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I just started studying different types of PDEs and solving them with various boundary and initial conditions. Generally, when working on class assignments the professors will somewhat lead us to the answer by breaking a single question (solving a PDE) into parts and starting with things like: $(a)$ start by finding the steady-state solution, $(b)$ ....

My question, however, is rather general. I want to know under what conditions does a steady-state solution exist for a PDE. If possible, I want to keep the question general, but I know it helps to have a reference sometimes so consider the following PDE: $$ \frac{\partial c_1}{\partial t} = \frac{\partial^2}{\partial z^2}\left[\frac{\partial^2 c_1}{\partial z^2}-\frac{\partial g(c_1)}{\partial c}\right]+F $$ where it's known that $ c_1 = c_1(z,t)$ only. Is it reasonable to try to solve for a steady-state solution and let $c_1(z,t) = s(z) +v(z,t)$? Would it depend on the associated boundary conditions? Is the existence of such a solution only dependent on the properties of the PDE? like it being linear? Any help, links, or comments would be much appreciated.

Again, if the example is ignored that's totally fine, the answer desired is the most general one possible!

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    $\begingroup$ This question can be very hard, and yes, boundary conditions are important here. For some important pdes, it is still unknown how to answer. I remember a course of P. Raphael in which he says that we are doing "really pathetic" in some cases. So no, there's no answer to your question. $\endgroup$ Oct 13, 2021 at 22:14
  • $\begingroup$ Appreciate the response. That explains why I couldn't find much on the internet when I typed the question into google. Is it a good starting point to try to find such a solution for a given PDE? or is it almost never actually possible when dealing with more difficult PDEs than say the heat equation? $\endgroup$
    – Mjoseph
    Oct 13, 2021 at 22:18
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    $\begingroup$ I wouldn't say that it is "never" possible, but yes, it can be difficult. There are many papers on such questions for the nonlinear heat equation $\partial_t u = \Delta u \pm u^m$, and the analogous nonlinear versions of the wave and of the Schrödinger equations. Typically the standing solutions, if it exists, is a distinguished one for the dynamics of the whole system; PDE people speak of "threshold" solution, etc... By this I mean: you opened a massive can of worms. I recommend closing it, just keep going. $\endgroup$ Oct 13, 2021 at 22:31
  • $\begingroup$ I appreciate your candor $\endgroup$
    – Mjoseph
    Oct 14, 2021 at 13:33

2 Answers 2

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A steady state for $$ u_t = f(u,u_x,\ldots) $$ exists if there is a solution to $$ 0 = f(u,u_x,\ldots), $$ which is an ordinary differential equation when there is only onoe space variable $x$. That solution needs to satisfy any given boundary conditions to be correct. But to be useful we hope it has the property that other solutions are attracted to it as time increases. The function $f$ need not be linear.

In your example, maybe you meant to write $\frac{\partial g(c_1)}{\partial c_1}$in which case that is a possibly nonlinear function of $c_1$. Any constant $c_1$ that makes that zero is a steady state solution if also $F = 0$. Whether it attracts other solutions would take a bit of study.

It is not usually going to work to look for $c_1 = s(z)+v(z,t)$ unless the equation is linear. Boundary conditions do matter, but the main point of steady states, when attractive, is that initial conditions don't matter in the long run.

This question leads to other ideas too. For example a forced damped oscillator $$ y''(t) +hy'(t)+ky(t) = f(t) $$ doesn't have a steady state, but has the property that the difference of any two solutions tends to zero as $t$ goes to infinity. That is rather significant, and also says that initial conditions don't matter in the long run.

Another related idea is the traveling wave solutions, $$ c_1(z,t) = f(z+kt)$$ that some PDEs have, where $k$ is a constant wave speed to be determined, and $f$ satisfies on ordinary differential equation.

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When does a PDE have a steady-state solution?

My conjecture, offered without proof, is that if each and every root of the characteristic equation are in the left-hand plane of the complex plane, then there will be a bounded steady-state solution.

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