# Thread: Consider the initial value problem

1. ## Consider the initial value problem

Hi everyone,
This is final time, so I am reviewing all chapters ready for final exam. I get 1 problem that I do not know how to figure out. Could you guys please help. I appreciate it.

Problem:

Consider the initial value problem: y' = 5.y^(4/5) ; y(1) = 0

Consider the two functions f1: R --> R and f2: R --> R such that
f1(x) ≡ 0 , f2(x) = (x -1)^5

a/ Show that f1 and f2 are both solutions to initial value problem.

b/ From part a, we see that the initial value problem has more than one solution. why does this not violate the statement of local existence and uniqueness theorem?

2. $\int y^{\frac{-4}{5}}dy=5\int dx \rightarrow 5y^{\frac{1}{5}}=5[x+c_1]\rightarrow y=(x+c_1)^5$

$y(1)=(1+c_1)^5=0\rightarrow 1+c_1=0\rightarrow c_1=-1$

$y=(x-1)^5$

I am struggling with obtaining y=0; however, Maple found it no problem.

3. here is y' = 5y^(4/5) not y' = 5y^(-4/5)

anyways, thanks dwsmith

4. Originally Posted by taro
here is y' = 5y^(4/5) not y' = 5y^(-4/5)

anyways, thanks dwsmith
I used separation of variables that is why I have a negative. Made a mistake with the constant 5 though.

$\frac{dy}{dx}=5y^{\frac{4}{5}}\rightarrow\frac{1}{ y^{\frac{4}{5}}}\frac{dy}{dx}=\frac{1}{y^{\frac{4} {5}}}5y^{\frac{4}{5}}\rightarrow\frac{dy}{y^{\frac {4}{5}}}=5dx$

$\int y^{\frac{-4}{5}}=5\int dx$ because $\frac{1}{y^{\frac{4}{5}}}=y^\frac{-4}{5}$

I am going to fix my original post due to the error with the 5.

5. Oh, I understand now. Thanks alot

6. Sorry, I got a typing mistake. The right problem is:

Problem:

Consider the initial value problem: y' = 5.y^(4/5) ; y(1) = 0

Consider the two functions f1: R --> R and f2: R --> R such that
f1(x) ≡ 0 , f2(x) = (x -1)^5

a/ Show that f1 and f2 are both solutions to initial value problem.

b/ From part a, we see that the initial value problem has more than one solution. why does this not violate the statement of local existence and uniqueness theorem?
f1(x) ≡ 0 not f1(x) = 0

7. Originally Posted by taro
Sorry, I got a typing mistake. The right problem is:

f1(x) ≡ 0 not f1(x) = 0

Modulo what?

8. Originally Posted by taro
Hi everyone,
This is final time, so I am reviewing all chapters ready for final exam. I get 1 problem that I do not know how to figure out. Could you guys please help. I appreciate it.

Problem:

Consider the initial value problem: y' = 5.y^(4/5) ; y(1) = 0

Consider the two functions f1: R --> R and f2: R --> R such that
f1(x) ≡ 0 , f2(x) = (x -1)^5

a/ Show that f1 and f2 are both solutions to initial value problem.

b/ From part a, we see that the initial value problem has more than one solution. why does this not violate the statement of local existence and uniqueness theorem?
Writing the ODE in 'standard form'...

$y^{'} = f(x,y)$ , $y(x_{0})=y_{0}$ (1)

... it has one and anly one solution if $f(x,y)$ is 'Lipschitz continous' in $(x_{0},y_{0})$. In Your case is $f(x,y)=\frac{1}{2}\cdot y^{\frac{4}{5}}$ which is not Lipschitz continous in $y_{0}=0$ because its partial derivative is unbounded. Furthemore in this case $f(x,y)$ is function of the y alone and that means that if $\phi (x)$ is solution of...

$y^{'} = \frac{1}{2}\cdot y^{\frac{4}{5}}$ (2)

... the $\phi(x-\xi)$ with $\xi >0$ arbitrary is also solution. In this case...

$\phi(x)=\left\{\begin{array}{ll}(x-1)^{5} ,\,\,x > 1\\{}\\0 ,\,\, x\le 1\end{array}\right.$ (3)

... satisfies (2) with initial condition $y(1)=0$ and that means that $\phi (x-\xi)$ with $\xi >0$ arbitrary also satisfies (2) with the same initial condition...

Kind regards

$\chi$ $\sigma$

9. Originally Posted by taro
Hi everyone,
This is final time, so I am reviewing all chapters ready for final exam. I get 1 problem that I do not know how to figure out. Could you guys please help. I appreciate it.

Problem:

Consider the initial value problem: y' = 5.y^(4/5) ; y(1) = 0

Consider the two functions f1: R --> R and f2: R --> R such that
f1(x) ≡ 0 , f2(x) = (x -1)^5

a/ Show that f1 and f2 are both solutions to initial value problem.
To "show" that a function satisfies a d.e. it is NOT necessary to integrate the equation. If y= f1 is identically 0 then, of course, it is constant and so y'= 0. The equation becomes 0= 5(0)^(4/5)= 0 which is true, and the initial condition, y(1)= f1(1)= 0 is true.
If y= f2= (x- 1)^5, then y'= 5(x- 1)^(4)= 5((x- 1)^5)(4/5) is true and y(1)= (1- 1)^5= 0 is true.

b/ From part a, we see that the initial value problem has more than one solution. why does this not violate the statement of local existence and uniqueness theorem?
Well, what exactly does the "local existence and uniqueness theorem" say? Especially the hypotheses. Note that, with y'= F(x,y)= 5y^(4/5), F_y= 4y^{-1/5} which is not defined at x= 0.

10. Thank you so much. I got it.