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In ODEs (or even SDEs), there is this theorem that for a given ODE that depends on a parameter, the solution of the equation depends even smoothly on the parameter (given that the function used in defining the ODE is sufficiently nice).

I wonder whether there exists something like this for certain classes of PDEs (eg. linear PDEs). For simple PDEs and simple parameters, we do get nice dependencies, eg. if we scale the parameter $c$ of the transport equation $$ \partial_t u(t,x) = c \partial_x u(t,x), $$ because the solutions are given by $$ f(x + ct), $$ where $f$ ranges over the class of continuously differentiable functions. In this case, we also have a continuous dependency on the initial datum.

Cloudscape
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  • For ODE's there is some information in https://math.stackexchange.com/questions/2809095/find-derivatives-with-respect-to-parameter-or-initial-condition/2809220#2809220 – Cesareo Jul 01 '18 at 08:25
  • Everyone's familiar with the ODE theory. I'm asking only about PDEs. – Cloudscape Jul 01 '18 at 10:55
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    I am not sure. But I think that we have smoothness for the solutions of linear homogeneous first-order PDEs, since they are equivalent, in some sense, to their characteristic system, which is an ODE system. – A. J. Pan-Collantes Mar 26 '22 at 16:29
  • Also, here https://www.avhandlingar.se/avhandling/10c8d24828/ in page 20 it is shown that any system of linear homogeneous first-order PDEs can be reduced to a single linear first-order homogeneous PDE, so I think that this is valid also for systems... – A. J. Pan-Collantes Mar 31 '22 at 06:57

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"Linear PDEs" is already a very large class, but if you restrict to a well-understood class of problems with established estimates then you can often get as-smooth-as-possible dependence on the coefficients. Here's a very rough sketch of how to proceed for a simple example:

If $u,v$ are solutions to the elliptic boundary value problems $$a^{ij}(x,c_0)\frac{\partial^2 u}{\partial x^i \partial x^j} = 0\\ a^{ij}(x,c_1)\frac{\partial^2 v}{\partial x^i \partial x^j} = 0$$ satisfying the same boundary condition $u|_{\partial\Omega} = v|_{\partial\Omega},$ then their difference $w=u-v$ satisfies the elliptic equation $$a^{ij}(x,c_0)\frac{\partial^2 w}{\partial x^i \partial x^j} + (a^{ij}(x,c_0)-a^{ij}(x,c_1))\frac{\partial^2 v}{\partial x^i \partial x^j}=0\tag1$$ with zero boundary condition. If we treat this second term as a source (let's call it $f$), then we can estimate its norm in terms of $|a^{ij}(c_0)-a^{ij}(c_1)|$ and $|D^2 v|,$ and note that (choosing the right norms, and having the right assumptions on $a^{ij}$) we have $|f| \to 0$ as $c_0 \to c_1.$ Since applying standard elliptic estimates to $(1)$ bounds $w$ in terms of $f$ (again, in some norms I'm being deliberately vague about), we should be able to conclude $|w| \to 0$ as $c_0 \to c_1,$ i.e. the problem is continuously dependent upon the parameter $c.$

The important idea to take from this is that once we have estimates for the norm of solutions of general problems in terms of their data, we can get continuity results by applying these estimates to differences of solutions.

Anthony Carapetis
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  • I'll have to book up on elliptic equations a bit in order to make this precise, but tricks like this is what I was looking for. – Cloudscape Jul 01 '18 at 13:09