Applying the Finite Difference Method to Boundary-Value Problems 4 |

In many problems, boundary conditions are specified in terms of the derivative of the function rather than the function itself. For example, consider the following boundary conditions:

$\displaystyle v'(a) = \alpha, v'(b) = \beta.$

This corresponds to the scenario discussed in “Applying Finite Difference Method to Bounday-Value Problems 1” where $p(x)=1, q(x)=0, r(x)=-c(x)$ and $s(x)=d(x)$ . Additionaly, the constants are: $\alpha_0 = \alpha_1 =0, \beta_0 = \beta_1=1,\gamma_0 =\alpha$ and $\gamma_1=\beta$ .

An alternative approach introduces grid points $x_{-1} = a-h$ and $x_{n+1} = b+h$ outside the interval $[a,b]$ . As a result, the discretized differential equation at the boundaries becomes

$\displaystyle -v_{-1} + (2+c_0h^2)v_0 -v_1 = -h^2d_0,\quad (i=0)\quad\quad\quad(4-1)$

$\displaystyle -v_{n-1} + (2 + c_nh^2)v_n -v_{n+1}=- h^2d_n, \quad(i=n)\quad\quad\quad(4-2)$

and for the interior points:

$\displaystyle -v_{i-1} + (2 + c_i h^2) v_i -v_{i+1} = -h^2d_i, \quad 1 \le i \le n-1.\quad\quad\quad(4-3)$

To approximate the boundary condition $v'(x_0=a) = \alpha$ more accurately, we use

$\displaystyle \frac{v_{1}-v_{-1}}{2h} = \alpha,$

which gives

$\displaystyle v_{-1} = v_1 - 2\alpha h.$

Substituting $v_{-1}$ into (4-1) yields

$\displaystyle -v_1+2\alpha h +(2+c_0 h^2)v_0 -v_1 = -d_0 h^2$

which is

$\displaystyle (2+c_0h^2)v_0-2v_1 = - h^2d_0 - 2\alpha h.\quad\quad\quad(4-4)$

Similarly, by eliminating $v_{n+1}$ from

$\displaystyle \begin{cases} -v_{n-1} + (2 + c_nh^2)v_n -v_{n+1}=- h^2d_n\\ \beta = \frac{v_{n+1}-v_{n-1}}{2h},\end{cases}$

we obtain

$\displaystyle -2v_{n-1} + (2+ c_n h^2)v_n = - h^2d_n - 2\beta h.\quad\quad\quad(4-5)$

Using (4-4), (4-5) and (4-3), we arrive at the following system of $n+1$ equations for the $n+1$ unknowns $v_0, ..., v_{n}$ :

$\displaystyle \begin{cases} (2+c_0h^2)v_0 - 2v_1 = -h^2d_0 - 2\alpha h\\ \\ -v_{i-1} + (2+c_i h^2)v_i - v_{i+1} = - h^2 d_i, \quad 1 \le i \le n-1 \\ \\ -2v_{n-1} + (2+c_n h^2)v_{n} = - h^2 d_{n} - 2\beta h.\end{cases}$

The coefficient matrix of this system is

$\displaystyle \begin{pmatrix} 2+c_0 h^2 & -2 & & \\ \\ -1 & 2+c_1h^2 & -1 & \\ \\ & \ddots & \ddots & \ddots \\ \\ & & -2 & 2+c_nh^2\end{pmatrix}$

According to “A Page in My Notebook [5]”, if $c_i > 0$ for $i=0, ..., n$ , the matrix $A$ is nonsingular. However, if all $c_i$ are zero, the matrix becomes singular. This singularity signifies the nonuniqueness of solutions to the boundary-value problem. Specifically, when $c(x) \equiv 0$ and $v$ is a solution, any shift $v + r$ , for any constant $r$ , is also a solution.

Another type of boundary condition that leads to nonunique solutions is when

$\displaystyle v(a) = v(b).$

In this case, $v_0$ and $v_n$ are both unknown, but only one of them needs to be included in the system of equations. Additionaly, if the solution is extended periodically outside of $[a, b]$ , we can assume that $v_{-1} = v_{n-1}$ , leading to the equation:

$\displaystyle -v_{-1} + (2 + c_0 h^2) v_0 -v_{1} = -h^2d_0, \quad (i=0)$

$\displaystyle \overset{v_{-1} = v_{n-1}}{\implies} (2 + c_0 h^2) v_0 -v_{1}-v_{n-1}= -h^2d_0.\quad\quad\quad(4-6)$

Similarly,

$\displaystyle -v_{n-2} + (2 + c_{n-1} h^2) v_{n-1} -v_{n} = -h^2d_{n-1}, \quad (i=n-1)$

$\displaystyle \overset{v(a) = v(b)\; : \;v_0=v_{n}}{\implies}-v_0 -v_{n-2} + (2 + c_n h^2) v_{n-1} = -h^2d_{n-1}.\quad\quad\quad(4-7)$

By combining (4-6) and (4-7) with the equation for the interior:

$\displaystyle -v_{i-1} + (2 + c_i h^2) v_i -v_{i+1} = -h^2d_i, 1 \le i \le n-2,$

we form a system of $n$ linear equations with $n$ unknowns $v_0, ..., v_{n-1}$ . The corresponding coefficient matrix is

$\displaystyle A = \begin{pmatrix} 2+c_0 h^2 & -1 & & -1 \\ \\ -1 & 2+c_1h^2 & -1 & \\ \\ & \ddots & \ddots & \ddots \\ \\ -1 & & -1 & 2+c_nh^2\end{pmatrix}$

The matrix $A$ remains nonsingular if all $c_i$ are positive, but becomes singular if all $c_i$ are zero. This singularity, once again, signifies the nonuniqueness of the solution.

Exercise-1 Explain “To approximate the boundary condition $v'(x_0=a) = \alpha$ more accurately, we use $\displaystyle \frac{v_{1}-v_{-1}}{2h} = \alpha$ ”

Exercise-2 Show that “when $c(x) \equiv 0$ and $v$ is a solution, any shift $v + r$ , for any constant $r$ , is also a solution.”

Applying the Finite Difference Method to Boundary-Value Problems 4 |

Most Popular

Google Forms for Formative Assessment in Math Class

2nd Grade Subtraction Worksheet | Subtraction of 2-Digit Numbers

S01 overview – Intellectual Mathematics

Digital SAT Math Problems and Solutions (Part

More from Author

Google Forms for Formative Assessment in Math Class

2nd Grade Subtraction Worksheet | Subtraction of 2-Digit Numbers

S01 overview – Intellectual Mathematics

Digital SAT Math Problems and Solutions (Part

Read Now

Google Forms for Formative Assessment in Math Class

2nd Grade Subtraction Worksheet | Subtraction of 2-Digit Numbers

S01 overview – Intellectual Mathematics

Digital SAT Math Problems and Solutions (Part

How To Find The Factors Of 20: A Simple Way

Addition & Subtraction Together |Combination of addition & subtraction

Two Back to School Ideas for Digital Classrooms

Digital SAT Math Problems and Solutions (Part

Percent of Increase Word Problems

Societal role of geometry in early civilisations – Intellectual Mathematics

Sub Plans for High School Math Class – Webquests are Perfect!

Digital SAT Math Problems and Solutions (Part