Analyzing IVP Solutions: Applying Lipschitz Conditions for Uniqueness

This assignment explores the application of the Lipschitz condition in determining the uniqueness of solutions for initial value problems (IVPs). It covers several problems, including verifying the Lipschitz condition for a given function, determining the interval of existence for a solution, and analyzi...Read More

Ans 1)
f ͼ C(s)
S = [ξ-a, ξ+a] x R
S satisfies Lipscitz Condition with respect to y in
S : |f(x,y) – f(x,z)| <= L|y-z|
Let y’ = f(t,x), for ξ<=x<= ξ+a, y(ξ)= η
Let C(J), J = [ξ-a, ξ+a]
||y|| = maxxͼJ |y(x) e-α|x-ξ||
According to Lipscitz Condition;
|f(t,y1) – f(t,y2)| <= L|y1 – y2|
Whenever (t, y1), (t,y2) are in D and D ͼ R2
Lipscitz condition is sufficient for uniqueness
||y|| = max|y(x) e- α|x-ξ| |
If e-α|x- ξ| = 0 then ||y|| can be maximized.
Now this is a modulus function, so x- ξ is always positive.
If x= ξ then e- α|x-ξ| = maximum
So, ||y|| = y(x)
|y’| = y’(x) and y(ξ) = η from equation that we got from S;
So we can say that in interval of J there exist an unique solution to this IVP.
Ans 2. Dy/dx=y 2
y 2 dy=dx
y 3 /3= x + c
If
y(0)=1

1/3=c
y 3 /3= x + 1/3
y 3 = 3x + 1
x=(y 3 -1)/3 &lt;= a
y-1 = cube root(3x+1) -1 &lt;= b
Solution exits on interval (y 3 -1)/3 to cube root(3x+1) -1.
Ans3)
f(x,y) satisfies a local Lipscitz condition; wrt y
D ϲ= R;
If A ϲ D is compact and f bounded on A then f satisfies a Lipscitz condition
w.r.t. y in A.
➔ f(x,y) is bounded on A; - (1)
➔ A c D -(2)
➔ And f(x,y) satisfies Lipscitz condition w.r.t. y in D where D c= R ----(3)
From analysing these three equations in order 3>1>2 then we can say that
actual domain of f(x,y) lies in A while satisfying the Lipscitz equation.
Ans 4)
Lipscitz condition is :
|f(x,y) – f(x,z)| <= L(y-z) w.r.t. y;
v ϵ ([a,b]) and w ϵ ([a,b]) and x ϵ [a,b];
If we want to establish the relation w.r.t. x then x <= [a,b]
V <= C[a,b] and w<= [a,b]
So Lipscitz condition is

|f(x, v(x)) – f(x, w(x))| <= L|v(x) – w(x)|
Ans 5. y = xy + 2/6 ( 1+y 2 ) 3/2 + c
y(0) = 1
1 = 0 + 2/6 (1) + c
3 = c
y = xy + 2/6 ( 1+y 2 ) 3/2 + 3 Lower Solution
y = xy + 2/6 (( 1+y 2 ) 3/2 )/y + c
y(0) = 1
y= 0 + 2/6 +c
y = xy + 2/6 (( 1+y 2 ) 3/2 )/y + 3 Upper Solution