# Solving PDEs using the Ritz method on variational calculus problem (Student questions)

I'm reading the book "Conduction Heat Transfer" by Vedat S. Arpaci. I'm currently at chapter 8 (I didn't read the rest of the book, though), which talks about the Variational Formulation - Solution by Approximate Profiles.

I have no background in Variational Calculus, but I think I understand the concept. The book shows an example of solving the following functional in order to obtain a function that minimizes the distance between two points:

$$\int_{a}^{b}F(x, y, y^\prime) dx = \int_{a}^{b}(1 + y^{\prime^2})^{1/2}dx$$

And then the book proceeds explaining how to get to the Euler equation associated with the variational problem:

$$\frac{\partial F}{\partial y} - \frac{d}{dx}\Bigg{(}\frac{\partial F}{\partial y^\prime}\Bigg{)} = \frac{d}{dx}\Bigg{(}\frac{y^\prime}{(1 + y^{\prime^2})^{1/2}}\Bigg{)} = 0$$

Finally, the book explains the difficulty on applying this method directly on physical problems. He proceeds explaining that, usually, what is done is the reverse: We consider the differential equation from our physical problem to be the Euler equation associated with the variational problem. So, for example, when solving the ODE:

$$\frac{d^2 \theta}{dx^2} - m^2 \theta = 0$$

Leas us to the following variational formulation:

$$\delta \int_0^L \Bigg{(} \frac{d \theta}{d x} \Bigg{)} + m^2\theta^2 dx = 0$$

Ok, so, as I understand, we now use the Ritz method to approximate a solution for $\theta$, by selecting an arbitrary function $y$ defined by

$$y(x) = \sum_{n=0}^N a_n \cdot \phi(x)$$

And for this particular problem, the book chooses

$$\frac{\theta(\xi)}{\theta_0} = 1 - (1 - \xi^2)(a_0 + a_1\xi^2 + a_2\xi^4 + ...)$$

1) First of all, searching on the internet, I found this so called "Rayleigh–Ritz" method. Do anyone knows if it is the same method as the one I'm studying?

Ans.: As I found out, it seems that yes, they are related. Each author explains some things sightly different, though

2) For this Ritz method, how to choose a function $\phi(x)$ in general? How do I know if my choice is good?

Ans.: As I understand, any ortogonal basis should do. It is common to use polynomials.

3) Is there any relation between this method and the FEM or FEA?

Ans.: They seem related, but there are still more math to be done. It seems that FEM can be viewed as the Ritz method applied with some pre-defined $\phi$ functions. It is a different perspective from the weights in the Galerkins method.

I managed to find out some information about the above questions. I'm still having trouble with the bellow one, though:

4) I was trying a different example by solving the classic unidimensional steady state heat transfer equation using this method. I started with

$$\frac{d^2u}{dx^2} = 0$$

with boundary conditions

$$u(0) = u_0, u(L) = u_L$$

with $u_0$ and $u_L$ constants. And then I said that the variational problem associated with this problem is as follows (I can show my steps if necessary):

$$\int_0^L \Bigg{(} \frac{du}{dx} \Bigg{)}^2 dx = 0$$

My doubt here is if my approach is correct so far, and now should be the point where I should choose a trial function. I don't know which function should I choose, so I don't know how to proceed to use the Ritz method here... Any help would be appreciated.

I think the idea is that find your functional, which you have found as $$I = \int_0^L \left( \frac{du}{dx}\right)^2 dx .$$ Then for Ritz, you try a trial function and try to minimize $I$ with respect to that. You are free to choose whatever trial function you want, so you could try for example a 3-rd order polynomial, $$\tilde{u}(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 .$$ Then you apply the boundary conditions. The first, $\tilde{u}(0) = u_0$ gives you $c_0 = u_0$. Then $\tilde{u}(L) = u_L$ leads to $$u_0 + c_1 L + c_2 L^2 + c_3 L^3 = u_L\\ \Rightarrow c_1 = \frac{u_L-u_0}{L} - c_2 L - c_3 L^2 .$$ Thus your trial function with the correct boundary conditions is now $$\tilde{u}(x) = u_0 + \frac{u_L-u_0}{L}x + c_2(x^2-L x) + c_3(x^3 - L^2 x) .$$ This you need to plug into $I$. After a whole bunch of calculations, you should get $$I = \left(\frac{u_L-u_0}{L}\right)^2 L + \frac{1}{3} L^3 c_2^2+\frac{4}{5}L^5 c_3^2 + L^4 c_2 c_3 .$$ Now for the minimizing part, we want to find the values for which $I$ is minimal. So we need to differentiate with respect to $c_2$ and $c_3$ and set that to zero, leading to a system of equations to solve. $$\frac{\partial I}{\partial c_2} = \frac{2}{3} L^3 c_2 + L^4 c_3 = 0\\ \frac{\partial I}{\partial c_3} = L^4 c_2 + \frac{8}{5} L^5 c_3 = 0 .$$ The solution to this is $c_2 = c_3 = 0$. So if you plug that back into the expression for the trial function, we are left with $$\tilde{u}(x) = u_0 + \frac{u_L-u_0}{L}x ,$$ which is the exact solution for your steady-state heat problem.
It is exact because the solution happened to be a polynomial in the first place. If you would have picked another trial function you would only have been able to approximate it. Similar to how a polynomial is not an exact solution to the $\theta''-m^2\theta=0$ problem but it approximates the exact solution as best as it can.