Answered: "Solution to x′=xsin⁡(πx) is uniqueI want to prove..."

p3sa1bynn

Answered question

2022-04-19

Solution to

x^{'} = x \sin (\frac{π}{x})

is unique
I want to prove that the only solution to the ODE

x^{'} = {\begin{cases} x \sin (\frac{π}{x}) if x \neq 0 \\ 0 else \end{cases}

obettyQuokeperg6

Beginner2022-04-20Added 14 answers

Step 1
If

x (t_{0}) \neq 0

for some

t_{0} \neq 0

.
We assume

t_{0} > 0

and

x (t_{0}) > 0

, the other cases is similar. Let

x_{0} \in (0, x (t_{0}))

so that

x_{0} \sin (\frac{π}{x_{0}}) = 0 .

Since

x (0) = 0

and

x (t_{0}) > x_{0}

, by continuity there is

t_{1} \in (0, t_{0})

so that

x (t_{1}) = x_{0}

. We also choose the smallest such

t_{1}

, so that

x (t) < x_{0}

for all

t < t_{1}

. Then the initial value problem

{\begin{cases} x^{'} = x \sin (\frac{π}{x}), \\ x (t_{1}) = x_{0} \end{cases}

has two solutions: x(t) and the constant solution

y (t) = x_{0}

. These two are different since

x (t) < x_{0}

for all

t < t_{1}

. It contradicts the uniqueness theorem, since

x \sin (\frac{π}{x})

is Locally Lipschitz around

(t_{1}, x_{0})

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?