Consider the boundary value problem
{
−
u
″
(
x
)
+
b
(
x
)
u
(
x
)
=
f
(
x
)
,
0
≤
x
≤
1
u
(
0
)
=
u
l
,
u
(
1
)
=
u
r
(
1
)
{\displaystyle {\begin{cases}-u''(x)+b(x)u(x)=f(x),0\leq x\leq 1\\u(0)=u_{l},u(1)=u_{r}\end{cases}}\qquad (1)\!\,}
b
(
x
)
,
f
(
x
)
∈
C
[
0
,
1
]
{\displaystyle b(x),f(x)\in C[0,1]\!\,}
b
(
x
)
≥
0
{\displaystyle b(x)\geq 0\!\,}
(
1
)
{\displaystyle (1)\!\,}
h
{\displaystyle h\!\,}
u
″
(
x
)
{\displaystyle u''(x)\!\,}
From Taylor expansion of
u
(
x
+
h
)
{\displaystyle u(x+h)\!\,}
u
(
x
−
h
)
{\displaystyle u(x-h)\!\,}
x
{\displaystyle x\!\,}
u
″
(
x
)
=
u
(
x
+
h
)
−
2
u
(
x
)
+
u
(
x
−
h
)
h
2
+
u
⁗
(
ξ
)
12
h
2
{\displaystyle u''(x)={\frac {u(x+h)-2u(x)+u(x-h)}{h^{2}}}+{\frac {u''''(\xi )}{12}}h^{2}\!\,}
P
=
{
x
i
}
i
=
0
N
+
1
{\displaystyle P=\{x_{i}\}_{i=0}^{N+1}\!\,}
[
0
,
1
]
{\displaystyle [0,1]\!\,}
h
{\displaystyle h\!\,}
i
=
2
,
…
,
N
−
1
{\displaystyle i=2,\ldots ,N-1\!\,}
[
−
1
h
2
2
h
2
+
b
(
x
i
)
−
1
h
2
]
[
u
i
−
1
u
i
u
i
+
1
]
=
f
(
x
i
)
{\displaystyle {\begin{bmatrix}-{\frac {1}{h^{2}}}&&&{\frac {2}{h^{2}}}+b(x_{i})&&&-{\frac {1}{h^{2}}}\end{bmatrix}}{\begin{bmatrix}u_{i-1}\\u_{i}\\u_{i+1}\end{bmatrix}}=f(x_{i})\!\,}
i
=
1
:
{\displaystyle i=1:\!\,}
[
2
h
2
+
b
(
x
1
)
−
1
h
2
]
[
u
1
u
2
]
=
f
(
x
1
)
+
u
l
h
2
{\displaystyle {\begin{bmatrix}{\frac {2}{h^{2}}}+b(x_{1})&&&-{\frac {1}{h^{2}}}\end{bmatrix}}{\begin{bmatrix}u_{1}\\u_{2}\end{bmatrix}}=f(x_{1})+{\frac {u_{l}}{h^{2}}}\!\,}
i
=
N
:
{\displaystyle i=N:\!\,}
[
−
1
h
2
2
h
2
+
b
(
x
N
)
]
[
u
N
−
1
u
N
]
=
f
(
x
N
)
+
u
r
h
2
{\displaystyle {\begin{bmatrix}-{\frac {1}{h^{2}}}&&&{\frac {2}{h^{2}}}+b(x_{N})\end{bmatrix}}{\begin{bmatrix}u_{N-1}\\u_{N}\end{bmatrix}}=f(x_{N})+{\frac {u_{r}}{h^{2}}}\!\,}
[
2
h
−
1
h
−
1
h
2
h
−
1
h
⋱
⋱
⋱
−
1
h
2
h
]
[
u
1
⋮
⋮
u
N
]
=
[
∫
0
1
f
ϕ
i
⋮
∫
0
1
f
ϕ
N
]
{\displaystyle {\begin{bmatrix}{\frac {2}{h}}&-{\frac {1}{h}}&&&\\-{\frac {1}{h}}&{\frac {2}{h}}&-{\frac {1}{h}}&&\\&\ddots &\ddots &\ddots &\\&&&-{\frac {1}{h}}&{\frac {2}{h}}\end{bmatrix}}{\begin{bmatrix}u_{1}\\\vdots \\\vdots \\u_{N}\end{bmatrix}}={\begin{bmatrix}\int _{0}^{1}f\phi _{i}\\\vdots \\\\\int _{0}^{1}f\phi _{N}\end{bmatrix}}\!\,}
∫
0
1
f
(
x
)
ϕ
i
(
x
)
≈
h
f
(
x
i
)
{\displaystyle \int _{0}^{1}f(x)\phi _{i}(x)\approx hf(x_{i})\!\,}
Show that the matrix in
(
b
)
{\displaystyle (b)\!\,}
Since the matrix is diagonally dominant, it is non-singular.
To show that the matrix has a non-zero determinant, 2n elementary row operation can be used to show that
[
2
h
−
1
h
−
1
h
2
h
−
1
h
⋱
⋱
−
1
h
2
h
]
{\displaystyle {\begin{bmatrix}{\frac {2}{h}}&-{\frac {1}{h}}&&\\-{\frac {1}{h}}&{\frac {2}{h}}&-{\frac {1}{h}}&\\&\ddots &\ddots &\\&&-{\frac {1}{h}}&{\frac {2}{h}}\end{bmatrix}}}
has the same determinant as
[
−
1
h
2
h
−
1
h
⋱
⋱
⋱
−
1
h
2
h
n
+
1
h
]
{\displaystyle {\begin{bmatrix}-{\frac {1}{h}}&{\frac {2}{h}}&-{\frac {1}{h}}&\\&\ddots &\ddots &\ddots &\\&&-{\frac {1}{h}}&{\frac {2}{h}}\\&&&{\frac {n+1}{h}}\\\end{bmatrix}}}
which is
n
+
1
h
n
≠
0
{\displaystyle {\frac {n+1}{h^{n}}}\neq 0}
Consider the following dissipative initial value problem,
{
y
′
+
f
(
y
)
=
0
,
0
≤
x
≤
1
y
(
0
)
=
y
0
(
2
)
{\displaystyle {\begin{cases}y'+f(y)=0,0\leq x\leq 1\\y(0)=y_{0}\end{cases}}\qquad (2)\!\,}
f
:
R
→
R
{\displaystyle f:R\rightarrow R\!\,}
0
≤
f
′
(
y
)
{\displaystyle 0\leq f'(y)\!\,}
Write the Backward Euler Method for (2). This gives rise to an algebraic equation. Explain how you would solve this equation.
Using Taylor Expansion we have
y
(
x
−
h
)
=
y
(
x
)
−
y
′
(
x
)
h
+
y
″
(
ξ
)
2
h
2
y
(
x
)
=
y
(
x
−
h
)
+
y
′
(
x
)
h
−
y
″
(
ξ
)
2
h
2
y
(
x
n
)
=
y
(
x
n
−
1
)
+
y
′
(
x
n
)
⏟
−
f
(
y
(
x
n
)
)
h
−
y
″
(
ξ
)
2
h
2
⏟
O
(
h
2
)
(
∗
)
{\displaystyle {\begin{aligned}y(x-h)&=y(x)-y'(x)h+{\frac {y''(\xi )}{2}}h^{2}\\y(x)&=y(x-h)+y'(x)h-{\frac {y''(\xi )}{2}}h^{2}\\y(x_{n})&=y(x_{n-1})+\underbrace {y'(x_{n})} _{-f(y(x_{n}))}h-\underbrace {{\frac {y''(\xi )}{2}}h^{2}} _{O(h^{2})}\qquad (*)\\\end{aligned}}}
y
n
=
y
n
−
1
−
h
f
(
y
n
)
(
∗
∗
)
{\displaystyle y_{n}=y_{n-1}-hf(y_{n})\qquad (**)\!\,}
e
n
≡
y
(
x
n
−
1
)
−
y
n
{\displaystyle e_{n}\equiv y(x_{n-1})-y_{n}\!\,}
Derive an error estimate of the form
|
y
(
x
n
)
−
y
n
|
≤
M
2
h
,
n
=
1
,
2
,
…
{\displaystyle |y(x_{n})-y_{n}|\leq {\frac {M}{2}}h,n=1,2,\ldots \!\,}
M
=
max
0
≤
x
≤
1
|
y
″
(
x
)
|
{\displaystyle M=\max _{0\leq x\leq 1}|y''(x)|\!\,}
Subtracting
(
∗
)
{\displaystyle (*)\!\,}
(
∗
∗
)
{\displaystyle (**)\!\,}
e
n
=
e
n
−
1
−
h
[
f
(
y
(
x
n
)
)
−
f
(
y
n
)
]
−
y
″
(
ξ
)
2
h
2
=
e
n
−
1
−
h
f
′
(
y
(
α
)
)
e
n
−
y
″
(
ξ
)
2
h
2
(by MVT)
e
n
2
=
e
n
−
1
e
n
−
h
f
′
(
y
α
)
⏟
<
0
e
n
2
−
y
″
(
ξ
)
2
h
2
e
n
≤
e
n
−
1
e
n
−
y
″
(
ξ
)
2
h
2
e
n
e
n
≤
e
n
−
1
−
y
″
(
ξ
)
2
h
2
‖
e
n
‖
≤
‖
e
n
−
1
‖
+
‖
y
″
‖
∞
2
h
2
‖
e
n
‖
≤
n
‖
y
″
‖
∞
2
h
2
≤
M
2
h
{\displaystyle {\begin{aligned}e_{n}&=e_{n-1}-h[f(y(x_{n}))-f(y_{n})]-{\frac {y''(\xi )}{2}}h^{2}\\&=e_{n-1}-hf'(y(\alpha ))e_{n}-{\frac {y''(\xi )}{2}}h^{2}\quad {\mbox{ (by MVT) }}\\e_{n}^{2}&=e_{n-1}e_{n}-h\underbrace {f'(y\alpha )} _{<0}e_{n}^{2}-{\frac {y''(\xi )}{2}}h^{2}e_{n}\\&\leq e_{n-1}e_{n}-{\frac {y''(\xi )}{2}}h^{2}e_{n}\\e_{n}&\leq e_{n-1}-{\frac {y''(\xi )}{2}}h^{2}\\\|e_{n}\|&\leq \|e_{n-1}\|+{\frac {\|y''\|_{\infty }}{2}}h^{2}\\\|e_{n}\|&\leq n{\frac {\|y''\|_{\infty }}{2}}h^{2}\\&\leq {\frac {M}{2}}h\end{aligned}}\!\,}
![{\displaystyle b(x),f(x)\in C[0,1]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a8785561b5f615148c84ecaf8aaaeb0c6ea5373)
![{\displaystyle [0,1]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b26fc5fee30dada6f15cc080115cae582606d7f3)
![{\displaystyle {\begin{aligned}e_{n}&=e_{n-1}-h[f(y(x_{n}))-f(y_{n})]-{\frac {y''(\xi )}{2}}h^{2}\\&=e_{n-1}-hf'(y(\alpha ))e_{n}-{\frac {y''(\xi )}{2}}h^{2}\quad {\mbox{ (by MVT) }}\\e_{n}^{2}&=e_{n-1}e_{n}-h\underbrace {f'(y\alpha )} _{<0}e_{n}^{2}-{\frac {y''(\xi )}{2}}h^{2}e_{n}\\&\leq e_{n-1}e_{n}-{\frac {y''(\xi )}{2}}h^{2}e_{n}\\e_{n}&\leq e_{n-1}-{\frac {y''(\xi )}{2}}h^{2}\\\|e_{n}\|&\leq \|e_{n-1}\|+{\frac {\|y''\|_{\infty }}{2}}h^{2}\\\|e_{n}\|&\leq n{\frac {\|y''\|_{\infty }}{2}}h^{2}\\&\leq {\frac {M}{2}}h\end{aligned}}\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/121831f773784548e823e73120ae8dae79959fa9)