Consider the following two boundary value problems

Boundary value problem 1: $$f''\left(x\right)+3f'\left(x\right)+2f\left(x\right) = \exp\left(-3x\right)$$ $$f\left(0\right) = 0$$ $$f'\left(1\right) = 0$$ Boundary value problem 2: $$g''\left(x\right)+3g'\left(x\right)+2g\left(x\right) = \exp\left(-3x\right)$$ $$g\left(0\right) = 0$$ $$\lim_{x \to +\infty}\left(g'\left(x\right)\right) = 0$$

Problem 1 has a unique solution $$f\left(x\right) = \frac{1}{2}\exp\left(-3x\right)+\frac{3\exp\left(-3\right)-\exp\left(-1\right)}{2\left(\exp\left(-1\right)-2\exp\left(-2\right)\right)}\exp\left(-2x\right)-\frac{3\exp\left(-3\right)-2\exp\left(-2\right)}{2\left(\exp\left(-1\right)-2\exp\left(-2\right)\right)}\exp\left(-x\right)$$

However, problem 2 doesn't have a unique solution. It has uncountably many solutions, indexed by the value of the arbitrary constant $G$ in $$g\left(x\right) = \frac{1}{2}\exp\left(-3x\right)+G\exp\left(-2x\right)-\left(G+\frac{1}{2}\right)\exp\left(-x\right)$$

Is there a name for this form of non-uniqueness that problem 2 exhibits and problem 1 doesn't, please? I've been walking round saying "problem 2 is ill-posed" (and indeed it does fit within Hadamard's definition of "ill-posed"), but someone I need to keep happy thinks my use of the phrase "ill-posed" is not sufficiently precisely defined.

  • 1
    $\begingroup$ I would not have said the problem is ill-posed. Instead I would just say the solution is non-unique. Many problems have an infinite set of solutions. Ill-posedness from my understanding usually results from the inability to actually construct a solution with the given set of conditions (e.g. backward heat equation $u_t + u_{xx} = 0$) $\endgroup$
    – Gregory
    Commented Jun 26, 2023 at 15:33
  • 2
    $\begingroup$ For your problem, more formally, I believe the details you are looking for can be found here: en.wikipedia.org/wiki/Fredholm_alternative Essentially, your linear operator can admit one solution, infinitely many solutions or no solutions based on the behavior of the inner product of the adjoint with the "RHS" of your equation. $\endgroup$
    – Gregory
    Commented Jun 26, 2023 at 15:36
  • $\begingroup$ Thanks @Gregory. I think I see the relevance of the Fredholm alternative: the homogeneous version of problem 1 doesn't have a nontrivial solution, so problem 1 must have a unique solution; the homogeneous version of problem 2 does have a nontrivial solution, so problem 2 is allowed, but not required, to fail to have a unique solution. The inner product thing is a way of predicting whether an inhomogeneous problem that is thus allowed to fail to have a solution actually does have a solution, right? $\endgroup$ Commented Jun 26, 2023 at 17:33
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    $\begingroup$ That is correct. Though I would not use the terminology "predict" - rather the statements are equivalent: $b = Ax$ has a (possibly infinite set) solution iff for all $y$ such that $A^T y = 0$ we have $y^T b = 0$. Uniqueness follows if we further restrict that the only solution of $A^T y = 0$ is $y \equiv 0$. $\endgroup$
    – Gregory
    Commented Jun 26, 2023 at 20:04
  • $\begingroup$ @Gregory Sorry to keep pulling at this thread, but I make the homogeneous adjoint equation for problem 2 $h''\left(x\right)-3h'\left(x\right)+2h\left(x\right)=0$, $h\left(0\right) = 0$, $\lim_{x \to +\infty}\left(3h\left(x\right)-h'\left(x\right)\right) = 0$. Does that sound right? $\endgroup$ Commented Jul 2, 2023 at 16:59


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