Chapter 1 · Section 1.3

1.3 Initial-Value Problems, Boundary-Value Problems, and Existence of Solutions

Section 1.2 showed that a differential equation usually has infinitely many solutions: a whole family. An engineer, however, wants one answer. A circuit has one current, a beam has one deflection curve. What singles out that answer is extra information about the system, such as its state at the moment we start watching it. In this section we add such conditions to an ODE, and then ask two deep questions: does the resulting problem have a solution at all, and if so, is there only one?

Initial-value problems

Let us start with the simplest possible case.

Example 1.3.1

Find the solution of $y^{'} = 6 x - 4$ whose graph passes through the point $(1, 5)$ .

Solution

All solutions. Integrate both sides: $y = 3 x^{2} - 4 x + c$ . This is a one-parameter family.
Use the condition. "Passes through $(1, 5)$ " means $y = 5$ when $x = 1$ . Substitute: $55 c = 3 (1)^{2} - 4 (1) + c = - 1 + c = 6.$
Answer. $y = 3 x^{2} - 4 x + 6$ , valid for all real $x$ . Check: $y^{'} = 6 x - 4$ and $y (1) = 3 - 4 + 6 = 5$ . $✓$

The condition " $y = 5$ when $x = 1$ " is written briefly as $y (1) = 5$ . The whole problem is then written as

y^{'} = 6 x - 4, y (1) = 5.

Read it as: find a function $y$ that satisfies the equation and takes the value $5$ at $x = 1$ . In engineering, conditions like this one, all imposed at a single starting point, are what we meet most often, because a system is usually described by its state at the moment we start observing it.

Definition 1.3.1 (Initial-value problem)

An initial-value problem (IVP) of order $n$ consists of an ODE of order $n$ together with $n$ conditions on the solution and its first $n - 1$ derivatives, all given at one point $x_{0}$ :

y (x_{0}) = y_{0}, y^{'} (x_{0}) = y_{1}, \dots, y^{(n - 1)} (x_{0}) = y_{n - 1} .

The numbers $y_{0}, \dots, y_{n - 1}$ are the initial values, and the conditions are the initial conditions. A solution of the IVP is a solution of the ODE on some interval containing $x_{0}$ that satisfies all the initial conditions.

For a first-order equation $y^{'} = f (x, y)$ there is one initial condition, $y (x_{0}) = y_{0}$ . Geometrically, we are looking for the integral curve that passes through the point $(x_{0}, y_{0})$ . For a second-order equation we prescribe both $y (x_{0})$ and $y^{'} (x_{0})$ : the curve must pass through a given point with a given slope. In mechanics this is natural. Newton's law $m x^{''} = F$ is second order, and to predict the motion we need the starting position $x (0)$ and the starting velocity $x^{'} (0)$ .

Method in a nutshell: IVP from a known family

Write down the $n$ -parameter family of solutions (and, if needed, its derivatives).
Substitute $x = x_{0}$ and the initial values. This gives $n$ equations for the $n$ constants.
Solve for the constants and put them back into the family.
State the largest interval containing $x_{0}$ on which the result is a solution.

Example 1.3.2

In Section 1.2 we saw that $y = x + 1 + c e^{x}$ solves $y^{'} = y - x$ for every constant $c$ . Solve the IVPs (a) $y (0) = 3$ and (b) $y (0) = 1$ .

Solution

(a) Put $x = 0$ , $y = 3$ : $3 = 0 + 1 + c e^{0} = 1 + c$ , so $c = 2$ . The solution is $y = x + 1 + 2 e^{x}$ on $(- \infty, \infty)$ .
(b) Put $x = 0$ , $y = 1$ : $1 = 1 + c$ , so $c = 0$ . The solution is the straight line $y = x + 1$ , the special integral curve we found in the direction field of Section 1.2.

Figure 1.3.1. The family

y = x + 1 + c e^{x}

(grey) fills the plane. An initial condition chooses the one curve through the given point:

y (0) = 3

gives

c = 2

, and

y (0) = 1

gives the straight line

c = 0

Example 1.3.3 (A second-order IVP)

Every solution of $y^{''} + 2 y^{'} + 17 y = 0$ has the form $y = e^{- x} (c_{1} cos 4 x + c_{2} sin 4 x)$ (you will learn why in Chapter 4). Solve the IVP

y^{''} + 2 y^{'} + 17 y = 0, y (0) = 2, y^{'} (0) = 6.

Solution

Derivative of the family. By the product rule, $y^{'} = - e^{- x} (c_{1} cos 4 x + c_{2} sin 4 x) + e^{- x} (- 4 c_{1} sin 4 x + 4 c_{2} cos 4 x) = e^{- x} [(- c_{1} + 4 c_{2}) cos 4 x + (- 4 c_{1} - c_{2}) sin 4 x] .$
First condition. At $x = 0$ : $cos 0 = 1$ , $sin 0 = 0$ , $e^{0} = 1$ . So $y (0) = c_{1} = 2$ .
Second condition. $y^{'} (0) = - c_{1} + 4 c_{2} = 6$ . With $c_{1} = 2$ : $4 c_{2} = 8$ , so $c_{2} = 2$ .
Answer. $y = 2 e^{- x} (cos 4 x + sin 4 x), - \infty < x < \infty.$ This is a damped oscillation: the factor $e^{- x}$ shrinks the amplitude while the trigonometric part oscillates.

Example 1.3.4 (Where the family is not enough)

The ODE $y^{'} = 2 x y^{2}$ has the one-parameter family $y = \frac{1}{c - x ^{2}}$ and also the extra solution $y \equiv 0$ (Section 1.2). Solve the IVPs (a) $y (1) = \frac{1}{3}$ , (b) $y (0) = - 1$ , (c) $y (2) = 0$ . Give the interval of each solution.

Solution

(a) $\frac{1}{c - 1} = \frac{1}{3}$ gives $c - 1 = 3$ , so $c = 4$ and $y = \frac{1}{4 - x ^{2}}$ . This is undefined at $x = \pm 2$ . The interval must contain $x_{0} = 1$ and avoid $\pm 2$ , so the solution lives on $(- 2, 2)$ .
(b) $\frac{1}{c - 0} = - 1$ gives $c = - 1$ and $y = \frac{1}{- 1 - x ^{2}} = - \frac{1}{1 + x ^{2}}$ . The denominator is never zero, so this solution lives on $(- \infty, \infty)$ .
(c) We would need $\frac{1}{c - 4} = 0$ , which is impossible. No member of the family works. But $y \equiv 0$ satisfies both the equation and $y (2) = 0$ . The solution is $y \equiv 0$ on $(- \infty, \infty)$ . Lesson: a family that misses some solutions can lead you to the wrong conclusion "no solution".

Boundary-value problems

Sometimes the extra conditions are given at two different points. A beam is clamped at both ends; a rod has its two ends held at given temperatures. This leads to a different kind of problem.

Definition 1.3.2 (Boundary-value problem)

A boundary-value problem (BVP) consists of an ODE together with conditions on the solution (or its derivatives) given at two or more different values of $x$ . The most common case is the two-point BVP, with conditions at the ends $x = a$ and $x = b$ of an interval.

BVPs behave very differently from IVPs. As the next example shows, the same equation with slightly different boundary data can have exactly one solution, no solution, or infinitely many.

Example 1.3.5 (Three BVPs, three outcomes)

Every solution of $y^{''} + 16 y = 0$ has the form $y = c_{1} cos 4 x + c_{2} sin 4 x$ . Solve each BVP.

(a) $y (0) = 2$ , $y (π /8) = 3$ (b) $y (0) = 2$ , $y (π /4) = - 2$ (c) $y (0) = 2$ , $y (π /4) = 5$

Solution

The left end, all cases. $y (0) = c_{1} cos 0 + c_{2} sin 0 = c_{1}$ . So $c_{1} = 2$ in every case, and $y = 2 cos 4 x + c_{2} sin 4 x$ .
(a) At $x = π /8$ : $4 x = π /2$ , $cos (π /2) = 0$ , $sin (π /2) = 1$ . So $y (π /8) = c_{2} = 3$ . Exactly one solution: $y = 2 cos 4 x + 3 sin 4 x$ .
(b) At $x = π /4$ : $4 x = π$ , $cos π = - 1$ , $sin π = 0$ . So $y (π /4) = - 2 + 0 \cdot c_{2} = - 2$ , whatever $c_{2}$ is. The condition is automatically satisfied. Infinitely many solutions: $y = 2 cos 4 x + c_{2} sin 4 x$ for every real $c_{2}$ .
(c) As in (b), $y (π /4) = - 2$ for every $c_{2}$ . It can never equal $5$ . No solution.

Why the difference?

Every solution of $y^{''} + 16 y = 0$ with $y (0) = 2$ is forced to come back to the value $- 2$ at $x = π /4$ , because $sin 4 x$ vanishes there. Prescribing the value at $x = π /4$ therefore either repeats information we already have, as in (b), or contradicts it, as in (c). In (a) the second point is "generic" and pins down $c_{2}$ . BVPs like these are the starting point for vibration modes and eigenvalue problems, which you will meet in later courses.

Figure 1.3.2. The three BVPs of Example 1.3.5. (a) Exactly one curve of the family passes through both

(0, 2)

and

(π /8, 3)

. (b) Every curve

y = 2 cos 4 x + c_{2} sin 4 x

(five are shown) passes through

(π /4, - 2)

. (c) For the same reason, none of them can reach

(π /4, 5)

Existence and uniqueness

Before trying to solve a problem, it is worth asking whether it can be solved. Example 1.3.5(c) shows that a BVP may have no solution. What about IVPs? Two questions arise.

Existence: does the IVP have at least one solution?
Uniqueness: does it have at most one solution?

For engineers these are not abstract questions. If a model has no solution, it is wrong. If it has several, the model does not predict what the system will do. In both cases a computer will still happily print numbers, which makes the questions even more important.

The following example shows that uniqueness can really fail, even for a harmless-looking first-order equation.

Example 1.3.6 (Infinitely many solutions)

Consider the IVP

y^{'} = 5 ∣ y ∣^{4/5}, y (0) = 0.

(Note that $∣ y ∣^{4/5} = 5 y^{4}$ , so the right-hand side is defined and continuous for every real $y$ .) Show that the IVP has infinitely many solutions.

Solution

The zero solution. $y \equiv 0$ gives $y^{'} = 0$ and $5∣0 ∣^{4/5} = 0$ . It satisfies $y (0) = 0$ . So $y \equiv 0$ is a solution.
A second solution. Try $y = x^{5}$ . Then $y^{'} = 5 x^{4}$ , and $5∣ x^{5} ∣^{4/5} = 5∣ x ∣^{4} = 5 x^{4} .$ The two sides agree for every $x$ , and $y (0) = 0$ . So $y = x^{5}$ is a different solution of the same IVP.
A whole family through the origin. Pick any two numbers $a \leq 0 \leq b$ and define $φ (x) = ⎩ ⎨ ⎧ (x - a)^{5}, 0, (x - b)^{5}, x < a, a \leq x \leq b, x > b .$ The same computation as in Step 2 (with $x$ replaced by $x - a$ or $x - b$ ) shows that each outer piece satisfies the ODE. The middle piece is the zero solution.
Do the pieces join smoothly? At $x = b$ the right piece has value $(b - b)^{5} = 0$ and derivative $5 (b - b)^{4} = 0$ , matching the zero piece. The same holds at $x = a$ . So $φ$ is differentiable everywhere and satisfies the ODE at every $x$ , including the joining points (where both sides equal $0$ ).
Conclusion. Since $φ (0) = 0$ , every choice of $a \leq 0 \leq b$ gives a solution of the IVP. There are infinitely many. Picture a curve resting on the $x$ -axis that can "take off" upward at any moment $b \geq 0$ and, going back in time, leave from below at any moment $a \leq 0$ .

Figure 1.3.3. Five of the infinitely many solutions of

y^{'} = 5∣ y ∣^{4/5}

y (0) = 0

: the zero solution,

y = x^{5}

, and three functions

φ

that rest on the

x

-axis for

a \leq x \leq b

and leave it smoothly on either side. All of them pass through the origin.

What goes wrong here? The function $f (y) = 5∣ y ∣^{4/5}$ is continuous, but it is very steep near $y = 0$ : its derivative $4 y /∣ y ∣^{6/5}$ (which equals $4 y^{- 1/5}$ for $y > 0$ ) becomes infinite as $y \to 0$ . The next theorem shows that this steepness is exactly what we must rule out.

Theorem 1.3.1 (Existence and uniqueness, first order)

Consider the IVP

y^{'} = f (x, y), y (x_{0}) = y_{0} .

Suppose that

$f (x, y)$ is continuous on an open rectangle $R = {(x, y) : a < x < b, c < y < d}$ ,
the partial derivative $\frac{\partial f}{\partial y} (x, y)$ is also continuous on $R$ , and
the initial point $(x_{0}, y_{0})$ lies in $R$ .

Then there is a number $h > 0$ such that the IVP has a solution $y = φ (x)$ defined on the interval $x_{0} - h < x < x_{0} + h$ , and this solution is unique: any solution of the IVP defined on an interval containing $x_{0}$ agrees with $φ$ wherever both are defined within $x_{0} - h < x < x_{0} + h$ .

Every part of this statement matters. Here is how to use it, and what it does and does not promise.

What you must check. Look at $f$ and $\partial f / \partial y$ near the starting point. If both are continuous on some open rectangle with $(x_{0}, y_{0})$ inside it, the theorem applies. The rectangle can be tiny.
What you get. In the direction field, one integral curve, and only one, goes through $(x_{0}, y_{0})$ , at least for $x$ close to $x_{0}$ .
What you do not get. A formula. The theorem is a guarantee, not a recipe: a search for the solution, by hand or by computer, has a target, and whatever solution you find is the one.
It is local. The number $h$ may be small, and the theorem does not say how small. In Example 1.3.4(a), $f = 2 x y^{2}$ and $\partial f / \partial y = 4 x y$ are continuous on the whole plane, yet the solution $y = 1/ (4 - x^{2})$ through $(1, \frac{1}{3})$ only exists on $(- 2, 2)$ : it blows up at $x = 2$ . Nice functions $f$ do not guarantee solutions that live forever.
Sufficient, not necessary. If the hypotheses fail, the theorem says nothing. The IVP might still have exactly one solution, or several, or none. Failure of the hypotheses is a warning sign, not a verdict.

Figure 1.3.4. Theorem 1.3.1 for Example 1.3.4(a). Since

f = 2 x y^{2}

and

\partial f / \partial y = 4 x y

are continuous everywhere, any rectangle

R

around

(1, \frac{1}{3})

will do. The theorem promises a unique solution on some interval

x_{0} - h < x < x_{0} + h

(short bar), without saying how large

h

is. The actual solution

y = 1/ (4 - x^{2})

lives on

(- 2, 2)

and blows up at both ends.

Existence versus uniqueness

A finer result (Peano's theorem) shows that continuity of $f$ alone already guarantees that at least one solution exists. The extra hypothesis on $\partial f / \partial y$ is what buys uniqueness. Example 1.3.6 fits this picture perfectly: $f = 5∣ y ∣^{4/5}$ is continuous, so solutions exist, but $\partial f / \partial y$ is not continuous at $y = 0$ , and uniqueness is lost.

Method in a nutshell: applying Theorem 1.3.1

Write the ODE in normal form $y^{'} = f (x, y)$ and read off $f$ .
Compute $\partial f / \partial y$ .
Find the set of points where both $f$ and $\partial f / \partial y$ are continuous.
If $(x_{0}, y_{0})$ lies inside that set (so that a small open rectangle around it fits inside), the IVP has a unique solution near $x_{0}$ . If not, the theorem gives no information.

Example 1.3.7 (The theorem applies everywhere)

Show that the IVP $y^{'} = x sin y + e^{x} y$ , $y (x_{0}) = y_{0}$ , has a unique solution near $x_{0}$ for every initial point $(x_{0}, y_{0})$ .

Solution

Read off $f$ . $f (x, y) = x sin y + e^{x} y$ .
Partial derivative. Treat $x$ as a constant: $\frac{\partial f}{\partial y} = x cos y + e^{x}$ .
Continuity. Both are built from $x$ , $sin y$ , $cos y$ , $e^{x}$ and $y$ by sums and products, so both are continuous on the whole plane. Any initial point lies inside some open rectangle (for example, a square of side $2$ centred at it).
Conclusion. By Theorem 1.3.1 there is exactly one solution through each point, on some interval around $x_{0}$ . We have no formula for it, but we know it is there and that it is the only one. A numerical method (Chapter 8) can now be used with confidence.

Example 1.3.8 (Same equation, different points)

For $y^{'} = 5∣ y ∣^{4/5}$ from Example 1.3.6, decide where Theorem 1.3.1 applies. Then solve the IVP with $y (1) = 1$ .

Solution

Continuity of $f$ . $f (x, y) = 5∣ y ∣^{4/5}$ is continuous on the whole plane.
The partial derivative. For $y > 0$ , $f = 5 y^{4/5}$ and $\frac{\partial f}{\partial y} = 4 y^{- 1/5}$ . For $y < 0$ , $f = 5 (- y)^{4/5}$ and $\frac{\partial f}{\partial y} = - 4 (- y)^{- 1/5}$ . Both are continuous away from $y = 0$ , but both grow without bound as $y \to 0$ ; at $y = 0$ the partial derivative does not exist.
Where the theorem applies. At any point with $y_{0} \neq = 0$ we can choose a rectangle that stays away from the $x$ -axis, so the IVP has a unique solution near $x_{0}$ . At points with $y_{0} = 0$ , no rectangle around the point avoids the $x$ -axis, so the theorem says nothing. (Example 1.3.6 showed that uniqueness really fails there.)
The IVP $y (1) = 1$ . From Example 1.3.6, $y = x^{5}$ solves the ODE, and $1^{5} = 1$ . By the theorem it is the only solution near $x = 1$ ; for instance on $(0, 2)$ , where it stays positive.
A subtle point. The function $y = x^{5}$ reaches the $x$ -axis at $x = 0$ . A solution of this IVP could follow $x^{5}$ for $x > 0$ but switch to the zero solution at the origin and stay there for a while (as in Example 1.3.6). Uniqueness is guaranteed only near $x_{0}$ , not forever.

Example 1.3.9 (Finding the good region)

For $y^{'} = \frac{y}{x - 4}$ , find all points $(x_{0}, y_{0})$ for which Theorem 1.3.1 guarantees a unique solution. Then show that uniqueness actually fails at the point $(5, 0)$ .

Solution

Where is $f$ continuous? $f (x, y) = y / (x - 4)$ needs $y \geq 0$ (square root) and $x \neq = 4$ (denominator).
Partial derivative. For $y > 0$ and $x \neq = 4$ , $\frac{\partial f}{\partial y} = \frac{1}{x - 4} \cdot \frac{1}{2 y} = \frac{1}{2 y ( x - 4 )},$ which is continuous there and undefined at $y = 0$ .
The region. An open rectangle on which both functions are continuous exists around $(x_{0}, y_{0})$ exactly when $y_{0} > 0$ and $x_{0} \neq = 4$ . So the theorem guarantees a unique solution at every point of the upper half-plane except on the vertical line $x = 4$ . It says nothing on the $x$ -axis or on the line $x = 4$ .
Uniqueness fails at $(5, 0)$ . Clearly $y_{1} \equiv 0$ is a solution with $y_{1} (5) = 0$ . Now consider, on $4 < x < \infty$ , $y_{2} (x) = {0, \frac{1}{4} (ln (x - 4))^{2}, 4 < x \leq 5, x > 5.$ For $x > 5$ we have $ln (x - 4) > 0$ , so $y_{2} = \frac{1}{2} ln (x - 4)$ and $y_{2}^{'} = \frac{1}{4} \cdot 2 ln (x - 4) \cdot \frac{1}{x - 4} = \frac{ln ( x - 4 )}{2 ( x - 4 )} = \frac{y _{2}}{x - 4} . ✓$ At $x = 5$ both pieces have value $0$ and slope $0$ (since $ln 1 = 0$ ), so they join smoothly. Thus $y_{2}$ is a second solution through $(5, 0)$ .

Figure 1.3.5. Example 1.3.9. In the shaded upper half-plane, away from the line

x = 4

, the theorem guarantees a unique solution. On the

x

-axis it says nothing, and indeed two solutions pass through

(5, 0)

y_{1} \equiv 0

, and

y_{2}

, which stays at

0

up to

x = 5

and then rises as

\frac{1}{4} [ln (x - 4)]^{2}

Try the theorem yourself. The checker below examines the hypotheses at a given initial point. It starts with Example 1.3.6 at the origin; move the point to $(0, 1)$ , or enter $f$ from Example 1.3.9, and see how the verdict changes.

The direction field of the same equation makes the loss of uniqueness visible. Click on the $x$ -axis and at nearby points: integral curves above and below the axis flatten out as they approach it, and can merge into the zero solution.

The linear case: a stronger theorem

For linear equations of any order, the situation is much better. We state the result now and use it throughout Chapter 4.

Theorem 1.3.2 (Existence and uniqueness, linear equations of order n)

Consider the linear IVP

a_{0} (x) y^{(n)} + a_{1} (x) y^{(n - 1)} + \dots + a_{n - 1} (x) y^{'} + a_{n} (x) y = b (x), y (x_{0}) = y_{0}, y^{'} (x_{0}) = y_{1}, \dots, y^{(n - 1)} (x_{0}) = y_{n - 1} .

Suppose that $a_{0}, a_{1}, \dots, a_{n}$ and $b$ are continuous on an interval $I$ , that $a_{0} (x) \neq = 0$ for every $x$ in $I$ , and that $x_{0}$ is a point of $I$ . Then for any real numbers $y_{0}, y_{1}, \dots, y_{n - 1}$ the IVP has exactly one solution, and this solution is defined on the whole interval $I$ .

Compare this with Theorem 1.3.1. There, the solution was only guaranteed near $x_{0}$ . Here, the solution exists on all of $I$ , wherever the coefficients behave well. The dangers are now easy to spot: points where some coefficient or the right-hand side is discontinuous, and points where the leading coefficient $a_{0} (x)$ vanishes.

A useful consequence: if $b \equiv 0$ and all initial values are zero, then $y \equiv 0$ is obviously a solution, so by uniqueness it is the only one.

Example 1.3.10

Find the largest interval on which Theorem 1.3.2 guarantees a unique solution of

(x - 5) y^{''} + x y^{'} + y = \frac{1}{x - 2}, y (3) = 1, y^{'} (3) = 0.

Solution

Identify the pieces. $a_{0} (x) = x - 5$ , $a_{1} (x) = x$ , $a_{2} (x) = 1$ , $b (x) = \frac{1}{x - 2}$ , and $x_{0} = 3$ .
List the trouble spots.
- $a_{0} (x) = 0$ at $x = 5$ .
- $x$ is continuous only for $x \geq 0$ .
- $b (x)$ is discontinuous at $x = 2$ .
Choose the interval. Mark $0$ , $2$ and $5$ on the number line. The initial point $x_{0} = 3$ lies between $2$ and $5$ . On $(2, 5)$ every coefficient is continuous and $a_{0} \neq = 0$ .
Answer. The IVP has exactly one solution on $(2, 5)$ . If the initial point were $x_{0} = 6$ instead, the interval would be $(5, \infty)$ .

Figure 1.3.6. Example 1.3.10 on the number line. The trouble spots

0

2

and

5

cut the line into pieces, and the guaranteed interval is the piece that contains

x_{0}

(2, 5)

for

x_{0} = 3

, and

(5, \infty)

for

x_{0} = 6

Exercises

Solve the IVP $y^{'} = 3 x^{2} + 2$ , $y (0) = - 1$ .
Answer
$y = x^{3} + 2 x - 1$
Every solution of $y^{'} + 2 y = 6$ has the form $y = c e^{- 2 x} + 3$ . Solve the IVPs (a) $y (0) = 7$ , (b) $y (1) = 3$ .
Answer
(a) $y = 4 e^{- 2 x} + 3$ ; (b) $y \equiv 3$
Every solution of $y^{''} - y^{'} - 20 y = 0$ has the form $y = c_{1} e^{5 x} + c_{2} e^{- 4 x}$ . Solve the IVP $y (0) = 2$ , $y^{'} (0) = 1$ .
Answer
$y = e^{5 x} + e^{- 4 x}$
Every solution of $x^{2} y^{''} - 5 x y^{'} + 5 y = 0$ on $(0, \infty)$ has the form $y = c_{1} x + c_{2} x^{5}$ . Solve the IVP $y (1) = 2$ , $y^{'} (1) = - 2$ , and state the interval that Theorem 1.3.2 guarantees.
Answer
$y = 3 x - x^{5}$ ; interval $(0, \infty)$ (since $a_{0} = x^{2}$ vanishes at $0$ ).
Every solution of $y^{''} + 25 y = 0$ has the form $y = c_{1} cos 5 x + c_{2} sin 5 x$ . Solve, or show there is no solution: (a) $y (0) = 1$ , $y (π /10) = 2$ ; (b) $y (0) = 1$ , $y (π /5) = - 1$ ; (c) $y (0) = 1$ , $y (π /5) = 4$ .
Answer
(a) $y = cos 5 x + 2 sin 5 x$ ; (b) $y = cos 5 x + c sin 5 x$ for every $c$ (infinitely many); (c) no solution.
Solve the BVP $y^{''} + 25 y = 0$ , $y^{'} (0) = 10$ , $y (π /20) = 22$ .
Answer
$y = 2 cos 5 x + 2 sin 5 x$
Does Theorem 1.3.1 guarantee a unique solution near $x_{0}$ ? (a) $y^{'} = y^{2} sin x + e^{x}$ , $y (0) = 3$ ; (b) $y^{'} = \frac{x + y}{x + 3}$ , $y (- 3) = 1$ ; (c) $y^{'} = y - 2$ , $y (0) = 2$ ; (d) $y^{'} = y - 2$ , $y (0) = 6$ .
Answer
(a) yes; (b) no guarantee ( $f$ undefined on $x = - 3$ ); (c) no guarantee ( $f$ undefined for $y < 2$ , $\partial f / \partial y$ unbounded near $y = 2$ ); (d) yes.
Find all points $(x_{0}, y_{0})$ at which Theorem 1.3.1 guarantees a unique solution of $y^{'} = \frac{ln y}{x ^{2} - 1}$ .
Answer
All points with $y_{0} > 0$ and $x_{0} \neq = \pm 1$ .
Show that $y \equiv 0$ and $y = {0, x^{4}, x \leq 0 x > 0$ are both solutions of $y^{'} = 4∣ y ∣^{3/4}$ , $y (0) = 0$ . Why does this not contradict Theorem 1.3.1?
Answer
For $x > 0$ : $y^{'} = 4 x^{3} = 4 (x^{4})^{3/4}$ ; the pieces join with value and slope $0$ . $\partial f / \partial y = 3 y /∣ y ∣^{5/4}$ is not continuous (it is undefined) at $y = 0$ , so the theorem does not apply at $(0, 0)$ .
Find the largest interval on which Theorem 1.3.2 guarantees a unique solution of $(x^{2} - 9) y^{''} + \frac{1}{x} y^{'} + y = ln (x + 5)$ with (a) $y (1) = 0$ , $y^{'} (1) = 2$ ; (b) $y (- 4) = 1$ , $y^{'} (- 4) = 0$ .
Answer
(a) $(0, 3)$ ; (b) $(- 5, - 3)$
Explain why $y \equiv 0$ is the only solution of $y^{''} + x^{2} y^{'} + e^{x} y = 0$ , $y (0) = 0$ , $y^{'} (0) = 0$ .
Answer
$y \equiv 0$ is a solution; the coefficients are continuous on $(- \infty, \infty)$ and $a_{0} = 1 \neq = 0$ , so by Theorem 1.3.2 the solution is unique.
Verify that $y = \frac{2}{1 - 2 x}$ solves $y^{'} = y^{2}$ , $y (0) = 2$ . On what interval does it exist? Is this consistent with Theorem 1.3.1?
Answer
$y^{'} = 4/ (1 - 2 x)^{2} = y^{2}$ ; interval $(- \infty, \frac{1}{2})$ . Yes: the theorem only promises a solution on some interval around $x_{0} = 0$ .