Uniqueness of differential equation (generalisation)

Question

What are the necessary conditions for the existance of a solution to a differential equation? What conditions must be added in order to have a unique solution? $$$$ I know $f(y_0)\neq0$ is needed to rule out $f(y)=0$ (for uniqueness) but I've also heard that a function must be Lipschitz continuous.

Answer 1

Consider the differential equation of the first order IVP

$y'=f(x,y)$ with $y(x_0) = y_0$ . . . . .$(1)$

Existence theorem: Suppose that $f(x, y)$ is continuous function in some region $$R=\{(x,y):|x-x_0|\le a,|y-y_0|\le b\}, (a,b\gt0)$$ Since $f$ is continuous in a closed and bounded domain, it is necessarily bounded in $R$, i.e., there exists $k > 0$ such that $|f(x, y)| ≤ k$, $∀(x, y) ∈ R$. Then the IVP $(1)$ has at least one solution $y = y(x)$ defined in the interval $|x − x_0| ≤ α$ where $\alpha=min\{a,\frac{b}{k}\}$.

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Uniquness theorem: Suppose that $f$ and $\frac{∂f}{∂y}$ are continuous function in $R$ (defined in the existence theorem). Hence, both the $f$ and $\frac{∂f}{∂y}$ are bounded in $R$, i.e.,

$1.$ $|f(x, y)| ≤ K$, $∀(x, y) ∈ R$

$2.$ $|\frac{∂f}{∂y}|≤ M$, $∀(x, y) ∈ R$

Then the IVP $(1)$ has at most one solution $y = y(x)$ defined in the interval $|x − x_0| ≤ α$ where $\alpha=min\{a,\frac{b}{k}\}$.

Combining with existence thereom, the IVP $(1)$ has unique solution $y = y(x)$ defined in the interval $|x − x_0| ≤ α$.

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Note: Condition $|\frac{∂f}{∂y}|≤ M$, $∀(x, y) ∈ R$ can be replaced by a weaker condition which is known as Lipschitz condition. Thus, instead of continuity of $\frac{∂f}{∂y}$, we require

$|f(x, y_1) − f(x, y_2)| ≤ L|y_1 − y_2|$ $ ∀(x, y_i) ∈ R$.

If $\frac{∂f}{∂y}$ exists and is bounded, then it necessarily satisfies Lipschitz condition. On the other hand, a function $f(x, y)$ may be Lipschitz continuous but $\frac{∂f}{∂y}$ may not exists.

For example $f(x, y) = x^2|y|$, $|x| ≤ 1$, $|y| ≤ 1$ is Lipschitz continuous in $y$ but $\frac{∂f}{∂y}$ does not exist at $(x, 0)$.

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Consider the second order initial value problem

$y''(t)+p(t)y'(t)+q(t)y=g(t)$, with $y(t_0) = y_0, y′(t_0) = y^′_0.$ . . . . . $(2)$

If the functions $p(t)$, $q(t)$, and $g(t)$ are continuous on the interval $I: α < t < β$ containing the point $t = t_0$. Then there exists a unique solution $y = φ(t)$ of the problem $(2)$, and that this solution exists throughout the interval $I$.

That is, the theorem guarantees that the given initial value problem $(2)$ will always have (existence of) exactly one (uniqueness) twice-differentiable solution, on any interval containing $t_0$ as long as all three functions $p(t), q(t),$ and $g(t)$ are continuous on the same interval. Conversely, neither existence nor uniqueness of a solution is guaranteed at a discontinuity of $p(t), q(t),$ or $g(t)$.

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For more information and examples you can find the following :

$1.$ 'Differential Equations Theory, Technique and Practice' by G. F. Simmons & S. G. Krantz (McGraw Hill Higher Education)

$2.$ "Differential Equations" by Shepley L. Ross

$3.$ http://home.iitk.ac.in/~sghorai/TEACHING/MTH203/ode5.pdf

$4.$ http://www.math.ucsd.edu/~y1zhao/2013UCSDFallQuarterMath20D/Files/Section3.2.pdf

$5.$ http://www.ltcconline.net/greenl/courses/204/ConstantCoeff/uniquenessExistence.htm