# solution of difference equation

I am trying to solve the following difference equation:

$$-\frac{\epsilon}{h^2}U_{n+1}+\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)U_{n}-\left(\frac{\epsilon}{h^2}+\frac{1}{h}\right)U_{n-1}=0,\mbox{ }\mbox{ }\mbox{ }\mbox{ }U_0=1,\mbox{ }U_1=0.$$

I try $U_{n}=Aw^n$ then I get

$$w_{1,2}=\frac{\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)\pm\sqrt{\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)^2-4\frac{\epsilon}{h^2}\left(\frac{\epsilon}{h^2}+\frac{1}{h}\right)}}{2\frac{\epsilon}{h^2}}.$$

This seems a bit far from what I want to get. I am trying to verify that the solution is

$$U_n=\dfrac{1-(1+\rho)^{n-N}}{1-(1+\rho)^{-N}},$$

where $0\leq n\leq N$ and $\rho=h/\epsilon$.

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In your third term in your difference equation, did you mean $U_{n-1}$? – Cameron Williams Jul 4 '13 at 18:19
If you want to verify a solution you already have, just substitute it in the equation and check it is true, and also check the initial conditions. – Mlazhinka Shung Gronzalez LeWy Jul 4 '13 at 18:39
Note that the homogeneous equation has solution $U_n=1$, equivalently you have a solution $w_1=1$. This makes life a lot easier. – Mark Bennet Jul 4 '13 at 18:39
@CameronWilliams, yes and thanks for that. – Vaolter Jul 5 '13 at 7:40
@MarkBennet, could you elaborate. – Vaolter Jul 5 '13 at 7:42

First of all note that, inside the square root, we have $$\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)^2-4\frac{\epsilon}{h^2}\left(\frac{\epsilon}{h^2}+\frac{1}{h}\right) = \frac{1}{h^2}$$ So the answers read $$w_{1,2}=\frac{\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)\pm \frac{1}{h}}{2\frac{\epsilon}{h^2}}. \Longrightarrow w_1=1, w_2=1+\frac{h}{\epsilon}$$ This means that the solution reads $U_n=A+B(1+\rho)^n$. Now let us impose boundary conditions . Since $U_0=1$, we get that $A+B=1$. Also, since $U_1=0$, we get that $A+B(1+\rho)=0$ or equivalently, $A=-(1+\rho)B$. Plugging this into the first equation, we get that $B=\frac{-1}{\rho}$, and that $A=\frac{1+\rho}{\rho}$, leading to a solution of the form $$U_n=\frac{1+\rho}{\rho} \bigg[ 1-(1+\rho)^{n-1} \bigg].$$ We can readily check that for $n=1$, we get $U_1=0$, and $n=0$, we get $U_0=1$.

bBut this is not what you are trying to recover. Your desired answer has $N$ in it, which is defined nowhere in the statement of the problem. My conjecture is that, $n$ is limited to the range $0,N$.

Let us now consider the following set of boundary conditions: $$U_0=1,U_N=0$$ Imposing the first condition on $U_n=A+B(1+\rho)^n$, we get $A+B=1$. Imposing the second condition, we get $A+B(1+\rho)^N=0$, which can be equivalently expressed as $A=-B(1+\rho)^N$. Plugging this into the first equation, we obtain $B=\frac{1}{1-(1+\rho)^N}$ . Also for $A$ we obtain $A=\frac{ -(1+\rho)^N}{1-(1+\rho)^N}$.

For the solution, we use the values of $A,B$ obtained above, and arrive at

$$U_n=\frac{(1+\rho)^n-(1+\rho)^N}{1-(1+\rho)^N}$$

Dividing both the numerator and the denominator by $-(1+\rho)^N$, we arrive at

$$U_n=\frac{1-(1+\rho)^{n-N}}{1-(1+\rho)^{-N}}$$

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