1
$\begingroup$

I've been studying linear parabolic equations from Evan's PDE text (chapter 7) and I seem to have shown something that's too strong. I will list the setup here:

Let $U \subset \mathbb R^n$ be a bounded domain, $T>0$ and set $U_T = U \times (0,T).$ Consider the divergence form parabolic PDE of the form,

$$ u_t + Lu = u_t - \sum_{i,j=1}^n \left(a^{ij}u_{x_i}\right)_{x_j} + \sum_{i=1}^n b^iu_{x_i} + cu = f $$

on $U_T,$ where $a^{ij},b^i,c \in C^1(\overline U_T)$ with $a^{ij}$ uniformly elliptic ($\sum_{i,j}a^{ij}\xi_i\xi_j \geq \theta|\xi|^2$) and $f \in L^2(U_T),$ subject to boundary conditions $u = \psi$ on $U \times \{0\}$ for some $\psi \in L^2(U)$ and $u=0$ on $\partial U \times (0,T).$

When showing existence via the Galerkin method, we fix a sequence $\{\phi_j\}_{j=1}^n$ in $H^1_0(U)$ which forms an orthonormal basis of $L^2(U)$ and is orthogonal in $H^1(U).$ We can then (assuming $f, \psi$ are sufficiently regular) construct approximate solutions,

$$ u^N(x,t) = \sum_{i=1}^N d_i^N(t) \phi(x), $$

such that $u^N_t + Lu^N = f_N$ weakly on $U_T$ and $u^N = \psi_N$ on $U \times \{0\},$ where $f_N$ and $\psi_N$ are orthogonal projections onto the relevant subspaces spanned by $\langle\phi_1,\dots,\phi_N\rangle.$ That is,

$$ f_N(x,t) = \sum_{i=1}^N \langle f(\cdot,t), \phi_N \rangle_{L^2(U)} \phi_N(x), \quad \psi_N(x) = \sum_{i=1}^N \langle \psi,\phi_N\rangle_{L^2(U)} \phi_N. $$

The idea is to show that as $N \rightarrow \infty,$ the sequence $u^N$ converges to a weak solution.


When doing this, we prove an energy estimate of the form,

$$ \max_{0\leq t\leq T} \lVert u^N(t) \rVert_{L^2(U)} + \lVert u^N \rVert_{L^2(0,T,H^1(U))} \leq C \left(\lVert f_N\rVert_{L^2(U_T)} + \lVert \psi_N \rVert_{L^2(U)} \right).$$

This is theorem 2 in 7.1 of Evans (p354). Although Evans has $f$ and $\psi$ on the RHS, the analogous argument also proves this estimate. If we instead take $\tilde f = f_N - f_M$ and $\tilde \psi = \psi_N - \psi_M$ however, by uniqueness of weak solutions (theorem 4) we get the unique weak solution must be $u_N - u_M.$ Hence we get the estimate,

$$ \max_{0\leq t\leq T} \lVert u^N(t)-u^M(t) \rVert_{L^2(U)} + \lVert u^N-u^M \rVert_{L^2(0,T,H^1(U))} \leq C \left(\lVert f_N-f_M\rVert_{L^2(U_T)} + \lVert \psi_N - \psi_M \rVert_{L^2(U)} \right).$$

Now if we let $N,M \rightarrow \infty,$ the right hand side vanishes. So our sequence $u^N$ is actually Cauchy in $C([0,T],L^2(U))$ and $L^2(0,T;H^1_0(U)),$ so it converges strongly to some $u$ in both spaces.

My question: What I have done wrong in the above?

When proving existence theorems, Evans instead observes the sequence is uniformly bounded in the spaces and hence extracts a weak limit. This however is not only a bit more delicate to work with, but it only guarantees the limit $u$ is in $L^{\infty}(0,T;L^2(U)).$

Precisely because my above argument isn't mentioned in the text, I have a strong suspicion that it is incorrect. I cannot find any fault in my answer however, despite having looked over it several times.

Note: I assumed earlier that $f,\psi$ are actually regular when proving the existence, but that was only to apply Picard-Lindelöf. For the actual energy estimates, I don't need to assume this. Hence taking $\tilde f, \tilde \psi$ as I did isn't an issue. Also once we have proved existence, we can remove this hypothesis by a density argument. I can elaborate on this if necessary, but I don't believe there is any issue here.

$\endgroup$
0
$\begingroup$

As it turns out this is correct, and there's no issues with this kind of reasoning. Since we have an a-priori estimate and the equation is linear, nothing goes wrong.

There are two points worth noting here however:

  • This kind of reasoning completely breaks down when dealing with nonlinear equations. Often in that setting we find suitable weak solutions by taking a sequence of approximate solutions, establishing a uniform estimate and taking limits. In this setting the best we can hope for is weak(*) convergence in the suitable spaces. I presume Evans opted for this argument to illustrate this technique of passing to weak limits.

  • Even in the nonlinear setting, one usually can still establish $L^2$-continuity in time. It is proven in section 5.9, theorem 3 that if $u \in L^2(0,T;H^1_0(U))$ with $\partial_t u \in L^2(0,T;H^{-1}(U)),$ then $u \in C([0,T];L^2(U)).$ Since these spaces are reflexive, if we can bound our approximating sequences with respect to these norms (note: this applies to the setup of my question also) then we obtain $L^2$-continuity in time.

$\endgroup$
  • $\begingroup$ You seem knowledgable in this field. I would like to ask a question and it would be very appreciated if you could answer it. Can we still define a weak solution to the parabolic equation with a non-zero boundary condition? In the elliptic case, one can deal with this by introducing a function that exactly matches the boundary. I am wondering if this is the case for the parabolic. If so, how can we guarantee the existence of such function? $\endgroup$ – induction601 Dec 19 '19 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.