In the previous sections, we have focused on initial value problems. This naturally raises the question: does every initial value problem admit a solution for a given ordinary differential equation, and if a solution exists, is it necessarily unique?
To illustrate this issue, consider the differential equation
$$ \frac{dy}{dx} = \frac{1}{x}. $$
Since the equation is separable, we may write
$$ \int 1, dy = \int \frac{1}{x}, dx, $$
which yields the general solution
$$ y = \ln |x| + C. $$
Now suppose we are given the initial condition $y(0) = 1$. This initial condition cannot be satisfied by the general solution, since the function $\ln |x|$ is undefined at $x = 0$. Consequently, although the differential equation itself is well defined for $x \neq 0$, there exists no solution that satisfies the given initial condition.
This example demonstrates that an initial value problem may fail to have a solution if the initial condition lies outside the domain of definition of the differential equation or its solutions.
Consider the Initial Value Problem (IVP):
$$ \begin{aligned} \frac{dy}{dt} &= F(t, y) \ y(t_0) &= y_0 \end{aligned} $$
Theorem Statement Suppose that $F(t, y)$ and $\frac{\partial F}{\partial y}$ are continuous on a rectangle containing the point $(t_0, y_0)$.
Then, the IVP has a unique solution near $(t_0, y_0) \in \mathbb{R}^2$.
Geometric Interpretation: If we consider a rectangle in the $ty$-plane containing $(t_0, y_0)$, and if the continuity conditions are met within this rectangle, we are guaranteed a unique solution curve passing through $(t_0, y_0)$ within some interval near $t_0$.
Linear vs Nonlinear ODEs and the Existence-Uniqueness Theorem
Example 1: Linear ODE
Problem: Find the largest interval on which the initial value problem
$$ (4-t^2)y^{\prime} - 2ty = \arctan t, \quad y(-1) = 3 $$
has a unique solution.
Solution:
First, rewrite the differential equation in the standard linear form $y^{\prime} + p(t)y = q(t)$:
$$ y^{\prime} - \frac{2t}{4-t^2}y = \frac{\arctan t}{4-t^2} $$
Here, we identify:
$$ p(t) = -\frac{2t}{4-t^2} \quad \text{and} \quad q(t) = \frac{\arctan t}{4-t^2} $$
According to the Existence-Uniqueness Theorem for linear equations, we need to find where $p(t)$ and $q(t)$ are continuous. The term $(4-t^2)$ appears in the denominator, so we must ensure it is not zero:
$$ 4-t^2 \neq 0 \implies t^2 \neq 4 \implies t \neq \pm 2 $$
The domain is divided into three intervals:
$$ (-\infty, -2) \cup (-2, 2) \cup (2, \infty) $$
We are given the initial condition $y(-1) = 3$. The initial point is $t_0 = -1$. We must choose the interval that contains $t_0 = -1$.
Since $-1 \in (-2, 2)$, the largest interval on which a unique solution exists is:
$$ (-2, 2) $$
Example 2: Non-Linear ODE (Regular Case)
Problem: Is there a unique solution to the IVP:
$$ y^{\prime} = y^{1/3}, \quad y(0) = 1 $$
near the point $(0, 1)$?
Solution:
Here, $F(t, y) = y^{1/3}$. This is a non-linear, 1st order ODE.
- Check continuity of $F(t,y)$: $F(t, y) = y^{1/3}$ is continuous near $(0, 1)$.
- Check continuity of $\frac{\partial F}{\partial y}$: Calculate the partial derivative:
$$ \frac{\partial F}{\partial y} = \frac{1}{3}y^{-2/3} = \frac{1}{3y^{2/3}} $$
We evaluate this near the point $(0, 1)$. At $y=1$:
$$ \frac{1}{3(1)^{2/3}} = \frac{1}{3} $$
The partial derivative is defined and continuous near $y=1$.
Conclusion: Since both $F$ and $\frac{\partial F}{\partial y}$ are continuous on a rectangle containing $(0, 1)$, by the Existence-Uniqueness Theorem, there is a unique solution to the ODE near $(0, 1)$.
Example 3: Non-Linear ODE (Singular Case)
Problem: Is there a unique solution to the IVP:
$$ y^{\prime} = y^{1/3}, \quad y(0) = 0 $$
near the point $(0, 0)$?
Analysis:
Check the conditions of the Existence-Uniqueness Theorem:
$$ \frac{\partial F}{\partial y} = \frac{1}{3y^{2/3}} $$
At the initial point $(0, 0)$, $y=0$. The derivative involves division by zero, so it is not continuous (undefined) at $y=0$.
Conclusion: The Existence-Uniqueness Theorem does NOT apply near $(0, 0)$.
Solving the ODE:
Let’s try to solve it to see what happens. $$ \frac{dy}{dt} = y^{1/3} $$ Using separation of variables ($y \neq 0$):
$$ \begin{aligned} \int y^{-1/3} , dy &= \int dt \\ \frac{y^{2/3}}{2/3} &= t + C \ \implies \frac{3}{2}y^{2/3} &= t + C \end{aligned} $$
Apply the initial condition $y(0) = 0$:
$$ \frac{3}{2}(0)^{2/3} = 0 + C \implies C = 0 $$
So,
$$ \frac{3}{2}y^{2/3} = t \implies y^{2/3} = \frac{2}{3}t $$
Raising both sides to the power $\frac{3}{2}$:
$$ y^2 = \left(\frac{2t}{3}\right)^3 \implies y(t) = \pm \sqrt{\left(\frac{2t}{3}\right)^3} $$
This gives us two solutions for $t \ge 0$. Additionally, by inspection, the trivial solution $y(t) \equiv 0$ also satisfies $y^{\prime} = y^{1/3}$ and $y(0)=0$.
Result: Thus, there are three solutions:
- $y(t) = \sqrt{\left(\frac{2t}{3}\right)^3}$
- $y(t) = -\sqrt{\left(\frac{2t}{3}\right)^3}$
- $y(t) = 0$
That is, the solution exists, but is not unique.