# Time discretisation after splitting a 4th order equation

Suppose we have a fourth-order parabolic PDE

$$\partial_t u + a(t)\Delta( \Delta u) - div( b(x,t)\nabla u) =0$$

We split the equation into two second-order equations by introducing $$w = \Delta u$$ thus

$$\partial_t u + a(t) \Delta w - div(b(x,t)\nabla u) =0$$ $$-\Delta u + w =0$$

I discretise the first equation as $$\dfrac{u^{n+1} - u^n}{\Delta t} + a(t^n) \Delta w^{n+1} - div(b(x,t^n) \nabla u^{n+1} =0 \ \ \ \ \ \ \text{(0 )}$$ My issue is with the second equation. Should it be $$$$-\Delta u^{n+1} + w^{n+1} =0 \ \ \ \ \ \ \text{(1)}$$$$ or this one $$$$-\Delta u^{n} + w^{n+1} =0 \ \ \ \ \ \ \text{(2)}$$$$ I was thinking of an algorithm for the numerical simulations and I think equation (2) makes more sense. Since we first solve it to obtain $$w^{n+1}$$ then we use $$w^{n+1}$$ in equation (0) to find $$u^{n+1}$$. If we use equation (1) I do not see how we would determine $$w^{n+1}$$ since $$u^{n+1}$$ is also not known.

Any help would be appreciated.

• There are no initial and boundary conditions specified, so the current formulation is ill-posed. Also you can directly discretise the 4th order formulation if you wish, i.e. you may avoid the substitution if you want. Aug 22, 2023 at 20:24

The introduction of $$w$$ is just a reformulation of your initial problem.
Alternatively, you can consider having the following viewpoint on your problem. First, you discretise in space to obtain a set of coupled ODEs $$u'=f(u)$$ where f contains the spatial discretisation of the RHS (4-th order diffusion term and divergence). Then, you can apply any temporal discretisation scheme you want directly on $$f$$, e.g. implicit Euler: $$\dfrac{u^{n+1}- u_{n}}{\Delta t} = f(u^{n+1})$$
Or you can split the RHS: $$u' = f_1(u) + f_2(u)$$, with $$f_1$$ being the diffusion term, and $$f_2$$ the divergence. Then you can apply implicit-explicit (IMEX) schemes, or other approaches, e.g. splitting. Note that what you did is simply inserting $$w(u)=\Delta u$$ in $$f_1$$, i.e. you obtain a formulation of the form $$u' = f_1^*(w(u)) + f_2(u)$$.
• Thank you! This was helpful. However, if we choose equation (1), how can we solve for $w^{n+1}$ since we have no previous step of it, while $u^{n+1}$ is also still unknown? Aug 22, 2023 at 11:50
• This depends on the solution method you use to solve (0). I guess your issue is more on the technical implementation of the scheme. Usually, the overall discretised ODEs (0) is solved via a Newton method, which iterates on $u^{n+1}$, each iteration requiring (usually) one call of $f$, until (0) is satisfied. Then, at each such call of $f$, the solver gives you its current guess of $u^{n+1}$ from which you can compute $w^{n+1}$. Here $w$ is purely an intermediate variable, e.g. some internal "cuisine" for your scheme, i.e. it should only exist within $f$. Aug 22, 2023 at 16:34