# Using the Existence and uniqueness theorem to find the interval of validity of a unique solution in a Linear ODE

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

• Your solution is not right. Possibly you made some mistake when plugging into the initial condition. – KittyL Jan 20 '15 at 10:14
• @abel No. In your case, $y(3) = 9$. – 5xum Jan 20 '15 at 10:26
• @ys wong: Do you know what the interval of validity mean? – Mhenni Benghorbal Jan 20 '15 at 14:40

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.

• But i thought i must make use of the existness and uniquness theorem to find the interval of validality? After all the topic is on existness and uniquness theorem – ys wong Jan 20 '15 at 10:22
• not necessarily. at $x = \sqrt 5, y = 1$ has no unique solution. once the solution reaches there you have problem. – abel Jan 20 '15 at 10:27

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.

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.

• I just edited my solution. – ys wong Jan 20 '15 at 10:19
• @yswong And it's still wrong. If $$y(x) = 1 + (x^2-5)^{1.5},$$ then $y(3) = 1 + 4^{1.5} = 1 + \sqrt{64} = 9\neq 7$. – 5xum Jan 20 '15 at 10:22
• yes i made a mistake. i have edited my answer. i need to watch out for these errors. – abel Jan 20 '15 at 10:28

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 $x\in (\sqrt{5}, \infty )$.

• please read the question. – abel Jan 20 '15 at 10:56
• your is wrong too. the right solution is $y = 1 - (x^2 - 5)^{3/2}$ – abel Jan 20 '15 at 11:20
• neither the solution nor the interval of maximal existence is correct. – abel Jan 20 '15 at 13:10
• why bother posting wrong answers? – abel Jan 20 '15 at 13:19
• Your final answer looks very little like an interval. – mrf Jan 20 '15 at 17:57