How to Prove Uniqueness in Ritz-Galerkin Approximations?

[greek9000]

Hi there I'm having a trouble with the following proof.I have the exam soon and he has asked this question every year so I have no doubt that he will ask it again.I have no idea how to do it?If you can help me I would be greatly appreciative.

Question:
Consider the 2-point boundary value problem
-u'' = f(x) on (0,1)
subject to the boundary conditions u(0) and u'(1)=0. The associated weak formulation reads:Find u element of V such that a(u,v) = l(v) for all v is an element of V,where a(u,v) = ∫(0 to 1) u'v'dx and l(v) = ∫(0 to 1) f(x)v(x)dx

Then:
(1)Define the suitable function space V(1)
(2) prove that if f is an element of C[0,1] and u is an element of C^2[0,1] satisfies the variational formulation,then u solves the differential equation.(7)
(3)Define the Ritz-Galerkin approximation to the variational statement.(2)
(4)Prove that if f is an element of L(subscript 2 at the bottom)(0,1),then the solution u(subscript h at the bottom) is an element of Vh(subscript h at the bottom of V) such that a(uh,vh) = (f,vh) for all vh is an element of Vh is unique? (8)

I need help with question 4 asap please guys.I'm desperate.Stressing out big time. Thanks very much in advance.

 

[Valkarie]

Solution: 1) The suitable function space V is defined as the space of all continuous functions that have continuous derivatives up to second order on the interval [0,1]. 2) To prove that if f is an element of C[0,1] and u is an element of C^2[0,1] satisfies the variational formulation,then u solves the differential equation,we must first show that a(u,v) = l(v) for all v is an element of V. This can be done by integrating by parts: a(u,v) = ∫(0 to 1) u'v'dx = [u'v - ∫(0 to 1) u''v dx] = [u'v - f(x)v] = [u'v - l(v)] = l(v) Therefore, a(u,v) = l(v) for all v is an element of V, which implies that u solves the differential equation. 3) The Ritz-Galerkin approximation to the variational statement is defined as follows: Find uh element of Vh such that a(uh,vh) = l(vh) for all vh is an element of Vh. Here, Vh is a finite-dimensional subspace of V and l(vh) is defined as l(vh)=∫(0 to 1) f(x)vh(x)dx. 4) To prove that if f is an element of L2(0,1), then the solution uh is an element of Vh such that a(uh,vh) = (f,vh) for all vh is an element of Vh is unique, we must first show that a(uh,vh) = (f,vh). This can be done by noting that, since f is an element of L2(0,1), we have a(uh,vh) = ∫(0 to 1) uh'vh'dx = [uh'vh - ∫(0 to 1) uh''vh dx]

 

[Mene]

Hi there,

I understand that you are under a lot of pressure to prepare for your exam and that this question has been a recurring one. I will try my best to provide a clear and helpful explanation for question 4.

First, let's define the Ritz-Galerkin approximation to the variational statement. This is a numerical method used to approximate the solution to a differential equation. It involves choosing a finite-dimensional subspace Vh of the function space V and finding the solution uh in this subspace that minimizes the error in the variational formulation. In other words, uh is the best approximation to the exact solution u in Vh.

Now, let's move on to the proof that if f is an element of L2(0,1), then the solution uh is unique. To prove this, we need to show that if there exists another solution uh' in Vh, then uh = uh'. This can be done using the uniqueness theorem for variational problems, which states that if a(u,v) = l(v) for all v in V, then u is the unique solution to the variational problem. Since uh and uh' both satisfy the variational formulation, it follows that uh = uh'.

To show that uh is in fact an element of Vh, we need to show that it satisfies the boundary conditions and the differential equation. First, we know that uh satisfies the boundary condition u(0) = 0. This is because uh is the best approximation to the exact solution u in Vh, and u(0) = 0 is a condition that all functions in Vh must satisfy.

Next, we need to show that uh satisfies the differential equation -u'' = f(x). This can be done by substituting uh into the variational formulation and using the properties of the inner product. Since uh satisfies the variational formulation, we have a(uh,vh) = l(vh) for all vh in Vh. If we substitute uh and f into this equation, we get a(uh,vh) = ∫(0 to 1) uh'vh'dx = (f,vh) = ∫(0 to 1) f(x)vh(x)dx. This shows that uh satisfies the differential equation -u'' = f(x).

In conclusion, we have shown that if f is an element of L2(0,1), then the solution uh is unique and