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For the optimization problem $\underset{\mathbf{x}\in \mathbb{R}^n}{\operatorname{argmin}} f(\mathbf{x})$, we can use the following standard nonlinear conjugate gradient method to find the solution:

  1. $\mathbf{d}_0 = -\nabla f(\mathbf{x}_0),$
  2. $\mathbf{x}_{i+1}=\mathbf{x}_i+\alpha_i\mathbf{d}_i,$ where $\alpha_i$ is found by line search,
  3. $\mathbf{g}_{i+1} = \nabla f(\mathbf{x}_{i+1})$,
  4. $\mathbf{d}_{i+1}= -\mathbf{g}_{i+1} + \beta_i\mathbf{d}_i$.

I wonder for the problem $\underset{\mathbf{x}\in \mathbb{R}^n, \; \mathbf{y}\in \mathbb{R}^n}{\operatorname{argmin}} f(\mathbf{x},\mathbf{y})$, where $f$ is a function of 2 variables, whether the corresponding nonlinear conjugate gradient method is:

  1. $\mathbf{d}_0^x = -\nabla_{\mathbf{x}} f(\mathbf{x}_0, \mathbf{y}_0), \mathbf{d}_0^y = -\nabla_{\mathbf{y}} f(\mathbf{x}_0, \mathbf{y}_0)$, where $\nabla_{\mathbf{x}}f$ and $\nabla_{\mathbf{y}}f$ are the derivatives of $f$ with respect to $\mathbf{x}$ and $\mathbf{y}$,
  2. $\mathbf{x}_{i+1}=\mathbf{x}_i+\alpha_i\mathbf{d}_i^x$, $\mathbf{y}_{i+1}=\mathbf{y}_i+\alpha_i\mathbf{d}_i^y$,
  3. $\mathbf{g}_{i+1}^x = \nabla_{\mathbf{x}} f(\mathbf{x}_{i+1},\mathbf{y}_{i+1})$, $\mathbf{g}_{i+1}^y = \nabla_{\mathbf{y}} f(\mathbf{x}_{i+1},\mathbf{y}_{i+1})$,
  4. $\mathbf{d}_{i+1}^x= - \mathbf{g}_{i+1}^x + \beta_i^x\mathbf{d}_i^x$, $\mathbf{d}_{i+1}^y= - \mathbf{g}_{i+1}^y + \beta_i^y\mathbf{d}_i^y$.

Also, I am not sure how to compute $\beta_i^x$, $\beta_i^y$. If $\beta_i^x,\beta_i^y$ are computed by the Fletcher–Reeves method, is $\beta_i^x=\beta_i^y=\frac{\|\mathbf{g}_{i+1}\|^2}{\|\mathbf{g}_{i}\|^2}$ (where $\mathbf{g}_{i} = [\mathbf{g}_{i}^x, \mathbf{g}_{i}^y]$), or $\beta_i^x=\frac{\|\mathbf{g}_{i+1}^x\|^2}{\|\mathbf{g}_{i}^x\|^2}$, $\beta_i^y=\frac{\|\mathbf{g}_{i+1}^y\|^2}{\|\mathbf{g}_{i}^y\|^2}$?

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up vote 7 down vote accepted

If you introduce a variable $\mathbf z=(\mathbf x^T, \mathbf y^T)^T \in \mathbb{R}^{2n}$, then you can write $f(\mathbf x, \mathbf y)=f(\mathbf z)$ and it fits the exact format you wrote in your outline of the first method and it will all work as described. You just have to match things like $\nabla_z f(\mathbf z) = (\nabla_x f(\mathbf x,\mathbf y)^T, \nabla_y f(\mathbf x,\mathbf y)^T)^T$ etc.

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Is this equivalent to my second method, where $\beta_i^x=\beta_i^y=\frac{\|\mathbf{g}_{i+1}\|^2}{\|\mathbf{g}_{i}\|^2}$? – chaohuang Jan 12 '13 at 4:37
Yes. The two betas must be the same (the CG method doesn't know anything about $x,y$, it only knows of one vector $z$). – Wolfgang Bangerth Jan 12 '13 at 12:49

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