# ODEUET1

Consider the initial value problem

${\dot {x}}=|x|^{{2/q}},x(0)=0\,$

where $\,q$ is a positive integer. Find all values of $\,q$ for which there is a unique solution. In the latter case, justify your answer by using an appropriate theorem.

Let $q=1\,$.

${\dot {x}}=|x|^{2}=x^{2},$

Check the Lipschitz condition.

$|f(x,t)-f(y,t)|=|x^{2}-y^{2}|\leq |x+y||x-y|=L|x-y|\,$

So $f\,$ is Lipschitz if $x\,$ is bounded.

Solving the DE,

${\dot {x}}=x^{2}\,$

${\frac {dx}{dt}}=x^{2}\,$

${\frac {dx}{x^{2}}}=dt\implies {\frac {-1}{x}}=t+c\,$

$x={\frac {-1}{t+c}}\,$

$x(0)={\frac {-1}{c}}=0\,$

There is no solution to this DE.

Let $q=2\,$.

${\dot {x}}=|x|\,$

Check Lipschitz.

$|f(x,t)-f(y,t)|=||x|-|y||\leq L|x-y|\,$ is true.

Solve the DE.

${\dot {x}}=|x|,x(0)=0\,$

if $x<0\,$,

${\frac {dx}{dt}}=-x\,$

${\frac {dx}{-x}}=dt\,$

$-\ln x=t+c_{1}\,$

$x(t)=-e^{{t+c_{1}}}\,=c_{2}e^{t}$

$x(0)=c_{2}=0\implies c_{2}=0\implies x=0\,$ is the unique solution.

If $x=0\,$,

${\frac {dx}{dt}}=0\,$

$x(t)=c_{3}\,$

$x(0)=0\implies c_{3}=0\implies x=0\,$ is the unique solution.

If $x>0\,$,

${\frac {dx}{dt}}=x\,$

${\frac {dx}{x}}=dt\,$

$\ln x=t+c_{4}\,$

$x(t)=e^{{t+c_{4}}}\,=c_{5}e^{t}$

$x(0)=c_{5}=0\implies c_{5}=0\implies x=0\,$ is the unique solution.

Let $q=n\geq 3\,$.

${\dot {x}}=|x|^{{2/n}}=x^{{2/n}}\,$

Test Lipschitz.

$|x^{{2/n}}-y^{{2/n}}|=|(x^{{1/n}}+y^{{1/n}})(x^{{1/n}}-y^{{1/n}})|\leq |x^{{1/n}}+y^{{1/n}}||x-y|\,$

But $(x^{{1/n}}+y^{{1/n}})\,$ is not bounded by a constant so $x\,$ is not Lipschitz and therefore there is not a unique solution.

Solve the DE.

${\dot {x}}=|x|^{{2/n}}=x^{{2/n}}\,$

$x^{{-2/n}}dx=dt\,$

${\frac {nx^{{\frac {n-2}{n}}}}{n-2}}=t+c_{1}\,$

$x^{{\frac {n-2}{n}}}={\frac {n-2}{n}}(t+c_{1})\,$

$x(t)=\left[{\frac {n-2}{n}}(t+c_{1})\right]^{{\frac {n}{n-2}}}\,$

$x(0)=\left[{\frac {n-2}{n}}c_{1}\right]^{{\frac {n}{n-2}}}=0\implies c_{1}=0\,$

So $x(t)=\left[{\frac {n-2}{n}}t\right]^{{\frac {n}{n-2}}}\,$

There are infinitely many solutions because the equation holds for all $n\geq 3\,$.