Science:Math Exam Resources/Courses/MATH215/December 2013/Question 01 (c)

MATH215 December 2013

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  

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  •  April 2013  •  December 2011  •  December 2013  •  

Question 01 (c)
$y_{1}(x)=x$ and $y_{2}(x)=-x$ are both solutions of the first-order initial value problem: $yy'=x,\;\;y(0)=0$. Does this contradict the existence and uniqueness theorem (`Picard's theorem' or `Cauchy-Lipschitz theorem')? Explain.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The point of a mathematical theorem is that it always holds true provided the conditions are satisfied. Hence this example does not contradict the theorem, but instead does not satisfy the conditions of the theorem.

Hint 2
Picard's theorem ensures existence and uniqueness of a solution provided the differential equation is of a certain form. What form?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Picard's theorem (at least one version of it) states that initial value problems of the form ${\frac {dy}{dx}}=f(x,y),\quad y(x_{0})=y_{0}$ have a unique solution over some interval I containing $x_{0}$ provided f is continuous at $(x_{0},y_{0})$. If we rearrange the ODE problem $y{\frac {dy}{dx}}=x,\quad y(0)=0$, we get ${\frac {dy}{dx}}={\frac {x}{y}},\quad y(0)=0$. When $x=0$, $y=0$, but $\displaystyle {\frac {x}{y}}$ is not continuous at (0,0) because it is not defined. Therefore the theorem cannot be used and uniqueness is not guaranteed.