Let be a continuous function defined over a rectangle where . If exists and is continuous on , then satisfies the Lipschitz condition with respect to in and the Lipschitz constant is given by
Proof
Since is continuous on a closed rectangle, it is bounded.
supremum exists.
Let .
Let and .
Then, by the Fundamental Theorem of Calculus (FTC),
Note: The converse need not be true.
Example: in .
Feb 4, Lect - 12
Theorem
Let be a continuous function defined over a rectangle where . If exists and is continuous on , then satisfies the Lipschitz condition with respect to in and the Lipschitz constant is given by
Proof
Since is continuous on a closed rectangle, it is bounded, which implies the supremum exists.
Let,
Let and . Then by the Mean Value Theorem,
where is between and .
Remark: The converse need not be true.
Example: in .
Problems
Check whether the following functions satisfy the Lipschitz condition (L.C.) and if so, find .
Gronwall Inequality
Theorem
Let and be two non-negative continuous functions for . Let be any non-negative constant. Then the inequality
implies
Proof
Given that and are continuous and
Let,
Then . Since and are non-negative, . If , then .
Also,
We have from (1),
Since ,
Integrating both sides from to :
From (1), we have . Therefore,
Hence the result.
Feb 4, Lect - 12
Theorem
Let be a continuous function defined over a rectangle where . If exists and is continuous on , then satisfies the Lipschitz condition with respect to in and the Lipschitz constant is given by
Proof
Since is continuous on a closed rectangle, it is bounded, which implies the supremum exists.
Let,
Let and . Then by the Mean Value Theorem,
where is between and .
Remark: The converse need not be true.
Example: in .
Problems
Check whether the following functions satisfy the Lipschitz condition (L.C.) and if so, find .
Gronwall Inequality
Theorem
Let and be two non-negative continuous functions for . Let be any non-negative constant. Then the inequality
implies
Proof
Given that and are continuous and
Let,
Then . Since and are non-negative, . If , then .
Also,
We have from (1),
Since ,
Integrating both sides from to :
From (1), we have . Therefore,
Hence the result.
Corollary
If
Then .
Proof
Given that
By Gronwall’s inequality,
Since is arbitrary, we have .
Picard’s Approximation Scheme
Start with an initial approximation and successively find the approximations of as follows:
If , then will be a solution of the integral equation.
Problems
Solve,
Example 1.
, using Picard’s approximation scheme.
Example 2.
Find the first 4 approximations of .
Example 3.
Find 4 approximations of when .
Solution 1.
We have,
Solution 2.
Feb 5, Lect - 13
Theorem
The sequence of successive approximations
with exists as continuous functions on where and where,
and for each and satisfies
Proof
We prove this by induction.
When , the result is obvious.
When ,
is continuous on a closed rectangle and hence it is bounded. Therefore, there exists such that on .
Let us assume that the result is true for and we will prove it for .
We have,
, and is continuous.
We have
Since is continuous on , there exists such that .
Picard’s Existence and Uniqueness Theorem
Let be a continuous real-valued function defined on a closed rectangle
and let . Further, suppose that satisfies the Lipschitz condition,
with Lipschitz constant . Then the successive approximation,
converges in the interval
to a unique solution of the IVP on .
Proof
We prove the theorem in three steps.
Step 1: The sequence converges.
Step 2: The limit is a solution of the IVP.
Step 3: The solution is unique.
Step 1:
We have,
We can write,
is the partial sum of the series,
If we are able to show that the series converges, then the sequence will converge.
By the previously proven theorem, we have,
Without loss of generality, let us assume that .
Consider,
We claim that,
Proof is by induction.
The result is true for . We assume that it is true for and will prove for .
The RHS is a series of positive terms, which converges to:
Thus the series is dominated by a convergent series of positive terms. Hence by the Weierstrass M-test, the series converges uniformly to .
Since ‘s are continuous and uniformly, is continuous too.
Step 2: is a solution of IVP.
It is enough to prove that is the solution of the integral equation:
We claim that converges uniformly to .
We know that uniformly; given , such that:
uniformly.
is the solution of the integral equation and hence it is a solution of the IVP.