Convergence of Steepest Descent

Copyright (C) 2026 Andreas Kloeckner

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import sympy as sp

lam1, lam2 = sp.symbols("lambda_1, lambda_2", positive=True)
u, v = sp.symbols("u,v", positive=True)

def grad(expr, vec):
    return sp.Matrix([expr.diff(vec[0]), expr.diff(vec[1])]) 

A = sp.Matrix([[lam1, 0], [0, lam2]])
A

$\displaystyle \left[\begin{matrix}\lambda_{1} & 0\\0 & \lambda_{2}\end{matrix}\right]$

x = sp.Matrix([u,v])
x

$\displaystyle \left[\begin{matrix}u\\v\end{matrix}\right]$

objective = (sp.Rational(1, 2) * x.T @ A @ x)[0]
objective

$\displaystyle \frac{\lambda_{1} u^{2}}{2} + \frac{\lambda_{2} v^{2}}{2}$

Question: What is the minimizer we're looking for?

grad_objective = grad(objective, x)
grad_objective

$\displaystyle \left[\begin{matrix}\lambda_{1} u\\\lambda_{2} v\end{matrix}\right]$

Question: What does this coincide with? Can you prove it?

Set up the line search as line and line_objective:

descent_direction = - grad_objective

alpha = sp.Symbol("alpha")
line = x + alpha * descent_direction
line_objective = objective.subs(u, line[0]).subs(v, line[1])
line_objective

$\displaystyle \frac{\lambda_{1} \left(- \alpha \lambda_{1} u + u\right)^{2}}{2} + \frac{\lambda_{2} \left(- \alpha \lambda_{2} v + v\right)^{2}}{2}$

And find the optimal $\alpha$:

alpha_opt = sp.solve(line_objective.diff(alpha), alpha)[0]
alpha_opt

$\displaystyle \frac{\lambda_{1}^{2} u^{2} + \lambda_{2}^{2} v^{2}}{\lambda_{1}^{3} u^{2} + \lambda_{2}^{3} v^{2}}$

Question: What is this in general? Can you prove it?

Next, find the next iterate:

x_next = line.subs(alpha, alpha_opt)
x_next

$\displaystyle \left[\begin{matrix}- \frac{\lambda_{1} u \left(\lambda_{1}^{2} u^{2} + \lambda_{2}^{2} v^{2}\right)}{\lambda_{1}^{3} u^{2} + \lambda_{2}^{3} v^{2}} + u\\- \frac{\lambda_{2} v \left(\lambda_{1}^{2} u^{2} + \lambda_{2}^{2} v^{2}\right)}{\lambda_{1}^{3} u^{2} + \lambda_{2}^{3} v^{2}} + v\end{matrix}\right]$

Next, consider the decrease in energy error:

def energy_error_squared(vec):
    return (vec.T @ A @ vec)[0]

ratio = sp.factor(energy_error_squared(x_next)/energy_error_squared(x))
ratio

$\displaystyle \frac{\lambda_{1} \lambda_{2} u^{2} v^{2} \left(\lambda_{1} - \lambda_{2}\right)^{2}}{\left(\lambda_{1} u^{2} + \lambda_{2} v^{2}\right) \left(\lambda_{1}^{3} u^{2} + \lambda_{2}^{3} v^{2}\right)}$

Take gradient, find critical point:

bad_soln, = sp.solve(grad(ratio, x), x)
x_bad = sp.Matrix(bad_soln)
x_bad

$\displaystyle \left[\begin{matrix}\frac{\lambda_{2} v}{\lambda_{1}}\\v\end{matrix}\right]$

sp.factor(ratio.subs(u, x_bad[0]).subs(v, x_bad[1]))

$\displaystyle \frac{\left(\lambda_{1} - \lambda_{2}\right)^{2}}{\left(\lambda_{1} + \lambda_{2}\right)^{2}}$