How Does the Ritz-Galerkin Method Ensure Uniqueness in Finite Element Solutions?

[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.

 

[lippyka]

Solution:To prove that the solution $u_h \in V_h$ is unique, we will use the uniqueness of minimizers of a functional. Let $F(u)$ be a functional defined as:$$F(u) = \frac{1}{2}\int_0^1 (u'(x))^2 \ dx - \int_0^1 f(x)u(x) \ dx$$We will show that the solution $u_h$ is the unique minimizer of $F(u)$ in the space $V_h$. First, we observe that $F(u)$ is convex since the integrand $(u'(x))^2$ is non-negative and the second term $\int_0^1 f(x)u(x) \ dx$ is linear. Hence, it suffices to show that $u_h$ satisfies the necessary conditions for a minimizer: For any $v_h \in V_h$,$$F(u_h) \leq F(u_h + v_h)$$ Expanding the left hand side, we get $$\frac{1}{2}\int_0^1 (u_h'(x))^2 \ dx - \int_0^1 f(x)u_h(x) \ dx \leq \frac{1}{2}\int_0^1 (u_h'(x) + v_h'(x))^2 \ dx - \int_0^1 f(x)(u_h(x) + v_h(x)) \ dx$$ Using the definition of $a(u_h,v_h)$ and $l(v_h)$, we have$$a(u_h,v_h) + l(v_h) \leq a(u_h+v_h,v_h+v_h)$$Since $u_h$ satisfies the variational formulation, we have $$a(u_h,v_h) = l(v_h).$$ Hence, $$l(v_h) \leq a(u_h+v_h,