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

If x=0\,,

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


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(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}}\,


{\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\,.

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