Theorem 2.2.1 states that if f and are continuous on a bounded rectangle R in the ty-plane, and the point () is inside R, then any solution to the IVP y' = f(t,y), with y() = can be extended (forwards and backwards in time) to the boundary of the rectangle R. The solution's extension need NOT leave R through any particular sides (left, right, top or bottom) and in particular, both the forward part of the extension and the backward part, may leave through the top of R.

Solutions which "blow up" (or down) to infinity in a finite amount of time are said to have a finite escape time.

A solution which has been extended as far as possible in the t direction, is said to be a maximally extended solution.

Here is problem 1 b, which investigates these issues:

1b) Find a solution formula for the maximally extended solution, find the t interval on which it is defined, sketch a solution curve and then use a numeric solver to check your work.

I'll do this problem sideways, using the extension theorem to guess about R, a numeric solver (over R) to make guesses about the answers, then verifying my guesses with the real solution formula. Here is the function f(t,y), and its partial wrt y:

> map(with,{DEtools, plots}):

> f:=y->-y^3/2;
diff(f(y),y);

Both are continuous everywhere, so any rectangle R containing (0, -1) will do.

> DEplot(2*diff(y(t),t) + y(t)^3 = 0, y(t),
t=-5..5,y=-5..5);

The direction field suggests that we won't be able to extend the unique solution through (0, -1) very far backwards. The slope lines are very steep and traveling backwards will cause y to drop very fast. It seems very unlikely that the solution will even make it back to t = -1. Traveling forwards appears as if we'll be attracted to the equilibrium solution y = 0, which we cannot cross, so it seems as if our solution will be negative, and approach zero as . Place your cursor at the end of the question and press [ENTER], then type your answer (in the Answer "style").

Make an analysis, similar to the one above about the solution with y(0) = 0.4, in particular make a guess as to how far backwards in time this solution will survive.

To find a formula, dsolve will work fine, but we can handle this one "by hand". It is separable, and we have -dy/y^3 = dt/2.

> step1:=int(-1*y^(-3), y) = int(.5, t)+c;

Take care of the constant:

> step2:=subs({y=-1, t=0},step1);

Put that back in:

> step3:=subs({c=.5},step1);

and solve for y

> solve(step3,{y});

Only the negative solution works (y(0) = -1), and we see why our solution can't make it very far backwards. No matter which rectangle (bounded, that is) we put around (0, -1), moving backwards will force the solution to escape through the bottom of the rectangle before t reaches -1. If the top of our rectangle is above y = 0, the forward extension will leave the right side of R. If the top of R is below y = 0, we can leave (in the forward direction) through either the top or right side of R. Here is a plot (all the way to infinity) of the solution:

> plot(-1/sqrt(t+1), t=-2..infinity);

The plot shows that our solution is indeed negative and doesn't reach y = 0 until t = infinity! Here is a safer way to see that y approaches zero as t approaches infinity.

> limit(-1/sqrt(t+1),t=infinity);

The t interval on which our solution is defined (maximally) is -1 < t < infinity, and here is a numeric solver check

> DEplot({2*diff(y(t),t) + y(t)^3 = 0}, y(t),
t=-.97..10,[[y(0)=-1]],y=-10..2,linecolor=blue,
stepsize=0.05);

Do problem 1a from the text.

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Autonomous differential equations, y' = f(y), where no t appears, have equilibrium solutions (constant solutions y(t) = c, where f(c) = 0) and direction fields which are "time invariant". The latter expression means that if y(t) is a solution, so is y(t + k) for any constant k. The addition of k inside the parenthesis simply shifts the graph of y(t) k units left (right if k is negative). Shown below is a DEplot of an autonomous IVP. If you click on this plot, it will become enclosed in an anchored frame and your toolbar will change to display the animation buttons. Click on the "play single frame" button, which looks like an arrow hitting a wall. Each press of this button will simply shift the solution left or right. Notice how all "shifts" fit the direction field arrows. The Maple code to make this "animation" is hidden inside the subsection below. It is NOT important, but feel free to take a peak at it.

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Maple code

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A solution to an autonomous differential equation y' = f(y) is called bounded if it does not approach plus or minus infinity anywhere (even as t approaches plus or minus infinity). Bounded solutions to autonomous differential equations, with both f(y) and continuous, must approach some equilibrium solution as t approaches infinity and as t approaches minus infinity. Which equilibrium solution a bounded (non-equilibrium) solution approaches is determined by the sign (positive or negative) of y' at any point on the solution. y' cannot be negative for one value of t and positive for another value of t, or y' would be zero somewhere. However, y' = f(y), which is only zero at equilibrium solutions, which cannot cross or intersect other solutions. These facts give rise to a qualitative analysis called sign analysis. The sign of the derivative (for non-equilibrium solutions trapped between two equilibrium solutions) determines the long term behavior of y(t). If y' is negative anywhere, the solution will decrease to the equilibrium solution below it, as t approaches infinity, and will increase (decreasing forwards is the same as increasing backwards) to the equilibrium solution above it, as t goes to minus infinity. Consider y' = y(y-1)^2*(y + 3). The equilibrium solutions are y = 0, 1 and -3. Solutions which pass through a point where -3 < y(t) < 1 are trapped by the equilibrium solutions above and below them. Here is a plot of y' versus y

> plot(y*(y-1)^2*(y+3), y= -3.2 ..2,
labels=[`y`,`dy/dt`]);

The graph above continues to rise forever, on both ends, i.e. as or , y' becomes very large and positive. Any solution with a point where y(t) > 1, will be increasing everywhere and go to infinity as t does. Any solution with a point where y(t) is larger than zero, and less than one, will have y' positive, and end up approaching the equilibrium solution y = 1, as t goes to infinity, while approaching y = 0 as t goes to minus infinity. If the y coordinant of a solution is ever between -3 and 0, the solution must decrease everywhere. Thus, it will approach the equilibrium solution y = -3 as t goes to infinity and y = 0 as t goes to negative infinity.

Make a DEplot of the ODE above and designate 4 initial conditions, one in each of the regions spoke of above ( y > 1, 0 < y < 1, -3 < y < 0 and y < -3), to verify graphically the statements made above.

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