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Show that each solution $(x(t), y(t))$ of the initial value problem $$ \left\{\begin{array}{cc}x' =&x^2+y \\ y' =&y^2+x \end{array}\right.\qquad \left\{\begin{array}{cc}x(0) =&x_0 \\ y(0) =&y_0 \end{array}\right. $$ with $x_0>0$ and $y_0>0$ cannot exist on an interval of the form $[0,\infty)$.

I have been learning about different theorems to show existence but I am not sure how I would show this DNE. I put it into matlab using this code and got no solutions but is there a way to show this algebraically or some other way?

syms x(t) y(t) x0 y0

ode1 = diff(x) == x^2+y;
ode2 = diff(y) == y^2+x;
odes = [ode1; ode2]
S = dsolve(odes)

xSol(t) = S.x
ySol(t) = S.y

[xSol(t), ySol(t)] = dsolve(odes)

cond1 = x(0) == x0;
cond2 = y(0) == y0;
conds = [cond1; cond2];
[uSol(t), vSol(t)] = dsolve(odes,conds)

fplot(xSol)
hold on
fplot(ySol)
grid on
legend('xSol','ySol','Location','best')
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    $\begingroup$ HINT: Show that the mapping $$ \begin{array}{ccc} f: \mathbb{R}^2 &\to& \mathbb{R}^2\\ (x,y) &\mapsto & (x^2 + y, y^2 + x) \end{array} $$ is not Lipschitz continuos $\endgroup$ – caverac Oct 12 '17 at 1:31
  • $\begingroup$ thank you! I will give that a try. :) $\endgroup$ – MathIsHard Oct 12 '17 at 1:56
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Since the right hand sides of your system are even analytic functions of $x$ and $y$ it follows that for any initial point $(x_0,y_0)\in\bigl({\mathbb R}_{>0}\bigr)^2$there is an $h>0$ and a solution $$t\mapsto\bigl(x(t),y(t)\bigr)\quad (0\leq t<h),\qquad x(0)=x_0, \ y(0)=y_0$$ of the given IVP. But such a solution cannot live forever, because it explodes in finite time.

Proof. By inspection one verifies that $x(t)>0$, $y(t)>0$ for all $t\geq0$, hence $$x'(t)> x^2(t)\qquad(t\geq0)\ .$$ It follows that $$\int_0^t{x'(\tau)\over x^2(\tau)}\>d\tau>\int_0^t 1\>d\tau=t\ ,$$ which expands to $${1\over x_0}-{1\over x(t)}>t\ ,$$ or $$x(t)>{1\over{1\over x_0}-t}\ .$$ This shows that the solution will explode before time $T:={1\over x_0}$.

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Given the system

$x' = x^2 + y, \tag 1$

$y' = y^2 + x, \tag 2$

we define the vector field

$\vec X(x, y) = \begin{bmatrix} x^2 + y \\ y^2 + x \end{bmatrix}, \tag 3$

which is defined on all of $\Bbb R^2$. Consider the set

$\Omega = \{ (x, y) \in \Bbb R^2 \mid x \ge x_0, y \ge y_0 \} \subset \Bbb R^2; \tag 3$

the boundary $\partial \Omega$ of $\Omega$ is

$\partial \Omega = \{ (x, y_0) \mid x \ge x_0 \} \cup \{ (x_0, y) \mid y \ge y_0 \}; \tag 4$

we see that $\partial \Omega$ consists of two half-lines emanating from $(x_0, y_0)$, each one parallel to one of the $x$- and $y$-axes, and each extending indefinitely in the positive direction.

On $\partial \Omega$, we have

$x' = x^2 + y \ge x_0^2 + y_0 > 0, \tag 5$

and

$y' = y^2 + x \ge y_0^2 + x_0 > 0; \tag 6$

it follows from (5) and (6) that $X(x, y)$ points into $\Omega$ on $\partial \Omega$, and from this observation we conclude that any trajectory initialized on $\partial \Omega$ enters $\Omega$ immediately, and never leaves it; that is, if $x(t_0), y(t_0) \in \partial \Omega$, then $(x(t), y(t)) \in \Omega$ for all $t > t_0$; likewise if $(x(t_0), y(t_0)) \in \Omega$, then $(x(t), y(t)) \in \Omega$ for all $ t > t_0$ as well.

Now let $(x_1, y_1) \in \Omega$, and consider the integral curve $\gamma(t) = (x(t), y(t))$ where $(x(t_0), y(t_0)) = (x_1, y_1)$; we have seen that $\gamma(t) \in \Omega$ for all $t \ge t_0$: thus, on $\gamma(t)$,

$x' = x^2 + y \ge x^2 + y_0, \tag 7$

from which

$\dfrac{x'}{x^2 + y_0} \ge 1; \tag 8$

thus, in a spirit similar to the arguments given by Christian Blatter and AVK,

$\displaystyle \int_{t_0}^t \dfrac{x'(s)}{x^2(s) + y_0}ds \ge \int_{t_0}^t ds = (t - t_0); \tag 9$

we have

$\displaystyle \int_{t_0}^t \dfrac{x'(s)}{x^2(s) + y_0}ds = \dfrac{1}{\sqrt{y_0}}\tan^{-1} \dfrac{x(t)}{\sqrt{y_0}} - \dfrac{1}{\sqrt{y_0}}\tan^{-1} \dfrac{x(t_0)}{\sqrt{y_0}}$ $= \dfrac{1}{\sqrt{y_0}}(\tan^{-1} \dfrac{x(t)}{\sqrt{y_0}} - \tan^{-1} \dfrac{x(t_0)}{\sqrt{y_0}}), \tag{10}$

and so, via (9),

$\dfrac{1}{\sqrt{y_0}}(\tan^{-1} \dfrac{x(t)}{\sqrt{y_0}} - \tan^{-1} \dfrac{x(t_0)}{\sqrt{y_0}}) \ge t - t_0, \tag{11}$

and thus

$\tan^{-1} \dfrac{x(t)}{\sqrt{y_0}} - \tan^{-1} \dfrac{x(t_0)}{\sqrt{y_0}} \ge \sqrt{y_0}(t - t_0), \tag{12}$

whence

$\tan^{-1} \dfrac{x(t)}{\sqrt{y_0}} \ge \sqrt{y_0}(t - t_0) + \tan^{-1} \dfrac{x(t_0)}{\sqrt{y_0}} \tag{13}$

and thus

$x(t) \ge \sqrt{y_0} \tan(\sqrt{y_0}(t - t_0) + \tan^{-1} \dfrac{x(t_0)}{\sqrt{y_0}}), \tag{14}$

the direction of the inequalities being preserved in (11)-(14) since $\tan$, $\tan^{-1}$, etc., are monotonically increasing functions of their respective arguments.

We note that for $x(t_0) \ge x_0$,

$0 < \tan^{-1} \dfrac{x(t_0)}{\sqrt{y_0}} < \dfrac{\pi}{2}, \tag{15}$

thus if we set

$t_{max} = t_0 + \dfrac{1}{\sqrt{y_0}}(\dfrac{\pi}{2} - \tan^{-1} \dfrac{x(t_0)}{\sqrt{y_0}}) \tag{16}$

we see that the solution $x(t)$ cannot possibly be continued as far as $t_{max}$, since from (14)

$x(t) \ge \sqrt{y_0} \tan(\sqrt{y_0}(t - t_0) + \tan^{-1} \dfrac{x(t_0)}{\sqrt{y_0}}) \to \infty \; \text{as} \; t \to t_{max}^-; \tag{17}$

the solution "blows up" in finite time.

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Firstly, the MATLAB program outputs the message

Warning: Explicit solution could not be found.

Basically it means that an attempt to find a solution as a closed-form expression was unsuccessful. It doesn't mean that the solution does not exist; moreover, for some value $\epsilon$ > 0, there exists a unique solution on the interval $t\in[-\epsilon,\epsilon]$ (by the Picard–Lindelöf theorem).

Consider the initial value problem \begin{equation}\tag{1} \left\{\begin{array}{lll} \dot x&=&x^2\\ \dot y&=&y^2,\\ \end{array} \right.\qquad \left\{\begin{array}{l} x(0)=x_0\\ y(0)=y_0, \end{array} \right. \end{equation} $x_0>0$, $y_0>0$. Its solution is $$ x_{(1)}(t)=-\frac{1}{t-\frac{1}{x_0}},\qquad y_{(1)}(t)=-\frac{1}{t-\frac{1}{y_0}}. $$ This solution does not exist on $[0,\infty)$ because it tends to $+\infty$ at $t\to1/x_0^{-}$ and $t\to1/y_0^{-}$.

The solution $x_{(2)}(t)$, $y_{(2)}(t)$ of the original initial value problem \begin{equation}\tag{2} \left\{\begin{array}{lll} \dot x&=&x^2+y\\ \dot y&=&y^2+x,\\ \end{array} \right.\qquad \left\{\begin{array}{l} x(0)=x_0\\ y(0)=y_0 \end{array} \right. \end{equation} can not be expressed as a closed-form expression, but we know that in some interval $(0,T)$ (before it tends to +$\infty$) $$ \dot x_{(2)}(t)=x^2+y>x^2=\dot x_{(1)}(t)>0, $$ $$ \dot y_{(2)}(t)=y^2+x>y^2=\dot y_{(1)}(t)>0, $$ ** Update **

($x(t)>0$ and $y(t)>0$ because they are greater than the solution of the initial value problem $\dot x=y$, $\dot y= x$, $x(0)=x_0$, $y(0)=y_0$ which is $x(t)= x_0\cosh t+y_0\sinh t$, $y(t)= x_0\sinh t+y_0\cosh t)$

** End of update **

thus, $$ x_{(2)}(t)= x_0+\int_0^t \dot x_{(2)}(t)\,dt>x_0+\int_0^t \dot x_{(1)}(t)\,dt= x_{(1)}(t)= -\frac{1}{t-\frac{1}{x_0}} $$ $$ y_{(2)}(t)= y_0+\int_0^t \dot y_{(2)}(t)\,dt>y_0+\int_0^t \dot y_{(1)}(t)\,dt= y_{(1)}(t)= -\frac{1}{t-\frac{1}{y_0}}. $$ The components of the solution of the problem (2) are greater than the functions that tend to $+\infty$ in finite time, therefore, the solution doesn't exist on $[0,\infty)$.

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    $\begingroup$ Thank you. How do we know that $x^2+y>x^2$? Why is $x>0$? $\endgroup$ – MathIsHard Oct 13 '17 at 2:59
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    $\begingroup$ @Math4Life I have updated the answer $\endgroup$ – AVK Oct 13 '17 at 11:00
  • $\begingroup$ @Math4Life: you know $x^2 + y > x^2$ because $y \ge y_0 > 0$ for all $t > 0$, as is explained in my answer below. Essentially, the solutions are trapped in the region $\{x \ge x_0, y \ge y_0 \}$. $\endgroup$ – Robert Lewis Oct 13 '17 at 17:46

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