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Consider the continuity equation on $[0,T] \times \mathbb{R}$ $$ \partial_{t}u + \partial_{x}(bu)=0$$with initial data $u(0,x) =u_{0}(x)$ which we assume to be smooth for now (also assume $b$ is smooth). I have read in papers that the formal adjoint to the forward continuity equation is the backward transport equation, i.e. $$\partial_{t}v + b\partial_{x}v=0 $$ with final data $v(T,x) = v^{T}(x)$. I would like to know if there is a precise way to show this? In other words how can we make sense of the backward transport equation as the adjoint of the forward continuity equation?

Here is my attempt but I am not totally convinced by it. Consider $u,v \in C^{\infty}_{c}(\mathbb{R})$ for convenience and define the operator $L$ by $Lu = \partial_{t} u + \partial_{x}(bu)$. Then we look for an operator $L^{*}$ (the adjoint) which satisfies $$ \langle Lu, v \rangle = \langle u, L^{*}v \rangle$$ where the duality bracket can be assumed to be $L^{2}([0,T]; L^{2}(\mathbb{R}))$ (i'm not actually sure why we can assume this). Then integrating by parts we see that \begin{aligned}\langle Lu, v \rangle &= \int_{t}\int_{x} v (\partial_{t}u + \partial_{x}(bu))~dxdt \\ &= - \int_{t}\int_{x} u(\partial_{t}v+b\partial_{x}v)~dxdt + \int_{\mathbb{R}} u(x,T)v(x,T)-u(x,0)v(x,0)~dxdt. \end{aligned} So formally the forward continuity equation will be adjoint to the transport equation if the boundary term vanishes, for which we can simply require that $u(x,t):=v(x,T-t)$, i.e. take $u$ to be the time-reversal of $v$. So the two equations will be adjoint if one is posed backwards in time and the other one is posed forwards in time.

But this doesn't really make sense since I am defining $u$ in terms of $v$ and the duality equality should hold for any $u,v$.

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  • $\begingroup$ There is a difference between the adjoint and the formal adjoint. You can see this answer, for example. I highlight the following : "In the case of the formal adjoint, it is left unspecified: indeed, the formula only really hold for sufficiently smooth function decaying sufficiently fast at infinity, and not in general for arbitrary functions $u,v \in \mathbb L^2$". Similarly, the "formal" adjoint is the backward continuity operator because the above relation does hold for all $u,v$ such that $u(0)=v(0)=u(T)=v(T)=0$. $\endgroup$ Jul 2, 2023 at 15:36
  • $\begingroup$ Here's another answer expressing the same sentiments. The formal adjoint is basically a transformation of a differential operator without considering the actual space that it's living on (and can be specified by a formula depending only on the operator, see here), while the actual adjoint requires the respective domains and hence goes further. $\endgroup$ Jul 2, 2023 at 15:42
  • $\begingroup$ @SarveshRavichandranIyer I see, so there's a difference between formal adjoint and adjoint. But with the boundary conditions you picked, we could argue that the adjoint of the forward continuity eqn is also the forward continuity equation. Wouldn't it be more suitable to say that the relation holds for all $u,v$ with $u(0) = v(T)$ and $v(0)=u(T)$? Anyway if you wish to post what you wrote as an answer I will be happy to accept it, thank you $\endgroup$
    – duelspace
    Jul 2, 2023 at 21:10
  • $\begingroup$ @dualspace With the boundary conditions I picked, we have that $\langle Lu,v \rangle = \langle u,L'v \rangle$ for all $u,v$ belonging to suitable subspace of the space in question, where $L'$ is the backward operator. The reason why I'm not writing an answer, at least right now, is because the $u,v$ in your space are maps to function spaces themselves, rather than to $\mathbb R$. The posts I attached all devoted themselves to real function spaces, whereas the functions here are function-valued themselves. Perhaps there is a multidimensional variation which I will find and place in my answer. $\endgroup$ Jul 3, 2023 at 1:42

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