In the study of ordinary differential equations (ODEs), we've learned how to solve equations analytically—especially when the equations are linear or separable. These methods involve transforming the equation into a simpler form and then solving it. However, not all ODEs are so easily managed. Sometimes, the equations are too sophisticated, or the solution methods give us answers that are difficult to interpret. This is where other approaches, like graphical and numerical methods, come into play.

It's important to understand the concept of well-posed problems and the existence and uniqueness of solutions.

1. What Makes a Problem Well-Posed?

In science and engineering, a well-posed problem is essential because it allows us to predict or explain something reliably. A problem is considered well-posed if it meets three criteria:

1. The problem has a solution.

Without a solution, we can't predict anything.

2. The solution is unique.

If there are too many solutions, it’s hard to know which one to use.

3. The solution depends continuously on the problem.

This means that if we make small changes to the initial conditions or the coefficients in our equation, the solution shouldn't change drastically. This is crucial because real-world models aren't perfect, and small errors or changes should lead to small differences in the outcome, not big ones.

When a problem is well-posed, it’s easier to approximate its solution using various techniques. But if it’s not well-posed, finding a reliable solution becomes very difficult, if not impossible.

2. Existence and Uniqueness of Solutions

Let's focus on the existence and uniqueness of solutions to initial-value problems, which are common in differential equations. An initial-value problem involves solving a differential equation with a given starting value.

Consider the general initial-value problem:

\frac{dy}{dt} = f(t, y), \quad y(t_0) = y_0

Here, $y(t)$ is the solution we’re looking for, and $y(t_0) = y_0$ gives us the initial condition, meaning the value of $y$ at the starting time $t_0$ .

2.1. The Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem is a key result that tells us when a solution exists and whether it's unique. Here’s the theorem in simple terms:

If the function $f(t, y)$ meets these conditions:

- Continuous over a certain region in the $t$ - $y$ plane.

- Differentiable with respect to $y$ in that region.

- The derivative with respect to $y$ , $\frac{\partial f}{\partial y}$ , is continuous over that region.

Then, for any initial point $(t_0, y_0)$ in that region, there exists a unique solution $y(t)$ that works for some time interval around $t_0$ .

This means that if our function $f(t, y)$ behaves nicely (i.e., it’s continuous and differentiable in the right way), we’re guaranteed to find one and only one solution for our problem, at least for a short time after $t_0$ .

2.2. Examples to Illustrate the Theorem

Example 1: A Non-Solvable ODE

Consider the equation:

\frac{dy}{dt} = \frac{\log(|y|)}{1 + t^2 - y^2}

with the initial condition $y(t_0) = y_0$ .

Here, $f(t, y) = \frac{\log(|y|)}{1 + t^2 - y^2}$ is defined everywhere except where $1 + t^2 - y^2 = 0$ (which corresponds to $y = \pm\sqrt{1 + t^2}$ ) or $y = 0$ . Therefore, we can define a region $S$ in the $t$ - $y$ plane that excludes these problematic points. Within this region, $f(t, y)$ is continuous, differentiable, and the partial derivative with respect to $y$ , $\frac{\partial f}{\partial y}$ , is also continuous.

Since these conditions are met, the theorem guarantees that for any initial point $(t_0, y_0)$ within this region, there exists a unique solution $y(t)$ that is valid for some time interval around $t_0$ .

1. Boundary curves and regions:

- The curves $y = 0$ , $y = -\sqrt{1 + t^2}$ , and $y = \sqrt{1 + t^2}$ divide the $(t, y)$ -plane into three distinct regions.

2. Region-restricted solutions:

- The solutions are confined to the specific region in which the initial value lies. That is, solutions starting in a region remain in that region.

3. Solution boundaries:

- Solutions cannot cross or extend beyond the boundaries defined by $y = \pm\sqrt{1 + t^2}$ . These boundary curves act as natural limits, and the solution will stop if it reaches a boundary.

4. Initial value dependence:

- The behavior and trajectory of the solution are uniquely determined by the initial value. Depending on where the initial condition falls, the solution remains confined to its respective region.

By breaking the plane into these regions, the behavior of solutions becomes predictable and determined by the starting point, with clear boundaries that cannot be crossed.

Example 2: A Function Defined Everywhere

Now consider:

\frac{dy}{dt} = \sqrt{1 + t^2 + y^2}

with the initial condition $y(t_0) = y_0$ .

In this case, $f(t, y) = \sqrt{1 + t^2 + y^2}$ is defined and continuous everywhere in the $t$ - $y$ plane. The partial derivative $\frac{\partial f}{\partial y} = \frac{y}{\sqrt{1 + t^2 + y^2}}$ is also continuous everywhere.

Since the function meets all the criteria across the entire plane, the theorem tells us that there’s a unique solution for any initial point $(t_0, y_0)$ , and this solution exists for all time.

2.3. Applying the Theorem to Specific Cases

Example 1: Linear ODE

Consider a linear ODE:

\frac{dy}{dt} = f(t) - a(t)y, \quad y(t_0) = y_0

where $a(t)$ and $f(t)$ are continuous functions.

Here, $f(t) - a(t)y$ is continuous and differentiable with respect to $y$ , and $\frac{\partial f}{\partial y} = -a(t)$ is continuous. According to the theorem, there exists a unique solution for this ODE, which is defined for the largest possible interval around $t_0$ .

Example 2: Separable ODE

Now consider a separable ODE:

\frac{dy}{dt} = f(t)g(y), \quad y(t_0) = y_0

where $f(t)$ is continuous over some interval, and $g(y)$ is differentiable with a continuous derivative $g'(y)$ .

In this case, the function $f(t)g(y)$ is continuous, and its derivative with respect to $y$ is also continuous. Therefore, the theorem assures us that there is a unique solution around the initial point $(t_0, y_0)$ .

2.4. Examples:

Example 1: Simple Linear ODE

Problem: Solve the initial-value problem

\frac{dy}{dt} = 2y, \quad y(0) = 1

Solution:

1. Identify the function: Here, $f(t, y) = 2y$ .

2. Check continuity: $f(t, y)$ is continuous and differentiable with respect to $y$ over the entire $t$ - $y$ plane. Also, the partial derivative with respect to $y$ is $\frac{\partial f}{\partial y} = 2$ , which is constant and continuous.

3. Apply the Existence and Uniqueness Theorem: The theorem applies, guaranteeing a unique solution for all $t$ .

4. Solve the equation: The ODE is separable:

\frac{dy}{y} = 2 \, dt

Integrating both sides:

\ln|y| = 2t + C

Exponentiate to solve for $y$ :

y(t) = Ce^{2t}

Use the initial condition $y(0) = 1$ to find $C$ :

1 = Ce^0 \quad \Rightarrow \quad C = 1

Final solution:

y(t) = e^{2t}

Example 2: Constant Function ODE

Problem: Solve the initial-value problem

\frac{dy}{dt} = 0, \quad y(0) = 5

Solution:

1. Identify the function: Here, $f(t, y) = 0$ .

2. Check continuity: The function is constant, so it's continuous and differentiable everywhere.

3. Apply the Existence and Uniqueness Theorem: Since $f(t, y) = 0$ satisfies the theorem's conditions, the solution is guaranteed to be unique.

4. Solve the equation: Integrate both sides:

y(t) = C

Use the initial condition $y(0) = 5$ to find $C = 5$ .

Final solution:

y(t) = 5

Example 3: Nonlinear Separable ODE

Problem: Solve the initial-value problem

\frac{dy}{dt} = y^2, \quad y(1) = 1

Solution:

1. Identify the function: Here, $f(t, y) = y^2$ .

2. Check continuity: The function $y^2$ is continuous and differentiable everywhere.

3. Apply the Existence and Uniqueness Theorem: Since $f(t, y) = y^2$ and its derivative $\frac{\partial f}{\partial y} = 2y$ are continuous, the theorem ensures a unique solution around $t = 1$ .

4. Solve the equation: The ODE is separable:

\frac{dy}{y^2} = dt

Integrate both sides:

-\frac{1}{y} = t + C

Solve for $y$ :

y(t) = -\frac{1}{t + C}

Use the initial condition $y(1) = 1$ to find $C$ :

1 = -\frac{1}{1 + C} \quad \Rightarrow \quad C = -2

Final solution:

y(t) = -\frac{1}{t - 2}

Now, what if the initial value was $y(1) = 0$ ? Then solving for $C$ using the initial value would lead to

0 = -\frac{1}{1 + C},

which cannot be solved. However, we know that for every $t$ and any initial value $y(1)$ , we should have a unique solution. Now, looking at the IVP $\frac{dy}{dt} = y^2$ with $y(1) = 0$ , it is easy to determine that $y(t) = 0$ is a solution to the IVP, which would then be the unique solution. So this theorem is very useful in this sense.

Example 4: Existence but No Uniqueness

Problem: Consider the initial-value problem

\frac{dy}{dt} = y^{1/3}, \quad y(0) = 0

Solution:

1. Identify the function: Here, $f(t, y) = y^{1/3}$ .

2. Check continuity: The function is continuous, but its derivative $\frac{\partial f}{\partial y} = \frac{1}{3}y^{-2/3}$ is not continuous at $y = 0$ .

3. Apply the Existence and Uniqueness Theorem: The theorem does not apply because the derivative is not continuous. This means that even though a solution exists, it might not be unique.

4. Find possible solutions: The equation is separable:

\frac{dy}{y^{1/3}} = dt

Integrate both sides:

\frac{3}{2}y^{2/3} = t + C

Solve for $y$ :

y(t) = \left(\frac{2}{3}(t + C)\right)^{3/2}

Using $y(0) = 0$ , $C$ can be 0. This leads to $y(t) = \left(\frac{2}{3}t\right)^{3/2}$ , but $y(t) = 0$ is also a solution. Thus, the solution is not unique.

Example 5: Initial Condition at a Point of Discontinuity

Problem: Solve the initial-value problem

\frac{dy}{dt} = \frac{1}{t}, \quad y(0) = 0

Solution:

1. Identify the function: Here, $f(t, y) = \frac{1}{t}$ .

2. Check continuity: The function $f(t, y) = \frac{1}{t}$ is not defined at $t = 0$ .

3. Apply the Existence and Uniqueness Theorem: The theorem cannot be applied because $f(t, y)$ is not continuous at $t = 0$ . Therefore, no solution exists that satisfies the initial condition $y(0) = 0$ .

In fact, by giving initial values at t = 0, the equation cannot be solved.

Example 6: First-Order Linear ODE

Problem: Solve the initial-value problem

\frac{dy}{dt} + 3y = 6, \quad y(0) = 2

Solution:

1. Identify the function: Here, $f(t, y) = 6 - 3y$ .

2. Check continuity: The function is linear, so it’s continuous and differentiable everywhere.

3. Apply the Existence and Uniqueness Theorem: Since $f(t, y) = 6 - 3y$ and its partial derivative $\frac{\partial f}{\partial y} = -3$ are continuous, the theorem applies, ensuring a unique solution.

4. Solve the equation: The ODE can be solved using an integrating factor:

\text{Integrating factor} = e^{\int 3 \, dt} = e^{3t}

Multiply both sides by the integrating factor:

\frac{d}{dt}(e^{3t}y) = 6e^{3t}

Integrate both sides:

e^{3t}y = 2e^{3t} + C

Solve for $y$ :

y(t) = 2 + Ce^{-3t}

Use the initial condition $y(0) = 2$ to find $C = 0$ .

Final solution:

y(t) = 2

Example 7: Separable ODE with Trigonometric Function

Problem: Solve the initial-value problem

\frac{dy}{dt} = \cos(t) \cdot y, \quad y\left(\frac{\pi}{2}\right) = 1

Solution:

1. Identify the function: Here, $f(t, y) = \cos(t) \cdot y$ .

2. Check continuity: The function is continuous and differentiable everywhere.

3. Apply the Existence and Uniqueness Theorem: The theorem applies, ensuring a unique solution.

4. Solve the equation: The ODE is separable:

\frac{dy}{y} = \cos(t) \, dt

Integrate both sides:

\ln|y| = \sin(t) + C

Exponentiate to solve for $y$ :

y(t) = Ce^{\sin(t)}

Use the initial condition $y\left(\frac{\pi}{2}\right) = 1$ to find $C = e^{-1}$ .

Final solution:

y(t) = e^{\sin(t) - 1}

Example 8: Nonlinear ODE with Quadratic Function

To solve the initial-value problem

\frac{dy}{dt} = t^2 + y^2, \quad y(0) = 0,

Let's follow these steps and explore the solution in more detail.

1. Identify the Function

The differential equation is:

\frac{dy}{dt} = t^2 + y^2.

2. Check Continuity

The function $f(t, y) = t^2 + y^2$ is continuous and differentiable everywhere in $t$ and $y$ . Thus, the conditions for applying the Existence and Uniqueness Theorem are satisfied.

3. Apply the Existence and Uniqueness Theorem

Since $f(t, y) = t^2 + y^2$ is continuous and its partial derivative with respect to $y$ , which is $\frac{\partial f}{\partial y} = 2y$ , is also continuous, the Existence and Uniqueness Theorem guarantees that there is a unique solution to this initial-value problem.

4. Solve the Equation

To solve the differential equation, we use the substitution method. Let’s try the substitution $y = ut$ where $u$ is a function of $t$ .

This differential equation is nonlinear and solving it analytically is quite complex. We can solve it numerically or use a software tool for exact solutions.

Example 9: Nonlinear ODE with an Absolute Value Function

\frac{dy}{dt} = |y|, \quad y(0) = 0,

1. Identify the Function

Here, the differential equation is

\frac{dy}{dt} = |y|.

2. Check Continuity

The function $f(t, y) = |y|$ is continuous everywhere. However, its derivative with respect to $y$ , which is the function $f_y(y) = \frac{d}{dy}|y|$ , is not defined at $y = 0$ . This means that the function $f(t, y)$ does not satisfy the condition of having a continuous derivative with respect to $y$ at $y = 0$ .

3. Apply the Existence and Uniqueness Theorem

The standard Existence and Uniqueness Theorem for ODEs states that if $f(t, y)$ and $\frac{\partial f}{\partial y}$ are continuous in some region containing the initial point, then there exists a unique solution to the ODE. In this case, because $\frac{\partial f}{\partial y}$ is not continuous at $y = 0$ , the theorem does not apply directly for the initial value y(0) = 0. Therefore, we cannot guarantee the uniqueness of the solution using this theorem alone.

4. Find Possible Solutions

We need to find possible solutions for the given initial-value problem. Let's solve the differential equation in different cases:

Case 1: $y(t) > 0$

When $y(t) > 0$ , $|y| = y$ . So, the differential equation becomes:

\frac{dy}{dt} = y.

This is a standard first-order linear differential equation. Its solution is:

y(t) = C e^t,

where $C$ is a constant. Using the initial condition $y(0) = 0$ :

0 = C e^0 \implies C = 0.

Thus, $y(t) = 0$ for $y(t) > 0$ .

Case 2: $y(t) < 0$

When $y(t) < 0$ , $|y| = -y$ . So, the differential equation becomes:

\frac{dy}{dt} = -y.

This is also a standard first-order linear differential equation. Its solution is:

y(t) = C e^{-t}.

Using the initial condition $y(0) = 0$ :

0 = C e^0 \implies C = 0.

Thus, $y(t) = 0$ for $y(t) < 0$ .

In both cases, we found that $y(t) = 0$ satisfies the differential equation and the initial condition. Therefore, the solution to the initial-value problem is:

y(t) = 0.

This is a valid solution. However, because $f(t, y) = |y|$ is not differentiable at $y = 0$ , other non-zero solutions might exist that are not covered by this standard analysis, but $y(t) = 0$ is a solution that satisfies the initial condition and the differential equation.

Example 10: Autonomous ODE with a Quadratic Nonlinearity

Problem: Solve the initial-value problem

\frac{dy}{dt} = y^2 - 4, \quad y(3) = -3

Solution:

1. Identify the function: Here, $f(t, y) = y^2 - 4$ .

2. Check continuity: The function is continuous and differentiable everywhere.

3. Apply the Existence and Uniqueness Theorem: The theorem applies, ensuring a unique solution.

4. Solve the equation: This is separable:

\frac{dy}{y^2 - 4} = dt

Factor the denominator and integrate:

\frac{1}{4} \ln\left|\frac{y-2}{y+2}\right| = t + C

Use the initial condition $y(3) = -3$ to solve for $C$ , leading to a specific form of the solution.

Note that if the initial value was either $y(3) = -2$ or $y(3) = 2$ , then solving for $C$ would not be possible with the derived form of the solution. However, the theorem guarantees that there will be a unique solution for either case.

We notice that $y(t) = -2$ and $y(t) = 2$ satisfy the two cases. By the theorem, we can guarantee that if $y(3) = -2$ , $y(t) = -2$ is the unique solution, and if $y(3) = 2$ , $y(t) = 2$ is the unique solution.

By understanding these concepts and working through examples, you'll gain a strong foundation in determining when and how solutions to differential equations exist and whether they are unique. This knowledge is crucial for solving ODEs, especially when more advanced methods or numerical approaches are needed.

Related Articles

11. Understanding Existence and Uniqueness in Ordinary Differential Equations

1. What Makes a Problem Well-Posed?

2. Existence and Uniqueness of Solutions

2.1. The Existence and Uniqueness Theorem

2.2. Examples to Illustrate the Theorem

2.3. Applying the Theorem to Specific Cases

2.4. Examples: