If y1 x) and y2 x) are two distinct solutions of the equation, then any combination ay1 x) + by2 x will also be a solution, called the general solution, for any constants a and b.
Finally, since (mathcal{N}^capmathcal{N}^=emptyset), we infer that (u_{0}^) and (u_{0}^) are two distinct solutions.
So, u ¯ 1 and u ¯ 2 are two distinct weak solutions of (P) and the proof is complete.
So, (bar{u}_{1}) and (bar{u}_{2}) are two distinct classical solutions of (1.2) and the proof is complete.
We will assume now that ( u i, v i, ω i, θ i ), i = 1, 2 are two distinct generalized solutions of the problem (1 - 11) in the domain Q T with the properties (12), (21), (23 - 24 23 - 24
Therefore, it follows from (2.7), (2.8), (2.9) and Lemma 2.2 that the operator T has two fixed points (u_{1}in Kcap overline{Omega}_{1}setminus{Omega }_{0})) and (u_{2}in Kcap overline{Omega}_{2}setminus{Omega}_{1})), which are two distinct positive solutions of problem (1.1).
Then, u 5 and u 6 are two distinct sign-changing solutions of BVP (1.1), (1.2).
Also it is shown that there are two distinct positive weak solutions to a perturbation of the equation.
There seem to be three distinct charges.
Assume now, contrary to the assertion, that f and g, which are not the solutions of M ˜ ( r, f ) = 0 and M ˜ ( r, g ) = 0, are two distinct zero-order transcendental entire solutions of (1.7), then we can write f n + M ˜ ( r, f ) = g n + M ˜ ( r, g ).
