Piecewise solution to Euler-Lagrange equations in effective field theory

I would like to consider a background for a quantum field theory made up by connecting continuously two different solutions of the Euler Lagrange equations.

The problem is one dimensional (let's call our coordinate $$r,\;\;r>0$$) and I would like to have a certain solution $$\phi_*(r)$$ defined as $$\;\phi_*(r):=\phi_0 (r)\;\;\forall\,r>r_*$$ and $$\phi_*(r):=\phi_c(r)\;\;\forall\,r\leq r_*$$.

In standard field theory I can do this as long as $$\phi_*(r_*)$$ is regular enough so that no operator of the Lagrangian is divergent (i.e. $$\phi_*\in C^k\;$$if the Lagrangian contains differential operators up to order $$k$$).

Now my question comes when we want to deal with an effective field theory, so that we need to be able to neglect operators with arbitrary number of derivatives. My conclusion would be that I can only accept such a solution if the junction between the two behaviors $$\phi_0\,,\,\phi_c$$ is $$C^{\infty}$$.

Is there any way to escape such conditions? Or are the conditions even stronger?(analicity of the background fields for instance?) I heard some sort of "piecewise" solution is used in the context of spontaneous scalarization in theoretical cosmology and I imagine that this kind of solutions might be studied in Electromagnetism and Mathematical Physics.

Can such solutions be made consistent with the effective field theory approach without discarding any globally non $$C^{\infty}$$ result?

• I'm confused. Effective field theory Lagrangian only has some low derivative operators. In that case $k$ associated with your Lagrangian will be small, like $k=2$ for $O(N)$ $\sigma-$model. Then why do you need $C^\infty$ functions and not $C^k$ functions with the corresponding finite k? – nGlacTOwnS May 16 at 10:59
• The fact of effective field theory is that while you write only some operators (with few derivatives) you aknowledge the existence of infinite others. That is the source of the problem. You can neglect those operators as long as they are small, that is as long as derivatives of the fields are smaller than the UV cutoff scale. If some gradient diverges then the effective field theory breaks and possibly infinite operators (all those containing a divergent term) will become dominant over the low derivative terms – AoZora May 16 at 11:40
• So you can ask this question for quantum field theories only at non trivial fixed points of RG flow? – nGlacTOwnS May 29 at 21:59
• I don't know how you get to this conclusion, but the fact is that any non $C^{\infty}$ behaviour implies UV modes (spikes, steps, in some high enough derivative), breaking your EFT approximation. If you get a background which is non infinitively derivable by solving equations of motion from a EFT lagrangian then this background is non consistent, because it contains UV modes itself. – AoZora May 29 at 22:42
• I should have been more elaborate, my apologies. What I meant is that the only way you can possibly allow $\mathcal{C}^k$ fields with finite $k$ is for non-trivial fixed points of RG flow which have only finite number of terms in the Lagrangian. I say this because for any other field theory RG flow will necessarily generate all irrelevant deformations. It might be true that you could possible argue, using some symmetry arguments that arbitrary irrelevant terms aren't allowed. But that will be presumably some very special cases. – nGlacTOwnS May 29 at 22:57