# Proof of uniqueness and existence theorem for first order ordinary differential equations

Suppose we have a first order ODE of the form y′=f(x,y)y^\prime = f(x,y) with initial conditions y(x0)=y0y(x_0) = y_0. The existence and uniqueness theorem states that if ff and its partial derivative with respect to yy are continuous in some rectangular region {(x,y);|x−x0|⩽a,|y−y0|⩽b}\{(x, y); |x – x_0| \leqslant a,|y – y_0| \leqslant b\} then there exists a unique solution of the ODE in the closed interval [x_0 – h, x_0 + h][x_0 – h, x_0 + h] where h < ah < a. I am familiar with the above definition and the various linked topics, such as Picard Iterations, Leibniz integral rule etc but was wondering if anyone could provide a proof of this theorem. =================      Is this about getting a local Lipschitz constant from a bound of f_yf_y? – LutzL 2 days ago      Well I was just after a quick-ish proof (if there is one!) – wrb98 2 days ago 2   Does the knowledge of Picard iterations extend to the theorem of Picard-Lindelöf? – LutzL 2 days ago      I second that. You want the Picard-Lindelöf theorem. – Michael Lee 2 days ago      Yes it does. Would anyone be willing to provide a proof of the theorem based on Picard-Lindelöf? – wrb98 2 days ago ================= 1 Answer 1 ================= The fundamental theorem tells you that f(x,y)-f(x,z)=\int_{[y,z]}f_ydy=\int_0^1f_y(x,y+t(z-y))آ·(z-y)\,dt. f(x,y)-f(x,z)=\int_{[y,z]}f_ydy=\int_0^1f_y(x,y+t(z-y))آ·(z-y)\,dt. The region \{(x,y);|xâˆ’x_0|â©½a,\|yâˆ’y_0\|â©½b\}\{(x,y);|xâˆ’x_0|â©½a,\|yâˆ’y_0\|â©½b\} is a compact set, f_yf_y is continuous thus bounded, let LL be a bound. Then \|f(x,y)-f(x,z)\|\le\int_0^1Lآ·\|z-y\|\,dt=Lآ·\|z-y\| \|f(x,y)-f(x,z)\|\le\int_0^1Lآ·\|z-y\|\,dt=Lآ·\|z-y\| which provides the Lipschitz condition on ff. Now apply Picard-Lindelأ¶f.      Bu why specifically is \frac{\partial f}{\partial y}\frac{\partial f}{\partial y} a requirement for Picard-Lindelöf? – wrb98 2 days ago      *that should read: \frac{\partial f}{\partial y}\frac{\partial f}{\partial y} is continuous – wrb98 2 days ago      Sorry, just seen it in your response. – wrb98 2 days ago      Could you please explain where the second integral on the first line comes from? It appears to be a substitution. – wrb98 2 days ago      The first is a line integral over the segment [y,z][y,z]. Parametrizing the segment as \gamma(t)=y+t(z-y)\gamma(t)=y+t(z-y), t\in[0,1]t\in[0,1] gives the second integral. – LutzL 2 days ago