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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. 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) 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 ${f(0)>0}$.