Uniqueness and Existence Theorem Continuous Partial Derivatives

The existence and uniqueness theorem for initial value problems of ordinary differential equations implies the condition for the existence of a solution of linear or non-linear initial value problems and ensures the uniqueness of the obtained solution.

Learn Ordinary Differential Equations

Open Rectangle: An open rectangle R is a set of points (x, y) on a plane, such that for any fixed points a, b, c and d

a < x < b and c < y < d

It is called "open" because points a, b, c and d are not included in the region R.

Theorem Statement

Existence Theorem: If f is a continuous function in an open rectangle

R = {(x, y) | a < x < b and c < y < d }

that contains a point (x_o, y_o), then the initial value problem

y' = f(x, y), y(x_o) = y_o

has atleast a solution in some open sub-interval of (a, b) which contains the point x_o.

Uniqueness Theorem: If f and f_y are continuous function in an open rectangle

R = {(x, y) | a < x < b and c < y < d }

that contains a point (x_o, y_o), then the initial value problem

y' = f(x, y), y(x_o) = y_o

has a unique solution on some open sub-interval of (a, b) which contains the point x_o.

Some important points that the existence and uniqueness theorem directly implies:

• It provides information about the existence of the solution to the initial value problem but does not state how to find the solution or find which open interval.
• The existence theorem does not provide information about how many solutions that the initial value problem may have.
• The uniqueness theorem ensures the uniqueness of the solution if the interval (a, b) ≠is not so large.

Solved Examples on Existence and Uniqueness Theorem

Example 1:

Check the existence and uniqueness of the solution for the initial value problem

y' = x – y + 1, y(1) = 2.

Solution:

Given the initial value problem

y' = x – y + 1, y(1) = 2.

where f(x, y) = x – y + 1 and its partial derivative with respect to y, f_y = – 1, which is continuous in every real interval. Hence the existence and uniqueness theorem ensures that in some open interval centred at 1, the solution of the given ODE exists.

Now,

y' + y = x + 1 is a linear differential equation of the form y' + P(x)y = Q(x), where P = 1 and

Q = x + 1.

Hence, I.F. = e^{∫P dx} = e^{∫ 1 . dx} = e^x

The solution is

y e^x = ∫ (x + 1). e^x dx

⇒ y e^x = ∫ x.e^x dx + ∫ e^x dx + C

⇒ y = x + C e^–x

At x = 1 ⇒ y = 2, we get

2 = 1 + C e^–1 ⇒ C = e

Thus, the solution of given ODE is y = x + e ^{1 – x}, which exists for all x ∈ R.

Example 2:

Check the existence and uniqueness of the solution for the initial value problem

y' = y² , y(0) = 1.

Solution:

Given the initial value problem

y' = y² , y(0) = 1.

where f(x, y) = y² and its partial derivative with respect to y, f_y = 2y, which is continuous in every real interval. Hence the existence and uniqueness theorem ensures that in some open interval centred at 0, the solution of the given ODE exists.

Now, separating the variables

y ^{– 2} dy = dx

Integrating both sides, we get

∫ y ^{– 2} dy = ∫ dx + C₁

⇒ – 2/ y = x + C₁

⇒ y = – 2/ (x + C₁)

At x = 0, y = 1

1 = –2/C₁ or C = 1 {Let – 2/C₁ = C}

Thus, the solution of given ODE is y = 1/ (1 – x), which exists for all x ∈ ( – ∞, 1).

Practice Problems

• Check the existence and uniqueness of the solution for the initial value problem

y' = 1 + y² , y(0) = 0.

• Check the existence and uniqueness of the solution for the initial value problem

y' = y^1/3 , y(0) = 0.

Frequently Asked Questions – FAQs

How do you know if an initial value problem has a unique solution?

By the uniqueness theorem, if f and f_y are continuous functions in an open rectangle R
that contains a point (x_o, y_o), then for the initial value problem y' = f(x, y), y(x_o) = y_ohas a unique solution on some open sub-interval of (a, b) which contains the point x_o.

What is the condition for the existence of a solution for an initial value problem?

By the existence theorem, if f is a continuous function in an open rectangle R that contains a point (x_o, y_o), then the initial value problem y' = f(x, y), y(x_o) = y_o has atleast a solution in some open sub-interval of (a, b) which contains the point x_o.

What is meant by an open rectangle?

An open rectangle R is a set of points (x, y) on a plane, such that for any fixed points a, b, c and d, and a

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