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Existence, uniqueness and the interval of validity

What continuity of $f$ and of $\partial f/\partial y$ each promise about an initial value problem, why a linear equation gets a stronger guarantee, and how far a solution is entitled to reach.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to check the two hypotheses of the existence and uniqueness theorem at an initial point, say which of existence and uniqueness each one buys, read the guaranteed interval of a linear problem straight off its coefficients, find the interval of validity of a nonlinear solution from the solution itself, and say precisely what the theorem refuses to promise.

2. What you already have

You have solved separable, linear and exact equations, and in two of those lessons a solution turned out to exist only on part of the line — the blow-up of $y' = y^{2}$, and the point $x = 0$ that $xy' + y = x^{2}$ cannot be solved across. This lesson is the theory behind both observations, and it is what lets you know the interval before doing the work rather than after.

3. Words for this lesson

4. What is promised, and how much of it

The theorem. Consider $y' = f(x, y)$ with $y(x_{0}) = y_{0}$, and a rectangle around $(x_{0}, y_{0})$.

• If $f$ is continuous on that rectangle, a solution exists on some open interval around $x_{0}$.
• If $\partial f/\partial y$ is continuous there as well, that solution is unique.

Two hypotheses, two separate conclusions, and it is worth keeping them apart: the second buys nothing extra about existence and the first buys nothing at all about uniqueness.

Why uniqueness needs the extra hypothesis. Take $y' = 3y^{2/3}$ with $y(0) = 0$. The function $y = 0$ solves it, and so does $y = x^{3}$, and so does every function that sits on the axis for a while and then lifts off as a cubic. Here $f$ is continuous, so existence is safe; but $\partial f/\partial y = 2y^{-1/3}$ is unbounded as $y \to 0$, and the hypothesis fails exactly on the line where the solutions branch.

The linear case is stronger. For $y' + p(x)y = q(x)$ with $p$ and $q$ continuous on an interval $I$ containing $x_{0}$, there is exactly one solution on the whole of $I$. Nothing is local about it: the interval is read off the coefficients before any solving, and the solution cannot leave early.

The nonlinear case is not. $y' = y^{2}$ with $y(0) = 1$ satisfies both hypotheses beautifully and its solution $y = 1/(1 - x)$ still escapes at $x = 1$. Nothing in the equation announces that point, which is precisely why the theorem promises a neighbourhood and refuses to say how large.

5. Where the interval comes from, linear against nonlinear

For a linear equation the interval is a property of the equation. Put it in standard form, mark every point where $p$ or $q$ is discontinuous, and the answer is the gap between the two marks either side of $x_{0}$. It can be found in ten seconds and it is guaranteed.

For a nonlinear equation the interval is a property of the solution, and it can depend on the initial condition. Compare two problems for $y' = y^{2}$: starting at $y(0) = 1$ gives $y = 1/(1 - x)$ and an interval ending at $1$; starting at $y(0) = 10$ gives $y = 1/(0.1 - x)$ and an interval ending at $0.1$. Same equation, same theorem, different windows — and no amount of staring at $y' = y^{2}$ reveals either number.

This is the practical content of the lesson. For a linear problem, state the interval first. For a nonlinear one, solve and then look. A solution is a function together with an interval it is a solution on. The existence theorem promises only a neighbourhood, and a formula that blows up at a finite point stops being an answer there — so a solution written without its interval is an answer to a question nobody asked.

6. What the theorem does not say

It does not say a solution is impossible when a hypothesis fails. $y' = 3y^{2/3}$ through the origin has plenty of solutions; what it lacks is the promise of only one. A failed hypothesis withdraws a guarantee, and withdrawing a guarantee is not the same as asserting the opposite.

It does not give the interval. In the nonlinear case it says some interval, and the word carries no size with it.

It is about one point. Uniqueness through $(0, 0)$ says nothing about uniqueness through $(0, 1)$, and the hypotheses have to be checked where the initial condition puts you.

It does not require the equation to be solvable in formulas. This is the reason the theorem is worth having: most equations cannot be solved in closed form at all, and a numerical method that computes a solution is only meaningful if there is a solution to compute and only one of it. Every solver in unit 6 rests on this lesson.

7. The method, step by step, and how to check it

Every question in this lesson is answered by the same short procedure, and it is worth doing in this order, because the order is what keeps the linear and nonlinear cases apart.

1. Write the problem as $y' = f(x, y)$ with $y(x_{0}) = y_{0}$. If the equation has a coefficient in front of $y'$, divide by it, and write down where that coefficient is zero: those points are outside the theorem from the start.
2. Is $f$ defined and continuous near $(x_{0}, y_{0})$? Look for division by zero, logarithms of non-positive numbers, even roots of negative numbers. If $f$ is continuous on a rectangle around the point, a solution exists.
3. Is $\partial f/\partial y$ continuous there? Differentiate in $y$, holding $x$ fixed, and look for the same kinds of trouble. A power of $y$ below one, such as $y^{2/3}$, usually fails here at $y = 0$, and that is where solutions branch.
4. Is the equation linear? If so, write it as $y' + p(x)y = q(x)$, mark every point where $p$ or $q$ fails, and take the gap containing $x_{0}$: there is exactly one solution on that whole interval.
5. If it is nonlinear, the theorem gives only some interval around $x_{0}$. To know how far the solution runs, solve it and find where the formula fails.

The reason for step 4 is worth knowing. For a linear equation the solution is built from integrals of $p$ and $q$ (the integrating factor formula), and those integrals are finite wherever $p$ and $q$ are continuous, so the solution cannot blow up inside the interval. A nonlinear equation feeds $y$ back into its own rate: in $y' = y^{2}$ a large $y$ makes $y$ grow faster still, and that feedback is what carries a solution to infinity in finite time.

How to check the answer. For a linear problem, confirm the initial point is inside your interval and that each end is a genuine trouble point of $p$ or $q$, not just a point where something looks awkward. For a nonlinear problem, check your formula satisfies the equation and the initial condition, and then check the end of the interval by substituting a value close to it: the solution should grow without bound as you approach. If a hypothesis failed, say which one and what that withdraws: continuity of $f$ withdraws existence, continuity of $\partial f/\partial y$ withdraws uniqueness.

8. In the world: an empty tank cannot tell you when it emptied

Torricelli's law says a tank draining through a hole in its base has depth $h$ obeying $h' = -k\sqrt{h}$. Starting full, the solution is $\sqrt{h} = \sqrt{h_0} - \frac{k}{2}t$, and the tank empties at a definite time. So far the model is predictive.

Now run it backwards. You find the tank empty at noon and want to know when it was full. The initial condition is $h(12) = 0$, and at $h = 0$ the right-hand side $f(h) = -k\sqrt{h}$ is continuous but its derivative $f'(h) = -\frac{k}{2\sqrt{h}}$ is not: it is infinite there. The uniqueness hypothesis fails, and it fails in practice. The constant solution $h = 0$ satisfies the condition, and so does every solution that emptied at some earlier time $t_1$ and then stayed empty. An empty tank at noon is consistent with a tank that emptied at eleven, at ten, or was never filled: the question has no single answer, and the failed hypothesis says so.

This matters well beyond tanks. Investigators reconstructing an accident from its end state, and simulations run backwards to find initial conditions, are only sound when uniqueness holds along the way. Checking the hypotheses of this lesson is how an analyst knows the backwards question has one answer.

9. In the world: a model that predicts its own breakdown

A self-heating process, such as a pile of damp hay or a batch of reacting chemicals, can release heat faster the hotter it gets. A crude model for the temperature excess $y$ above the surroundings is $y' = ay^{2}$: the heat released grows like the square of the excess. The right-hand side is a polynomial, so existence and uniqueness hold everywhere, and yet the solution through $y(0) = y_0$ is $y = \frac{y_0}{1 - ay_0t}$, which is infinite at $t = \frac{1}{ay_0}$.

With $a = 0.01$ per degree per minute and a starting excess of $20$ degrees, the blow-up time is $\frac{1}{0.01 \times 20} = 5$ minutes; from an excess of $40$ degrees it is $2.5$ minutes. Doubling the start halves the time left, which is the signature of runaway.

No real temperature becomes infinite. What reaches infinity is the model, and the finite time is the useful prediction: it says the process runs away within about five minutes, which is the time an operator has to act. Chemical engineers call this thermal runaway and design cooling to keep the equation linear. The lesson is the one this lesson teaches: continuity of the right-hand side promises a solution near the start, and says nothing about how long it lasts.

10. Continuity of the right-hand side is not continuity of the solution's reach

The commonest over-reading of this theorem is to take a polynomial right-hand side — continuous everywhere, differentiable everywhere, nothing to complain about — and conclude that the solution must therefore exist everywhere. $y' = y^{2}$ is the standing refutation: as well-behaved as an equation can be, and its solutions still reach infinity in finite time. The hypotheses are about $f$ on a neighbourhood of the starting point; the conclusion is about a neighbourhood of the starting point; and nothing in the argument ever looks further than that. The mirror-image error is to read a failed hypothesis as bad news about the equation — to say that $y' = 3y^{2/3}$ through the origin has no solution, when in fact it has infinitely many. What has failed is a promise, and the honest report is that the theorem is silent here and the equation must be examined on its own.

11. A linear problem, interval first

1. Take $(x - 3)y' + y = \ln x$ with $y(1) = 2$. Divide every term by $x - 3$ to reach standard form.

   $\dfrac{(x - 3)y'}{x - 3} + \dfrac{y}{x - 3} = \dfrac{\ln x}{x - 3} \quad\Rightarrow\quad y' + \dfrac{1}{x - 3}y = \dfrac{\ln x}{x - 3}$

   Divide before looking for trouble: the coefficients must be read from the standard form, where $y'$ has coefficient $1$.

2. Find where $p$ fails to be continuous.

   $p(x) = \dfrac{1}{x - 3}: \quad \text{undefined at } x = 3$

   A quotient is continuous wherever its denominator is not zero.

3. Find where $q$ fails to be continuous.

   $q(x) = \dfrac{\ln x}{x - 3}: \quad \text{needs } x > 0 \text{ and } x \ne 3$

   The logarithm is defined only for positive $x$, and the division again rules out $3$.

4. Mark the trouble points on the line and find the gap containing the initial point $x = 1$.

   $\text{trouble at } 0 \text{ and } 3; \qquad 0 < 1 < 3 \quad\Rightarrow\quad (0, 3)$

   The interval must contain $x_{0}$ and no trouble point.

5. State what the theorem promises.

   $\text{exactly one solution on the whole of } (0, 3)$

   For a linear equation the interval is settled before anything is integrated, and the solution exists on all of it. Beyond $0$ and $3$ nothing is promised.

12. A nonlinear problem, interval last

1. Take $y' = y^{2}$ with $y(0) = 2$. Check both hypotheses.

   $f = y^{2}, \quad \dfrac{\partial f}{\partial y} = 2y: \text{ both continuous everywhere}$

   Both hold, so there is exactly one solution near $x = 0$, but how far it runs is not promised.

2. Separate and integrate.

   $\displaystyle\int y^{-2}\,dy = \int dx \quad\Rightarrow\quad -\frac{1}{y} = x + C$

   Divide by $y^{2}$, multiply by $dx$, and use the power rule on the left.

3. Substitute $x = 0$, $y = 2$ to find $C$.

   $-\dfrac{1}{2} = 0 + C \quad\Rightarrow\quad C = -\dfrac{1}{2}$

   At $x = 0$ only the constant is left.

4. Multiply both sides by $-1$ and take reciprocals.

   $\dfrac{1}{y} = \dfrac{1}{2} - x \quad\Rightarrow\quad y = \dfrac{1}{\frac{1}{2} - x} = \dfrac{2}{1 - 2x}$

   Multiplying top and bottom by $2$ tidies the fraction.

5. Read the interval off the solution.

   $1 - 2x = 0 \text{ at } x = \tfrac{1}{2} \quad\Rightarrow\quad \left(-\infty, \tfrac{1}{2}\right)$

   The number $\tfrac{1}{2}$ came from the initial condition, not the equation, so the theorem could never have predicted it.

6. Compare with a larger start, $y(0) = 10$.

   $C = -\tfrac{1}{10}, \qquad y = \dfrac{10}{1 - 10x}, \qquad \left(-\infty, \tfrac{1}{10}\right)$

   Same equation, a different window: the larger the start, the sooner the solution escapes.

13. When uniqueness fails

1. Take $y' = 3y^{2/3}$ with $y(0) = 0$. Check the first hypothesis.

   $f = 3y^{2/3} = 3\left(\sqrt[3]{y}\right)^{2}: \quad \text{continuous everywhere}$

   Cube roots are defined for every real $y$, so a solution exists.

2. Check the second hypothesis.

   $\dfrac{\partial f}{\partial y} = 3 \cdot \tfrac{2}{3}y^{-1/3} = \dfrac{2}{\sqrt[3]{y}}: \quad \text{undefined at } y = 0$

   The initial point sits on $y = 0$, exactly where this fails, so uniqueness is not promised.

3. Check the constant function $y = 0$.

   $y = 0: \quad y' = 0, \qquad 3 \cdot 0^{2/3} = 0$

   Both sides are zero, so $y = 0$ is one solution.

4. Separate for $y \ne 0$ and integrate.

   $\displaystyle\int y^{-2/3}\,dy = \int 3\,dx \quad\Rightarrow\quad 3y^{1/3} = 3x + C$

   The power rule: $\int y^{-2/3}\,dy = \frac{y^{1/3}}{1/3} = 3y^{1/3}$.

5. Apply $y(0) = 0$ and solve for $y$.

   $0 = 0 + C \;\Rightarrow\; C = 0, \qquad y^{1/3} = x \;\Rightarrow\; y = x^{3}$

   Cubing both sides undoes the cube root.

6. Check $y = x^{3}$ in the equation.

   $y' = 3x^{2}, \qquad 3\left(x^{3}\right)^{2/3} = 3x^{2}$

   Both sides agree, so there are now two different solutions through the same point.

7. Build infinitely many more: stay on the axis until any $a \ge 0$, then lift off.

   $y = \begin{cases} 0, & x \le a \\ (x - a)^{3}, & x > a \end{cases}$

   Each piece solves the equation, and they join with matching value and slope $0$ at $x = a$. A failed hypothesis withdrew the promise of one solution, and there are infinitely many.

14. Your turn: what is promised for $y' = \dfrac{y}{x - 2}$ with $y(5) = 1$?

1. Classify the equation and read off its coefficients.

   $y' - \dfrac{1}{x - 2}y = 0: \quad p = -\dfrac{1}{x - 2}, \quad q = 0$

   It is linear, so the interval comes from where $p$ and $q$ are continuous.

2. Your turn: work this step out. Its working is at the end of the packet.

   Find the failures and take the gap containing $x = 5$.

3. Your turn: work this step out. Its working is at the end of the packet.

   State the promise, and compare it with the formula.

15. Guided practice

Match each initial value problem to what the existence and uniqueness theorem promises it.

16. Guided practice

Complete the worked solution: where is a solution of $y' + \dfrac{y}{x - (-3)} = \dfrac{1}{x - 8}$ with $y(1) = 6$ guaranteed, and how much room is there?

1. Read off $p$ and $q$ from the standard form.

   $p(x) = \dfrac{1}{x - (-3)}, \qquad q(x) = \dfrac{1}{x - 8}$

   For a linear equation the guarantee depends only on where $p$ and $q$ are continuous.

2. Find where each fails, and take the gap containing the starting point.

   $p \text{ fails at } -3, \quad q \text{ fails at } 8, \quad -3 < 1 < 8 \quad\Rightarrow\quad (-3, 8)$

   The guarantee runs outward from the starting point to the nearest failure on each side.

3. Subtract the left end from the right end for the length.

   $8 - (-3) =$ _l

   Subtracting a negative number adds its size.

4. Measure from the starting point to each end, and take the smaller.

   $\min\left(1 - (-3),\ 8 - 1\right) =$ _r

   The nearer failure is how far the solution can be followed in both directions.

17. Guided practice

What does the existence and uniqueness theorem promise for $y' = x + y$ through $(0, 1)$?

• Exactly one solution near the point
• At least one solution, but uniqueness is not promised
• Nothing: the equation is not defined at the point

18. Practice

For $y' + \dfrac{y}{x - (-4)} = \dfrac{1}{x - 3}$ with $y(0) = 4$, on which interval is a solution guaranteed?

This task has no paper form; do it on a device.

19. Practice

$y' + \dfrac{y}{x - (-2)} = \dfrac{1}{x - 4}$ with $y(0) = 8$. How long is the interval on which a solution is guaranteed?

Answer:

20. Practice

A self-catalysing reaction is heated so that its rate constant rises steadily with time, and the concentration $y$ of its product obeys $y' = 2xy^{2}$, with $x$ the time. The model predicts that $y$ becomes infinite in finite time, which is where it stops describing the vessel. Starting from $y(0) = 4$, the solution exists on an interval $(-T, T)$ and no further. Give that interval.

This task has no paper form; do it on a device.

21. Somewhere new

For $y' = 4y^{2}$ with $y(0) = 6$, both $f$ and its derivative in $y$ are continuous everywhere. Mark the two statements the theorem actually supports.

This task has no paper form; do it on a device.

22. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

23. Test question

A self-catalysing reaction is heated so that its rate constant rises steadily with time, and the concentration $y$ of its product obeys $y' = 2xy^{2}$, with $x$ the time. The model predicts that $y$ becomes infinite in finite time, which is where it stops describing the vessel. Starting from $y(0) = 25$, the solution exists on an interval $(-T, T)$ and no further. Give that interval.

This task has no paper form; do it on a device.

24. What you can do now

You can check the hypotheses at an initial point, name the guaranteed interval for a linear problem and the interval of validity for a nonlinear one. Say in your own words why a failed hypothesis is not the same as no solution.

Working for the steps left to you

14. Your turn: what is promised for $y' = \dfrac{y}{x - 2}$ with $y(5) = 1$?, step 2

14. Your turn: what is promised for $y' = \dfrac{y}{x - 2}$ with $y(5) = 1$?, step 3