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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 $\mathbb{R} \rightarrow \mathbb{R^+}$?

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

1 Answer 1

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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.

To obtain this error, you need that the first step has the same order or better one order more, so the third order Heun method, or even an RK4 step would be appropriate.


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.

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