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Cauchy Problem with two solutions

Suppose f:\mathbb{R}^2 \to \Bbb{R} is continuous and t_0, x_0 \in \Bbb{R}. Prove that if the Cauchy Problem \begin{cases} \dot{x}=f(t,x) \\ x(t_0)=x_0 \end{cases} has two distinct solutions then it has infinitely many solutions.

Proof: Without loss of generality we may assume that x_0=t_0=0. Then there exist two solution x_1,x_2 of the Cauchy problem such that they are different in a point a which we may assume is greater than 0. Therfore assume x_1(a) < x_2(a) and denote P(a,h) a point on the segment a \times (x_1(a),x_2(a)) we can extend x towards 0. From Cauchy’s existence theorem, we can see that there exists a solution x around P of the differential equation \begin{cases} \dot{x}=f(t,x) \\ x(a)=h \end{cases}. Denote by K the compact determined by x_1,x_2 and the line x=a. Since x is in that compact in a left neighborhood of a, by the compact extension theorem, it can be extended until it reaches the boundary of K. From the intersection point of the graph of x with the boundary of K we can go on the graph of x_1 or x_2 until we reach (0,0) and by the corollary of Lagrange’s theorem the graph we choose is the graph of a solution to the initial differential equation.

Therefore, we have found a solution x_h for all h \in (x_1(a),x_2(a)), and therefore the intial equation has uncountably infinitely many solutions.

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