Show that the two-step method
is of order 2 but does not satisfy the root condition.
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Method finds an approximation for such that .
Let be the th point of evaluation where is the starting point and is the step size.
Taylor Expansion of y' about a_0[edit | edit source]
Substituting into the second term on the right hand side of and simplifying yields
Taylor Expansion of y about a_0[edit | edit source]
Since , we also take Taylor Expansion of about
Substituting and simplifying yields,
Take Difference of Taylor Expansions[edit | edit source]
Hence shows that (1) is a method of order 2.
Does Not Satisfy Root Condition[edit | edit source]
The Characteristic equation of (1) is
Giving the roots
clearly does not satisfy
Give an example to show that the method (1) need not converge when solving .
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Let . Then . We have the difference equation
which has general solution (use the roots)
If , then as
Hence, . Therefore if , then .
Consider the boundary value problem
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Prove that has at most one solution
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Let and be solutions. Let .
By subtracting the two equations and their conditions we have
Multiplying by test function and integrating by parts from 0 to 1, we want to find such that for all
Let . Then, we have
Since , , and are all , . Hence .
Discretize the problem. Take a uniform partition of
Use the three point difference formula for and the simplest difference formula for the boundary condition at . Write the resulting system as a matrix vector equation where
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The three point difference formula for is given by
Substituting into with our difference formula we have in matrix formulation
We can eliminate the variable by using the approximation
which implies
Using this relationship and the three difference formula, we have
Since , we can eliminate the variable by substituting into the n-1 equation.
Prove that the equation found in has a unique solution
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Since the matrix is diagonally dominant, the system has unique solution.
Transform the problem into an equivalent problem with homogeneous boundary conditions.
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Let ,a solution of the boundary value problem, be represented as the sum of solutions to two different boundary value problems i.e.
where
Suppose . Then
and which implies and hence
Substituting into , we then have
which implies
Since , we have
Therefore an equivalent boundary value problem with hemogenous boundary conditions is given by
Variational Formulation and Sobolev space[edit | edit source]
Using the problem's notation, we want to find such that for all , we have
The above comes from integrating by parts and applying the boundary conditions.
To show that is unique, we show that the hypothesis of the Lax-Milgram Theorem are met.
a(v,w) bounded and continuous[edit | edit source]
Consider the approximation of by piecewise linear finite elements. Define precisely the piecewise linear finite element subspace (use the partition (3)). Show that the finite element problem has a unique solution.
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Let be a nonlinear function with zero :
- , .
Consider the iteration
- , .
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Prove (4) is locally convergent.
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is a fixed point iteration. By the contraction mapping theorem, if is a contraction in some neighborhood of then the iteration converges at least linearly.
We have to show there exists such that .
By the mean value theorem we have that , that is for some in our neighborhood of .
In particular, , implying that is a contraction and that the iterative method converges at least linearly.
Show that the convergence is at least quadratic.
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where satisfies when .
Then, we obtain .
Write the Newton iteration and compare it with (4)
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The Newton iteration looks like this:
Where B is the inverse of the Jacobian of f.
That is, in the Newton Iteration gives (4).