coercivity
When can you guarantee that an equation has a solution and that the solution is unique? In the variational world the answer hinges on one property of the bilinear form, called coercivity. Intuitively it says the form has a definite 'positive grip': it can never let a nonzero function slip through with arbitrarily small energy. That grip is what stops solutions from collapsing or multiplying, and it is the make-or-break hypothesis of Lax-Milgram.
Precisely, a bilinear form a on a Hilbert space H is coercive if there is a constant alpha > 0 with a(u, u) at least alpha ||u||^2 for every u in H. In words: when you feed the same function into both slots, the result is not just positive but bounded below by a fixed multiple of the function's squared size — the form cannot shrink toward zero unless the function itself does. For the Dirichlet Laplacian, a(u, u) = integral of |grad u|^2, and coercivity on H^1_0 is exactly the Poincare inequality (the gradient energy alone controls the full H^1 norm). Coercivity is the functional-analytic echo of two finite-dimensional ideas at once: it is positive-definiteness of a symmetric form (eigenvalues bounded away from zero) and, for the operator behind the form, the property of being bounded below — both of which force invertibility.
Why it is decisive: coercivity is precisely what makes the solution operator invertible and stable. If a(u, u) is at least alpha ||u||^2, then a solution of a(u, v) = L(v) cannot be large unless L is — you get the a priori bound ||u|| at most ||L||/alpha for free, which is continuous dependence on the data, the third leg of well-posedness. Lose coercivity and you lose all of this: an indefinite problem (the form a(u, u) can be zero or negative for some nonzero u) may have no solution, or infinitely many, and you are forced into the Fredholm alternative where solvability becomes conditional. This is exactly what happens for the Helmholtz equation minus Laplacian u minus k^2 u = f at a resonant k: the negative term destroys coercivity, and existence depends delicately on whether k^2 is an eigenvalue. Coercivity, in short, is the dividing line between 'always uniquely solvable' and 'solvable only sometimes'.
Compare two forms on H^1_0(0, 1). The form a(u, u) = integral of (u')^2 is coercive (Poincare gives a(u, u) at least a positive constant times ||u||_{H^1}^2), so minus u'' = f is uniquely solvable. But b(u, u) = integral of ((u')^2 - 10 u^2) is NOT coercive — for u = sin(pi x) the gradient energy pi^2 is overwhelmed by the 10 u^2 term, making b(u, u) negative — and indeed minus u'' minus 10 u = f need not be uniquely solvable (10 sits past the first eigenvalue pi^2).
A reaction term with the wrong sign can destroy coercivity and with it unique solvability.
Coercivity is stronger than mere positivity a(u, u) > 0: the latter can hold while a(u, u) creeps toward zero for a sequence of unit-norm functions, which is exactly the loophole that breaks existence. The uniform lower bound alpha > 0 is the whole point. Note also that coercivity is relative to the NORM you use — a form coercive in H^1_0 (thanks to Poincare) need not be coercive in L^2.