2
$\begingroup$

Consider the following difference equation: $$-(1+a)u_{i-1} + 2 u_i - (1-a) u_{i+1} - \frac{h^2}{\epsilon} = 0, \; i = 1, ..., N, \tag{1}$$ $$u_0 = 0, \; u_N = 1.$$

Can someone explain me how to solve this equation?

I read somewhere the following method:

1) Solve the homogeneous equation;

2) Find a particular solution of inhomogeneous equation (1);

3)The general solution of (1) is the sum of the particular solution and the solution of homogeneous equation.

Is this method right?

I can't find a particular solution of equation (1). Can someone help me?

Thank you!

$\endgroup$
2
$\begingroup$

Note that (1) can be written as $$(1+a)(u_i-u_{i-1})- (1-a)(u_{i+1}-u_i) = \frac{h^2}{\epsilon}. \tag{2}$$ Let $v_i=u_i-u_{i-1}$ and then (2) becomes $$(1+a)v_i- (1-a)v_{i+1} = \frac{h^2}{\epsilon} = 0$$ or $$(1+a)\left(v_i-\frac{h^2}{2a\epsilon}\right)=(1-a)\left(v_{i+1}-\frac{h^2}{2a\epsilon}\right)$$ or $$v_{i+1}-\frac{h^2}{2a\epsilon}=\frac{1+a}{1-a}\left(v_i-\frac{h^2}{2a\epsilon}\right)\tag{3}$$ From (3), one obtains $$ v_i-\frac{h^2}{2a\epsilon}=\left(v_1-\frac{h^2}{2a\epsilon}\right)\left(\frac{1+a}{1-a}\right)^{i-1}=\left(u_1-\frac{h^2}{2a\epsilon}\right)\left(\frac{1+a}{1-a}\right)^{i-1}.$$ Thus \begin{eqnarray} u_i&=&\sum_{k=1}^i(u_k-u_{k-1})+u_0\\ &=&\sum_{k=1}^iv_i\\ &=&\sum_{k=1}^i\left[\frac{h^2}{2a\epsilon}+\left(u_1-\frac{h^2}{2a\epsilon}\right)\left(\frac{1+a}{1-a}\right)^{i-1}\right]\\ &=&\frac{h^2}{2a\epsilon}i+\left(u_1-\frac{h^2}{2a\epsilon}\right)\frac{1-\left(\frac{1+a}{1-a}\right)^{i}}{1-\frac{1+a}{1-a}}\\ &=&\frac{h^2}{2a\epsilon}i-\frac{1-a}{2a}\left(u_1-\frac{h^2}{2a\epsilon}\right)\left[1-\left(\frac{1+a}{1-a}\right)^{i}\right]. \end{eqnarray} Using $u_N=1$, one can determine $u_1$ and I omit the detail.

$\endgroup$
  • $\begingroup$ I didn't understand why does $$ v_i-\frac{h^2}{2a\epsilon}=\left(v_1-\frac{h^2}{2a\epsilon}\right)\left(\frac{1+a}{1-a}\right)^{i-1}=\left(u_1-\frac{h^2}{2a\epsilon}\right)\left(\frac{1+a}{1-a}\right)^{i-1}.$$ result from (3). Can you give me more details? Thank you! $\endgroup$ – g.pomegranate Nov 30 '16 at 18:18
  • $\begingroup$ @g.pomegranate, this is because, if $a_{i+1}=qa_i$, then $a_i=a_1q^{i-1}$. $q$ is called the common ratio. $\endgroup$ – xpaul Nov 30 '16 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.