Regularization Of Solutions From Transport-Diffusion Equation Let $\epsilon > 0$ be Real. Let $u \in L^\infty_{loc}(\mathbb{R}_+,L^2(\mathbb{R^2})^2) \cap L^2_{loc}(\mathbb{R}_+, \dot{H}^1(\mathbb{R^2})^2)$ be a divergence free vector field, and $w_0 \in L^2(\mathbb{R^2})$. I have successfully proved the existence of a solution $w \in L^\infty(\mathbb{R}_+, L^2(\mathbb{R^2}) \cap L^2(\mathbb{R}_+, \dot{H}^1(\mathbb{R^2}) \cap C_w(\mathbb{R}_+, L^2(\mathbb{R^2}) $ of the following PDE :
$$\partial_tw + (u\cdot\nabla) w - \epsilon\Delta w = 0 \ \text{in} \ ]0,+\infty[ \times \mathbb{R^2}\\ w_{|t=0} = w_0 \in L^2(\mathbb{R^2})  
$$
Where we define $w \in C_w(\mathbb{R}_+, L^2(\mathbb{R^2}) $ if and only if $w \in L^\infty(\mathbb{R}_+, L^2(\mathbb{R^2}) $ and for any $\phi \in C_c^\infty(\mathbb{R}_+ \times L^2(\mathbb{R^2})) $ the map :
$$ t \mapsto \int_\mathbb{R^2} w(t,x)\phi(t,x)dx $$
is continuous. So $C_w$ stands for weak continuity. Also, $\dot{H}$ denotes the homogeneous Sobolev space and $u \cdot \nabla$ is the following differential operator : $$ (u \cdot \nabla) w = \sum_{i=1}^2 u_i \partial_i w$$
As $u$ is fixed, this term is linear.
I am trying to prove uniqueness of the solutions. In order to do that, as the PDE is linear, I am taking the weak formulation of the PDE, for $w_0 = 0$ wich is, for any $T > 0$ :
$$\int_\mathbb{R^2} w(T,x)\phi(T,x) = \int_0^T \int_\mathbb{R^2} w(t,x)[\partial_t\phi + u \cdot\nabla\phi + \epsilon\Delta\phi](t,x) dxdt$$
I'm trying to prove that for any $T>0$ :
$$ \int_\mathbb{R^2} w(T,x)\psi(x)dx = 0$$
for any function $\psi \in C_c^\infty(\mathbb{R^2})$.
So I'm looking for a solution that I know exists of the following PDE :
$$  \partial_t\phi + u \cdot\nabla\phi + \epsilon\Delta\phi = 0 \\ \phi(T,x) = \psi(x) \in C_c^\infty(\mathbb{R^2}) \subset L^2(\mathbb{R^2})
$$
If the solution $\phi$ was regular enough and compact supported that would prove my point. Unfortunatly $\phi$ does not have the regularity required to be taken as a test function in the weak formulation.
In order to fix this problem, I can regularize $\phi$ by convolution. and we then denote :
$$\phi_\delta = \phi \ast \rho_\delta$$
We obtain :
$$  \partial_t\phi_\delta + (u \cdot\nabla\phi) \ast \rho_\delta + \epsilon\Delta\phi_\delta = 0 \\ 
$$
My question is :

How do we modify the term $ (u \cdot\nabla\phi) \ast \rho_\delta$, in order the get a similar PDE for $\phi_\delta$  ?

Thank you very much.
 A: I have found the solution :
We have : $\partial_t\phi_\delta + (u \cdot\nabla\phi) \ast \rho_\delta + \epsilon\Delta\phi_\delta = 0 $
So : $$\partial_t\phi_\delta + u \cdot\nabla\phi_\delta+ \epsilon\Delta\phi_\delta = r_\delta
$$
where $r_\delta = u \cdot\nabla\phi_\delta - (u \cdot\nabla\phi) \ast \rho_\delta$
One may show the following statement (wich is a variation of the Di Perna-Lions Lemma shown in the proof of uniqueness in the Di Perna-Lions Theorem) :
$$ r_\delta \rightarrow 0$$ in $L^1([0,T] \times \mathbb{R^2})$ and $L^2([0,T], H^{-1}(\mathbb{R^2}))$
Let $\chi \in C_c^1\mathbb(R^2)$ be a cut-off function wich value is $1$ in $B(0,1)$. We define $\chi_R= \chi(\frac{\cdot}{R})$. We take $\Psi = \phi_\delta\chi_R \in C_c^1(\mathbb{R^2})$ as a test function in the weak formulation.
We have :
\begin{align}
\int_\mathbb{R^2} w(T,x)(\psi \ast \phi_\delta)\chi_Rdx = \int_0^T \int_\mathbb{R^2} w(t,x)[\partial_t(\phi_\delta\chi_R) + u \cdot\nabla(\phi_\delta\chi_R) + \epsilon\Delta(\phi_\delta\chi_R)](t,x) dxdt
= \int_0^T \int_\mathbb{R^2} w(t,x)[\partial_t\phi_\delta + u \cdot\nabla\phi_\delta + \epsilon\Delta\phi_\delta]\chi_R(t,x) dxdt
 + \int_0^T \int_\mathbb{R^2} \phi_\delta \partial_t \chi_R + (ect...)
= \int_0^T \int_\mathbb{R^2} w(t,x)r_\delta\chi_R +  
\int_0^T \int_\mathbb{R^2} \phi_\delta \partial_t \chi_R + (ect...)
\end{align}
So we finally have :
\begin{align}
\int_\mathbb{R^2} w(T,x)\psi(x) = \int_\mathbb{R^2} w(T,x)[\psi - (\psi \ast \phi_\delta)\chi_R]dx + \int_0^T \int_\mathbb{R^2} w(t,x)r_\delta\chi_R +  
\int_0^T \int_\mathbb{R^2} \phi_\delta \partial_t \chi_R + (ect...)\\ = 
\end{align}
One can easily majorate the term $| \int_\mathbb{R^2} w(T,x)\psi(x) |$ by a constant $C$ that depends on $\delta$ and $R$ :
$$| \int_\mathbb{R^2} w(T,x)\psi(x) | \leq C(||w|| ||r_\delta||_{L^1}, \delta, R)$$
Then, as $\delta \rightarrow 0$ and $R \rightarrow \infty$ we conclude that :
$$\int_\mathbb{R^2} w(T,x)\psi(x) = 0$$
for any $\psi \in C_c^1(\mathbb{R^2})$ and $w = 0$ for any $T > 0$.
That gives us the Uniqueness.
