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Qualitative study of a differential equation

Consider the following Cauchy problem:

\displaystyle x'(t)=\sqrt{1+x(t)-x^4(t)},\ x(0)=0.

Prove that any solution {x} of the above problem cannot be bounded on {\Bbb{R}}.

Solution: First note that {x} is increasing, since {x'(t) \geq 0} for every {t \in \Bbb{R}}. If we neglect the initial condition, we may wonder if the differential equation has some constant solutions. Plugging {x \equiv c} we get

\displaystyle 0 = 1+c-c^4.

Define {g(x)=1+x-x^4} and note that {g'(x)=1-4x^3}, whcih is strictly decreasing with a unique zero at {x_0=4^{(-1/3)}}. It is easy to see that {g(x_0)>0} and {\lim_{x \rightarrow \pm \infty}g(x)=-\infty}. This implies that {g} has exactly two zeros {m,M} and since {g(0)=1>0} we have {m<0<M}. Thus the only constant solutions to the differential equation are {x\equiv m} and {x\equiv M}.

Let’s come back to the problem with initial condition {x(0)=0}. Suppose that {x} is bounded as {t \rightarrow +\infty}. This means that there exists {\lim_{t \rightarrow \infty} x(t)=L\geq 0}. The differential equation then tells us that there exists {\lim_{t \rightarrow \infty} x'(t)=\sqrt{1+L-L^4}}. Using the mean value theorem on {[n,n+1]} we find a {c_n \in (n,n+1)} such that {f'(c_n)=f(n+1)-f(n)}. Taking {n \rightarrow \infty} we get {\lim_{n \rightarrow \infty} f'(c_n)=0}, so we must have

\displaystyle 0 = 1+L-L^4.

This coupled with {L \geq 0} shows that {L=M}.

We have obtained that {\lim_{ t \rightarrow \infty} x(t)=M} and {x(t)<M} for {t \geq 0}. Let’s now use the particularities of this problem, and note that if we differentiate the two members of the differential equation we get

\displaystyle x''(t)=\frac{x'(t)-4x^3(t)x'(t)}{2\sqrt{1+x(t)-x^4(t)}}=\frac{1}{2}(1-4x^3(t))

Using the study made above, we can see that as {t \rightarrow +\infty} we have

\displaystyle \lim_{t \rightarrow \infty} x''(t) = \frac{1}{2}(1-4M^3)<0,

which implies in particular that {x''(t)\leq c<0} for every {t\geq t_0}. Define {h(t)= \int_{t_0}^t x''(t)dt}. On one hand we have {h(t) \leq c(t-t_0)} and on the other {h(t)=x'(t)-x'(t_0)}. So we have

\displaystyle -x'(t_0) = \lim_{t \rightarrow \infty} h(t)\leq \lim_{t \rightarrow \infty} c(t-t_0) = -\infty,

since {c<0}. This contradiction implies that the assumption that {x} was bounded towards {+\infty} is false. We could make a similar reasoning towards {-\infty}.

The same reasoning could be applied to any problem of the type {x'(t)=\sqrt{f(x(t))}, x(0)=0}, where {f:\Bbb{R} \rightarrow \Bbb{R}} is a function such that {f'} is decreasing with a unique zero, {\lim_{t \rightarrow \pm \infty}f(t)<0} and {f(0)>0}.

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