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It's a shame that, though I've taken the "Equations of Mathematical Physics" class for one semester and solved numbers of PDEs with Mathematica, I'm still unclear about how many initial conditions(ICs) or boundary conditions(BCs) are needed for getting the determine solution of a PDE or a set of PDEs: I never met a book which mentioned that.

I roughly know that there's a theorem for ODEs which tells us that for a n-th order ordinary differential equation, we need n ICs or BCs which have lower order than the ODE to get the determine solution, and usually it's the principle I followed when I tried to solve PDEs numerically with NDSolve in Mathematica (Of course in this case the number of IC or BC is considered seperately for every argument), but it's inaccurate, right? A popular case is the d'Alembert's formula, the 1D wave equation

$${ \partial^2 u \over \partial t^2 } = c^2 { \partial^2 u \over \partial x^2 } $$ with only 2 initial conditions $$u(x,0)=f(x) $$ $$u_t(x,0)=g(x) $$

gives the determine solution

$$u(x,t) = \frac{f(x-ct) + f(x+ct)}{2} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds$$

while with my "principle" we need 4 ICs or BCs(2 for x, 2 for t).

I also encountered several equations that don't follow my "principle" when I wandered in Mathematica.SE, for example, it seems that this set of equations needs 6 BCs, but it can be solved with only 5 BCs in fact:

https://mathematica.stackexchange.com/questions/13795/problem-with-ndsolve-for-a-system-of-equations

And this one, which actually needs only 4 BCs in total, no matter which variables the BCs are given to, while with my "principle", we need 1 for ρg, 1 for u, 2 for te:

https://mathematica.stackexchange.com/questions/9277/i-failed-to-solve-a-set-of-one-dimension-fluid-mechanics-pdes-with-ndsolve

So, as my title said, is there a theorem or something for the decision of the number of IC and BC? Any help would be appreciated.

Community
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xzczd
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    For equations that can be shown to be hyperbolic, you can use Cauchy-Kowalevski. But in general the number depends on the "type" of the equation, and there are very few general rules. For evolution equations a good rule of thumb (though not always accurate) is to pretend your equation is actually an ODE on infinite dimensional space and apply your intuition from ODE theory. – Willie Wong Jul 24 '13 at 07:58
  • @WillieWong Er… what's "an ODE on infinite dimensional space"? I failed to find an informal explanation for that. – xzczd Jul 24 '13 at 09:32
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    At each instant of time, the solution belongs to a function space, which is infinite dimensional. But I do not have the time to explain in more detail. For your question of IC and BC, see, perhaps, pages 5 through 7 of this document. – Willie Wong Jul 24 '13 at 11:41
  • @WillieWong The exact position is the top of page 10. This document is pretty good! it's so… straight-out! It confirms several guess of mine, thanks very much! – xzczd Jul 24 '13 at 14:04
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    (BTW, that is from Olver's upcoming new book Introduction to Partial Differential Equations. http://www.math.umn.edu/~olver/pdn.html ) – Willie Wong Jul 24 '13 at 15:35
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    See this extreme example you will understand for the solution(s) of a PDE exist, the proportion between "B.C.s" and "I.C.s" can even be not restricted, the nature of the solution(s) are only depending on the number of conditions. – doraemonpaul Jul 24 '13 at 23:47
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    @doraemonpaul In fact I was also planning to ask a question which might be titled as "Is there a mathematical distinction between IC and BC?", but now it seems to be not that necessary :) – xzczd Jul 25 '13 at 07:36
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    @xzczd: In fact "IC"(initial condition) and "BC"(boundary condition) are physical terms. "IC" specifically refers the condition at the starting time while "BC" specifically refers the condition at the boundary. When the PDE does not come from physics, it should not have the concepts about "time independent variable" and "dimension independent variable(s)". Unfortunately the modern PDE analyses are often very dictatorial. They often treat all PDEs should come from physics, make us have not enough free rooms for studying the non-physical properties of PDEs. – doraemonpaul Jul 28 '13 at 04:11
  • @doraemonpaul: Don't understand what you're complaining about. You're free to come up with any non-physical PDE that you want and any "pure mathematical" treatment that you want. Nobody forbids you to create those "free rooms". – Han de Bruijn Mar 21 '14 at 12:41

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I believe that your intuition that you need two boundary/initial conditions per derivation degree and variable is actually quite helpful for elliptic and parabolic PDEs. However, in the case of hyperbolic PDEs in an infinite domain, you needn't set BCs. Intuitively, this is related to the fact that in hyperbolic systems local perturbations are "felt" or propagated only locally, but in elliptic or parabolic systems they are propagated instantaneously to the whole domain, so local hyperbolic solutions "don't care" about what is happening non-locally.

However, note that in a finite domain, the wave equation does require BCs in addition to the ICs you wrote (as stated in the Cauchy problem). For example, $u(0,t)=u(L,t)=0$ is typical, and it would comply with your rule. This is done by using the d'Alembert formula and judiciously extending f and g outside the solution domain. This example insinuates that intuitions break down at infinity.

marsarius
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