MHB Approximating Equations for Unknowns: How to Justify the Form of $U_0$?

[evinda]

Hello! (Wave)

Given the problem $$-u''(x)+q(x)u(x)=f(x), 0 \leq x \leq 1, \\ u'(0)=u(0), \ \ u(1)=0$$ where $f,g$ are continuous functions on $[0,1]$ with $q(x) \geq q_0>0, x \in [0,1]$. Let $U_j$ be the approximations of $u(x_j)$ at the points $x_j=jh, j=0, 1, \dots , N+1$, where $(N+1)h=1$, that gives the finite difference method $$-\frac{1}{h^2}\left (U_{j-1}-2U_j+U_{j+1}\right )+q(x_j)U_j=f(x_j), \ \ 1 \leq j \leq N \\ \frac{1}{h}(U_1-U_0)-U_0=\frac{1}{2}h\left (q(x_0)U_0-f(x_0)\right )$$ where $U_{N+1}=0$.

I have to justify the form of the equation for the unknown $U_0$. We have that the approximation of the first derivative $u'(x_j)$ is $$u'(x_j) \approx \frac{u(x_{i+1})-u(x_{i-1})}{2h}$$

so from $u'(0)=u(0)$ we have $$\frac{U_1-U_0}{h}=U_0 \Rightarrow \frac{1}{h}(U_1-U_0)-U_0=0$$ but this is not the desired result.

What have I done wrong? How do we get $\frac{1}{h}(U_1-U_0)-U_0=\frac{1}{2}h\left (q(x_0)U_0-f(x_0)\right )$ ? (Thinking)

 

[I like Serena]

Hey evinda! (Smile)

I believe we're including the second order correction:
$$u(h) = u(0) + hu'(0) +\frac 12 h^2 u''(0)$$
Thus
$$u(h) =u(0) + hu(0) +\frac 12 h^2\Big(q(0)u(0)-f(0)\Big) \\
\Rightarrow U_1 = U_0 + hU_0 +\frac 12h^2\Big(q(0)U_0-f(0)\Big)
$$
(Thinking)

 

[evinda]

Hey evinda! (Smile)

I believe we're including the second order correction:
$$u(h) = u(0) + hu'(0) +\frac 12 h^2 u''(0)$$
Thus
$$u(h) =u(0) + hu(0) +\frac 12 h^2\Big(q(0)u(0)-f(0)\Big) \\
\Rightarrow U_1 = U_0 + hU_0 +\frac 12h^2\Big(q(0)U_0-f(0)\Big)
$$
(Thinking)

I see... So do we suppose that at the Taylor expansion there is no error? (Thinking)

- - - Updated - - -

Also how could we show that the matrix of coefficients

$A=\begin{bmatrix}
-\frac{1}{h^2}+\frac{1}{h}+\frac{q(x_0)}{2} & -\frac{1}{h^2} & 0 & 0 & \cdots& 0\\
-\frac{1}{h^2} & \frac{2}{h^2}+q(x_1) & -\frac{1}{h^2} & 0 & \cdots & 0 \\
0 & -\frac{1}{h^2}& \frac{2}{h^2}+q(x_2) & -\frac{1}{h^2} & & 0\\
& & & & \ddots & 0 \\
& & & & & -\frac{1}{h^2}\\
& & & & -\frac{1}{h^2} & \frac{2}{h^2}+q(x_N)
\end{bmatrix}$

is invertible? (Thinking)

 

[I like Serena]

I see... So do we suppose that at the Taylor expansion there is no error? (Thinking)

There will still be an error, just an order of magnitude less.
Not bad eh? (Mmm)

Also how could we show that the matrix of coefficients

$A=\begin{bmatrix}
-\frac{1}{h^2}+\frac{1}{h}+\frac{q(x_0)}{2} & -\frac{1}{h^2} & 0 & 0 & \cdots& 0\\
-\frac{1}{h^2} & \frac{2}{h^2}+q(x_1) & -\frac{1}{h^2} & 0 & \cdots & 0 \\
0 & -\frac{1}{h^2}& \frac{2}{h^2}+q(x_2) & -\frac{1}{h^2} & & 0\\
& & & & \ddots & 0 \\
& & & & & -\frac{1}{h^2}\\
& & & & -\frac{1}{h^2} & \frac{2}{h^2}+q(x_N)
\end{bmatrix}$

is invertible? (Thinking)

We won't be able to guarantee that it's invertible for any $h$ and any function $q$.
I think that for any $h$ there will be a function $q$ such that the matrix is not invertible.

However, we can write $A$ as:
$$A=\frac 1{h^2}\begin{bmatrix}
-1+h+h^2\frac{q(x_0)}{2} & -1 & 0 & 0 & \cdots& 0\\
-1 & 2+h^2q(x_1) & -1 & 0 & \cdots & 0 \\
0 & -1& 2+h^2q(x_2) & -1 & & 0\\
& & & & \ddots & 0 \\
& & & & & -1\\
& & & & -1 & 2+h^2q(x_N)
\end{bmatrix}$$
And if $h$ is small enough, it approaches:
$$A \approx \frac 1{h^2}\begin{bmatrix}
-1 & -1 & 0 & 0 & \cdots& 0\\
-1 & 2 & -1 & 0 & \cdots & 0 \\
0 & -1& 2 & -1 & & 0\\
& & & & \ddots & 0 \\
& & & & & -1\\
& & & & -1 & 2
\end{bmatrix} $$
Would that be invertible? (Wondering)

 

[Ackbach]

There will still be an error, just an order of magnitude less.
Not bad eh? (Mmm)

We won't be able to guarantee that it's invertible for any $h$ and any function $q$.
I think that for any $h$ there will be a function $q$ such that the matrix is not invertible.

However, we can write $A$ as:
$$A=\frac 1{h^2}\begin{bmatrix}
-1+h+h^2\frac{q(x_0)}{2} & -1 & 0 & 0 & \cdots& 0\\
-1 & 2+h^2q(x_1) & -1 & 0 & \cdots & 0 \\
0 & -1& 2+h^2q(x_2) & -1 & & 0\\
& & & & \ddots & 0 \\
& & & & & -1\\
& & & & -1 & 2+h^2q(x_N)
\end{bmatrix}$$
And if $h$ is small enough, it approaches:
$$A \approx \frac 1{h^2}\begin{bmatrix}
-1 & -1 & 0 & 0 & \cdots& 0\\
-1 & 2 & -1 & 0 & \cdots & 0 \\
0 & -1& 2 & -1 & & 0\\
& & & & \ddots & 0 \\
& & & & & -1\\
& & & & -1 & 2
\end{bmatrix} $$
Would that be invertible? (Wondering)

From the wiki on tridiagonal matrices, it would appear so, since this particular matrix has all the off-diagonal elements equal (it's also Toeplitz, but that's more general). However, these results have only been obtained around 1996 or 1997 - fairly recently. Not sure they've made their way into a lot of textbooks yet.

 

[evinda]

There will still be an error, just an order of magnitude less.
Not bad eh? (Mmm)

We would have an error of order $O(h^3)$. Can we ignore it since it converges to $0$?
If so, how could we justify it formally? (Thinking)