I'm trying to solve the following ODE using the Ritz method:

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

With the boundary conditions

$$\frac{d\theta}{dx}\Bigg{|}_{x=0} = 0$$

$$\theta(1) = \theta_0$$

I'm trying to follow the book "Conduction Heat Transfer" by Vedat S. Arpaci (chap. 8). So this is what I did so far:

1) Transform the problem into a variational problem:

$$ \int_o^1 \Bigg{(}\frac{d^2 \theta}{d x^2} - m^2\theta\Bigg{)} \delta\theta\ dx = 0 $$

2) (Ritz Method) Select a convergent sequence of functions such that

$$ y(x) = \sum_{n=0}^Na_n\phi_n(x) $$

I choose the following $y(x)$:

$$y(x) = \sum_{n=0}^N\theta_0(1 - (1 - x^2)(a_0 + a_1x^2 + a_2x^4 + ...))$$

Which yields the approximation

$$ y(x) = \phi_0(x) = \theta_0(1 - (1 - x^2)a_0) $$

3) Apply (2) in (1) in order to obtain $a_0$

$$ \int_o^1 \Bigg{(}\frac{d^2 \theta}{d x^2} - m^2\theta\Bigg{)} \delta\theta\ dx = $$ $$ \theta_0\int_o^1 \Bigg{(}2a_0 - m^2(1 - (1-x^2)a_0) \Bigg{)} \delta\theta\ dx = 0 $$

Where $\delta F$ is the variation of the functional $F(x, y, y^\prime)$, that is,

$$ \delta F = \frac{\partial F}{\partial y}\delta y + \frac{\partial F}{\partial y^\prime}\delta y^\prime $$

And this is where I got stuck. The book suggest the following:

$$ \theta_0\int_o^1 \Bigg{[}2a_0 - m^2(1 - (1-x^2)a_0) \Bigg{]} \delta\theta\ dx = $$ $$ \theta_0\int_o^1 \Bigg{[}2a_0 - m^2(1 - (1-x^2)a_0) \Bigg{]} \Bigg{[} -(1-x^2)\delta a_0 \Bigg{]} dx = 0 $$

But I don't understand the application of $\delta \theta$ in this case, because I though $a_0$ was meant to be a constant (Should it really?), then $\theta$ is not a functional in relation to it.

Assuming $a_0$ is a function, and the step above is correct (Which I don't understand why), I would also be stuck in the next step:

$$ \theta_0\int_o^1 \Bigg{[}2a_0 - m^2(1 - (1-x^2)a_0) \Bigg{]} \Bigg{[} -(1-x^2)\delta a_0 \Bigg{]} dx = 0 $$

For this step, the book suggests to use the following identity, which I was unable to apply here:

$$ \int_0^l(l^2-x^2)^mdx = \frac{(2m)!!}{(2m+1)!!}l^{2m+1} $$

I would be grateful for any suggestion at this point.

  • $\begingroup$ Is your use of $\delta$ different than $\partial$? $\endgroup$ – gen-z ready to perish Nov 29 '17 at 23:01
  • 1
    $\begingroup$ Thanks for noticing. I messed up the symbols. They're now fixed. $\delta$ means the variation of a functional. It is similar to differentiation, but for function of functions (a functional). $\endgroup$ – Tarcisio Fischer Nov 30 '17 at 0:13

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