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# 2.2 Existence and uniqueness of solutions for initial value problems

If we are to solve an initial value problem of the type in Equation (2.2), we must first be sure that it has a solution and that the solution is unique. Such conditions are guaranteed by the following criteria:

The criteria for existence and uniqueness

Consider the domain $$\mathcal{D}=\{(y,x): |y-a| \leq c, x_0 \leq x \leq x_1\}$$, with $$c$$ an arbitrary constant. If $$f(y,x)$$ is Lipschitz continuous in $$y$$ and continuous in $$x$$ over $$\mathcal{D}$$, then there is a unique solution to the initial value problem (2.2)-(2.2) at least up to time $$X^*=min(x_1,x_0+c/S)$$, where \begin{align} S = \max_{(y,x)\in\mathcal{D}} |f(y,x)|. \nonumber \end{align}

We say that $$f(y,x)$$ is Lipschitz continuous in $$y$$ over some domain $$\mathcal{D}$$ if there exists some constant $$L \geq 0$$ so that \begin{align} |f(y,x)-f(y^*,x)| \leq L |y-y^*| \nonumber \end{align} for all $$(y,x)$$ and $$(y^*,x)$$ in $$\mathcal{D}$$. Note that this requirement is more restrictive than plain continuity, which only requires that $$|f(y,x)-f(y^*,x)| \to 0$$ as $$y \to y^*$$. In fact, Lipschitz continuity requires that $$|f(y,t)-f(y^*,t)| \to \mathcal{O}(|y-y^*|)$$ as $$y \to y^*$$. To give a definition for $$L$$, consider that $$f(y,x)$$ is differentiable with respect to $$y$$ in $$\mathcal{D}$$ and that its derivative $$\frac{\partial f(y,x)}{\partial y}$$ is bounded, then we can take \begin{align} L = \max_{(y,x)\in\mathcal{D}} |\frac{\partial f(y,x)}{\partial y}|, \tag{2.12} \end{align} since \begin{align} f(y,x) = f(y^*,x)+ \frac{\partial f(v,x)}{\partial y} (y-y^*) \nonumber \end{align} for some value $$v$$ between $$y$$ and $$y^*$$.

For (2.4), or higher order ODEs in general, one has to consider that such equations can be written as systems of first order ODEs. Similar criteria apply in that case to define the existence and uniqueness of solutions. However, special attention must be paid to the definition of Lipschitz continuity, in particular to the definition of appropriate bounds in terms of a given norm. This aspect will not be covered in this notes.

Fulfillment of the criteria for existence and uniqueness

Consider the initial value problem \begin{align} y'(x) = (y(x))^2\,,\quad y(0) = a > 0.\nonumber \end{align} It is clear that $$f(y)=y^2$$ is independent of $$x$$ and Lipschitz continuous in $$y$$ over any finite interval $$|y-a| \leq c$$ with $$L = 2 (a+c)$$ and $$S=(a+c)^2$$. In this case, we can say that there is unique solution at least up to time $$c/(a+c)^2$$ and then take the value of $$c$$ that maximizes that time, i.e. $$c=a$$, yielding a time of $$1/(4a)$$.

In this simple problem we can verify our considerations by looking at the exact solution \begin{align} y(x) = \frac{1}{a^{-1}-x}.\nonumber \end{align} It can be easily seen that $$y(x) \to \infty$$ as $$x \to 1/a$$, so that there is no solution beyond time $$1/a$$.

Violation of the criteria for existence and uniqueness

Consider the initial value problem \begin{align} y'(x) = \sqrt{y(x)}\,,\quad y(0) = 0.\nonumber \end{align} In this case we have that $$f(y)=\sqrt{y}$$ is not Lipschitz continuous near $$y=0$$ since $$f'(y)=1/(2\sqrt{y}) \to \infty$$ as $$y \to 0$$. Therefore, we can not find a constant $$L$$ so that the bound in (2.12) holds for $$y$$ and $$y^*$$ near $$0$$. This consideration implies that this initial value problem does not have an unique solution. In fact, it has two solutions: \begin{align} y(x) = 0\nonumber \end{align} and \begin{align} y(x) = \frac{1}{4} x^2.\nonumber \end{align}