I am trying to implement a stiff ODE solver that uses step doubling with the backward Euler method but I want to make a modification and I am unsure how to incorporate it.

I am trying to solve $y'=f(y), y(a)=d$ and I am using an approximation to $f$ such that $y'\approx F(y),\;t\in (T,T+dt), y(T)=Y$ where

$$F(y)=f(y)+Df(Y)(y(t) - Y).$$

I understand how to implement the backward Euler method when we simply have $y'=f(t,y)$ but I am having trouble writing a program to solve the problem here because $F(y)$ is dependent on $y(t)$, which is obviously unknown so the usual formulation of $\Delta x=Df^{-1}(x)f(x)$ doesn't seem to work. I was wondering how I could go about writing a method for this?

  • 1
    $\begingroup$ Don't you mean $F(y) = f(Y) + Df(Y)(y - Y)$ since the Taylor Series is about $Y$? Assuming that, your problem becomes $y' = (f(Y) - AY) + Ay$, where $A = Df(Y)$. The first term is independent of $y$, making your equations linear with respect to $y$ since the latter term is linear. Should be pretty straight forward to do implicit integration then! $\endgroup$
    – spektr
    Dec 5, 2016 at 23:12
  • $\begingroup$ Thanks! That helped but honestly I am still a bit confused on how to do the implicit integration? The way I calculated I got that $Df(Y)\Delta y=-f(y)$ so I solve that using Newton-Raphson and then do $\Delta y + y_t=y_{t+1}$ but that seems to be giving me the wrong answer. Is there anything I am missing? Sorry, I am very new to this haha. $\endgroup$
    – Twis7ed
    Dec 7, 2016 at 3:30
  • $\begingroup$ Sorry, meant to say that $\Delta y(I-Df(Y)\Delta t)=\Delta t f(Y)$ $\endgroup$
    – Twis7ed
    Dec 7, 2016 at 3:37

1 Answer 1


Let's assume you have some differential equation that can be written as the following:

$$ \frac{dy}{dt} = f(t,y)$$

for some initial condition $y(a) = d$. We then want to integrate this quantity using Backward Euler, with the fundamental formula being derived in the following way:

$$ \begin{align} \frac{dy}{dt} &= f(t_{n+1},y_{n+1})\\ \frac{y_{n+1} - y_{n}}{\Delta t} &= f(t_{n+1},y_{n+1})\\ y_{n+1} &= y_{n} + \Delta t f(t_{n+1},y_{n+1}) \end{align}$$

Now we might want to make things a little easier than solving nonlinear equations for $y_{n+1}$, if we're willing to add in a little bit of error, by linearizing $f(t_{n+1},y_{n+1})$ around $y_{n}$. We can show the linearization for $f(t_{n+1},y_{n+1})$ about $y_{n}$ is:

$$ f(t_{n+1},y_{n+1}) = f(t_{n+1},y_{n} + (y_{n+1} - y_{n})) \approx f(t_{n+1},y_{n}) + \left.\frac{\partial f}{\partial y}\right|_{(t_{n+1},y_{n})}\left(y_{n+1} - y_{n}\right)$$

If we substitute this into the Backward Euler equation, and set $A_n = \left.\frac{\partial f}{\partial y}\right|_{(t_{n+1},y_{n})}$, we can work out the following: $$ \begin{align} y_{n+1} &= y_{n} + \Delta t f(t_{n+1},y_{n}) + \Delta tA_n\left(y_{n+1} - y_{n}\right) \\ (I + \Delta t A_n)y_{n+1} &= (I - \Delta t A_n)y_{n} + \Delta t f(t_{n+1},y_{n}) \\ y_{n+1} &= y_{n} + \Delta t (I - \Delta t A_n)^{-1}f(t_{n+1},y_{n}) \end{align}$$

The final formula above should be all you need, assuming you can compute $A_n$.

  • $\begingroup$ Thank you so much! This was extremely helpful and helped me learn why the motivation behind the modification. $\endgroup$
    – Twis7ed
    Dec 9, 2016 at 4:07
  • $\begingroup$ @Twis7ed no problem! Glad it helped! $\endgroup$
    – spektr
    Dec 9, 2016 at 4:11

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