Good day,

Let's take this initial-boundary value parabolic PDE

\begin{align} \partial_t u + Lu &=f \text{ in } \Omega_T=\Omega \times (0,T] \\ u&=0 \text{ on } \partial \Omega \times (0,T) \\ u(x,0)&=u_0(x) \text{ for } x \in \Omega \end{align} where $Lu:=-\text{div}(A^T(x,t) \nabla u)+\langle b(x,t),\nabla u \rangle+c(x,t)u $

and its weak form

\begin{align} \frac{d}{dt} (u(t),v)_{L^2}+a(u(t),v;t)&=\langle f(t),v \rangle_{H^{-1},H_0^1} \text{ for all } v \in H_0^1=:V \\ u(0)&=u_0\end{align} where $a(w,v;t):=\int \langle \nabla w(x), A(x,t) \nabla v(x) \rangle + \langle b(x,t),\nabla w(x)\rangle v(x) + c(t) w(x) v(x) \text{ d}x $

Proving the existence of a weak solution in $L^2(0,T;H_0^1)$ for such a system can be done via Galerkin approximation where we test this system with finite-dimensional test functions. We get an ODE that has a unique solution to Cauchy-Lipschitz and one can show that this solution has a subsequence that converges weakly to the solution of our parabolic system.

Using Galerkin we get

\begin{align}(u_n'(t),v)_{L^2} + a(u_n(t),v;t)&=\langle f(t),v \rangle \text{ for all } v \in V_n \text{, almost all } t \in (0,T] \\ u_n(0)&=u_{0n} \end{align}

and since $v \in V_n:=\text{span} \{ w_1,...,w_n \}$ and $v$ is linear in this equation we can test with $w_j$ for $j=1,...,n$. Further we look for $u_n(t)\in V_n$ i.e. $u_n(t)=\sum_{k=1}^n c_{nk}(t) w_k$ where $c_{nk}(t)$ are some 'coefficients'. Therefore most authors conclude $\color{red}{u_n'(t)=\sum_{k=1}^n c'_{nk}(t) w_k}$ and get by plugging in:

\begin{align}\sum_{k=1}^n \color{red}{c'_{nk}(t)}(w_k,w_i)_{L^2} + \sum_{k=1}^n c_{nk}(t) a(w_k,w_i;t)&=\langle f(t),v \rangle, ~ 1 \leq i \leq n \\ c_{nk}(0)&=\alpha_{nk} \end{align}

This approach can be seen in several PDE/Finite Element books that treat weak solutions for parabolic type PDEs. Take for example 'Galerkin Finite Element Methods for Parabolic Problems' by V. Thomée.

Now my question: In this last Galerkin equation we have the time derivative of these coefficients $c_{nk}(t)$ (see the red equation above). But how do we even know that these derivatives exist? It seems like we just assume that so we can solve the ODE. But this is no real explanation to me. Why do the $u_n(t)$ have such a representation with these coefficients that are differentiable? Can somebody please help me with this?

Thanks a lot, Marvin


The differentiability of the functions $c_{nk}$ is built into the very fabric of the Galerkin scheme.

For the sake of simplicity let me freeze $n$ so that I can get away with one index. We pick a basis $\{w_k\}_{k=1}^\infty \subseteq H^1_0$ and then consider $W = \text{span}(w_1,\dotsc,w_n)$. The start of the Galerkin process is to assume that we have a finite dimensional approximation $$ u_n(x,t) = \sum_{k=1}^n c_k(t) w_k(x) $$ where $c_k :[0,T] \to \mathbb{R}$ is a differentiable function. We want the weak form to hold for all of the $v \in W$, which is equivalent, via linearity, to $$ \frac{d}{dt} (u_n,w_j) + a(u_n,w_j) = <f,w_j> \text{ for each }j=1,\dotsc,n. $$ Expanding and using linearity, this is equivalent to $$ \sum_k (w_k,w_j) \dot{c}_k(t) + \sum_k c_k a(w_k,w_j) = <f,w_j>. $$ In turn, if we define the matrices $M$ and $N(t)$ via $$ M_{kj} = (w_k,w_j) \text{ and } N_{kj} = a(w_k,w_j) $$ and the vector $F(t)$ with components $F_j = <f,w_j>$ then the previous system is equivalent to $$ M \dot{c}(t) + N(t) c = F(t), $$ where here we view $c$ as the vector with components $c_k$. A bit of work shows that $M$ is invertible, and so again we have the equivalence to $$ \dot{c}(t) + M^{-1} N(t) c(t) = M^{-1} F(t). $$ This is a linear ODE with continuous coefficients and a some reasonable forcing (you didn't specify, but let's say $f \in C^0 H^{-1}$), so we can construct a global-in-time solution once we specify the data by expanding $u(\cdot,0)$ in $W$. The solution $c(t)$ then is of class $C^1$ and we have that $$ u_n'(t) = \sum_k c_k'(t) w_k $$
as you desire.

Now, in principle we might not want to quite assume as much as $f \in C^0 H^{-1}$ or we might want to assume less about the coefficients $N$ (which are related to assumptions on $A$ I didn't state precisely above). In this case we won't be able to apply completely classical ODE results to produce $c$ but we will be able to produce solutions in $1D$ Sobolev spaces, and we'll get at the very least that $c$ is absolutely continuous, which means that it will be differentiable a.e. in time, and so the formula $$ u_n'(t) = \sum_k c_k'(t) w_k $$
will still hold for a.e. $t$.

  • $\begingroup$ Thanks a lot for your detailed answer, it is pretty helpful to me. Just to be clear, you first assume that we have such a representation with differentiable $c_{k}$ and then we get to the ODE $$ \dot{c}(t) = -M^{-1} N(t) c(t) + M^{-1} F(t).$$ Now you say that the right hand side exists and therefore $\dot{c}(t)$ exists and because of that $c$ has to be differentiable. Have I got that that right? Otherwise I am not sure why we have a solution $c(t)$ of class $C^1$. (By the way I think we just have $f \in L^2(0,T;H^{-1})$ so $f(t) \in H^{-1}$; therefore thanks for your last paragraph.) $\endgroup$ – Fritz Nov 29 '16 at 15:11
  • 1
    $\begingroup$ No, I'm saying that the finite dimensional approximation of the weak formulation is equivalent to the given ODE for $c$, and this ODE has a solution due to the classical theory of ODEs. It's not that the RHS of your equation exists and so $c$ is differentiable. It's rather that this equation is one for the unknown $c$, and ODE theory (Cauchy-Lipschitz) provides us with such a $c$. $\endgroup$ – Glitch Nov 29 '16 at 16:28

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