The Rayleigh-Ritz method


This method starts straight from the weak form of a differential equation. It follows by taking the same functional space for the solution functions and the weight functions. This makes it similar to the straight Galerkin method – indeed it is often equivalent, as we will discuss in a future article.

The variational point of view

In a linear problem, the RR method is quite simple. In this case, our weak problem can be written as


where B is a bilinear operator, and l is a linear one.

In our example our final expression in the weak form was:

ex. \int\left[ a\frac{du}{dx}\frac{d w }{dx}-wf\right]-w(L)Q_0=0

So we can readily identify

ex. B(u,w)=\int\left[a\frac{du}{dx}\frac{dw}{dx}-wf\right]  and  l(w)=wf+w(L)Q_0.

It is quite interesting physically that w can be considered a variation from the true solution u^*, so we may write w=\delta u, and solve

B(u,\delta u)-l(\delta u)=0.

For these (bi)linear operators, this can be written, with a little functional analysis, as

\delta \left( \frac{1}{2} B(u,u)-l(u)\right)=0,

IF B is symmetric, on top of bilinear.

So, indeed, we are solving a variational problem in the spirit of classical mechanics. One would call the quantity B(u,u)/2-l(u) the potential energy, or the Lagrangian perhaps. Sometimes, such a physical name is not appropriate, but clearly the quantity is still useful as the thing that is being minimized (or, rather, extremized).

The Rayleigh-Ritz method

Since in this method we take the same functional space for the solution functions and the weight functions, we end up with N equations:

B(\phi_i,\sum c_j\phi_j + \phi_0 )-l(\phi_i)=0

Given the (bi)linearity, we can arrive at a standard linear algebra problem:

\sum_j B_{ij}c_j=F_i,

where the coefficient matrix has components


and the vector is


Notice B is bilinear, but doesn’t have to be symmetric. Hence the matrix does not have to be symmetric either.

Boundary conditions

We recall that NBCs are already included in a proper weak formulation. Hence, our solution function u_N=\sum c_j\phi_j+\phi_0 must only comply with the EBCs. This is why we keep dragging the \phi_0: this function usually satisfies the non-homogeneous BCs that may be, while the rest satisfies homogeneous BCs. If the BCs are homogeneous to start with, \phi_0 is simply zero, and we don’t even have to write it down (as is done as an easy case in some books). Don’t forget that the \phi_j are also used for w in the RR method, which is fine: the weight should indeed comply with homogeneous BC.


Let us consider this case:

A(u)=-\frac{d}{dx}(au')-cu\qquad f(x)=-x^2.

The weak form from which we begin is:

\int_0^L (au'w'-cuw+x^2w)dx=0 + boundary terms

We don’t write down the boundary terms since we are going to consider two sets. Also, we will set a=c=L=1, so

\int_0^1 (u'w'-uw+x^2w)dx=0 + boundary terms

Set 1

Homogeneous Dirichlet BCs: u(0)=u(1)=0. These are clearly two EBCs. They therefore impose w(0)=w(1)=0. The weak form is, in this case simply

\int_0^1 (u'w'-uw+x^2w)dx=0

Also, we may set \phi_0=0 since they are homogeneous. Hence our linear algebra problem is

\sum_j B_{ij}c_j=F_i,


B_{ij}=\int_0^1(\phi'_i\phi'_j-\phi_i\phi_j)dx\qquad F_i=-\int_0^1x^2\phi_idx.

The set of functions must comply with the EBCs. For example, we may try the following:

\phi_i=x^i (x-1) .

The results are trivial but really tedious to copy (see Reddy, p. 45).

A fair convergence is obtained with only N=3. This kind of polynomials is, nevertheless, known to behave badly as N increases, since they are “not too orthogonal”. By the way, the exact solution to this problem is


Set 2

Homogeneous Dirichlet BC on 0 only: u(0)=0, plus Neumann on L: u'(1)=1.

There only one EBC, the first one, which imposes w(0)=0. The weak form is, in this case

\int_0^1 (u'w'-uw+x^2w)dx=w(1)

(The Neumann BC is not a EBC since it is w which appears in the weak form, not w‘.)

We may set \phi_0=0 again since the EBC is homogeneous. Hence our linear algebra problem is

\sum_j B_{ij}c_j=F_i,

with the same B_{ij} as before, but now


The set of functions must comply with the EBC. For example, we may try:


so that \phi_i(1)=1.

Now the coefficients are easier:

B_{ij}=\frac{ij}{i+j-1}-\frac{1}{i+j+1}\qquad F_i=1-\frac{1}{i+3}.

Again, fair convergence is obtained with only N=3, but these simple polynomials are known to be treacherous. By the way, the exact solution to this problem is




One thought on “The Rayleigh-Ritz method

  1. Pingback: Approximation for weighted residuals « Daniel Duque Campayo

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s