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For an algebraic system of equations or a system of ordinary differential equations the following rule holds:(right?)

the total number of unknown variables must be equal to the number of equations (and also the same number of boundary conditions are needed, but that's not my question)

Is it generally correct for a system of PDEs too?

I asked this specific question on physics.SE, where one of the users presented the following example in comments saing that it is not the case for a system of PDEs : $$\partial_x f=0 \,\,\,\,\,\partial_y f=0 \,\,\,\,\text{(two equations)}$$ $$\to f(x,y)=0 \,\,\,\text{(unique solution)}$$ (I've seen this question (without answer), which asks about number of needed boundary conditions; my question is about the number of independent equations needed)

Community
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user215721
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  • With spatial problems, the discretization of a PDE often results in a system of equations. Then the fact that the total number of unknown variables must be equal to the number of equations can be employed the other way around: to establish what the proper Boundary Conditions must be for the original PDE. (I think it's sort of illegal to think of that, but I've done it.) With the discretization of a time-like PDE you have be certain (numerical/analytical) that the previous step completely determines the next one, which makes Initial Conditions PDE's an entirely different sort of problem. – Han de Bruijn Mar 21 '14 at 12:58

1 Answers1

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First, it is not true for algebraic systems that you need the same number of unknowns and equations to have a unique solutions. $$(x,y)\in\mathbb{R}^2,\ x^2 + y^2 =0 \Rightarrow x=y=0.$$

Second, it is not obvious how to characterize the dimension of the solution space of a PDE system. See chapter 8 of Seiler's book, "Involution: The Formal Theory of Differential Equations and its Applications".

Sergio Parreiras
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