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Problem 4 (Failure of uniqueness). Show that the following initial value problem has two solutions y = y(t) defined for t > 0: { y' = Vy, y(0) = 0. Why does the uniqueness theorem not apply in this case?

Problem 4 (Failure of uniqueness). Show that the following initial value problem has two solutions y = y(t) defined for t > 0: { y' = Vy, y(0) = 0. Why does the uniqueness theorem not apply in this case?

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Problem 4 (Failure of uniqueness). Show that the following initial value problem has two solutions y = y(t) defined for t > 0: { y' Vy, y(0) = 0. Why does the uniqueness theorem not apply in this case?
Transcribed Image Text:Problem 4 (Failure of uniqueness). Show that the following initial value problem has two solutions y = y(t) defined for t > 0: { y' Vy, y(0) = 0. Why does the uniqueness theorem not apply in this case?
Expert Solution
Step 1

Here, y'=y

⇒f(y,t)=y

also, y(0)=0

⇒dfdy=-12y

We can see that, dfdy is not continuous at y=0.

Therefore, dfdy is not continuous at (0,0)

It would not necessarily have a unique solution,

y'=y⇒1ydy=dt ⇒2y=t+c

y(0)=0 ⇒0=0+c⇒c=0

⇒2y=t ⇒y=t2

⇒y=t24 and y=0 are functions that satisfy the given IVP.  

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