leq h}\underbrace {(x-x_{i})} _{\leq h}]dx\\&\leq (u^{(2)}(\xi _{i})h)^{2}\end{aligned}}\!\,}
Therefore the error on the entire interval is given by
|
u
−
I
h
u
|
1
2
≤
∑
i
=
1
n
(
u
(
2
)
(
ξ
i
)
h
)
2
≤
max
0
<
ξ
<
1
(
u
(
2
)
(
ξ
)
h
)
2
⋅
n
{\displaystyle {\begin{aligned}|u-I_{h}u|_{1}^{2}&\leq \sum _{i=1}^{n}(u^{(2)}(\xi _{i})h)^{2}\\&\leq \max _{0<\xi <1}(u^{(2)}(\xi )h)^{2}\cdot n\end{aligned}}\!\,}
which implies
|
u
−
I
h
u
|
1
≤
max
0
<
ξ
<
1
u
(
2
)
(
ξ
i
)
n
h
{\displaystyle |u-I_{h}u|_{1}\leq \max _{0<\xi <1}u^{(2)}(\xi _{i}){\sqrt {n}}h\!\,}
u
{\displaystyle u\!\,}
needs to be twice differentiable.