-1

$$(y_{n+1}-2y_n+y_{n-1})-\frac{h}{2}(y_{n+1}-y_{n-1})-6h^2y_n=0$$

Show that solution is approximately $(1-2h)^n$ when $n$ tends to infinity and $0<h<<1$, given initial condition $y_0=1$

I have tried to substitute $y_n=C_1\lambda_1^n+C_2\lambda_2^n$, it gives me characteristic equation $\lambda^2-2\lambda+1-\frac{h}{2}(\lambda^2-1)-6h^2\lambda=0$.

Solving gives $\lambda_{1,2}$ as $$\lambda_{1,2}=\frac{2+6h^2 \pm6h\sqrt{h^2+25/36}}{2-h}$$ which does not seem like the solution to the equation $y_n$, and hard to simplify

Any help is appreciated

José Carlos Santos
  • 427,504
Steve
  • 868
  • Sorry but how is the (almost instantly) accepted answer below, even addressing your question? – Did Jul 25 '17 at 09:38

1 Answers1

1

$$y_{n+1}(1-h/2)+y_n(-2-6h^2)+y_{n-1}(1+h/2)=0.$$

So the char. equation is $$x^2(1-h/2)+x(-2-6h^2)+(1+h/2)=0.$$

If $h=2$ the term in $x^2$ vanishes and $x=1/13$ and $y_n=y_{n-1}/13.$

If $h\ne 2$ the discriminant of the char. eq. is $$(-2-6h^2)^2-4(1-h/2)^2=(2+6h^2)^2-(2-h)^2=(h)(6h+1)(6h^2-h+4).$$

If $h\ne 2$ and the discriminant is $0$ there is one solution for $x,$ and $y_n=(A+nB)x^n$ with constants $A,B$.

DanielWainfleet
  • 57,985
  • Sorry I still don't understand when the question asks to show that solution is bounded by $(1-2h)^n$ when n tends to infinty and $0<h<<1$, given initial condition $y_0=1$ – Steve Nov 28 '16 at 23:10
  • You had the wrong char. eq. Thought I'd help with that. – DanielWainfleet Nov 29 '16 at 01:56