Exactly one solution of $y'=-xy^{1/3}, y(0)=0$

Question

I want to show that the IVP $$y'=-xy^{1/3}, y(0)=0$$ has an unique solution given by $y\equiv 0$. One cannot apply the Picard-Lindelöf theorem because $f(x,y)=-xy^{1/3}$ is not Lipschitz in any neighbourhood of (0,0). Futhermore, the sufficent condition that $$\int_0^{0+\alpha} s^{-1/3} ds$$ is not convergent is not useful. So, how does one prove that there exists a unique solution for the IVP? Thanks for your answers!

Answer 1

A more theoretical solution. By Gronwall’s inequality every solution exists in the entire real line. Assume there exists $x_0>0$ such that $y(x_0)>0$. Then you can find a maximal interval $I$ containing $x_0$ where $y>0$. The interval $I$ is bounded from below by zero. Since $y$ is strictly positive in $I$ using the differential equation you get that $y’<0$ in $I$, which means that $y$ is strictly decreasing in $I$. Now let $a\ge 0$ be the left endpoint of $I$. By continuity of $y$ and the maximality of $I$ you have $y(a)=0$, which is a contradiction since $y(a)\ge y(x_0)>0$. If $y(x_0)<0$ you can reason in a similar way using increasing instead of decreasing. This proves that $y=0$ for $x$ positive. A similar reasoning works for $x$ negative. If the function is positive at some negative point, then it is strictly increasing in a maximal interval but then it can never go down to zero.

Answer 2

I would first check whether there is really only one solution. For this we could solve the ODE e.g.: $$ \begin{align*} \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} &= -x \cdot \left( y\left( x \right) \right)^{\frac{1}{3}} \quad\mid\quad \text{multiply by differential}\\ \operatorname{d}y\left( x \right) &= -x \cdot \left( y\left( x \right) \right)^{\frac{1}{3}}\,\operatorname{d}x \quad\mid\quad \div\left( y\left( x \right) \right)^{\frac{1}{3}}\\ \frac{1}{\left( y\left( x \right) \right)^{\frac{1}{3}}}\, \operatorname{d}y\left( x \right) &= -x\,\operatorname{d}x\\ \left( y\left( x \right) \right)^{-\frac{1}{3}}\, \operatorname{d}y\left( x \right) &= -x\,\operatorname{d}x\\ \int \left( y\left( x \right) \right)^{-\frac{1}{3}}\, \operatorname{d}y\left( x \right) &= \int -x\,\operatorname{d}x\\ \frac{3}{2} \cdot \left( y\left( x \right) \right)^{\frac{2}{3}} &= -\frac{1}{2} \cdot x^{2} + c_{1}\\ \left( y\left( x \right) \right)^{\frac{2}{3}} &= -\frac{1}{3} \cdot x^{2} + c\\ y\left( x \right) &= \pm\left( -\frac{1}{3} \cdot x^{2} + c \right)^{\frac{3}{2}}\\ y\left( x \right) &= \pm\sqrt{\left( -\frac{1}{3} \cdot x^{2} + c \right)^{3}}\\ \end{align*} $$

$$\fbox{$ \begin{align*} y\left( x \right) = \pm\sqrt{\left( -\frac{1}{3} \cdot x^{2} + c \right)^{3}} \vee y\left( x \right) = 0\\ \end{align*} $} \tag{1}$$

If we set the constant $c = 0$: We can already see here that there is one real solution ($y\left( x \right) = 0$ <= The solution was lost when deviding by $\sqrt[3]{y}$.), three complex solutions ($y\left( x \right) = 0$, $y\left( x \right) = \sqrt{\left( -\frac{1}{3} \cdot x^{2} + c \right)^{3}}$ and $y\left( x \right) = -\sqrt{\left( -\frac{1}{3} \cdot x^{2} + c \right)^{3}}$) and infinitely many hypercomplex solutions.

Now let's use the boundary conditions: $$ \begin{align*} 0 &= \pm\sqrt{\left( -\frac{1}{3} \cdot 0^{2} + c \right)^{3}}\\ 0 &= \pm\sqrt{c^{3}}\\ 0 &= c\\ \end{align*} $$

So our Solution is: $$\fbox{$ \begin{align*} y\left( x \right) = \pm\sqrt{\left( -\frac{1}{3} \cdot x^{2} \right)^{3}} \vee y\left( x \right) = 0\\ \end{align*} $} \tag{2}$$

We can simplify this even further: $$ \begin{align*} y\left( x \right) &= \pm\sqrt{\left( -\frac{1}{3} \cdot x^{2} \right)^{3}}\\ y\left( x \right) &= \pm\sqrt{-\frac{1}{27} \cdot x^{6}}\\ y\left( x \right) &= \pm\sqrt{-1} \cdot \sqrt{\frac{1}{27} \cdot x^{6}}\\ y\left( x \right) &= \pm i \cdot \sqrt{\frac{1}{27} \cdot x^{6}}\\ y\left( x \right) &= \pm \sqrt{\frac{1}{27} \cdot x^{6}} \cdot i\\ y\left( x \right) &= \pm \frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i\\ \end{align*} $$

Thus, our three final complex solutions are: $$\fbox{$\fbox{$ \begin{align*} y\left( x \right) = \frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i \vee y\left( x \right) = -\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i \vee y\left( x \right) = 0\\ \end{align*} $}$} \tag{3}$$

Whether there is a unique solution depends on the number range in which this solution should be. In real terms, as shown, there is only one solution. In complexes there are already three. In hypercomplexes there are still infinitely many. ${;)}$

Note: My calculation steps are not valid for general hypercomplex algebras.

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Let's check $y = 0$: $$ \begin{align*} y' &= -x \cdot y^{1/3}\\ 0' & =-x \cdot 0^{1/3}\\ 0 &= 0 \quad \square\\ \end{align*} $$

Let's check $y = \pm\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i$: $$ \begin{align*} y' &= -x \cdot y^{1/3} \quad\mid\quad \div x\\ y' \cdot \frac{1}{x} &= -y^{1/3} \quad\mid\quad \left( \cdot \right)^{3}\\ \left( y' \cdot \frac{1}{x} \right)^{3} &= -y\\ \left( \left( \pm\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i \right)' \cdot \frac{1}{x} \right)^{3} &= -\pm\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i\\ \left( \pm\frac{3}{3 \cdot \sqrt{3}} \cdot x^{2} \cdot i \cdot \frac{1}{x} \right)^{3} &= -\pm\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i\\ \left( \pm\frac{1}{\sqrt{3}} \cdot x^{2} \cdot i \cdot \frac{1}{x} \right)^{3} &= \mp\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i\\ \left( \pm\frac{1}{\sqrt{3}} \cdot x \cdot i \right)^{3} &= \mp\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i\\ \pm \left( \frac{1}{\sqrt{3}} \right)^{3} \cdot x^{3} \cdot i^{3} &= \mp\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i\\ \pm\frac{1}{\left( \sqrt{3} \right)^{2} \cdot \sqrt{3}} \cdot x^{3} \cdot (-i) &= \mp\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i\\ \mp\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i &= \mp\frac{1}{3 \cdot \sqrt{3}} \cdot x^{3} \cdot i \quad \square\\ \end{align*} $$