Linear BVP | Sahithyan's Notes

A boundary value problem is linear iff

y'' = p(x)y' + q(x)y + r(x), \quad a \leq x \leq b, \quad y(a) = \alpha,\ y(b) = \beta

for continuous functions $p$ , $q$ , and $r$ .

Linear Shooting Method

Solves a linear BVP by combining solutions to 2 initial value problems.

A unique solution exists if

$p$ , $q$ , and $r$ are continuous on $[a,b]$
$q(x) > 0$ for $x \in [a,b]$

Superposition Idea

If $y_1$ solves

y_1'' = p(x)y_1' + q(x)y_1 + r(x)

and $y_2$ solves the homogeneous equation

y_2'' = p(x)y_2' + q(x)y_2

then $y = y_1 + c\,y_2$ solves the ODE for any constant $c$ .

Auxiliary IVPs

Fix

y_1(a) = \alpha,\ y_1'(a) = 0, \qquad y_2(a) = 0,\ y_2'(a) = 1

Solve

y_1'' = p(x)y_1' + q(x)y_1 + r(x), \quad y_1(a) = \alpha,\ y_1'(a) = 0 \\[8pt] y_2'' = p(x)y_2' + q(x)y_2, \quad y_2(a) = 0,\ y_2'(a) = 1

with order 4 Runge-Kutta.

Combining Solutions

Choose $c$ so that $y(b) = \beta$ .

c = \frac{\beta - y_1(b)}{y_2(b)}, \qquad y(x) = y_1(x) + \frac{\beta - y_1(b)}{y_2(b)}\,y_2(x)

Linear Finite-Difference Method

Replace derivatives with centered differences at mesh points and solve the resulting linear system.

Mesh Points

Partition $[a,b]$ into $N+1$ subintervals of length $h = \dfrac{b-a}{N+1}$ .

x_i = a + ih, \quad i = 0, 1, \ldots, N+1

Centered Difference Formulas

y''(x_i) = \frac{y(x_{i+1}) - 2y(x_i) + y(x_{i-1})}{h^2} + O(h^2), \qquad y'(x_i) = \frac{y(x_{i+1}) - y(x_{i-1})}{2h} + O(h^2)

Linear Difference Equation

With $y(x_i) \approx w_i$ , $w_0 = \alpha$ , and $w_{N+1} = \beta$ ,

\left(1 + \frac{h}{2}p(x_i)\right)w_{i-1} - \big(2 + h^2 q(x_i)\big)w_i + \left(1 - \frac{h}{2}p(x_i)\right)w_{i+1} = h^2 r(x_i)

for $i = 1, \ldots, N$ , with truncation error $O(h^2)$ .

Matrix Form

Aw = b

where $A$ is tridiagonal, with

a_i = -1 - \frac{h}{2}p(x_i), \quad d_i = 2 + h^2q(x_i), \quad c_i = -1 + \frac{h}{2}p(x_i) \\[8pt] A = \begin{pmatrix} d_1 & c_1 & & \\ a_2 & d_2 & c_2 & \\ & \ddots & \ddots & \ddots \\ & & a_N & d_N \end{pmatrix} \\[8pt] w = \begin{pmatrix}w_1\\w_2\\\vdots\\w_N\end{pmatrix}, \quad b = \begin{pmatrix}-h^2r(x_1)-a_1\alpha\\-h^2r(x_2)\\\vdots\\-h^2r(x_N)-c_N\beta\end{pmatrix}

Solve for $w$ with LU decomposition or another linear solver.

If $p$ , $q$ , and $r$ are continuous on $[a,b]$ and $q(x) \geq 0$ , the linear system has a unique solution whenever $h < \dfrac{2}{L}$ , where $L = \max_{a \leq x \leq b}|p(x)|$ .

Worked Examples

Example 1

Solve $y'' = y$ , $0 \leq x \leq 1$ , $y(0) = 0$ , and $y(1) = 1$ with the linear shooting method.

Solution Reveal Hide

$p = 0$ , $q = 1$ , and $r = 0$ .

For the shooting method, solve the auxiliary IVPs

y_1'' = y_1, \quad y_1(0) = 0,\ y_1'(0) = 0

and

y_2'' = y_2, \quad y_2(0) = 0,\ y_2'(0) = 1

The first IVP has the zero solution, so $y_1(x) = 0$ . The second IVP has the solution $y_2(x) = \sinh x$ . Therefore,

c = \frac{1-y_1(1)}{y_2(1)} = \frac{1-0}{\sinh 1} = 0.8509

Combining the auxiliary solutions gives

y(x) = y_1(x) + c y_2(x) = 0 + \frac{\sinh x}{\sinh 1}

Example 2

Solve $y'' = y$ , $0 \leq x \leq 1$ , $y(0) = 0$ , and $y(1) = 1$ with the linear finite-difference method for $N = 3$ .

Solution Reveal Hide

For the finite-difference method,

h = \frac{b-a}{N+1} = \frac{1-0}{3+1} = 0.25, \qquad x_1=0.25,\ x_2=0.50,\ x_3=0.75

Since $p = r = 0$ and $q = 1$ ,

-w_{i-1} + (2+h^2)w_i-w_{i+1}=0, \qquad 2+h^2=2+0.25^2=2.0625

Using $w_0=0$ and $w_4=1$ gives

\begin{pmatrix}2.0625&-1&0\\-1&2.0625&-1\\0&-1&2.0625\end{pmatrix}\begin{pmatrix}w_1\\w_2\\w_3\end{pmatrix}=\begin{pmatrix}0\\0\\1\end{pmatrix}

The 1st and 3rd equations give

w_1 = \frac{w_2}{2.0625}, \qquad w_3 = \frac{1+w_2}{2.0625}

Substitute these into the 2nd equation:

-\frac{w_2}{2.0625}+2.0625w_2-\frac{1+w_2}{2.0625}=0

\left(2.0625-\frac{2}{2.0625}\right)w_2=\frac{1}{2.0625} \implies w_2=0.4437

Back-substitution gives

w_1=\frac{0.4437}{2.0625}=0.2151, \qquad w_3=\frac{1+0.4437}{2.0625}=0.6999

$x_i$	$w_i$	Exact value	Error
$0.25$	$0.2151$	$0.2150$	$0.0002$
$0.50$	$0.4437$	$0.4434$	$0.0003$
$0.75$	$0.6999$	$0.6997$	$0.0002$

Example 3

Solve $y'' = 6x$ , $0 \leq x \leq 1$ , $y(0) = 0$ , $y(1) = 1$ with the linear finite-difference method for $N = 1$ .

Solution Reveal Hide

Here $p = q = 0$ and $r(x)=6x$ . Calculate the mesh size and interior point:

h=\frac{b-a}{N+1}=\frac{1-0}{1+1}=0.5, \qquad x_1=a+h=0.5

Evaluate the forcing term at the interior point:

r(x_1)=r(0.5)=6(0.5)=3

Since $p=q=0$ , the difference equation is

-w_0+2w_1-w_2=-h^2r(x_1)=-(0.5)^2(3)=-0.75

Apply the boundary values $w_0=y(0)=0$ and $w_2=y(1)=1$ :

-0+2w_1-1=-0.75

2w_1=0.25 \implies w_1=0.125

The exact solution is $y=x^3$ , so

y(0.5)=(0.5)^3=0.125

The finite-difference method gives the exact value at the interior point.