Banach space-valued test functions in the definition of a weak solution of a PDE problem

In the literature about PDEs it is easy to find books that talk about weak solutions of a partial differential equations. A short reminder of the usual definition is given bellow. More information could be found, for example, in Dafermos's book.

Let's consider a problem:

$$\hspace{1.1cm} u(x,t)_t + f(u(x,t))_x = g(u(x,t)), \; x \in \mathbb{R}, t \in [0,T],\label{1}\tag{1}$$ $$\hspace{1cm} u(x,0)=u_{0} (x), \; x \in \mathbb{R}. \label{2}\tag{2}$$

The weak solution $$u$$ of problem \eqref{1}-\eqref{2} is given with:

$$\int_{0}^T \int_{\mathbb{R}} [u \psi_{t} + f(u) \psi_{x}] \; dx dt + \int_{\mathbb{R}} u_{0}(x) \psi (x,0) \; dx = - \int_{0}^T \int_{\mathbb{R}} g(u) \psi \; dx dt,\label{3}\tag{3}$$

for every test function $$\psi(x,t) \in C_{c}^{\infty}(\mathbb{R} \times [0,T))$$. So here the test functions are smooth real valued functions with compact support.

A few days ago I was reading a paper (Sueur) and I stumbled upon some interesting technique. That paper talks about strong solutions of Euler system and weak solution of Navier-Stokes system and their connection.

The sentence that made me think was:"One may wish to apply the weak formulation of the Navier-Stokes equations with the solution of the Euler equations as a test function."

Using a (relatively smooth) solution of one system as a test function for the other system sounds unusual and amazing at the same time. I then remembered the problem that I was working on a few years ago. So I have two questions.

1. Instead of the usual real valued test functions $$\psi(x,t) \in C_{c}^{\infty}(\mathbb{R} \times [0,T))$$ could we use some Banach space-valued test functions such as $$\psi(t)(x) \in C^{\infty}([0,T);H^m(\mathbb{R}))$$ or $$\psi(t)(x) \in C^{2}([0,T);H^m(\mathbb{R}))$$ or $$\psi(t)(x) \in C([0,T);H^m(\mathbb{R}))$$? The last one is the one I am the most interested in and those would be "smooth solutions". Here $$H^m(\mathbb{R})$$ represents Sobolev space $$W^{m,2}(\mathbb{R})$$ - those were the one I was working on a few years ago.

2. What would be the rules based on which we could decide whether some space of test functions is acceptable or not? For example could we use $$\psi(t)(x) \in L^{\infty}([0,T);H^m(\mathbb{R}))$$ or $$\psi(t)(x) \in L^{1}([0,T);H^m(\mathbb{R}))$$ or whatever else? I am sure that the answer would be NO because of that the t derivative maybe doesn't even exist (just see \eqref{3}). But I am maybe wrong too.

Any help with this would be great (whether it is some reference in the literature or good old fashioned way by writting the answer).

Writing your equation (3) as $$a(u,\psi)=0$$, it is indeed common to call $$u\in L^1_\mathrm{loc}$$ a `weak solution' to your problem if and only it satisfies

$$a(u,\psi) = 0 \mbox{ for all } \psi \in C_c^\infty(\mathbb{R}\times [0,T))$$

However, this is a definition which implicitly encodes $$C^\infty_c(\mathbb{R}\times [0,T))$$ as the base set of 'test functions' for your problem. I mean this in the sense that only once you have verified that $$a(u,\psi)=0$$ for all $$\psi\in C_c^\infty(\mathbb{R}\times [0,T))$$ can you declare $$u$$ to be a (weak) 'solution' your problem.

In the other direction, this notion of 'solution' gives us a very powerful mechanism for learning things about anything we know to be a solution. Namely, we know that any weak solution satisfies $$a(u,\psi)= 0$$, no matter what smooth compactly supported function we choose as $$\psi$$. By making clever particular choices of $$\psi$$ (which I think is what Sueur is proposing), one can learn a great deal about functions $$u$$ which are 'solutions' in the above sense.

People do sometimes use 'test function' in slightly looser senses as well (see Why is this test function admissible? [Paper explanation]), but that's a separate issue.

• Even if your putative solutions only a priori belong a 'big' space which includes lots of coarse functions, it is often the case that being a member of this big space and being a weak or distributional solution to your PDE actually implies much greater smoothness. As you may be aware, this idea is 'regularity theory'. I don't have a reference to hand, but I'd imagine there are existing regularity theorems which are applicable to your PDE. Once you know your solutions are actually smooth, you can multiply them by smooth cutoff functions to get genuine test functions, at which point you can test them against whatever you like.

• It's often possible to make some headway by approximating a problem of interest by a sequence of approximate problems (see the proof of theorem 4 in the paper discussed in the post associated with the answer I linked to above).

• I know you're quite keen to see things in contexts other than that of the Navier Stokes equations, but the notes http://homepages.warwick.ac.uk/~masdh/Campinas.pdf discuss what looks at first glance like a slightly different notion of 'weak solution' from the one discussed above (here I'm thinking particularly of definition 1.2 and the discussion leading up to it). The two notions end up being the same, but this takes a little bit of deduction (the argument is sketched in the notes).

• I should also say: you are perfectly free to define a notion of 'solution' in terms of whatever set of 'test functions' you choose, but this will generally give you a notion of 'solution' which is different from the usual ones.
– DCM
Jun 6 '19 at 20:16
• Thanks for the answer. In your other post you said: " 'test function' often just means 'thing we're free to choose which will fit into the other slot of the pairing from what we already have'." and that explains it. Sueur talks about weak-strong uniqueness where one compares two solutions of the same equation (or two different equations - in his case Euler and Navier-Stokes) only one of which being smooth. IF I have a solution that is in the $L^{\infty}(0,T,L^{\infty}(\mathbb{R}))$ or in $L^{\infty}(0,T,L_{loc}^{1}(\mathbb{R}))$, I could use test function from $C([0,T];H^m(\mathbb{R}))$?Yes?
– Mark
Jun 7 '19 at 8:43
• If you accept the standard definition of 'weak solution' you have a pairing $a: L^1_\mathrm{loc}\times C^\infty_c\to \mathbb{R}$. As long as you have a $C^\infty_c$ function in one slot, it doesn't really matter which one you refer to as the 'test function'.
– DCM
Jun 7 '19 at 18:43
• Re. "If I have a solution in ..., I could use a test function from ... ?": no, since you'd want at least slot of your pairing to be filled by something in $C^\infty_c$. I agree with your statement that "Sueur talks about weak-strong uniqueness where one compares two solutions of the same equation (or two different equations - in his case Euler and Navier-Stokes) only one of which being smooth", but the bit about "one of which being smooth" is important if you wish to stick to the standard notion of 'weak solution'.
– DCM
Jun 7 '19 at 19:03
• Testing a something in $L^\infty([0,T],H^m(\mathbb{R})$ against something in $C([0,T],H^m(\mathbb{R})$ doesn't tell you anything about whether either is a weak solution of your problem (since the whole notion of 'weak solution' is encoded in terms of $C^\infty_c$), but I'm sure whatever argument you have in mind will work if you replace one of them by a suitable $C^\infty_c$ approximation.
– DCM
Jun 7 '19 at 19:07

More (too long) a comment than an answer. It is somewhat paradoxical that almost 90 years after Serguei Sobolev introduced the notion of weak solutions of PDE, also almost 70 years after Laurent Schwartz wrote his treatise on Distribution theory, we keep writing weak solutions using test functions. For the example of the IVP for the Incompressible Navier-Stokes System in the whole $$\mathbb R^3$$, we can write with $$w=H(t) v$$, $$H=\mathbf 1_{[0,+\infty)}$$ $$\frac{\partial w}{\partial t}+\mathbb P\text{div}(w\otimes w)+\nu\ \text{curl}^2 w= v_0\otimes \delta_0(t), \tag{1}$$ where $$v_0$$ is the initial datum (a vector field with null divergence), $$\delta_0$$ is the 1D Dirac mass at $$0$$ and $$\mathbb P$$ is the Leray projection $$=(-\Delta)^{-1}\text{curl}^2$$. The difficulty for that equation to make sense is that the tensor product $$w\otimes w$$ should have a meaning and a simple a priori assumption is to require that $$w$$ is locally in $$L^2$$. That way of writing the equation achieves two different goals: first using test functions you get the standard equation but you have also a weak version of the IVP and you may very well accept solutions which are not continuous in time valued in some Banach space. Now, once you have written $$(1)$$, you want at least to prove existence and one way to do this is to approximate $$w$$ by $$w_\epsilon$$, multiply the equation by $$w_\epsilon$$, justify the integration-by-parts and pass to the limit for the non-linear term. After this you are free to look for uniqueness in a specific class e.g. in $$L^p_tL^q_x,\quad \frac{2}{p}+\frac{3}{q}=1, \tag{2}$$ which is true (the endpoint $$q=3$$ is pretty hard). The main problem for that supercritical equation is that existence is proven in $$L^p_tL^q_x,\quad \frac{2}{p}+\frac{3}{q}=\frac 32, \tag{3}$$ and uniqueness is not known for that class. You have also to keep in mind that the main virtue of $$(2)$$ is the scaling invariance ($$\lambda v(\lambda^2 t, \lambda x)$$ is also a solution) whereas (3) is on the different scale of the energy inequality. The litmus test for the choice a space of solutions should be to see which kind of scaling invariance you get: if you have enough regularity to have (2), you are in good shape to get uniqueness (although it could be tough as noted above for $$L^\infty_tL^3_x$$). If your space is less regular, say $$\frac{2}{p}+\frac{3}{q}>1,$$ then uniqueness would be a breakthrough.

• Thanks for the long comment :). I didn't know that it passed 90 years from the first mentioning of weak solutions. And it surprises me to that today we still write test functions using weak solutions. I am working on a problem that is similar to this in some way, a passing to the limit in the non-linear term is the place I have the biggest problem (I can't find a good way to do it - directly or using some compactness lemma). Maybe I will use the litmus test for the choice of a space of solutions in some problem in the future - scaling invariance is useful.
– Mark
Jun 10 '19 at 12:15