Existence & Uniqueness Theorem, Ex1
HTML-код
- Опубликовано: 6 сен 2024
- Existence & Uniqueness Theorem, Ex1
Subscribe for more math for fun videos 👉 bit.ly/3o2fMNo
For more calculus & differential equation tutorials, check out @justcalculus
👉 / justcalculus
To learn how to solve different types of differential equations:
Check out the differential equation playlist: 👉 bit.ly/bprpdiffeq
Check out my Diff Eq Ultimate Study Guide: 👉 • 24 First-Order Differe...
💪 Support this channel, / blackpenredpen
#DifferentialEquations
we cannot ignore the fact that he can flip pens instantly.
: )
That part!
you are a wonderful human
Uo o p p
2 Years after your class and I'm still using your videos to get through college, thanks a lot Profesor (Matthew K., Calc 2 Fall 2019)
Hey Matthew! Hope all is well!
In the first problem, how is f(x,y) continuous since left hand limit does not exist(limit y tends to 3-)( and consequently, is not equal to f(4,3) and the right hand limit).Please explain sir! I'm very confused.
y=3 is an endpoint for f(x,y). Thus, f(x,y) is NOT continuous at the point (4,3). In other words, y-3 must be strictly greater than 0, NOT great than or equal to zero. This means we cannot even guarantee the existence of a solution to this differential equation. Good catch on the above video!
can you make a tutorial of how to change markers with one hand quickly.
This is a great video man. Simply and slowly explained like a highschooler could understand it. You're great at breaking it down to make it seem easy unlike so many other people on youtube, thanks!
It might be part of the theroem, but I want to ask why we take the partial of y (df/dy), not x ?
Did you found an answer?
Why is x sqrt(y-3) continuous around (4,3)? It's undefined when y = 2.999.
but y=2.999 is not in the domain of sqrt(y-3).
We say a function is continuous if it is cont. at the values in its domain.
so, sqrt(y-3) is cont. when y>=3
Thank you for your good explanation.
But as I know that for existence and uniqueness of solution, there should be a region surrounding the initial value y(x0)=y0 in which f(x,y) and df/dy are both continuous
R={(x,y); |x-x0|
anas Nassar Without reading much into it, is that a "fancy" way of saying Continuous Around vs Continuous At?
Yes, this is what I understand from the definition. the function must be continuous around this point (which also means that it's continuous at this point).
Look at this notes (page 1 theorem 2):
www.math.uiuc.edu/~tyson/existence.pdf
Because of limits we re talking about continuity
Thanks for saving me.
you're welcome
The Ivp dy/dx=3y^(2/3) ,y(0)=0 has two solutions or infinitely many solutions?
Please give reply sir
Notes integral of 1/√(y-3) exists if y is not = to 3 for the uncountable number of y values. The first answer is when y is not = to 3 for the uncomfortable number of x values. The second answer is when y=3 for the uncountable number of x values which is the case with the second answer.
When you squared both sides, I think you should add the condition x^2/4-4 >= 0.Because when x^2/4-4 < 0, y = (x^2/4-4)^2+3 is not solution of the equation.We have 1/4*(-16+x^2)*x = -1/4*(-16+x^2)*x.
i think because he wants to solve the ivp at x=4, he is not interested in taking restrictions, as 4^2/4 -4 =0 >= 0
Is your function not discontinuous given that y < 3 returns a nonreal answer? In otherwords, it's continuous at 3 itself but not around 3, no?
The condition that "partial derivative of f with y is bounded " is sufficient but NOT necessary. The condition fails does not guarantee that the equation has no unique solution. Please clarify.
Thank you for explaining this concept in detail.
Did you assume that Y cannot be 3 when you timed both sides by 1/sqrt(y-3)? But Y can be 3
you say it is a missing solution, but it is covered by your differential solution when ( 1 / 4 )^x^2 - 4 = 0 which means that x = + or - 4
Love your math videos! makes me exited about maths again after chucking my text book out the window!
i was just so happy when i found out that you have a video about this theorem sir thank you so much !
Thanku for this video.It helps me to understand the existence and uniqueness theorem in a proper way...May God bless u
May I know which book did you refer for this
Thankyou, my professor is teaching with zoom and a horrible microphone and he is maddeningly boring, thankfully he has no due dates and I can just teach myself with HW and help from videos like these.
if (x^2 /4 - 4)^2 +3 is a solution, then it must satisfy the initial condition. I don't see how it does! even its derivative does not!
pls explain... thank you
mathphschjhb ((4)^2/4-4)^2+3=3
I can’t help but laugh imagining a question going like
« Can you promise me there’s a unique solution? »
Love your explanation and how you smile eachvtime,you make it look very exciting .keep on 🤗
Love your pep and energy!!!
thanks a lot, really clear
廖逸如 you're welcome 😊
y=3 is not missing solution, it is included when x=4
and x = - 4 also
Thank you!
how did you determine at 1:41 that the function is continuous?
is it because we found a particular number by putting (4,3) to function? because we couldn't find any solution by putting (4,3) and we found it is discontinuous at 3:19
Yea, it's easier to find that where the function is not continuous. If there's no place the function is not continuous, then its continuous. Usually check if the denominator is zero or not.
Nice explanation👌🏻
Love from india❤
when we say dy/dx= x* square root of(y-3), why in the second step @0:46 we say f(x,y) = dy/dx= x* square root of(y-3). Should we take a integral first before make it equal to f(x,y)?
This is a year late but he's taking the derivative with respect to y so treat x as a constant. If you were using the product rule, it would have two functions with the same variable.
THAAANKS! My exam is tomorrow and I had no idea how to solve this
you uploded this on my brithday
This is a nice example to chew on. I'll be giving a lecture on the existence and uniqueness theorem in a few days. I'll probably steal this example!
But how did you jump up to the solution of y=3? Quite obvious that it satisfies all the conditions but didn't undertsand.
First : 5:28, you dont multiply the equation by dx, that is not correctly explained. Being a separable equation the member on the left becomes the derivative of a compound function and then you integrate on both sides, my teachers would give you 0 on that exercise because of that.
Second: in the end you didnt exactly prove that there are multiple solutions, you have to draw the graphs and study it
sqrt(y-3) is not continuous around y=3 as there is no positive constant, c, for which sqrt(y-3) exists when y=3-c
would like to mention that I love your videos, though
You're such a genius. Keep it up, but I would like to know how you were able know that y(x) = 3 is also a solution.
We have come to know that it is a solution but my question rests on how we are able to know that.
Could that proposition be based on any characteristic of the IVP? If so, could it be specified?
Nice explanation ! Plz tell me that, is dy/dX=√|y| gives unique solution about (0,0). Plz reply me fast.
You're missing solutions because you're forgetting that the expression you get in the separation of variables holds only for y>3 and there are solutions obtained by gluing the two "types" of solutions you found
appreciate this so much
My go-to math RUclipsr
What is interesting this equation for each point y>3 has two different solutions. The special solution y=3 is an envelope of the set of all solutions with C≤0. It touches every solution such solution, and doesn't touch any of the solution where C>0: imgur.com/aB8odbb
Where dose the 2 in x/2(y-3)^1/2 come from?
its the derivative of it
Nyc explanation. So we can say that it possess only two solution Or we say that it has infinitely many solution? Plz explain sir
Thank you
beastpenredpen coming in clutch yet again
really nice work, thanks!
I like how it looks like you're peeking from the right. I very much appreciate your explanation. Very helpful
how do you find the missing solution if you are not given initial points?????
That sleight of hand at 3:16 is probably why your channel is named blackpenredpen
Thanks
Can you please do some Cauchy problems with qualitative graph, Please help me
Sir Please, make conceptual video on this exact and uniqueness topic
why is the partial derivative of x (y-3)^-.5 not (x(y-3)^.5)/2
thank you so much. your teaching is very engaging and clear.
Thanks!
You break it down!! Thanks.
I am still confused with your final conclusion that y(x) = 3 for all x, is that means all element x to be the solution?
Thank you so much 😭❤️❤️
why only df/dy to check the uniqueness? How about df/dx? Is it enough only to check df/dy?
Great question
Great video, thankyou. Hope you are well
At 2:17 why did you take partial with only respect y? Why did you differentiate with x as well?
possibly because it wasnt needed as taking a partial derivative and seeing it was niy continuous proves the question
When does uniquenes of ivp fails
Very good explanations, thanks
Thanks GOAT!
Man, you save me in a differential equations
Thank you very much! You helped me clear a horrible confusion I had
Thank you my lord
Thank you professor.
Great explaining
Excellent explaining.
You are an amazing teacher!
thank you very much.. you helped me a lot!!
So clear to understand, thank you sir!
thank God
why we have missing solution y=3, where it is come from ?
ivan luthfi y(4) = 3 thats where it comes from.
good video thank u so much
Thaaaaaaaank You Maaaan ! God Bless You
You sir are great.
Awesome video!
Hey playlist doesn’t exists?
Great video. So neat. Thanks a lot! I really appreciate it :)
thank you sir
is good lecturar
lovely!
you're a legend mate.
Thanks!
How will get the initial condition
Prasad Palle : )
We can prove the solution isn’t unique as y=3 is a solution along with y=3+(1/4*x^2-4)^2
Legend
Great
Thank you =)
you're welcome
Fantastic
At first it looks weird because dy/dy=x*f(y-3) where f(0)=0 guaranties that y(x)=3 is always a solution. So? We always have two? Well, of course not: when the existence and uniqueness theorem are satisfied, the constant solution emerges from the separation of variables!
Funny video.
that mic 😍😍😍
W video
"does this have a unique solution? no. Can I still have one unique solution? yes." sigh...
I am abangladeshi thanks
Ha, funny enough when I solved the equation in my head I got y=0, which is always an annoying solution, but it was, a, solution