# Determining Internal Node Values Homework Help

**Variational Principles**:

For the differential equations that describe several physical systems the internal energy of the system is an integral. For example for the steady state heat equation

u_{xx}+ u_{yy}= g(x, y)

The internal energy is the integral:

R is region on which we are working it is able to be shown that u(x, u) is a solution of it and only if it is minimizer of I[u] in.

**The finite element solution**:

Evoke that a finite element solution is a linear combination of finite element functions

Where n is the number of nodes. To acquire the values at the internal nodes we will plug U(x, y) into the energy integral and minimize. That is we get the minimum of:

I[U]

For every choices of {C_{j}}^{m}_{j}=1 where m is the number of internal nodes. In this as with any other minimization problem the manner to find a possible minimum is to differentiate the quantity with respect to the variables and set the results to zero. In this circumstances the free variables are {C_{j}}^{m}_{j}=1. Therefore to find the minimizer we should try to solve:

∂I[U]/ ∂Cj= 0 for 1 ≤ j ≤ m.

We describe this set of equations the internal node equations. At this point we must ask whether the internal node equations can be solved as well as if so is the solution actually a minimizer (and not a maximizer). The subsequent two facts answer these questions. These facts create the finite element method practical

•For largely applications the internal node equations are linear.

• For generally applications the internal node equations give a minimizer.

We are able to demonstrate the first fact using an example.

**Application to steady state heat equation**:

If we plug the entrant finite element solution U(x, y) into energy integral for heat equation, we obtain

Differentiating with respect to Cjwe acquire the internal node equations

Now we have numerous simplifications first note that since:

Likewise ∂Uy/∂Cj= (Φ_{j})y. The integral subsequently becomes:

Next we utilize the fact that the region R is subdivided into triangles {Ti}pi=1 and the functions in question have different definitions on each triangle. The integral afterwards is a sum of the integrals:

At this time note that the function Φ_{j}(x, y) is linear on triangle Tiand so Φ_{ij}(x, y) = Φ_{j}|Ti (x, y) = a_{ij}+ b_{ij}x + c_{ij}y.

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