From the existness and uniqueness Theorem,the initial value problem
$$y'=3x(y-1)^{1/3} , y(3)=-7$$
has a unique solution on some open interval that contains $x=3$. Find the solution and determine the largest open interval on which it’s unique.
What i tried, First i tried to solve the equation by the seperable equation method to get,$$y=1+(x^2-5)^{1.5}$$. Then from here i used the existence and uniquness theorem to calculate $f(x,y)$ and $f_{y}$. From the calculations,when $y$ not equals to $1$, $f_{y}$ will be continuous, hence there will be a unique solution when $y$ not equals to $1$, according to the wxistness and uniquness theorem. However im stuck from here onwards as to finding the interval of validaty. Is my working correct. Could anyone explain. Thanks
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$\begingroup$once you have $y = 1 - (x^2 -5)^{3/2}$, you can see that $\sqrt 5 \le x < \infty$ is the maximal interval of existence.
$\endgroup$ 2 $\begingroup$I'm interpreting $u^{1/3}$ as $$u^{1/3}:={\rm sgn}(u)\>|u|^{1/3}\qquad(u\in{\mathbb R})\ .$$ The initial point $(-3,7)$ is below the line $y=1$; whence we change $(y-1)^{1/3}$ to $-(1-y)^{1/3}$, where now the radicand is positive in the neighborhood of the initial point.
Separation of variables leads to $$\int_{-7}^y -(1-\eta)^{-1/3}\ d\eta=\int_3^x 3\xi\>d\xi\ ,$$ so that we obtain $$y=1-(x^2-5)^{3/2}\qquad\bigl(x>\sqrt{5}\bigr)\ .\tag{1}$$ It is easily checked that $(1)$ satisfies the given ODE as well as the initial condition.
So far we have not needed the existence and uniqueness theorem. The right side of the given ODE is defined in all of ${\mathbb R}^2$ and satisfies the assumptions of this theorem everywhere, except on the line $y=1$ (which happens to be a solution). Since we already have found a solution it seems that we don't need the theorem at all. But there is still the question of uniqueness: There could be other solutions that cannot be found using separation of variables.
Note that the solution $(1)$ passes only through good points. If $x\mapsto\bar y(x)$ is a different solution then arguing in the neighborhood of the point $\bigl(\xi,y(\xi)\bigr)$ where $\xi=\inf\{x\>|\>\bar y(x)\ne y(x)\}$ would produce a contradiction. I omit the details.
$\endgroup$ $\begingroup$If $$y(x) = 1 + (x^2-13)^{1.5},$$
then $$y(3) = 1 + (-6)^{\frac32}$$ is most surely not equal to $-7$. Therefore, your solution is wrong.
$\endgroup$ 3 $\begingroup$The answer is correct and it gives the technique for finding the interval of validity. By the way it is the first answer to be posted solving the issue of the interval of validity.
I'll advance based on your solution. To find the interval of validity we need the solution of the ode
$$ y = 1+ (x^2-5)^{3/2} $$
which implies imposing the condition
$$ x^2-5 \geq 0 \implies x\in ( -\infty, -\sqrt{5} )\cup (\sqrt{5}, \infty ). $$
The above suggested that the interval of validity is $x\in (\sqrt{5}, \infty )$.
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