0
$\begingroup$

Let $X$ and $Y$ be Banach spaces and $D$ is a compact set in $X$. One can view a PDE as an operator and the solution should fall in the null space of it. Let $$ \mathcal{L}:X \mapsto Y, $$ where $\mathcal{L}$ is a PDE operator. And we are looking for $u \in X$ satisfying $\mathcal{L}[u] = 0$.

Suppose I am trying to solve a PDE, $\mathcal{L}[u]=0$, numerically and iteratively. Let $u^*$ be a solution and $u_k$ be a sequence of approximations. We can also assume that $\{u_k\}$ is in a compact set $D$.

In order for this framework to work, I think the PDE operator should be continuous. That is, for any $v_k \to v$, $$ \lim_{k\to \infty} \|\mathcal{L}[v_k] - \mathcal{L}[v]\|_Y = 0. $$ Unless, even though $u_k$ is close to $u^*$, $\mathcal{L}[u_k]$ will not be close to zero.

My question is that if the above argument is correct, is it true that all PDEs can be solved numerically (esp, iteratively) should be continuous operators? If it is the case, I would like to see a couple of examples showing a PDE is continuous under suitable choices of $X$ and $Y$.

Any comments/suggestions/answers will be very appreciated.

$\endgroup$
1
$\begingroup$

Typically, differential operators are not continuous. What you iterate is not the differential operator, but an integral operator related to it. Thus to solve $y' + y = f$ with $y(0)=0$, you wouldn't start with a guess $u_0$ and iterate $u_{n+1} = u_n' + u_n - f$, rather you would write the equation as $y(x) + \int_0^x y(t)\; dt = F(x)$ where $F(x) = \int_0^x f(t)\; dt$ and iterate $u_{n+1}(x) = u_n(x) + \int_0^x u_n(t)\; dt - F(x)$.

$\endgroup$

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.