Question

For each initial value problem, determine whether Picard's Theorem can be used to show the existence of a unique solution in an open interval containing t = 0. Justify your answer.
(a) y' = ty4/3, y(0) = 0
(b) y' = tył/3, y(0) = 0
(c) y' = tył/3, y(0) = 1

Answer 1

Answer:

Part a: $f , \, f_y$ is continuous at the initial value (0,0) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

Part b: $f_y$ is not continuous at the initial value (0,0) so due to Picardi theorem there does not exist an interval such that the IVP has a unique solution.

part c: $f , \, f_y$ is continuous at the initial value (0,1) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

Step-by-step explanation:

Part a

as $y^{' }=ty^{4/3}$

Let

$f(t,y)=ty^{4/3}$

Now derivative wrt y is given as

$f_y=\frac{4}{3}ty^{1/3}$

Finding continuity via the initial value

$f$ is continuous on $R^2$ also $f_y$ is also continuous on $R^2$

Also

$f , \, f_y$ is continuous at the initial value (0,0) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

Part b

as $y^{' }=ty^{1/3}$

Let

$f(t,y)=ty^{1/3}$

Now derivative wrt y is given as

$f_y=\frac{1}{3}ty^{-2/3}$

Finding continuity via the initial value

$f$ is continuous on $R^2$ also $f_y$ is also continuous on $R^2$

Also

$f_y$ is not continuous at the initial value (0,0) so due to Picardi theorem there does not exist an interval such that the IVP has a unique solution.

Part c

as $y^{' }=ty^{1/3}$

Let

$f(t,y)=ty^{1/3}$

Now derivative wrt y is given as

$f_y=\frac{1}{3}ty^{-2/3}$

Finding continuity via the initial value

$f$ is continuous on $R^2$ also $f_y$ is also continuous on $R^2$ when $y\neq 0$

Also

$f , \, f_y$ is continuous at the initial value (0,1) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.