Examples

Solution that blows up

Solution that blows up

Consider the IVP:

$$y^{\prime }(t)=y^2,y(0)=y_0>0.$$

This is a separable differential equation. Assuming $y\ne 0$, we can solve it: $-y^{-1}=t+c$. From the algebraic viewpoint we had been using, the solution to the ODE is the family of curves $(c-t)y=1$, and there is no particular problem with this family.

Now write the solution as a functional solution and solve for the constant using the initial condition $y(0)=y_0$:

$$y(t)=\frac{1}{-t+y_0^{-1}}.$$

Look what happens as time evolves: for $0\le t<y_0^{-1}$, the function $y(t)$ is well-behaved. But as $t\to y_0^{-1}$ we have $y(t)\to +\infty$. In other words, $y(t)$ blows up at $y_0^{-1}$. There is a solution to the IVP (and in fact it is unique when it exists), but the solution cannot be extended beyond the range $t\in (-\infty ,y_0^{-1})$.

This blow-up behavior is not weird algebraically, but functionally it is undesirable, since the solution function does not exist beyond a certain time $t=y_0^{-1}$.

Notice that if the ODE is replaced by $y^{\prime }(t)=-y^2$, then the solution is $y(t)=\frac{1}{t+y_0^{-1}}$, which does not blow up for $t\ge 0$. (Although it does blow up for $t<0$.)

Once-differentiable functions given piecewise, non-uniqueness

Once-differentiable functions given piecewise, non-uniqueness

Consider the IVP:

$$y^{\prime }(t)=2y^{1/2},y(0)=0.$$

This is a separable differential equation: $\frac{1}{2}{y}^{-1/2}dy=dt$ so $y^{1/2}=t+c$. Solving this explicitly: $y=(t+c{)}^2$, and $y=t^2$ is a solution satisfying $y(0)=0$.

But look! Here is a piecewise function, for each $t_0\in \mathbb{R}$, that also solves this IVP:

$$y(t)=\{\begin{array}{ll}(t-t_0{)}^2& t\ge t_0\\ 0& t<t_0.\end{array}$$

Clearly this $y(t)$ is a solution for $t\ne t_0$. Considering $t=t_0$, we have:

$$\begin{array}{rl}\underset{t\to t_0^{+}}{\lim }{y}^{\prime }(t)=\underset{t\to t_0^{+}}{\lim }2(t-t_0)& =0\\ \underset{t\to t_0^{-}}{\lim }{y}^{\prime }(t)=\underset{t\to t_0^{-}}{\lim }0& =0,\end{array}$$

and therefore $y^{\prime }(t_0)$ does exist, namely $y^{\prime }(t_0)=0$, and the ODE is satisfied everywhere.

Notice that $y^{\prime \prime }(t)$ does not exist at $t_0$!

Furthermore, the piecewise solutions are not given by a single equation with parameter, so they do not qualify as a “family of curves” according to our previous usage in the algebraic viewpoint.

Infinitely differentiable - not enough for uniqueness

Infinitely differentiable: not enough for uniqueness

Consider the IVP: $ydt-t^{2}dy=0,y(0)=0.$

This differential equation is separable, and the solution family is calculated to be $y=C{e}^{-1/t}$ for $C\in \mathbb{R}$. We can also check that $y=0$ is a solution too.

But look! Here is a piecewise function, for each $C\in \mathbb{R}$, that solves the IVP:

$$y(t)=\{\begin{array}{ll}C{e}^{-1/t}& t>0\\ 0& t\le 0.\end{array}$$

At $t=0$ we have $y^{(n)}(t)=0$ for any $n$. For example, from the right $y^{\prime }(t)=\frac{C}{t^2}{e}^{-1/t}$, and this limits to zero as $t\to 0^{+}$. (It is not trivial to prove this, but intuitively the exponential goes to zero much faster than $t^2$ does.)

All these functions are infinitely differentiable, they all satisfy the IVP, and yet they differ from each other for any $t>0$ (for differing $C$ values).

Picard-Lindelöf frequently not applicable

Picard-Lindelöf frequently not applicable

The IVP $y^{\prime }(t)=2y^{1/2},y(0)=0$ does not satisfy Picard-Lindelöf. We have $\frac{\partial f}{\partial y}=\frac{1}{\sqrt{y}}$, and this is neither continuous nor bounded on any interval containing $(0,0)$. So the “Variants” of Picard-Lindelöf do not apply. Furthermore, the difference quotient can be made arbitrarily large by letting it approximate this partial near $(0,0)$, so the original Picard-Lindelöf does not apply either.

Picard iteration can yield an exact solution

Picard iteration can yield an exact solution

Consider the IVP $y^{\prime }=2y,y(0)=1$. Change variables to $z=y-1$ and get the IVP $z^{\prime }=2(z+1)$ and $z(0)=0$. Set $z_0(t)=0$ and iterate:

$$\begin{array}{rl}{z}_1& ={\int }_0^{t}f(s,0)ds={\int }_0^{t}2ds=2t\\ z_2& ={\int }_0^{t}f(s,z_1)ds={\int }_0^{t}2(2s+1)ds=2t^2+2t\\ z_3& ={\int }_0^{t}f(s,z_2)ds={\int }_0^{t}2(2s^2+2s+1)ds=\frac{4}{3}{t}^3+2t^2+2t\\ & ⋮\\ z_n& ={\int }_0^{t}f(s,z_{n-1})ds={\int }_0^{t}2(z_{n-1}+1)ds.\end{array}$$

So we conclude that $z(t)$ is given by the power series

$$z(t)=2t+2t^2+\frac{4}{3}{t}^3+\frac{2}{3}{t}^4+\dots$$

and the coefficients are defined recursively by $a_1=2$, $a_n=\frac{2}{n}{a}_{n-1}$. This recursion relation is solved by the general term $a_n=\frac{2^n}{n!}$, so we have:

$$\begin{array}{rl}z(t)& =0+\frac{(2t)}{1!}+\frac{(2t{)}^2}{2!}+\frac{(2t{)}^3}{3!}+\dots \\ & =e^{2t}-1.\end{array}$$