# Implicit linear multistep method gives weird results on this ODE

Given a ODE with its initial value

$$u'(t) = 5u(t), u(0) = 0.5$$

Just for reference, the analytical solution would be $$u(t) = \frac{1}{2}e^{5t}$$

Given this implicit, multi-step formula:

$$u_{i+1} = \frac{4}{3}u_i - \frac{1}{3}u_{i-1} + \frac{2}{3}h\cdot f_{i+1}$$

As we can observer, we need $$u_1$$ for solving $$u_2 = \frac{4}{3}u_1 - \frac{1}{3}u_0 + \frac{2}{3}h\cdot f(x_2, u_2)$$.

So my plan was, to calculate one step of the implicit euler formula first: $$u_1 = u_0 + h\cdot f(x_1, u_1)$$ $$u_1 = \frac{1}{2} + \frac{1}{2}\cdot 5u_1$$ $$u_1 = \frac{1}{2} + \frac{5}{2}\cdot u_1$$ $$-\frac{3}{2}u_1 = \frac{1}{2}$$ $$u_1 = -\frac{1}{3}$$

So how can $$u_1$$ be $$< 0$$ if the analytical formula $$u(t) = \frac{1}{2}e^{5t}$$ is a function of type $$\mathbb{R} \rightarrow \mathbb{R^+}$$?

• You seem to lose $h$ at the very first step. Jul 11, 2019 at 21:46
• @user58697 $h = \frac{1}{2}$? Jul 12, 2019 at 6:40

Your method $$3u_{i+1}-4u_i+u_{i-1}=2hf_{i+1}$$ is based on the order 2 backwards differentiation formula and thus of consistency order 2. As the roots of the characteristic polynomial on the left side are $$1$$ and $$\frac13$$, it is also stable and thus order 2 convergent.
With step size $$h=\frac12$$ you get a linear recursion equation $$-2u_{i+1}-4u_i+u_{i-1}=0$$ where the characteristic roots are $$-1\pm\sqrt{\frac32}$$, which is far away from the exponential step factor $$e^{5h}=12.18249$$.
For general $$h$$ the characteristic roots of the recursion $$(3-10h)u_{i+1}-4u_i+u_{i-1}=0$$ are $$(2q-1)^2=(1+10h)q^2\implies q=\frac1{2\pm\sqrt{1+10h}}=\frac{2\mp\sqrt{1+10h}}{3-10h}$$ The root close to $$1$$ satisfies $$q=\frac1{1-5h+\frac54h^2- \frac{125}2h^3 +O(h^4)}=\exp\left(5h + \frac{125}3h^3+O(h^4) \right)$$ To get correct results, you need $$\frac{125}3h^2\ll 5$$ or $$h\ll\frac{\sqrt3}5=0.346410..$$, so $$h=0.2$$ could start to give useful results.