I am trying to work out a second order or higher order correction for the method of Spectral deferred Correction (SDC). Specifically using as a corrector a second order or third order multi-step.
In SDC, the timestep $[t_{n},t_{n+1}] $ is subdivided in m quadrature nodes $[t_{n,1},t_{n,2},...,t_{n,m-1},t_{n,m}] $.
The ODE problem is written in its integral form: $$ \begin{equation} \label{eq:ODEPicard} \begin{array}{ccl} \phi(t) & = & \phi_{0} + \displaystyle\int_{0}^{t} F(\tau,\phi(\tau))d\tau \end{array} \end{equation}$$
And later rewritten in function of the error.
$ \begin{equation} \begin{array}{ccl} \label{eq:CorrectionEmeasure} \delta(t) & = & \phi_{0} + \displaystyle\int_{0}^{t} [F(\tau,\tilde{\phi}+\delta(t)) - F(\tau,\tilde{\phi})]d\tau + E(t,\tilde{\phi}) \end{array} \end{equation}$
where $\tilde{\phi} $ denotes the approximate solution. This previous eq is discretized and based on the definitions of the error $\delta(t)$ and the residual $E(t,\tilde{\phi})$ can be discretize by a r-multi-step method as:
$ \begin{equation} \label{eq:CorrectionMultiStepDiff1} \begin{array}{ccl} \displaystyle\sum^{r}_{j=0}{\{\alpha_{j}\tilde{\phi}^{[k]}_{i-j} + \displaystyle\sum_{w=0}^{j} \alpha_{w} I^{i-j}_{i-j-1}(F(\tilde{\phi}^{[k-1]})) \}} - \{F(t_{i-1},\tilde{\phi}_{i-1}^{[k]} ) - F(t_{i-1},\tilde{\phi}_{i-1}^{[k-1]})\} = 0 \end{array} \end{equation} $ \ Being $I^{i+1}_{i}(F)$ an interpolatory quadrature rule $\begin{equation} \label{eq:QuadratureInterpolatory} \int^{t_{i+1}}_{t_{i}} F(\tau,\phi(\tau))d\tau \approx I^{i+1}_{i}(F) = \int^{t_{i+1}}_{t_{i}} p(\tau)\omega(\tau)d\tau = \displaystyle\sum_{j=1}^{n} \omega_{j}F(t_{j}) \end{equation}$
Well, this as every multistep method needs an initialization in the first step. What I have done is to resolve that initialization by RK2 or midpoint rule. But furthermore, the first subinterval $[t_{n,1},t_{n,2}]$ require values of the previous r-1 subnodes values. This is a problem, because as it is pointed out in Layton [1]. $\delta(t)$ is continous in the interval $[t_{n},t_{n+1}]$ but may be discontinous in the previous one. I do not totally understand why this happen, so if someone could explain me this I would be so grateful, I think it is because of the quadrature term but I don't still see it. Well, but she say that this produces a lack of smoothness in the solution and the order or correction is only 1 per step instead of r for an r-step. And later in that paper she proposes some solutions for this, (page 5), but it is not clear what she exactly mean by "fixed starting values" and "variable starting values".
What I have tried to keep the order of accuracy of the corrector (specially I have focused in Second order Adams Bashforth)
$ \displaystyle \beta_{1} = \frac{\Delta t_{i-1}^2}{2\Delta t_{i-2}} \\ \displaystyle \beta_{2} = \frac{2\Delta t_{i-1}\Delta t_{i-2} + \Delta t_{i-1}^2 }{2\Delta t_{i-2}} $
$ \begin{equation} \phi_{i}^{[k+1]} = \phi_{i-1}^{[k+1]} + \beta_{1}[F(t_{i-1},\phi_{i-1}^{[k+1]}) - F(t_{i-1},\phi_{i-1}^{[k]})] + \beta_{2}[F(t_{i-2},\phi_{i2}^{[k+1]}) - F(t_{i-2},\phi_{i-2}^{[k]})] +I^{i}_{i-1}(\phi^{[k]}) \end{equation} $
And the first I tried it was the use of midpoint rule in the first subinterval, but as it is pointed out in IDC [2] this lack of smoothness (this concept is also redefined in this paper), it has a degree of smoothness one, in the discrete sense of the numerical set. And later I have tried to use other initialization but also lacking of smoothness.
So my question is does anybody has an idea how to obtain a second or higher order smoothness in the discrete sense based on a combination of Multistep and other method (maybe multistage). Or how to initialize at each subinterval the multistep method not to suffer order reduction due to the lack of smoothness.
P.S. If someone could clarify me also the comment that Layton does in the page 5 of [1] about continuity of the error it would be great too!
