2
$\begingroup$

Consider the IVP: $$\frac{dy}{dt}=f(t,y)~,~y(0)=y_{0},$$ where $f(t,y)$ is a bounded function, continuous on the set $\Omega(t,y)=\{ |t| < a , |y-y_{0}| < b \}$ and satisfies a Lipschitz condition there with Lipschitz constant $K.$ Find a condition which will ensure that a unique solution will exist inside $\Omega$ and that the condition has a maximal area on $\Omega.$

My approach. The solution of the IVP is given by $$y(t)=y_{0}+\int_{0}^{t} f(s,y(s))~ds.$$ $$\implies | y(t) - y_{0} | \leq \int_{0}^{t} | f(s,y(s))|~ds \leq M \cdot |t|,~~\text{where}~M=\max_{s}|f(s,y(s))|.$$ Since in $\Omega,~|t| < a,$ we have $$| y(t) - y_{0} | < a \cdot M.$$ Also, incorporating the Lipschitz condition in the second argument yields $$|f(t,x)-f(t,y)| \leq K \cdot |x-y|.$$

I don't know how to continue from here onwards. Any help is much appreiciated.

$\endgroup$
0
$\begingroup$

Let $I(t) = \begin{cases} [t,0], & t<0 \\ [0,t], & t \ge 0 \end{cases}$. Let $u$ and $v$ be two solutions on $(-a,a)$. Then, as you correctly write,

$$u(t) = y_0 + \int \limits _{I(t)} f \big(s,u(s) \big) \ \Bbb d s \qquad \text{and} \qquad v(t) = y_0 + \int \limits _{I(t)} f \big(s,v(s) \big) \ \Bbb d s ,$$

so that

$$\sup _{|t|<a} |u(t) - v(t)| = \sup _{|t|<a} \left| \int \limits _{I(t)} f \big(s,u(s) \big) - f \big(s,v(s) \big) \ \Bbb d s \right| \le \sup _{|t|<a} \int \limits _{I(t)} \Big| f \big(s,u(s) \big) - f \big(s,v(s) \big) \Big| \ \Bbb d s \le \\ \sup _{|t|<a} \int \limits _{I(t)} K \ |u(s) - v(s)| \ \Bbb d s \le K \sup _{|t|<a} \sup _{s \in I(t)} |u(s) - v(s)| \int \limits _{I(t)} \Bbb d s = K \sup _{|t|<a} \sup _{s \in I(t)} |u(s) - v(s)| \ t \le \\ aK \sup _{|s|<a} |u(s) - v(s)| $$

Looking only at the endpoints of this chain of inequalities, and assuming now that $u \ne v$, it follows that $\sup _{|s|<a} |u(s) - v(s)| \ne 0$ so we may divide by it, thus obtaining $1 \le aK$. What we have obtained, then is the following implication: if the problem does not have a unique solution, then $aK \ge 1$. It follows then that if $aK < 1$ then the problem has a unique solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy