Computational Science Asked by kostas1335 on January 5, 2021

**The Problem**

The TR-BDF2 explained in this paper [1], is quite a popular numerical scheme used to integrate $dot{y} = f(t,y)$, consistent of the following two stages:

begin{align}

y_{n+gamma} & = y_n + gamma frac{h}{2}left( f_n + f_{n+gamma} right) tag{1}

%

y_{n+1} & = frac{1}{gamma(2-gamma)}y_{n+gamma} – frac{(1-gamma)^2}{gamma(2-gamma)}y_n +

frac{1-gamma}{2-gamma}hf_{n+1} tag{2}

end{align}

The above are two implicit algebraic equations that need to be solved in each step. This is accomplished by performing a simplified Newton (chord) iteration, where in accordance to Hosea & Shampine’s paper [1] is performed by firstly letting $z=hf(t,y)$. As a result by substituting this into equations $(1)$ and $(2)$ above for the $k^{text{th}}$ Newton iteration, we have:

begin{align}

y_{n+gamma}^k & = left(y_n + frac{gamma}{2}z_nright) + frac{ gamma}

{2}z_{n+gamma}^k tag{3}

%

y_{n+1}^k & = left(y_n +frac{sqrt{2}}{4}z_n + frac{sqrt{2}}{4}z_{n+gamma} right) + frac{gamma}{2}z_{n+1}^k tag{4}

end{align}

From the above, the paper goes straight away to say that the Newton step and Newton iteration are as follows:

begin{align}

left(I-hfrac{ gamma}{2}frac{partial f}{partial y}right)Delta^k = hf(t_{n+gamma},y^k_{n+gamma}) – z_{n+gamma}^k,

text{and} z_{n+gamma}^{k+1} = z_{n+gamma}^k + Delta^k text{for equation (3)} tag{5}

%

left(I-hfrac{ gamma}{2}frac{partial f}{partial y}right)Delta^k = hf(t_{n+1},y^k_{n+1}) – z_{n+1}^k,

text{and} z_{n+1}^{k+1} = z_{n+1}^k + Delta^k text{for equation (4)} tag{6}

end{align}

**The Question**

Therefore I do not understand how the last two equations above were derived, and what the quantity $z$ represents.

To elaborate, the paper says that $z=hf(t,y)$, so $z_{n+gamma}^k=hf(t_{n+gamma},y_{n+gamma}^k)$, but doesn’t that make the RHS of equations $(5)$ and $(6)$ zero? Also, the typical Newton iteration is of the form $u^{k+1} = u^k – (g’)^{-1}g(u^k)$. In this case then what is the function $g$ corresponding to for equations $(5)$ and $(6)$? It surely can’t be $g = hf(t_{n+gamma},y^k_{n+gamma}) – z_{n+gamma}^k$, since the Jacobian of this function is not $I-hfrac{ gamma}{2}frac{partial f}{partial y}$ as shown in equation $(5)$ for example.

Thanks for your help in advance.

**Bibliography**

*Hosea, M. E.; Shampine, L. F.*,**Analysis and implementation of TR-BDF2**, Appl. Numer. Math. 20, No. 1-2, 21-37 (1996). ZBL0859.65076.

Maybe it would be easier if you first consider the Newton method on the variable $y_{n+gamma}$. I'll only treat the first stage of the method which defines $y_{n+gamma}$, as the second one can be handled similarly. I'll also remove the dependence on time as it superfluous for the explanation. Let's recall this first stage:

$$y_{n+gamma} = y_n + gamma frac{h}{2}left( f_n + f_{n+gamma} right)$$

Let's introduce $y=y_{n+gamma}$ and reformulate as: $$ 0 = g(y) = y - frac{gamma h}{2} f(y) - underbrace{left(y_n + frac{gamma h}{2} f(y_n) right)}_{A}$$ where $A$ is constant.

To solve this, we apply the Newton method on $g$, starting from an initial guess $y^0$ for $y$, and the $k$-th Newton step reads: $$y^{k} = y^{k-1} - J(y_k)^{-1} g(y^k)$$

with the Jacobian $J(y^k)= left(partial_y gright)(y^k)$.

Replacing $J(y^k)$ by $J(y^0)$ yields the corresponding simplified Newton method. Otherwise the full Newton method would compute the Jacobian of $g$ at each iterate anew.

Now the authors of the reference you gave have introduced $z=f(y)$. The first stage can then be reformulated as: $$y = y_n + gamma frac{h}{2}left( z_n + z right)$$

That is not what we will solve, as it would not make sense on its own. Indeed we have introduced a new variable, hence an additional equation is required to solve the system. That is why the system to be solved is now: $$y = y_n + gamma frac{h}{2}left( z_n + z right)$$ $$z = f(y)$$ which is equivalent to $qleft( (y,z)^t right)=0$

You could write the Newton method to solve $q=0$. But here $y$ can be explicitly expressed from $z$ from the first equation. Hence we take the RHS of the first equation and use it in the second to obtain: $$z = fleft( y_n + gamma frac{h}{2}left( z_n + z right)right)$$ $$Leftrightarrow 0 = g_2(z)$$

You apply a Newton loop on $g_2$ as shown before for $g$, but this time the Jacobian is $partial_z g_2 = mathrm{I} - frac{gamma h}{2} partial_y f$. Thus we obtain Equations $(5)$ and $(6)$ from your post.

So, regarding your last question: you actually precisely want the RHS of $(5)$ and $(6)$ to got to 0 as your Newton method converges. Then, when convergence is achieved, you have $z=f(y)$, from which you can compute $y$ based on Equation $(5)$;

Correct answer by Laurent90 on January 5, 2021

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