# Deciding whether solution to ODE is unique

I have an ODE:

$$y' = y^{1/4}$$ for $y \ge 0$.

and $y' = 0$ for $y < 0$.

The initial condition is $y(a) = b$.

I'm trying to use Picard's theorem to decide whether there is a unique solution in the cases:

1. $(a,b) = (0,0)$
2. $(a,b) = (0,1)$
3. $(a,b) = (1,0)$.

For the first case I can see that the function is not Lipschitz around the origin and hence we are not guaranteed a unique solution. In fact we can see that $y = (\frac{4}{3}x)^{3/4}$ is a solution, as is $y=0$.

For the second case, the ODE satisfies all the conditions of Picard's theorem (provided we choose a suitable rectangle that stays away from the origin) and hence there is a unique solution.

Now I am struggling with the third case, to me it seems like it should be exactly the same as the 1st case, however I'm not sure what happens when we alter the initial value of $x$.

Thanks for any help

• You're right, it's just like case 1. The equation doesn't contain $x$ explicitly, so it is translation invariant. – Hans Lundmark Jun 18 '14 at 9:08
• Okay great, thanks for confirming – Wooster Jun 18 '14 at 9:10

For all $C\in ]-\infty, -1]$, consider the function $\varphi_C\colon \mathbb R\to \mathbb R$ given by $$\varphi_C(x)=\begin{cases} \left(\dfrac 3 4\right)^{4/3}(x+C)^{4/3}, &\text{ if }x\ge -C\\ 0, &\text{ if }x\leq -C\end{cases}$$

These are solutions to the IVP $y'=y^{1/4}, y(1)=0$.

First check that $\varphi _C$ is differentiable (at the eventually problematic point $-C$): $$\lim \limits_{x\to -C^+}\left[\dfrac{\varphi_C(x)-\varphi_C\left(-C\right)}{x+C}\right]=\lim \limits_{x\to -C^+}\left[\left(\dfrac 3 4\right)^{4/3}\dfrac{(x+C)^{4/3}}{x+C}\right]=0.$$

Since $C\leq -1$, the initial condition is verified.

Finally check that this function is a solution to the differential equation: $$\forall x\in \mathbb R\left[\varphi_C'(x)=\left(\dfrac 3 4\right)^{4/3}\dfrac 43(x+C)^{1/3}=\left(\dfrac 3 4\right)^{1/3}(x+C)^{1/3}=(\varphi_C(x))^{1/4}\right].$$

In fact with the abuse $C=-\infty$, these are all solutions.

Alternatively and in view of Hans Lundmark's comment take any two different solutions to the first IVP and use this answer.

You may have a typo.
If I apply the Lie group G[$x'=\lambda^\alpha x$ and $y'=\lambda^\beta y$] with the unitary transformation $\lambda_o=1$, we can set $\alpha$=1 w.o.l.o.g.
Thus $dx'=\lambda dx$, $dy'=\lambda^\beta dy$, $\dot{y}'=\frac{dy'}{dx'}=\frac{\lambda^\beta dy}{\lambda dx}=\lambda^{\beta -1}\dot{y}$, $\ddot{y}'=\lambda^{\beta -2}\ddot{y}$, etc.

Infintestimal tranformations for these new variables are $\bigg(\frac{\partial x'}{\partial\lambda}\bigg)_{\lambda_o =1}=x$, $\bigg(\frac{\partial y'}{\partial\lambda}\bigg)_{\lambda_o =1}=\beta y$, $\bigg(\frac{\partial \dot{y}'}{\partial\lambda}\bigg)_{\lambda_o =1}=(\beta-1)\dot{y}$, $\bigg(\frac{\partial \ddot{y}'}{\partial\lambda}\bigg)_{\lambda_o =1}=(\beta -2)\ddot{y}$ and so on.

Using the method of characteristics to find group stabilizers, $$d\lambda = \frac{dx}{x}=\frac{dy}{\beta y}=\frac{d\dot{y}}{(\beta -1)\dot{y}}=\frac{d\ddot{y}}{(\beta -2)\ddot{y}}=...$$

$\frac{dx}{x}=\frac{dy}{\beta y} \rightarrow ln x^\beta = ln y - ln \mu$ where $\mu$ is a constant of integration: a polynomial stabilizer which Sophus Lie referred to as a differential invariant. This leads to $\mu = \frac{y}{x^\beta}$, a first-order invariant. $\frac{dx}{x}=\frac{d\dot{y}}{(\beta -1)\dot{y}}$ produces a second-order stabilizer: $\nu=\frac{d\dot{y}}{(\beta -2)\dot{y}}$. As predicted by Lie's Thm. 4.4.1 from his Differential Invariant Paper of 1884, there are an infinite number of these differential invariants of higher order, all following the same pattern: $\eta=\frac{\ddot{y}}{x^{\beta -2}}$, $\xi=\frac{\dddot{y}}{x^{\beta -3}}$, and so on, which together represent a complete solution to the problem.

Example: your DEQ $\dot{y}=y^{\frac{1}{4}}$ Applying G, $\lambda^{\beta -1}\dot{y}=(\lambda ^{\frac{\beta}{4}})y^{\frac{1}{4}}$. For invariance, $\beta -1=\frac{\beta}{4} \rightarrow \beta=\frac{4}{3}$. Invariants for this DEQ are $\mu=\frac{y}{x^\frac{4}{3}}$, $\nu=\frac{\dot{y}}{x^\frac{1}{3}}$, $\eta=x^{\frac{2}{3}}\ddot{y}$ and so on. (To check these stabilizers compute $G(\mu)=\mu$, $G(\nu)=\nu$, $G(\eta)=\eta$, and so on.)

In the direction field of the Lie group stabilizers, $$\frac{d\nu}{d\mu}=\frac{x\frac{d\nu}{dx}}{x\frac{d\mu}{dx}}=\frac{\eta -(\beta -1)\nu}{\nu - \beta \mu}$$ In similar fashion, $$\frac{d\eta}{d\nu}=\frac{x\frac{d\eta}{dx}}{x\frac{d\nu}{dx}}=\frac{\xi -(\beta -2)\eta}{\eta - (\beta -1) \nu}$$ Invariant points such as singularities, saddle points and the separatrices that connect them are undefined points in the direction fields where slope is both zero and infinite simultaneously. This requires $\frac{d\mu}{dx}=0$, $\frac{d\nu}{dx}=0$, and so on for the rest. The algebra between the stabilizers is particularly simple at these points. $$\nu=\beta \mu$$ $$\eta=(\beta -1)\nu=\beta(\beta -1)\mu$$ $$\xi=(\beta-2)\eta=\beta(\beta -1)(\beta -2)\mu$$ If we multiply your DEQ by $\frac{1}{x^{\frac{1}{3}}}$ we get $\frac{\dot{y}}{x^{\frac{1}{3}}}=\bigg(\frac{y}{x^{\frac{4}{3}}}\bigg)^{\frac{1}{4}}$ or $\nu=\mu^{\frac{1}{4}}$.
Since $\nu=\frac{4}{3}\mu$, $\frac{4}{3}\mu^{\frac{3}{4}}=1$, and $\mu=\bigg(\frac{3}{4}\bigg)^{\frac{4}{3}}$. Since $\mu=\frac{y}{x^{\frac{4}{3}}}$, your special solution should be this: $$y=\bigg(\frac{3}{4}x\bigg)^{\frac{4}{3}}$$ (Okay, yeah, I could have just pointed out the typo, but where's the fun in that? Admittedly I didn't answer your specific question, but I had so much fun chasing the problem around I just had to share. Thanks!)

As a follow-up, to solve the general equation observe that $$x\frac{d}{dx}\mu =\nu-\beta \mu=\mu^{\frac{1}{4}}-\frac{4}{3}\mu$$ $$\frac{d\mu}{\mu^{\frac{1}{4}}-\frac{4}{3}\mu}=\frac{dx}{x}$$ $$-ln(3-4\mu^{\frac{3}{4}})=lnx+lnA$$ A is an integration constant. $$Ax(3-4\mu^{\frac{3}{4}})=1$$ For convenience, let $A=(4C)^{-1}$. $$\mu=\bigg(\frac{3}{4}-\frac{C}{x}\bigg)^{\frac{4}{3}}$$ This leads to the general solution: $$y=\bigg(\frac{3}{4}x-C\bigg)^{\frac{4}{3}}$$ Note that the general solution collapses to the special solution when C=0, as expected. Note also that if we had chosen $A=-(3C)^{-1}$ the general solution would be: $$y=\bigg(\frac{3}{4}\bigg)^{\frac{4}{3}}\bigg(x+C\bigg)^{\frac{4}{3}}$$