# Derivation of backward differentiation formulas(BDF)

I have been reading upon numerical techniques that are used to solve stiff ordinary differential equations.

From the description given here, I could follow the steps till equation (5).

I am finding it difficult to understand how equation 6 is obtained and the calculations involved in computing the coefficients of BDF.

Could someone recommend a reference in which these steps have been detailed?

• This is probably more suited to math.se – Nox Jan 6 '19 at 17:22
• I think there won’t be a reference because it’s just calculus: the lhs is a function of t, the rhs is a polynomial in t, so you just differentiate both to get a linear equation in $y_n$, solve for it and read off the coefficients of the other y’s. – Kirill Jan 6 '19 at 17:50
• Actually, there is no solve involved. One directly reads off the coefficients for all $y_{n-j}$s so that $f(t_n)=\dot y(t_n)$ is exactly approximated by the polynomial (up to order $k$). – Jan Jan 8 '19 at 14:03

In BDF schemes for $$\dot y = f$$, one uses $$f(t_n)=\dot y(t_n)$$ and tries to approximate $$\dot y(t_n)\approx \sum_{j=0}^k\alpha_k y_{n-j}$$ by the current value $$y_n$$ (that is to be computed) and the $$k$$ previously computed approximations.

In the presented approach, in $$(5)$$, $$y$$ is approximated as a polynomial $$p$$ in $$t$$ fitted to $$y_{n-j}$$, so that the time derivative of the polynomial at $$t_n$$ approximates $$\dot y(t_n)$$ as desired. With that, for a given approximation order $$k$$, one can read of the coefficients $$\alpha_j$$ by evaluating $$\dot p$$ at $$t_n$$:

For $$k=1$$:

$$\quad \dot y(t_n) \approx \dot p(t_n) = \frac{1}{h}(y_n - y_{n-1})$$ which gives that $$h\dot y(t_n)$$ is approximated by $$1\cdot y_n + (-1)\cdot y_{n-1}.$$

For $$k=2$$ the terms read: $$k=2: \quad \dot y(t_n) \approx \dot p(t_n) = \frac{1}{h}(y_n - y_{n-1})+\frac{1}{2h^2}[(t_n-t_n)\nabla^2y_n + (t_n-t_{n-1})\nabla^2y_n]$$ which, with $$t_n-t_{n-1}=h$$ and $$\nabla^2y_n = y_n - 2y_{n-1} + y_{n-2}$$ gives that $$h\dot y(t_n)$$ is approximated by $$\frac{3}{2}\cdot y_n + (-2)\cdot y_{n-1} + \frac{1}{2}y_{n-2}$$.

And so on...

• If my understanding is right, the interpolating polynomial is in Newton's form, $p(t)=y[t_1] + y[t_1,t_2](t-t_1)+y[t_1,t_2,...,t_k](t-t_1)(t-t_2)..(t-t_{k-1})$ – Natasha Jan 9 '19 at 4:19
• yes, if you say it. I wasn't sure and I did not check the definition. – Jan Jan 9 '19 at 7:22
• How does one choose the intial data in the best way? – Emil Jan 9 '19 at 19:19
• Maybe my comment was off topic, found this anyway: math.stackexchange.com/a/421107/68036 (predictor-corrector) – Emil Jan 9 '19 at 19:53