ODEUET1

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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.

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

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.

$|f(x,t)-f(y,t)| = ||x|-|y|| \le L |x-y|\,$ 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 $q=n\ge 3\,$ .

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

Test Lipschitz.

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

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 $n\ge 3\,$ .

Main Page : Ordinary Differential Equations

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