Why can't we say that all PDEs of a specified order require a fixed number of boundary conditions?

Question

For an $n$th order ODE we always need $n$ boundary conditions (right?). But, as I've seen somewhere, for 2nd order PDEs there are many possible situations and a general answer to the question of how/what boundary conditions are needed doesn't exist. For example, Dirichlet or Neumann boundary conditions give unique solution for elliptic equations, while they don't for hyperbolic equations.

Isn't it really possible for a PDE with some specified boundary conditions to tell easily (with a rule of thump, like for ODEs) if it has a unique solution?

Answer 1

With PDEs is not as simple as with ODEs.

In ODEs, if the equation is of $n$th order, then you need $n$ initial conditions, to guarantee uniqueness.

In the case of hyperbolic PDEs of $n$th order, $$ L_1\cdots L_n u=f, $$ where $L_j$ are first order linear operators, then you need $n$ initial condition, as in the case of the wave equation.

But in the case of elliptic equations, i.e., $L=\Delta^n$, then you need $n$ boundary conditions for $Lu=f$, with $L$ being an operator of order $2n$.

Answer 2

In fact for the "rule" that "An $n$th order ODE should need $n$ boundary conditions" is only true that for assuming the solutions which their continuous section(s) are maximized. Once allowing the solutions can be piecewise functions, the "rule" is no longer to be true.

For example for a simplest ODE $y'(x)=0$ , besides $y(x)=C$ are the solutions, e.g. $y(x)=Bu(x)+C$ , $y(x)=B[x]+C$ are also the solutions.

In fact allowing the solutions can be suitable types of piecewise functions, the ODEs can match more number(s) of the condition(s). So as for the PDEs.

In fact either ODEs and PDEs can have the concepts that "introducing suitable types of piecewise functions of the solutions so that the equations can match more number(s) of the condition(s)" , PDEs ones have much more discussions that ODEs ones only because of e.g. the real situations request.