Worked examples · Sheet 2

Cauchy problems for separable equations

Theory used: Separable equation Method of separation of variables Cauchy problem Local existence and uniqueness

Exercise 1

Solve the Cauchy problem and state the interval of existence of the solution:

{y^{'} = y^{2} y (0) = 1

General solution. The constant $y \equiv 0$ is a solution, but it does not satisfy the initial condition. For $y \neq = 0$ :

\int \frac{d y}{y ^{2}} = \int d x ⟹ - \frac{1}{y} = x + C ⟹ y = - \frac{1}{x + C} .

Initial condition.

y (0) = - \frac{1}{C} = 1 ⟹ C = - 1 ⟹ y = \frac{1}{1 - x} .

Interval of existence. The solution is defined on $(- \infty, 1)$ , the interval containing $x_{0} = 0$ on which the denominator does not vanish.

y = \dfrac{1}{1 - x}

y = \dfrac{1}{1 - x}

The formula

y = \frac{1}{1 - x}

has two branches: the solution of the problem is the branch in color, defined on

(- \infty, 1)

and passing through

(0, 1)

; the dashed branch does not contain the initial point

The function $f (x, y) = y^{2}$ and its partial derivative with respect to $y$ are continuous on the whole plane, so the local existence and uniqueness theorem applies and the solution is unique. The zero solution is discarded immediately, since at $x = 0$ it takes the value $0$ and the initial condition requires the value $1$ . Separating the variables and integrating gives the family $y = - \frac{1}{x + C}$ ; the initial condition determines $C = - 1$ . The formula $\frac{1}{1 - x}$ defines a function on two intervals separated by the point $x = 1$ , but the solution of the Cauchy problem is a function on an interval containing the initial point, therefore only $(- \infty, 1)$ is kept. As $x \to 1^{-}$ the solution tends to $+ \infty$ : the interval of existence is bounded on the right even though $f$ is defined everywhere.

Exercise 2

Solve the Cauchy problem and state the interval of existence of the solution:

⎩ ⎨ ⎧ y^{'} = - \frac{x}{y} y (0) = 2

\int y d y = \int (- x) d x ⟹ \frac{y ^{2}}{2} = - \frac{x ^{2}}{2} + C ⟹ y^{2} = 2 C - x^{2} .

Initial condition. $y (0)^{2} = 2 C = 4$ , so $y^{2} = 4 - x^{2}$ ; since $y (0) = 2 > 0$ the positive root is chosen:

y = 4 - x^{2} .

Interval of existence. The solution is defined on $(- 2, 2)$ .

y = \sqrt{4 - x^{2}}

y = -\sqrt{4 - x^{2}}

y^{2} = 4 - x^{2}

describes the circle with center at the origin and radius

2

: the solution is the upper semicircle (in color), passing through

(0, 2)

, with the endpoints

(- 2, 0)

and

(2, 0)

excluded

The equation is separable with $h (y) = \frac{1}{y}$ , defined for $y \neq = 0$ ; the initial condition assigns the value $2$ , so we work in the region $y > 0$ . Multiplying by $y$ and integrating gives the relation $y^{2} = 2 C - x^{2}$ , which describes a family of circles centered at the origin. The initial condition fixes $2 C = 4$ . Taking the square root produces two functions, one positive and one negative; since the initial value is positive, the solution is the positive root. At the endpoints $x = \pm 2$ the solution equals $0$ and the equation loses meaning, since the denominator $y$ vanishes and the derivative tends to infinity; therefore the interval of existence is the open interval $(- 2, 2)$ .

Exercise 3

Solve the Cauchy problem and state the interval of existence of the solution:

{y^{'} = (2 x + 1) e^{- y} y (0) = 0

\int e^{y} d y = \int (2 x + 1) d x ⟹ e^{y} = x^{2} + x + C .

Initial condition. $e^{0} = C = 1$ , so

e^{y} = x^{2} + x + 1 ⟹ y = ln (x^{2} + x + 1) .

Interval of existence. The discriminant of $x^{2} + x + 1$ is $1 - 4 = - 3 < 0$ , so the polynomial is always positive and the solution is defined on all of $R$ .

The equation is separable with $g (x) = 2 x + 1$ and $h (y) = e^{- y}$ ; the factor $e^{- y}$ never vanishes, so there are no constant solutions. Multiplying by $e^{y}$ and integrating gives $e^{y} = x^{2} + x + C$ , and the initial condition gives $C = 1$ . The logarithm can be applied only where the right-hand side is positive, therefore the interval of existence coincides with the set where $x^{2} + x + 1 > 0$ around the initial point. The discriminant of the trinomial is negative and the coefficient of $x^{2}$ is positive, so the trinomial is positive for every $x$ and the solution is defined on all of $R$ .