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elias001
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The quoted passages in (A) and (B) of Part 2 below are respectively taken from:
Advanced Mathematics An Invitation in Preparation for Graduate School by P Guidotti
and
Differential Equations An Introduction to Basic Concepts, Results and Applications by I I. Vrabie
Questions
I have a quick question about 4 Blow-up condition., in Part 1 in the Background section below. Specifically, in the hint to the question, I would like to know if ##R## should be defined as ##R := \liminf_{t \rightarrow T} |x(t)| < \infty##. I did a google search for ordinary differential equations whose function solutions is non globally lipschitz and maximal interval of existence. I found two references' expositions which doesn't assume the global liptschitz condition: I have reproduced them in Part 2 labeled respectively (A) and (B).
In (A) where it says: ##\limsup_{t \nearrow T(x_0)} |x(t, x_0)|_2 < \infty,## there exists a sequence ##(t_k)_{k \in \mathbb{N}}## with ##t_k \nearrow T(x_0)## as ##k \to \infty##... I am not sure with the way about why the right end point cannot be included, that it should be lim inf instead.
Alsp in (B), in the proof of Theorem 2.4.4, it assumed by way of contradiction that (highlighted in bold) ##\liminf_{t \uparrow b} \|x(t)\| < +\infty.## After looking at both sources, I couldn't decide if ##R## should be defined using upper or lower limits. So I decide it might be better if I bring it to the MSE community and ask the experts here.
Background
Part 1
Let ##f## be a vector field on ##\mathbb{R}^n##. In class (chapter 17 section 2, and class lectures), we proved a theorem on local Existence and Uniqueness of solutions to the system
$$x' = f(x). \qquad (1)$$
3 Globally Lipschitz vector fields. Suppose for the moment that there is a constant ##L## such that
$$|f(x) - f(y)| \leq L|x - y| \qquad \text{for all } x,y \in \mathbb{R}^n.$$
4 Blow-up condition. In general (if ##f## is not globally Lipschitz), fix a solution ##x(t)##, and let ##I_{\text{max}} = (T_{\text{min}}; T_{\text{max}}) \in \mathbb{R}## be its maximal time of existence. Prove that
* either ##T_{\text{max}} = +\infty##, that is, the solution exists for all time ##t > 0##,
* or ##|x(t)| \rightarrow \infty## as ##t \nearrow T_{\text{max}}##.
Hint: Prove the contrapositive: For any ##T > 0##,
$$R := \liminf_{t \rightarrow T} |x(t)| < \infty \implies T < T_{\text{max}}.$$
(Argue that whenever ##|x(s)| \leq R## for some ##s##, you can extend the solution by a fixed amount, ##\tau##, up to time ##s + \tau##. Use that the ball ##B := \{x \in \mathbb{R}^n : |x| \leq R\}## is compact.) So the only way a solution can cease to exist is by running off to infinity.
Part 2
(A)
Fact. Assume that ##F : \mathbb{R}^n \to \mathbb{R}^n## be locally Lipschitz. Then, given ##x_0 \in \mathbb{R}^n##, there is a unique solution ##x(\cdot, x_0)## of the initial value problem
$$\dot{x} = F(x), \quad x(0) = x_0,$$ on the interval ##[0, T]## for some ##T > 0##. This solution can be extended to a maximal interval of existence ##[0, T(x_0))## with ##T(x_0) > 0##. If ##T(x_0) < \infty##, then
$$\lim_{t \nearrow T(x_0)} |x(t, x_0)|_2 = \infty.$$
We shall use the Banach fixed-point theorem as we did for the linear equation. For fixed ##T > 0## (to be chosen later), define
##X = X(T, r, x_0) = \{x \in C([0, T], \mathbb{R}^n) \mid x(0) = x_0, |x(t) - x_0|_2 \le r, t \in [0, T]\},##
which is a complete subset of ##C([0, T], \overline{B}(x_0, r))## with respect to the supremum norm ##\|\cdot\|_\infty##. Define
$$\Phi(x)(t) = x_0 + \int_0^t F(x(\tau)) d\tau, \quad t \in [0, T], x \in X,$$
and observe that
$$|\Phi(x)(t) - x_0|_2 \le \int_0^t |F(x(\tau))|_2 d\tau, \quad t \in [0, T], x \in X.$$
Since continuous maps assume their maximum and minimum on compact sets, we have that
$$|F(z)|_2 \le M, \quad z \in \overline{B}(x_0, r),$$
for some ##M \ge 0##, and consequently, that
$$|\Phi(x)(t) - x_0|_2 \le MT, \quad t \in [0, T], x \in X.$$
By choosing ##T## sufficiently small, we can achieve that ##\Phi(x)(t) \in \overline{B}(x_0, r)## for ##t \in [0, T]##, which shows that ##\Phi## maps ##X## into itself. Notice that this is possible for any ##r > 0##. Local Lipschitz continuity yields ##r_0 > 0## and ##L \ge 0## such that
$$|F(y) - F(z)|_2 \le L|y - z|_2, \quad y, z \in \overline{B}(x_0, r_0).$$
Then it holds that
\begin{align*}
|\Phi(x)(t) - \Phi(\tilde{x})(t)|_2 &\le \int_0^T |F(x(\tau)) - F(\tilde{x}(\tau))|_2 d\tau \\
&\le LT\|x - \tilde{x}\|_\infty.
\end{align*}
This shows that
$$\|\Phi(x) - \Phi(\tilde{x})\|_\infty \le \frac{1}{2}\|x - \tilde{x}\|_\infty, \quad x, \tilde{x} \in X,$$
provided ##T \le T_0## and ##T_0## is chosen so that ##LT_0 \le \frac{1}{2}##. The Banach fixed-point theorem then yields a unique solution ##x(\cdot, x_0) \in X = X(T_0, r_0, x_0)##. The reader can verify that such a fixed point is a solution of the initial value problem and that any solution of the initial value problem must be a fixed point of ##\Phi##. The length of the existence interval ##[0, T_0]## depends on the local properties of ##F## near the point ##x_0##. Considering the equation
$$\dot{x} = F(x), \quad x(0) = x_1 = x(T_0, x_0),$$
the argument can be repeated to obtain a solution ##x(\cdot, x_1)## on some interval ##[0, T_1]## for ##T_1 > 0##. Glueing the solutions together, i. e., extending ##x(\cdot, x_0)## by setting
$$x(t, x_0) = x(t - T_0, x_1), \quad t \in [T_0, T_0 + T_1],$$
we obtain a (still unique) solution on ##[0, T_0 + T_1]##, and, repeating the argument, on an interval ##[0, T(x_0))## of length ##T(x_0) = \sum_{k \ge 0} T_k##, where ##T_k > 0## for ##k \ge 0##. Clearly, we either have ##T(x_0) = \infty## or ##T(x_0) < \infty##. In the latter case, if ##\limsup_{t \nearrow T(x_0)} |x(t, x_0)|_2 < \infty,## there exists a sequence ##(t_k)_{k \in \mathbb{N}}## with ##t_k \nearrow T(x_0)## as ##k \to \infty## such that ##x(t_k, x_0) \to x_\infty \in \mathbb{R}^n##, in which case the solution could be extended once more using ##x_\infty## as a new initial datum. This argument also shows that the maximal interval of existence cannot include the right-end point.
(B)
Saturated Solutions
Let ##\mathbb{I}## be a nontrivial interval in ##\mathbb{R}##, ##\Omega## a nonempty and open subset in ##\mathbb{R}^n##, let ##f : \mathbb{I} \times \Omega \to \mathbb{R}^n## be a given function, ##a \in \mathbb{I}## and ##\xi \in \Omega##. Let ##\mathcal{D} = (\mathbb{I}, \Omega, f, a, \xi)## and let us consider the Cauchy problem
\begin{equation*}
\begin{cases} x' = f(t, x) \\ x(a) = \xi. \end{cases} \tag{CP($\mathcal{D}$)}
\end{equation*}
We recall that a solution ##x : \mathbb{J} \to \Omega## of CP(##\mathcal{D}##) (CP##:=\text{The Cauchy Problem}##, ##\mathcal{D} = (\mathbb{I}, \Omega, f, a, \xi)##) is called continuable at the right (left) if there exists a right (left) solution ##y : \mathbb{K} \to \Omega## with ##\sup \mathbb{J} < \sup \mathbb{K}## (##\inf \mathbb{J} > \inf \mathbb{K}##) and such that ##x(t) = y(t)## for each ##t \in \mathbb{J} \cap \mathbb{K}##. We also recall that a solution ##x : \mathbb{J} \to \Omega## is called saturated at the right (left) if it is not continuable at the right (left). A right (left) solution ##x : \mathbb{J} \to \Omega## is called a global at the right (left) solution if ##\mathbb{J} = \{t \in \mathbb{I} : t \ge a\}## (##\mathbb{J} = \{t \in \mathbb{I} : t \le a\}##). Since we merely consider right solutions, in all that follows, by a "continuable", respectively "saturated" solution we shall mean a "continuable at the right", respectively "saturated at the right" solution.
Theorem 2.4.4 Let ##f : \mathbb{I} \times \Omega \to \mathbb{R}^n## be continuous on ##\mathbb{I} \times \Omega## and let us assume that it maps bounded subsets in ##\mathbb{I} \times \Omega## into bounded subsets in ##\mathbb{R}^n##. Let ##x : [a, b) \to \Omega## be a saturated solution of CP(##\mathcal{D}##). Then, either
Theorem 2.4.4. Let ##x : [a, b) \to \Omega## be a saturated solution of ##CP(\mathcal{D})##. Then, either:
##(i')## ##x## is unbounded on ##[a, b)## and, if ##b < +\infty##, there exists ##\lim_{t \uparrow b} \|x(t)\| = +\infty##, or
##(ii)## ##x## is bounded on ##[a, b)##, and, either ##x## is global, i.e., ##b = \sup \mathbb{I}##, or
##(iii')## ##x## is bounded on ##[a, b)##, and ##x## is not global, i.e., ##b < \sup \mathbb{I}##, and in this case there exists ##\lim_{t \uparrow b} x(t) = x^*## and ##x^*## lies on the boundary of ##\Omega##.
Proof. In view of Theorem 2.4.3, the only fact we have to prove here is that whenever (ii) does not hold, then one of the two conditions (i') or (iii') must hold. So, let us assume for the beginning that (i') and (ii) do not hold. Then, from Lemma 2.4.1 and Proposition 2.4.1, it follows that there exists
$$\lim_{t \uparrow b} x(t) = x^*,$$
while from (iii) in Theorem 2.4.3 we have ##x^* \in \partial \Omega##. So, (iii') holds true.
Let us assume now that (ii) and (iii') do not hold, and that ##b < +\infty##. To show that ##\lim_{t \uparrow b} \|x(t)\| = +\infty##, let us assume by contradiction that, under these circumstances, we have
$$\liminf_{t \uparrow b} \|x(t)\| < +\infty.$$
Equivalently, there exist at least one sequence ##(t_k)_{k \in \mathbb{N}}## in ##(a, b)##, converging to ##b##, and ##r > 0## such that ##\|x(t_k)\| \le r##, for each ##k \in \mathbb{N}##.
Let ##C = \{y \in \Omega : \|y\| \le r + 1\}##. Since ##f## maps bounded subsets in ##\mathbb{I} \times \Omega## into bounded subsets in ##\mathbb{R}^n##, ##b < \sup \mathbb{I}## and ##C## is bounded, there exists ##M > 0##, such that
\begin{equation*}
\|f(\tau, y)\| \le M, \tag{2.4.3}
\end{equation*}
for each ##(\tau, y) \in [a, b) \times C##. Now let us choose ##d > 0## satisfying
\begin{equation*}
dM < 1, \tag{2.4.4}
\end{equation*}
and fix ##k \in \mathbb{N}## such that ##b - d < t_k < b##. Since ##x## is unbounded on ##[a, b)##, it is necessarily unbounded on ##[t_k, b)##. Then, there exists ##t^* \in (t_k, b)## such that
\begin{equation*}
\|x(\tau)\| < r + 1, \tag{2.4.5}
\end{equation*}
Thank you in advance
Advanced Mathematics An Invitation in Preparation for Graduate School by P Guidotti
and
Differential Equations An Introduction to Basic Concepts, Results and Applications by I I. Vrabie
Questions
I have a quick question about 4 Blow-up condition., in Part 1 in the Background section below. Specifically, in the hint to the question, I would like to know if ##R## should be defined as ##R := \liminf_{t \rightarrow T} |x(t)| < \infty##. I did a google search for ordinary differential equations whose function solutions is non globally lipschitz and maximal interval of existence. I found two references' expositions which doesn't assume the global liptschitz condition: I have reproduced them in Part 2 labeled respectively (A) and (B).
In (A) where it says: ##\limsup_{t \nearrow T(x_0)} |x(t, x_0)|_2 < \infty,## there exists a sequence ##(t_k)_{k \in \mathbb{N}}## with ##t_k \nearrow T(x_0)## as ##k \to \infty##... I am not sure with the way about why the right end point cannot be included, that it should be lim inf instead.
Alsp in (B), in the proof of Theorem 2.4.4, it assumed by way of contradiction that (highlighted in bold) ##\liminf_{t \uparrow b} \|x(t)\| < +\infty.## After looking at both sources, I couldn't decide if ##R## should be defined using upper or lower limits. So I decide it might be better if I bring it to the MSE community and ask the experts here.
Background
Part 1
Let ##f## be a vector field on ##\mathbb{R}^n##. In class (chapter 17 section 2, and class lectures), we proved a theorem on local Existence and Uniqueness of solutions to the system
$$x' = f(x). \qquad (1)$$
3 Globally Lipschitz vector fields. Suppose for the moment that there is a constant ##L## such that
$$|f(x) - f(y)| \leq L|x - y| \qquad \text{for all } x,y \in \mathbb{R}^n.$$
4 Blow-up condition. In general (if ##f## is not globally Lipschitz), fix a solution ##x(t)##, and let ##I_{\text{max}} = (T_{\text{min}}; T_{\text{max}}) \in \mathbb{R}## be its maximal time of existence. Prove that
* either ##T_{\text{max}} = +\infty##, that is, the solution exists for all time ##t > 0##,
* or ##|x(t)| \rightarrow \infty## as ##t \nearrow T_{\text{max}}##.
Hint: Prove the contrapositive: For any ##T > 0##,
$$R := \liminf_{t \rightarrow T} |x(t)| < \infty \implies T < T_{\text{max}}.$$
(Argue that whenever ##|x(s)| \leq R## for some ##s##, you can extend the solution by a fixed amount, ##\tau##, up to time ##s + \tau##. Use that the ball ##B := \{x \in \mathbb{R}^n : |x| \leq R\}## is compact.) So the only way a solution can cease to exist is by running off to infinity.
Part 2
(A)
Fact. Assume that ##F : \mathbb{R}^n \to \mathbb{R}^n## be locally Lipschitz. Then, given ##x_0 \in \mathbb{R}^n##, there is a unique solution ##x(\cdot, x_0)## of the initial value problem
$$\dot{x} = F(x), \quad x(0) = x_0,$$ on the interval ##[0, T]## for some ##T > 0##. This solution can be extended to a maximal interval of existence ##[0, T(x_0))## with ##T(x_0) > 0##. If ##T(x_0) < \infty##, then
$$\lim_{t \nearrow T(x_0)} |x(t, x_0)|_2 = \infty.$$
We shall use the Banach fixed-point theorem as we did for the linear equation. For fixed ##T > 0## (to be chosen later), define
##X = X(T, r, x_0) = \{x \in C([0, T], \mathbb{R}^n) \mid x(0) = x_0, |x(t) - x_0|_2 \le r, t \in [0, T]\},##
which is a complete subset of ##C([0, T], \overline{B}(x_0, r))## with respect to the supremum norm ##\|\cdot\|_\infty##. Define
$$\Phi(x)(t) = x_0 + \int_0^t F(x(\tau)) d\tau, \quad t \in [0, T], x \in X,$$
and observe that
$$|\Phi(x)(t) - x_0|_2 \le \int_0^t |F(x(\tau))|_2 d\tau, \quad t \in [0, T], x \in X.$$
Since continuous maps assume their maximum and minimum on compact sets, we have that
$$|F(z)|_2 \le M, \quad z \in \overline{B}(x_0, r),$$
for some ##M \ge 0##, and consequently, that
$$|\Phi(x)(t) - x_0|_2 \le MT, \quad t \in [0, T], x \in X.$$
By choosing ##T## sufficiently small, we can achieve that ##\Phi(x)(t) \in \overline{B}(x_0, r)## for ##t \in [0, T]##, which shows that ##\Phi## maps ##X## into itself. Notice that this is possible for any ##r > 0##. Local Lipschitz continuity yields ##r_0 > 0## and ##L \ge 0## such that
$$|F(y) - F(z)|_2 \le L|y - z|_2, \quad y, z \in \overline{B}(x_0, r_0).$$
Then it holds that
\begin{align*}
|\Phi(x)(t) - \Phi(\tilde{x})(t)|_2 &\le \int_0^T |F(x(\tau)) - F(\tilde{x}(\tau))|_2 d\tau \\
&\le LT\|x - \tilde{x}\|_\infty.
\end{align*}
This shows that
$$\|\Phi(x) - \Phi(\tilde{x})\|_\infty \le \frac{1}{2}\|x - \tilde{x}\|_\infty, \quad x, \tilde{x} \in X,$$
provided ##T \le T_0## and ##T_0## is chosen so that ##LT_0 \le \frac{1}{2}##. The Banach fixed-point theorem then yields a unique solution ##x(\cdot, x_0) \in X = X(T_0, r_0, x_0)##. The reader can verify that such a fixed point is a solution of the initial value problem and that any solution of the initial value problem must be a fixed point of ##\Phi##. The length of the existence interval ##[0, T_0]## depends on the local properties of ##F## near the point ##x_0##. Considering the equation
$$\dot{x} = F(x), \quad x(0) = x_1 = x(T_0, x_0),$$
the argument can be repeated to obtain a solution ##x(\cdot, x_1)## on some interval ##[0, T_1]## for ##T_1 > 0##. Glueing the solutions together, i. e., extending ##x(\cdot, x_0)## by setting
$$x(t, x_0) = x(t - T_0, x_1), \quad t \in [T_0, T_0 + T_1],$$
we obtain a (still unique) solution on ##[0, T_0 + T_1]##, and, repeating the argument, on an interval ##[0, T(x_0))## of length ##T(x_0) = \sum_{k \ge 0} T_k##, where ##T_k > 0## for ##k \ge 0##. Clearly, we either have ##T(x_0) = \infty## or ##T(x_0) < \infty##. In the latter case, if ##\limsup_{t \nearrow T(x_0)} |x(t, x_0)|_2 < \infty,## there exists a sequence ##(t_k)_{k \in \mathbb{N}}## with ##t_k \nearrow T(x_0)## as ##k \to \infty## such that ##x(t_k, x_0) \to x_\infty \in \mathbb{R}^n##, in which case the solution could be extended once more using ##x_\infty## as a new initial datum. This argument also shows that the maximal interval of existence cannot include the right-end point.
(B)
Saturated Solutions
Let ##\mathbb{I}## be a nontrivial interval in ##\mathbb{R}##, ##\Omega## a nonempty and open subset in ##\mathbb{R}^n##, let ##f : \mathbb{I} \times \Omega \to \mathbb{R}^n## be a given function, ##a \in \mathbb{I}## and ##\xi \in \Omega##. Let ##\mathcal{D} = (\mathbb{I}, \Omega, f, a, \xi)## and let us consider the Cauchy problem
\begin{equation*}
\begin{cases} x' = f(t, x) \\ x(a) = \xi. \end{cases} \tag{CP($\mathcal{D}$)}
\end{equation*}
We recall that a solution ##x : \mathbb{J} \to \Omega## of CP(##\mathcal{D}##) (CP##:=\text{The Cauchy Problem}##, ##\mathcal{D} = (\mathbb{I}, \Omega, f, a, \xi)##) is called continuable at the right (left) if there exists a right (left) solution ##y : \mathbb{K} \to \Omega## with ##\sup \mathbb{J} < \sup \mathbb{K}## (##\inf \mathbb{J} > \inf \mathbb{K}##) and such that ##x(t) = y(t)## for each ##t \in \mathbb{J} \cap \mathbb{K}##. We also recall that a solution ##x : \mathbb{J} \to \Omega## is called saturated at the right (left) if it is not continuable at the right (left). A right (left) solution ##x : \mathbb{J} \to \Omega## is called a global at the right (left) solution if ##\mathbb{J} = \{t \in \mathbb{I} : t \ge a\}## (##\mathbb{J} = \{t \in \mathbb{I} : t \le a\}##). Since we merely consider right solutions, in all that follows, by a "continuable", respectively "saturated" solution we shall mean a "continuable at the right", respectively "saturated at the right" solution.
Theorem 2.4.4 Let ##f : \mathbb{I} \times \Omega \to \mathbb{R}^n## be continuous on ##\mathbb{I} \times \Omega## and let us assume that it maps bounded subsets in ##\mathbb{I} \times \Omega## into bounded subsets in ##\mathbb{R}^n##. Let ##x : [a, b) \to \Omega## be a saturated solution of CP(##\mathcal{D}##). Then, either
Theorem 2.4.4. Let ##x : [a, b) \to \Omega## be a saturated solution of ##CP(\mathcal{D})##. Then, either:
##(i')## ##x## is unbounded on ##[a, b)## and, if ##b < +\infty##, there exists ##\lim_{t \uparrow b} \|x(t)\| = +\infty##, or
##(ii)## ##x## is bounded on ##[a, b)##, and, either ##x## is global, i.e., ##b = \sup \mathbb{I}##, or
##(iii')## ##x## is bounded on ##[a, b)##, and ##x## is not global, i.e., ##b < \sup \mathbb{I}##, and in this case there exists ##\lim_{t \uparrow b} x(t) = x^*## and ##x^*## lies on the boundary of ##\Omega##.
Proof. In view of Theorem 2.4.3, the only fact we have to prove here is that whenever (ii) does not hold, then one of the two conditions (i') or (iii') must hold. So, let us assume for the beginning that (i') and (ii) do not hold. Then, from Lemma 2.4.1 and Proposition 2.4.1, it follows that there exists
$$\lim_{t \uparrow b} x(t) = x^*,$$
while from (iii) in Theorem 2.4.3 we have ##x^* \in \partial \Omega##. So, (iii') holds true.
Let us assume now that (ii) and (iii') do not hold, and that ##b < +\infty##. To show that ##\lim_{t \uparrow b} \|x(t)\| = +\infty##, let us assume by contradiction that, under these circumstances, we have
$$\liminf_{t \uparrow b} \|x(t)\| < +\infty.$$
Equivalently, there exist at least one sequence ##(t_k)_{k \in \mathbb{N}}## in ##(a, b)##, converging to ##b##, and ##r > 0## such that ##\|x(t_k)\| \le r##, for each ##k \in \mathbb{N}##.
Let ##C = \{y \in \Omega : \|y\| \le r + 1\}##. Since ##f## maps bounded subsets in ##\mathbb{I} \times \Omega## into bounded subsets in ##\mathbb{R}^n##, ##b < \sup \mathbb{I}## and ##C## is bounded, there exists ##M > 0##, such that
\begin{equation*}
\|f(\tau, y)\| \le M, \tag{2.4.3}
\end{equation*}
for each ##(\tau, y) \in [a, b) \times C##. Now let us choose ##d > 0## satisfying
\begin{equation*}
dM < 1, \tag{2.4.4}
\end{equation*}
and fix ##k \in \mathbb{N}## such that ##b - d < t_k < b##. Since ##x## is unbounded on ##[a, b)##, it is necessarily unbounded on ##[t_k, b)##. Then, there exists ##t^* \in (t_k, b)## such that
\begin{equation*}
\|x(\tau)\| < r + 1, \tag{2.4.5}
\end{equation*}
Thank you in advance
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