- #1
heman
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Can anybody help me with the proof of Existence and Uniqueness Theorems.
Existence and Uniqueness Theorem
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Discussion Overview
The discussion centers around the Existence and Uniqueness Theorem in the context of ordinary differential equations (ODEs). Participants explore the conditions under which unique solutions exist for given initial value problems, discussing various examples and mathematical properties relevant to the theorem.
Discussion Character
- Technical explanation
- Debate/contested
- Mathematical reasoning
- Homework-related
Main Points Raised
- Some participants discuss the criteria for the existence of a unique solution, specifically the continuity of the function f(x,y) and its partial derivative with respect to y.
- Others argue that continuity of the partial derivative is not necessary, proposing that the Lipschitz condition in y is sufficient for uniqueness.
- A participant mentions that the proof involves showing the initial value problem has a unique solution if the corresponding integral equation has a unique solution.
- There is a suggestion that continuity implies boundedness, which is relevant to the proof's convergence arguments.
- Some participants express a need for further clarification on the Lipschitz condition and its implications for existence and uniqueness.
- Multiple participants request links to resources or textbooks that provide detailed proofs of the theorem.
- Discussions include references to specific textbooks and personal experiences with them, indicating varying preferences for learning materials.
Areas of Agreement / Disagreement
Participants express differing views on the necessity of continuity of the partial derivative for uniqueness, with some asserting it is essential while others maintain that the Lipschitz condition suffices. The discussion remains unresolved regarding the best approach to understanding the proof and the implications of these conditions.
Contextual Notes
Participants mention various examples of ODEs that fail the criteria for uniqueness, indicating the complexity of the topic. There are also references to the Banach fixed point theorem and its application in proving the existence and uniqueness of solutions, but the details remain partially explored.
Who May Find This Useful
This discussion may be useful for students and practitioners in mathematics and engineering who are studying differential equations, particularly those interested in the theoretical underpinnings of existence and uniqueness theorems.
- #2
saltydog
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heman said:
Start with the basic first:
\frac{dy}{dx}=f(x,y)
A unique solution of this equation about the point (x_0,y_0) exists if f(x,y) and it's partial with respect to y is continuous about the point. So:
\frac{dy}{dx}=\frac{1}{x}
would fail this criteria along the x-axis. That is, any initial condition (0,y) would not be guaranteed to lead to a unique solution. How about:
\frac{dy}{dx}=\frac{1}{1+y}
Well, taking the partial with respect to y leads to \frac{-1}{(1+y)^2} and that's not continuous at y=-1 so again, any initial condition of the form (x,-1) would not be guaranteed to lead to a unique solution and in fact the IVP of y(0)=-1 leads to 2 solutions. Same dif with higher-ordered ODEs. Just break it up into a simultaneous system and apply this rule to the individual parts.
What about:
ty^{''}+(t-1)y^{'}+y=t^2;\quad y(0)=0,\quad y^{'}(0)=0
Come up with any ODE with a form that fails this criteria and there will be a problem with existence or uniqueness.
Or are you having problems with the specific details of the proof? Rainville and Bedient, "Elementary Differential Equations" is the reference I've used to study the proof and is one I think is nice to follow.
- #3
heman
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saltydog said:
Thanks for Amazing Clarification Salty but my doubt is this that i want to know why the happening of continuity of F(x,y) and Fy(x,y) ensures the uniqueness.!
- #4
saltydog
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heman said:
Well, I tell you what Heman. It's best to carefully go through the proof. But in general, continuity implies that it's bounded. And as you go through the proof, an expression is created which is bounded by a particular series which is show to converge. Later in the proof, a series of those expressions converge to the solution of the ODE with convergence guaranteed by the convergence of the series which bounds it.
- #5
heman
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Salty,
Can you give me an online link on Differential eqns. which stresses on proof!
Actually my textbook Kreyzig sucks!
Can you give me an online link on Differential eqns. which stresses on proof!
Actually my textbook Kreyzig sucks!
- #6
HallsofIvy
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Actually, continuity of Fy(x,y) is not necessary. What is necessary for uniqueness is that F be "Lipschitz" in y. A function of one variable is Lipschitz (on a set) if and only if |f(x)-f(y)|\leq C|x-y|. If a function has a continuous derivative on a set, then you can use the mean value theorem to show that it must be Lipschitz on that set so that "continuous derivative" is a sufficient condition.
The basic idea of the "uniqueness and existence" theorem is this: Show that the initial value problem \frac{dy}{dx}= F(x, y) y(x_0)= y_0 has a unique solution if and only if the integral equation y(x)= \int_{x_0}^x f(t,y(t))dt+ y_0 has a unique solution.
Define the the operator: A(y)= y_0+ \int_{x_0}^xf(t,y(t))dt from the set of functions continuous on some interval containing x0 to itself. Clearly, the integral equation has a unique solution if and only if that operator has a unique "fixed point". You then use continuity and Lipschitz property of F to show that, if necessary limiting the domain in R2, the operator is a "contraction operator" and so the Banach fixed point theorem can be applied.
The basic idea of the "uniqueness and existence" theorem is this: Show that the initial value problem \frac{dy}{dx}= F(x, y) y(x_0)= y_0 has a unique solution if and only if the integral equation y(x)= \int_{x_0}^x f(t,y(t))dt+ y_0 has a unique solution.
Define the the operator: A(y)= y_0+ \int_{x_0}^xf(t,y(t))dt from the set of functions continuous on some interval containing x0 to itself. Clearly, the integral equation has a unique solution if and only if that operator has a unique "fixed point". You then use continuity and Lipschitz property of F to show that, if necessary limiting the domain in R2, the operator is a "contraction operator" and so the Banach fixed point theorem can be applied.
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- #7
saltydog
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HallsofIvy said:
Hall, thanks. Is that a typo? Should it read:
|f(x)-f(y)|\leq K|x-y|
for some value K thus ensuring the function f is bounded.
- #8
saltydog
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heman said:
Heman, really I think only "text book in hand" is the best way to learn the proof. Go through each line carefully and make sure you understand every detail. Draw pictures, go back and review all the theorems that are quoted. Do examples.
Also, I have Kreyzig. It's a good book for Engineers but not for Mathematicians.
God I hope I don't get in trouble with the Engineers in here for saying that.
- #9
heman
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Salty,i have got the book Tenenbaum, Morris Today and i will see to it that i follow the proof.!
- #10
HallsofIvy
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Yes, saltydog, that was a typo! Thanks.
(I've edited it. I used "\le" instead of "\leq"!)
(I've edited it. I used "\le" instead of "\leq"!)
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- #11
heman
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Thanks Halls for the explanation! 
- #12
Feynman
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You can also think about athers existence theorem like LAX-Miligram, Hill-yosida-Lumeur-Phillips,...
- #13
Maxwell
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saltydog said:
YOU ARE IN TROUBLE!
I LOVE Kreyzig... lol. I still use that book as a reference for O.D.E's and Linear Algebra stuff. Then again, I am an engineer.
- #14
barathiviji
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can anyone explain about the lipschitz condition? with an example and how it is related to existence and uniqueness of solution...i am in problem...help me
- #15
matematikawan
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Me too required further explanation on the lipschitz condition.
It is not that we prove uniqueness by assuming there are two solutions y1(x) and y2(x) and then show that y1=y2.
But do not know how they go about proving the existence of solution.
It is not that we prove uniqueness by assuming there are two solutions y1(x) and y2(x) and then show that y1=y2.
But do not know how they go about proving the existence of solution.
- #16
barathiviji
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give a link for the proof of this uniqueness and existence theorem...and also give the link of differential equation which includes singular points, that is covering pg syllabus...
- #17
elenamk
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I also need a proof of this uniqueness and existence theorem!
I have to use the Banach Fixed point theorem...
I have to use the Banach Fixed point theorem...
- #18
HallsofIvy
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The existence and uniquess theorem says:
If f(x,y) is continuous in both variables in some neighborhood of (x0,y0) and satifies the Lipshchitz property in the variable y in that neighborhood, then there exist some interval about x0 such that there is a unique solution to the differential equation dy/dx= f(x,y) on that interval that satisfies y(x0)= y0.
I explained what "Lispschitz" means earlier in this thread.
Yes, you should use the Banach fixed point theorem. You would need to know (probably will not be required to prove it) that the set of all continuous functions on a given interval, with the metric (distance function) d(f, g)= max|f(x)- g(x)| where the max is taken over all points in the interval, is a "complete metric space" in order to apply the Banach fixed point theorem.
Since f(x,y) satisfies properties on some neighborhood of (x0, y0) there exists some 'interval' \left{(x, y)| |x- x_0|< \delta |y- y_0|< \epsilon\right} on which f satisfies those properties. Now consider the set of functions continuous on that interval, with the "metric" defined above. In order to use the Banach fixed point theorem you would need to prove that the operator,
\int_{x_0}^x f(t, y(t))dt+ y_0[/itex] maps that set of functions into itself. <br /> <br /> In fact, it <b>doesn't</b>! But you can reduce \delta, reducing the possible values for x, so that it does map this smaller interval into itself.<br /> <br /> Then, since the Banach fixed point theorem requires a "contraction map" you need to prove that this operator is a "contraction". That is, that, if f and g are two functions in the (smaller) interval, then, for some c< 1,<br /> max\left|\left(\int_{x_0}^x f(t,y(t))dt+ y_0\right)- \left\int_{x_0}^x g(t,y(t))dt+ y_0\right)\right|<br /> = max\left|\int_{x_0}^x f(t,y(t))- g(t,y(t))dt\right|\le c max\left|f((x,y(x))- g(x,y(x)\right|<br /> where the maximum is take over all x values in the interval.<br /> <br /> Again, this NOT generally true! You will have to again reduce the \delta value so that it is. Once you have done that, you can appeal to the Banach fixed point theorem to assert that there is a unique function y(x), on that interval, such that<br /> y(x)= \int_{x_0}^x f(t, y(t))dt+ y_0<br /> and so dy/dx= f(x,y), y(x<sub>0</sub>)= y<sub>0</sub>.
If f(x,y) is continuous in both variables in some neighborhood of (x0,y0) and satifies the Lipshchitz property in the variable y in that neighborhood, then there exist some interval about x0 such that there is a unique solution to the differential equation dy/dx= f(x,y) on that interval that satisfies y(x0)= y0.
I explained what "Lispschitz" means earlier in this thread.
Yes, you should use the Banach fixed point theorem. You would need to know (probably will not be required to prove it) that the set of all continuous functions on a given interval, with the metric (distance function) d(f, g)= max|f(x)- g(x)| where the max is taken over all points in the interval, is a "complete metric space" in order to apply the Banach fixed point theorem.
Since f(x,y) satisfies properties on some neighborhood of (x0, y0) there exists some 'interval' \left{(x, y)| |x- x_0|< \delta |y- y_0|< \epsilon\right} on which f satisfies those properties. Now consider the set of functions continuous on that interval, with the "metric" defined above. In order to use the Banach fixed point theorem you would need to prove that the operator,
\int_{x_0}^x f(t, y(t))dt+ y_0[/itex] maps that set of functions into itself. <br /> <br /> In fact, it <b>doesn't</b>! But you can reduce \delta, reducing the possible values for x, so that it does map this smaller interval into itself.<br /> <br /> Then, since the Banach fixed point theorem requires a "contraction map" you need to prove that this operator is a "contraction". That is, that, if f and g are two functions in the (smaller) interval, then, for some c< 1,<br /> max\left|\left(\int_{x_0}^x f(t,y(t))dt+ y_0\right)- \left\int_{x_0}^x g(t,y(t))dt+ y_0\right)\right|<br /> = max\left|\int_{x_0}^x f(t,y(t))- g(t,y(t))dt\right|\le c max\left|f((x,y(x))- g(x,y(x)\right|<br /> where the maximum is take over all x values in the interval.<br /> <br /> Again, this NOT generally true! You will have to again reduce the \delta value so that it is. Once you have done that, you can appeal to the Banach fixed point theorem to assert that there is a unique function y(x), on that interval, such that<br /> y(x)= \int_{x_0}^x f(t, y(t))dt+ y_0<br /> and so dy/dx= f(x,y), y(x<sub>0</sub>)= y<sub>0</sub>.
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- #19
elenamk
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Thank you , HallsofIvy...
You really helped me a lot...
You really helped me a lot...
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