Dr Jerzy J. Jasielec

Dirichlet boundary problem

PMiKNoM seminars - class 5

Introduction

• Second Fick's Law for stationary states and other examples of boundary problems.

The general equation

There are many applications, that require solution of second order differential equation with boundary conditions expressed by known values of the function at the boundary points (optionally its derivatives or combination of function value and derivative). An example of such problem is given as:

The problem defined with the above equations is known as Dirichlet boundary problem. The important thing is, that the values of the function `u(x)` at the boundaries are given.

Solution of Dirichlet boundary problem:

General equation:

• Divide the range `[x_L,x_R]` into `n+1` equal sections. Distance between each two points `x_i` and `x_{i+1}` equals `h=(x_R-x_L)/(n+1)`.
• Write the general equation for each `i`-th discrete point:
  `-\alpha(x_i) u \text{"} (x_i) + \beta(x_i) u \text{'} (x_i) + \gamma(x_i)u(x_i) = f(x_i)`
• Substitute the first and second derivative using finite differences:
  ` u \text{'}(x_i) = \frac{u_{i+1}-u_{i-1}}{2h} `
  ` u \text{"}(x_i) = \frac{u_{i+1}-2u_i+u_{i-1}}{h^2} `
• You obtain:
  `-\alpha(x_i) \frac{u_{i+1}-2u_i+u_{i-1}}{h^2} + \beta(x_i) \frac{u_{i+1}-u_{i-1}}{2h} + \gamma(x_i)u(x_i) = f(x_i)`
• Multiply both sides by `h^2` and rearrange the equation:
  `u_{i-1} ( - \alpha(x_i) - \frac{h}{2} \beta(x_i) ) + u_i (2 \alpha(x_i) + h^2 \gamma(x_i)) + u_{i+1} ( - \alpha(x_i) + \frac{h}{2} \beta(x_i) ) = h^2 f(x_i)`

Boundaries:

• Rewrite the equation above, substituting `i=1`:
  `u_0 ( - \alpha(x_1) - \frac{h}{2} \beta(x_1) ) + u_1 (2 \alpha(x_1) + h^2 \gamma(x_1)) + u_2 ( - \alpha(x_1) + \frac{h}{2} \beta(x_1) ) = h^2 f(x_1)`
• Having `u_0 = u(a) = u_L`, one can move it to the right-hand side of the equation:
  `u_1 (2 \alpha(x_1) + h^2 \gamma(x_1)) + u_{2} ( - \alpha(x_1) + \frac{h}{2} \beta(x_1) ) = h^2 f(x_1) + u_L ( \alpha(x_1) + \frac{h}{2} \beta(x_1) )`
• We perform similar substitution for the point `i=n`, obtaining:
  `u_{n-1} ( - \alpha(x_n) - \frac{h}{2} \beta(x_n) ) + u_n (2 \alpha(x_n) + h^2 \gamma(x_n)) + u_{n+1} ( - \alpha(x_n) + \frac{h}{2} \beta(x_n) ) = h^2 f(x_n)`
• Having `u_{n+1} = u(b) = u_R`, one can move it to the right-hand side of the equation:
  `u_{n-1} ( - \alpha(x_n) - \frac{h}{2} \beta(x_n) ) + u_n (2 \alpha(x_n) + h^2 \gamma(x_n)) = h^2 f(x_n) + u_R ( \alpha(x_n) - \frac{h}{2} \beta(x_n) )`

Tri-diagonal matrix:

• The set of equations:
  $$ \left\{ \begin{array}{l} u_1 (2 \alpha(x_1) + h^2 \gamma(x_1)) + u_{2} ( - \alpha(x_1) + \frac{h}{2} \beta(x_1) ) = h^2 f(x_1) + u_L ( \alpha(x_1) + \frac{h}{2} \beta(x_1) ) \\ u_{i-1} ( - \alpha(x_i) - \frac{h}{2} \beta(x_i) ) + u_i (2 \alpha(x_i) + h^2 \gamma(x_i)) + u_{i+1} ( - \alpha(x_i) + \frac{h}{2} \beta(x_i) ) = h^2 f(x_i) \text{, for } i=2..n-1 \\ u_{n-1} ( - \alpha(x_n) - \frac{h}{2} \beta(x_n) ) + u_n (2 \alpha(x_n) + h^2 \gamma(x_n)) = h^2 f(x_n) + u_R ( \alpha(x_n) - \frac{h}{2} \beta(x_n) ) \end{array} \right. $$
  forms the tri-diagonal matrix:
  $$ \left\{ \begin{array}{l} d_1 u_1 + c_1 u_2 = b_1 \\ a_{i-1} u_{i-1} + d_i u_i + c_i u_{i+1} = b_i \text{, for } i=2..n-1\\ a_{n-1} u_{n-1} + d_n u_n = b_n \end{array} \right. $$
• Hence, in order to solve the problem, first one has to assign values to the tri-diagonal matrix, i.e. `d_1 = (2 \alpha(x_1) + h^2 \gamma(x_1))`, `c_1 = ( - \alpha(x_1) + \frac{h}{2} \beta(x_1) )`, `b_1 = h^2 f(x_1) + u_L ( \alpha(x_1) + \frac{h}{2} \beta(x_1) )`, and so on...
• Solution of the tri-diagonal matrix is presented in class 4.

Additional materials:

• R. Filipek, K. Szyszkiewicz-Warzecha "Metody Matematyczne dla Ceramików", pages 128-147.