Initial Value Problems, Existence and Uniqueness

An initial value problem (IVP) pairs an equation $y' = f(t, y)$ with a starting value $y(t_0) = y_0$. In Direction Fields and Euler's Method you traced a curve from a point by following the slopes. It is natural to assume exactly one curve leaves each point. Usually that is true, but not always, and the exceptions are worth knowing.

Idea clave

Theorem. If $f$ and $\partial f/\partial y$ are continuous on a rectangle containing $(t_0, y_0)$, then the IVP has exactly one solution on some interval around $t_0$.
Linear case. For $y' + p(t)y = g(t)$, if $p$ and $g$ are continuous on an open interval $I$ containing $t_0$, the unique solution exists on all of $I$.

Por qué funciona

Suppose $y_1$ and $y_2$ both solve the IVP and let $w = y_1 - y_2$, so $w(t_0) = 0$. By the mean value theorem, $$w' = f(t, y_1) - f(t, y_2) = \frac{\partial f}{\partial y}(t, \xi)\,w$$ for some $\xi$ between $y_1$ and $y_2$. If $|\partial f/\partial y| \le K$, then $|w|$ can grow at most like $|w(t_0)|e^{K|t - t_0|} = 0$. A bounded $\partial f/\partial y$ keeps two solutions from peeling apart. When $\partial f/\partial y$ blows up, that protection is gone.

When uniqueness fails

Ejemplo resuelto: Two solutions from one point

Solve $y' = 3y^{2/3}$, $y(0) = 0$.

$y = 0$ works: both sides are 0.
Try $y = t^{3}$: $y' = 3t^{2}$ and $3y^{2/3} = 3(t^{3})^{2/3} = 3t^{2}$. It works too, with $y(0) = 0$.
In fact $y = 0$ until any time $c \ge 0$, followed by $y = (t - c)^{3}$, is also a solution: infinitely many.
Here $\partial f/\partial y = 2y^{-1/3}$, which is undefined at $y = 0$, so the theorem never promised uniqueness.

Respuesta: The IVP has more than one solution.

Error común

Assuming every initial value problem has exactly one solution. For $y' = \sqrt{y}$ with $y(0) = 0$, both $y = 0$ and $y = \frac{t^{2}}{4}$ (for $t \ge 0$) work. Before trusting a single answer, check the hypotheses: are $f$ and $\partial f/\partial y$ continuous at the starting point? Here $\partial f/\partial y = \frac{1}{2\sqrt{y}}$ is not.

How far a solution lasts

For a linear equation the answer comes from the coefficients alone: divide by the leading coefficient to reach standard form, list every $t$ where $p$ or $g$ is undefined or discontinuous, and take the gap that contains $t_0$.

Ejemplo resuelto: The interval for a linear IVP

On what interval is the solution of $(t - 3)y' + (\ln t)y = 2t$, $y(1) = 2$, guaranteed to exist and be unique?

Standard form: $y' + \frac{\ln t}{t - 3}y = \frac{2t}{t - 3}$.
$p = \frac{\ln t}{t - 3}$ needs $t > 0$ and $t \ne 3$; $g = \frac{2t}{t - 3}$ needs $t \ne 3$.
The trouble points split the line into $(0, 3)$ and $(3, \infty)$; $t_0 = 1$ lies in $(0, 3)$.

Respuesta: $(0, 3)$

Ejemplo resuelto: A short life without any trouble point

Does $y' = y^{2}$, $y(0) = 1$ have a unique solution, and for how long?

$f = y^{2}$ and $\partial f/\partial y = 2y$ are continuous everywhere, so there is exactly one solution near $t = 0$.
Separating variables gives $y = \frac{1}{1 - t}$.
It blows up as $t \to 1^{-}$, even though the equation looks perfectly smooth.

Respuesta: Unique, but it exists only for $t < 1$. For nonlinear equations the theorem promises just some interval.

Slopes of $y' = 3y^{2/3}$. The blue curve from $(0, 0)$ stays on the equilibrium $y = 0$, while the teal curves, shifted copies of $y = (t - c)^{3}$, leave the same line with slope 0. Several solutions touch at each point of $y = 0$.

When uniqueness fails

How far a solution lasts

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