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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 most usual definition is given bellow.

Let's consider a problem:

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

In order to get the weak solution, we take test function of two variables $\psi(x,t)$ that is smooth and has a compact support on $\mathbb{R} \times [0,T)$, i.e. $\psi(x,t) \in C_{c}^{\infty}(\mathbb{R} \times [0,T))$ and multiply it with the equation (1). Than we integrate everything and perform the integration by parts. At the end, we get that the weak solution $u$ of a problem (1)-(2) is given with:

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

for every test function $\psi(x,t) \in C_{c}^{\infty}(\mathbb{R} \times [0,T))$.

On the other hand, not so usual definition of the weak solution of the above problem (that can be found for example in the papers of E. Feireisl) is given by:

$$\int_{0}^T \phi_t \int_{\mathbb{R}} u \varphi \; dx dt + \int_{0}^T \phi \int_{\mathbb{R}} u \varphi_x \; dx dt + \phi (0) \int_{\mathbb{R}} u_{0} \varphi \; dx = - \int_{0}^T \phi \int_{\mathbb{R}} f \varphi \; dx dt,$$

for every test function $\phi$ that is smooth and has compact support on $[0,T)$ and for every test function $\varphi$ that is smooth and has a compact support on $\mathbb{R}$. So here we used two test functions of single variables $\phi(t)$ and $\varphi(x)$ instead of one test function of two variables $\psi (x,t)$.

I have two questions.

  1. What are the advantages/disadvantages of the second approach where we used two test functions $\phi(t)$ and $\varphi (x)$ comparing to the first approach where we used $\psi (x,t)$?

  2. Could I used just $\phi(t)$ or just $\varphi (x)$ (I am not sure if that $u$ than counts as a weak solution)?

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    $\begingroup$ The second definition is just like the first one, but with test functions restricted to the form $\psi(x,t)=\varphi(x)\phi(t)$. I'm pretty sure (though this is new to me, and I haven't checked it) that linear combinations of test functions on this form are dense in the space of all test functions, so the two approaches should be equivalent. The advantage of the second approach would be that it might be easier to check. Note also that Kružkov's famous doubling of variables proof only relies on test functions on product form. $\endgroup$ Commented Aug 19, 2018 at 8:26
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    $\begingroup$ Actually, I'm not familiar with the one-dimensional case. So my answer may not appropriate to your question. So I explain the one I know currently. The second definition of a weak solution has an advantage when we proceed with the Galerkin approximation with compactness argument such as Aubin-Lions lemma. In the case of the Navier-Stokes equation, see Lemma 2.1 and the proof of Theorem 2.6(in particular Step 3) in the following lecture note goo.gl/gLmEZp $\endgroup$
    – Will Kwon
    Commented Aug 22, 2018 at 2:06
  • $\begingroup$ @HaraldHanche-Olsen: Thanks for your comment. I agree that the second approach is a little bit easier to check. In the problem I have I do not use Kruzkov's doubling of variables (but it is a nice reminder). What do you think of question two? I am thinking of a PDE problem where I do not have a problem with a variable t (so I wouldn't use $\phi(t)$). I would use just $\varphi(x)$. Could I think then of a solution of a PDE as a weak solution too? Also sorry for a late answer (I was on two weaks vacation). $\endgroup$
    – Mark
    Commented Sep 2, 2018 at 9:31
  • $\begingroup$ @WillKwon: Thank you for your comment and for the lecture note. Faedo-Galerkin with Aubin-Lions is a nice example (also E. Feireisl use something similar in a few of his papers). I am not working on Navier-Stokes currently but this maybe could help me with something else I was working on. But somehow I think there is more to it with using different types of test functions. Hope I'll figure it out. $\endgroup$
    – Mark
    Commented Sep 2, 2018 at 11:33
  • $\begingroup$ I am not sure I understand your second question. But I'd say if you only use $\varphi(x)$ then you are not involving the $t$ variable at all, so how can it be a reasonable definition of weak solution? And if you don't have a $t$ variable, as you allude to in your followup comment, isn't your PDE then an ODE? Unless, of course, $x$ is multidimensional, in which case notions of weak solutions certainly do exist. $\endgroup$ Commented Sep 3, 2018 at 10:44

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