Proving existence results for some initial and boundary value problem, we usually find a corresponding integral equation first and then use some fixed point theorem to prove the existence of solution of differential equation. Why finding corresponding integral equation is not enough and it is important to use fixed point theorems to prove existence of solutions? some counter example will be helpful.
Dear Asma,
Consider the initial value problem dy/dt = f(t,y(t)), y(p) = q,
if f(t,y(t)) is continuous, one can't guarantee the uniqueness
of solution.
Example: Consider the initial value problem:
dy/dt = 2√t, y(0) = 0.
Direct integration shows that y=(4/3)t√t.
Is it a unique solution? The answer is no.
Consider the family of solutions:
ys = (t-s)², for t>s>0. ys (0)=0 for t
Dear Asma,
The main point is uniqueness.
Fixed points are not used to show the existence of solutions.
The counterexample shows the existence of infinitely many solutions.
In fact, if we define the operator
T(y)[x] =y(p) + ∫p x f(t,y(t))dt
Then any solution y of the IVP must satisfies satisfies T[y]= y ,
that's why uniqueness of the fixed point is equivalent to
the uniqueness of the solution.
T(y)[x] =y(p) + ∫p x f(t,y(t))dt
As an example:
dy/dt = y , y(0)=1.
We apply the same idea :
T(y)[x] =1 + ∫0 x y(t)dt and Ty = y (fixed point if exist)
We apply the iterations
yn+1 = 1 + ∫0 x yn dt
y1 = 1 + ∫0 x y0dt = 1 + ∫0 x 1.dt = 1+x
y2 = 1+ ∫0 x (1+x)dt = 1 + x + x2 /2
as n goes to infinity
the sequence {yn } converges to 1 + x + x2 /2 + x3 /3! +.... xn /n!+.... = ex
Indeed, it is the only solution is y(x) = ex that represents
the fixed point of the operator equation T(y) = y.
Best regards
dear
the corresponding integral equation is not enough because the such integral equation needs also to solution wrt x(t) specially when the function f is abstracted.
So ,we cannot consider it as solution of main problem but it's equivalent to it.
@ Sabri,
Your post is not clear.
Direct integration can determine the exact solution.
The previous answer shows the relation between the uniqueness of the solution and the fixed point theorem.
( Read the notes followed by Asma Question).
Issam> Example: Consider the initial value problem:
dy/dt = 2√t, y(0) = 0.
Direct integration shows that y=(4/3)t√t.
Is it a unique solution? The answer is no.
Consider the family of solutions [of the equation y'(t) = 2 \sqrt{t}:
ys = (t-s)², for t>s>0. ys (0)=0 for t0) are not defined for 0< t \le s, unless there should be: ys (t)=0 for 0\le t \le s, instead of ys (0)=0 for t
Added: In the case of y'= 2 \sqrt{y} the direct derivation leads to the unique solutions under assumption, that they positive. Otherwise, the intermediate step of dividing by \sqrt{y} is not justified. Combining with suitable additional analysis we get:
Corollary: y(t) = t2, t\ge 0, is the unique solutions of the ODE y'(t) = 2 \sqrt{y(t)}, t \ge 0, among solutions positive within the open interval t\in (0,\infty), satisfying the intial condition y(0)=0.
Regards, Joachim
Dear Joachim,
Thank you for the excellent answer.
In fact, I focused on the part that deals with :
[ Why finding corresponding integral equation is not enough and it is important to use fixed point theorems to prove existence of solutions? ]
Asma wonder if there is any relation between
1, The Uniqueness of the solution and 2. Fixed point theorem.
I have two answers, one of them includes a DE where the solution is not unique, and the other DE has a unique solution.
I show how the solution considered as a fixed point of some operator.
Best wishes
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