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

Thank you Issam.

But we use fixed point theorems to prove existence.

for uniqeuness we give separate proofs using lipschitz type conditions.

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

If you have the corresponding integral equation, then you need the differentiability to fullfill the differential equation, and this is not generally available. See: Jawad: Unique strong and strict solution for inhomogeneous hyperbolic differential equations in Banach space (revised revision). You can see there that the solution of the integral equation exists for f belongs to L^1 (t_1, t_2) but this solution don't satisfy the differential equation. It is proved there that one needs for that that f is of bounded variation. The proceeding via the fixed point theorems or via Picard iteration is a usually way to construct a solution for nonlinear problems under restrictions on the nonlinearity. Naturally these restrictions depend on the regularity of the solution you need.