Question

Consider the boundary-value problem $\mathit{y}\mathbf{"}\mathbf{+}\mathit{x}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$.

(a) Find the difference equation corresponding to the differential equation. Show that for$\mathit{i}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{n}\mathbf{-}\mathbf{1}$ , the difference equation yields equations in $\mathit{n}\mathbf{+}\mathbf{1}$unknows${\mathit{y}}_{\mathbf{-}\mathbf{1}}\mathbf{,}{\mathit{y}}_{\mathbf{0}}\mathbf{,}{\mathit{y}}_{\mathbf{1}}\mathbf{,}{\mathit{y}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{y}}_{\mathbf{n}\mathbf{-}\mathbf{1}}$ . Here${\mathit{y}}_{\mathbf{-}\mathbf{1}}$ and ${\mathit{y}}_{\mathbf{0}}$are unknowns, since ${\mathit{y}}_{\mathbf{-}\mathbf{1}}$represents an approximation$\mathit{y}$ to at the exterior point$\mathit{x}\mathbf{=}\mathbf{-}\mathit{h}$ and ${\mathit{y}}_{\mathbf{0}}$is not specified at$\mathit{x}\mathbf{=}\mathbf{0}$ .

(b) Use the central difference approximation (5) to show that${\mathit{y}}_{\mathbf{1}}\mathbf{-}{\mathit{y}}_{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{2}\mathit{h}$ . Use this equation to eliminate${\mathit{y}}_{\mathbf{-}\mathbf{1}}$ from the system in part (a).

(c) Use $\mathit{n}\mathbf{=}\mathbf{5}$and the system of equations found in parts (a) and (b) to approximate the solution of the original boundary-value problem.

Short answer

(a) So, the difference equation is$y_{i+1}+(-2+h^2{x}_i)y_i+y_{i-1}=0i=0,1,2,....n-1$ corresponding to the differential equation.

(b) So, the equation is $2y_1+(-2+h^2{x}_0)y_0-2h=0$.

(c) So, the approximate solutions are

$$\begin{gathered} y_0=-2.2755,y_1=-2.0755,y_2=-1.8589, \\ {y}_3=-1.6126,y_4=-1.3275 \end{gathered}$$

Step-by-step solution

Definition of finite difference

Finite difference methods are a class of numerical techniques for solving differential equations by approximating derivatives with finite differences.

Solve for finite difference(a)

Given boundary value problem is$y"+xy=0,y\left(0\right)=1,y\left(1\right)=-1$ .

This equation is same as (7) in the text except for the boundary conditions.

Since derivation of the difference equation (8) does not change.

Now comparing $y"+xy=0$with (7) we get

$p\left(x\right)=0$and$Q\left(x\right)=x,f\left(x\right)=0$

From (8) the finite difference equation is

$$\begin{gathered} \left(1+\frac{h}{2}.0\right)y_{i+1}+\left(-2+h^2.x_i\right)y_i+\left(1-\frac{h}{2}.0\right)y_{i-1}=0 \\ \Rightarrow y_{i+1}+\left(-2+h^2{x}_i\right)y_i+y_{i-1}=0i=0,1,2,....n-1 \end{gathered}$$

Solve for interior points

Thus for $i=0,1,2.....n-1$we get system as

$$\begin{gathered} y_1+\left(-2+h^2{x}_0\right)y_0+y_{-1}=0 \\ {y}_2+\left(-2+h^2{x}_1\right)y_1+y_0=0 \\ {y}_n+\left(-2+h^2{x}_{n-1}\right)y_{n-1}{y}_{n-2}=0 \end{gathered}$$

In the above system we know only $y_n$.

Thus, the system yield equations in$n+1$ unknowns namely $y_{-1,}{y}_0,.....y_{n-1.}$.

(b) Step 1: Solve for equation

We know $y_0=y\left(0\right)$and $y_{-1}=y(0-h),y_1=y(0+h)$Now form (5)$\frac{1}{2h}[y_1-y_{-1}]=y\left(0\right)$

But $y\left(0\right)=1$

So$$\begin{gathered} \frac{1}{2h}\left[y_1-y_{-1}\right]=y\left(0\right) \\ \Rightarrow y_1-y_{-1}=2h \\ \Rightarrow y_{-1}=y_1-2h \end{gathered}$$

So, the first equation in the system becomes

$$\begin{gathered} y_1+(-2+h^2{x}_0)y_0+y_1-2h=0 \\ \Rightarrow 2y_1+(-2+h^2{x}_0)y_0-2h=0 \end{gathered}$$

(c)Step 1: Solve for interior point

Now we $n$get equations in $n$variables.

$n=5\Rightarrow h=\frac{b-a}{n}=\frac{1-0}{5}=\frac{1}{5}$

Now the interior points are

$$\begin{gathered} x_1=0+\frac{1}{5}=\frac{1}{5},x_2=0+\frac{2}{5}=\frac{2}{5},x_3=0+\frac{3}{5}=\frac{3}{5} \\ {x}_4=0+\frac{4}{5}=\frac{4}{5},x_5=0+\frac{5}{5}=1 \end{gathered}$$

So, the system is

$$\begin{gathered} 2y_1+(-2+h^2{x}_0)y_0-2h=0 \\ {y}_2+(-2+h^2{x}_1)y_1+y_0=0 \\ {y}_3+(-2+h^2{x}_2)y_2+y_1=0 \\ {y}_4+(-2+h^2{x}_3)y_3+y_2=0 \\ {y}_5+(-2+h^2{x}_4)y_4+y_3=0 \end{gathered}$$

Use initial condition

Form the initial conditions $y_5=-1$so the system is

$$\begin{gathered} 2y_1+(-2y_0)-0.4=0 \\ {y}_2+(-1.992)y_1+y_0=0 \\ {y}_3+(-1.984)y_2+y_1=0 \\ {y}_4+(-1.976)y_3+y_2=0 \\ -1+(-1.968)y_4+y_3=0 \end{gathered}$$

Step 3:

Solving the system, we get

$$\begin{gathered} y_0=-2.2755 \\ {y}_1=-2.0755 \\ {y}_2=-1.8589 \\ {y}_3=-1.6126 \\ {y}_4=-1.3275 \end{gathered}$$

Thus, the required solution is

$$\begin{gathered} y_0=-2.2755,y_1=-2.0755,y_2=-1.8589, \\ {y}_3=-1.6126,y_4=-1.3275 \end{gathered}$$