## Astrophysics one year ago @ganeshie8 (O.D.E. Intervals)

1. Astrophysics

I have to find the interval where the solution exists depends on the initial value $$y_0$$. $y'=\frac{ t^2 }{ y(1+t^3) }$ where the initial condition is $$y(0)=y_0$$ I got $y = \left( \frac{ 2 }{ 3 }\ln|1+t^3| \right)^{1/2}+y_0$

2. Astrophysics

Oh wait..

3. anonymous

separation of variables?

4. Astrophysics

Yup

5. anonymous

$y \frac{ dy }{ dt }=\frac{ t^2 }{ 1+t^3 }$ $y \frac{ dy }{ dt }=t^2+\frac{ 1 }{ t }$

6. anonymous

$ydy=(t^2+\frac{ 1 }{ t })dt$

7. Astrophysics

No you cannot do that :P, but that's not what I need help with, I need to find the interval that depends on y_0 xD which I don't know how to do

8. anonymous

ohh

9. Astrophysics

$\frac{ t^2 }{ 1+t^3 } \neq t^2+ \frac{ 1 }{ t }$

10. Astrophysics

I mean if I look at $\frac{ dy }{ dt } = \frac{ t^2 }{ y(1+t^3) }$ we have t can't be -1 right

11. Astrophysics

But I have to look at the solution, I guess since we're focusing on the ivp

12. Astrophysics

initial condition

13. anonymous

mmm

14. anonymous

so are you allowed to solve it though?

15. Astrophysics

Yes

16. anonymous

like what you did in the top post?

17. Astrophysics

Yup, that's what I got when I solved it, I guess if you want to check that, would be nice :P

18. anonymous

hmm let me think

19. Empty

Yeah I thought you weren't supposed to solve this, more like take the derivative or something I forget.

20. Astrophysics

I had to solve it anyways, not sure how to deal with the interval stuff..

21. anonymous

we need to find y(o) in terms of t i rekn

22. anonymous

to do the interval stuff

23. anonymous

Graph it and see if there are any discontinuities

24. anonymous

i don't think its as easy as that..

25. anonymous

I'm on mobile and latex is not displayed properly..

26. anonymous

let me see how you solved y(t)

27. Astrophysics

$\frac{ dy }{ dt } = \frac{ t^2 }{ y(1+t^3) } \implies \int\limits y dy = \int\limits \frac{ t^2 }{ (1+t^3) }dt$ doing u sub yadi, yadi, yada haha.

28. anonymous

lol i'm out of ideas lel

29. Astrophysics

After it says to talk about the theorem of existence and uniqueness, so I think I may have to do that first, or it at least requires that

30. anonymous

you'd know more than me at this stage haha

31. Astrophysics

Haha, well thanks for trying, this interval stuff is pretty difficult, at least for me

32. anonymous

my math knowledge stops after 2nd sem eng maths 2nd yr. so i probs won't have much luck with this

33. Astrophysics

I don't think I really understand theorem of existence and uniqueness either, err

34. anonymous

haha i hope ganeshie comes and saves the day

35. Astrophysics

Well finding the subset D, yeah me to lol

36. anonymous

good luck

37. Astrophysics

Thanks haha

38. ganeshie8

wolfram says the solution is $y^2 = y_0^2 + \frac{2}{3}\log (1+t^3)$

39. ganeshie8

Clearly, the domain is a subset of $$(e^{-3/(2y_0^2)}-1, \infty )$$

40. ganeshie8

Also, depending on whether the intiial value, $$y_0$$ is "positive or negative", the solution curve is either "top one or bottom one" respectively : https://www.desmos.com/calculator/ugxiyenayy

41. Astrophysics

Oh I should've kept all the constants then, also how did you get that domain?

42. ganeshie8

$$y^2 = y_0^2 + \frac{2}{3}\log (1+t^3)$$ solve $$y_0^2 + \frac{2}{3}\log (1+t^3)\gt 0$$

43. Astrophysics

Ohhh wow, so if it depends on the initial value you just look at your solution then? I still thought I had to consider everything separately, but that makes a whole lot of sense

44. ganeshie8

notice that i was careful with what i put above, i didn't say that the domain is $$(e^{-3/(2y_0^2)}-1, \infty )$$ i only said that $$(-\infty , e^{-3/(2y_0^2)}-1]$$ must be excluded from the domain

45. ganeshie8

we can say that much by looking at the solution

46. Astrophysics

Ok I see, so that's where it would be unique?

47. ganeshie8

you need to analyze separately for uniqueness

48. ganeshie8

look at the graph, for each positive value of $$y_0$$, there exists an unique curve that passes through $$(0, y_0)$$ for each negative value of $$y_0$$, there exists an unique curve that passes through $$(0, y_0)$$ yes ?

49. ganeshie8

vary $$y_0$$ and convince yourself that an unique curve passes through $$(0,y_0)$$ for each of ur choice

50. Astrophysics

Yes, very interesting

51. ganeshie8

what can we say about the initial value $$(0,0)$$

52. ganeshie8

plugin x=0, you get y=0 but that makes the slope undefined, so...

53. Astrophysics

Oh oops it's (0,0) so it does not exist

54. Astrophysics

yes

55. ganeshie8

so can we say, when the initial value $$y_0 = 0$$, there is no solution ?

56. Astrophysics

Yes

57. ganeshie8

for all other values of $$y_0$$, there exists an "unique" solution ?

58. Astrophysics

I'd think so but, it doesn't look like it, since the negative and positive seem to give similar results right?

59. Astrophysics

Or does that not matter

60. ganeshie8

How ? (0, -2) gives you some bottom curve as solution (0, 2) gives you some top curve as solution

61. Astrophysics

Ah squared!

62. ganeshie8

you get different curves depending on whether $$y_0$$ is positive or negative

63. Astrophysics

Yes, ok I see

64. baru

so you didn't need to solve it right? existance and uniqueness fails if y' is undefined at at (x0,y0)

65. Astrophysics

I think so...you have to define a subset and take the partial derivative

66. baru

the existance and uniqness theorem for first order ode goes somthing like y' and d(y')/dy should be continuous around (x0,y0)...

67. Astrophysics

I think I'm just mixing up all of these theorems, there's three of them: Theorem of existence and unique for linear eq. General theorem of existence and uniqueness theorem (which I believe is this question) Theorem of existence and uniqueness, which is not limited to the linear eq. I think 1 and 3 are similar...

68. Astrophysics

Yes that's the general theorem of existence and uniqueness

69. baru

ohh..ok

70. baru

this eq isnt linear..so not 1?

71. Astrophysics

Yeah I guess not

72. Astrophysics

I don't really know how to define "unique" in each case though haha, like what exactly does it mean

73. baru

for the first order eq: there are infinite number of curves that satisfy a diff equation, but for a given point (x0,y0) only one curve passes through it, consequently no two solution curves can be tangent or intersect. thats my understanding :)

74. Astrophysics

That sounds pretty good, and this can only be satisfied with a ivp?

75. Astrophysics

I don't really care about the answers xD, I'm just trying to understand all of this, feel so dumb haha.

76. Astrophysics

am*

77. baru

yep..thats why its an ivp. essentially an ivp just supplies a point (x0,y0). if existence and uniqueness holds, i.e one and only one solution passes through a given point, then all the information i would need to single out a particular solution from the infinitely many solutions would be a point on the solution curve(i.e x0,y0 )

78. Astrophysics

I see, so can we say all I.V.P.'s are unique?

79. baru

yes...if existance and uniqueness holds.

80. baru

but it works a bit differently for higher order ODEs... infinite solutions pass through a given point(even if existance and uniqeness holds), so you would need more info than just a point (x0,y0)....(ignore if you havent reached higer order )

81. Astrophysics

Well we did start second order, but it seems the third theorem I mentioned would fall into the higher order category but I just realized/ looks like, "general theorem..." seems to be for IVPs. Yeah, I also think you would need more initial conditions then correct

82. Astrophysics

By the way thanks everyone, that was very helpful, I'm going to go over these definitions and try to figure out what exactly they are saying haha.

83. baru

do let me know if you gain any additional insight... :)

84. Astrophysics

Haha, alright, the difficult part will be understanding the subset stuff

85. Astrophysics

Actually, the initial conditions become a subset of of the domain, so that would make the subset unique..hmm haha, and I think specifically that's where the unique solution is, in that subset, ok I think I'm overthinking this lol.

86. Astrophysics

Having a "unique subset" haha I don't think even makes sense

87. baru

lol...i didnt go into that much detail

88. Astrophysics

Haha, yeah I end up overthinking not sure if that's a good thing or bad

89. Astrophysics

Usually when I see definitions in math I'm clueless, and don't really pay attention, but this is pretty interesting , probably because I actually need it

90. baru

yep, i ask myself the same thing...overthink too much and you end up in a mathematicians territory...

91. Astrophysics

Haha xD, now I understand why they find math so fascinating, have to pay attention to the definitions, and find the exact meaning of them!

92. baru

yep :)

93. ganeshie8

to understand existence and uniqueness thoroughly, you need to be an expert in real analysis let me also tag our analysis gurus @eliassaab @zzr0ck3r

94. Astrophysics

Ok thanks!

95. Astrophysics

Oh @ganeshie8 so the interval we have, is that a subset for the DE pretty much? It's like going backwards from the general and uniqueness theorem kind of right

96. Astrophysics

should be $I = ([e^{-3/2y_0^2}-1]^{1/3},\infty)$