Why can't I have x˙(t)&lt;0?I'm in the very beginning of "Ordinary Differential Equations and Dynamical Systems" by Gerald Teschl and on page 9 he starts with differential equations of the form:x˙=f(x),x(0)=x0,f∈C(R)To solve these he quickly goes to:∫0tx˙(s)dsf(x(s))=tLetting F(x)=∫x0xdyf(y) he points out that any solution must satisfy that F(x(t))=t. So, since F(x) is is strictly monotone in some neighborhood of x0, any solution ϕ will need to satisfyϕ(t)=F−1(t),ϕ(0)=F−1(0)=x0Assuming, f(x0)&gt;0 we can find a maximal interval (x1,x2)nix0 where f is positive in this interval as well. So we can define:T+=limx↑x2F(x)∈(0,+∞]T−=limx↓x2F(x)∈[−∞,0)So, ϕ∈C1(T−T+). So far, so good. But, where I get confused is the following statement. Teschl states that:In particular, ϕ is defined for all t&gt;0 if and only if:T+=∫x0x2dyf(y)=+∞

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Why can&#039;t I have x˙(t)&amp;lt;0?I&#039;m in the very beginning of &quot;Ordinary Differential Equations and Dynamical Systems&quot; by Gerald Teschl and on page 9 he starts with differential equations of the form:x˙=f(x),x(0)=x0,f∈C(R)To solve these he quickly goes to:∫0tx˙(s)dsf(x(s))=tLetting F(x)=∫x0xdyf(y) he points out that any solution must satisfy that F(x(t))=t. So, since F(x) is is strictly monotone in some neighborhood of x0, any solution ϕ will need to satisfyϕ(t)=F−1(t),ϕ(0)=F−1(0)=x0Assuming, f(x0)&amp;gt;0 we can find a maximal interval (x1,x2)nix0 where f is positive in this interval as well. So we can define:T+=limx↑x2F(x)∈(0,+∞]T−=limx↓x2F(x)∈[−∞,0)So, ϕ∈C1(T−T+). So far, so good. But, where I get confused is the following statement. Teschl states that:In particular, ϕ is defined for all t&amp;gt;0 if and only if:T+=∫x0x2dyf(y)=+∞

brelhadorqkh · Accepted Answer

Step 1The function f is strictly positive in the maximal interval (x1,x2); thus, by the fundamental theorem of Calculus, F is strictly monotone on (x1,x2) which contains x0. Moreover, F(x0)=0,F(x)&amp;gt;0 for x0&amp;lt;x&amp;lt;x2. It follows that F maps (x0,x2) to some open interval (0,β) (possible infinite) and the inverse F−1:(α,β)→(x1,x2) is also differentiable and increasing. As the OP pointed outx(t)=F−1(t)solves the initial value problem, and is defined on (α,β). This implies thatx(−β)=limt≠arrowβx(t)=limt≠arrowβF−1(t)=x2and t=F(x(t))=∫{x(0)}x(t)dyf(y)Step 2Letting t≠arrowβ givesβ=limt≠arrowβ∫{x(0)}x(t)dyf(y)Therefore, x is defined over (0,∞) (i.e. β=∞) iff∫{x(0)}x(β−)dyf(y)=∞

Why can't I have \(\displaystyle\dot{{{x}}}{\left({t}\right)}{ < }{0}\)? I'm

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