Optimizing fractions and Lagrange Multiplier

[member 428835]

Hi PF!

When minimizing some fraction $f(x)/g(x)$ can we use Lagrange multipliers and say we are trying to optimize $f$ subject to the constraint $g=1$?

Thanks

 

[S.G. Janssens]

No, I don't think so, if I understand your question correctly. Why would that be true? Consider $f(x) = e^{-x}$ and $g(x) = 1 + x^2$.

 

[member 428835]

Yea those are good counter examples. I am curious because I am reading about the Ritz method for approximating eigenvalues. My book says: consider the eigenvalue problem $Lu=\lambda Mu$ where $L$ and $M$ are linear operators, $u$ is a function and $\lambda$ is the eigenvalue. Then the smallest eigenvalue $\lambda_1$ is given by $$\lambda_1=\min\frac{(Lu,u)}{(Mu,u)}$$ or equivalently $$\lambda_1=\min\limits_{(Mu,u)=1}(Lu,u)$$ where $(f,g)=\int_a^bfg\,dx$.

 

[S.G. Janssens]

Ah ok, yes, but this problem has a special structure. Let's write
$$
\lambda_1 := \min_{(Mu,u) \neq 0}\frac{(Lu,u)}{(Mu,u)}
$$
and
$$
\mu_1 := \min_{(Mu,u)=1}(Lu,u)
$$
Your book now claims that $\lambda_1 = \mu_1$. How would you prove this directly (i.e. no Lagrange multipliers or something like that)?

 

[member 428835]

Ah ok, yes, but this problem has a special structure. Let's write
$$
\lambda_1 := \min_{(Mu,u) \neq 0}\frac{(Lu,u)}{(Mu,u)}
$$
and
$$
\mu_1 := \min_{(Mu,u)=1}(Lu,u)
$$
Your book now claims that $\lambda_1 = \mu_1$. How would you prove this directly (i.e. no Lagrange multipliers or something like that)?

Take a derivative of what we're trying to minimize w.r.t...hmmm well now I'm not so sure. Any ideas?

 

[S.G. Janssens]

Any ideas?

Yes, but I would like you to puzzle a little bit, too.

You have not given much of the precise context. Let's assume that

• $H$ is a real Hilbert space with inner product $(\cdot,\cdot)$,
• $L$ and $M$ are operators with domains $D(L) = D(M) = H$,
• $M$ is a symmetric and positive operator, i.e. $0 \le (Mu, u)$ for all $u \in H$.

To be safe, I also replace your minima by infima. Let's make the minimization problems a bit more explicit using set notation. Write
$$
\Lambda_1 := \left\{\frac{(Lu,u)}{(Mu,u)}\,:\, u \in H,\, (Mu,u) > 0 \right\}, \qquad M_1 := \left\{(Lv,v)\,:\, v \in H,\, (Mv,v) = 1 \right\},
$$
so $\lambda_1 = \inf{\Lambda_1}$ and $\mu_1 = \inf{M_1}$. (Note that the positivity of $M$ allowed me to replace the condition $(Mu,u) \neq 0$ by $(Mu,u) > 0$ in $\Lambda_1$.)

Now observe that $M_1 \subseteq \Lambda_1$. (This already gives $\lambda_1 \le \mu_1$.) With a little bit (but not much) more work, you also show the reverse inclusion:
$$
M_1 \supseteq \Lambda_1 \qquad (\ast)
$$
Once this is done, you have $\Lambda_1 = M_1$, so $\lambda_1 = \mu_1$ follows. If you like, try to deduce $(\ast)$ yourself.

(The essential property you will use, is that both the numerator and the denominator of the original function
$$
u \mapsto \frac{(Lu,u)}{(Mu,u)}
$$
are homogeneous of degree two, so any scaling of $u$ does not change the function value.)

 

[member 428835]

Now observe that $M_1 \subseteq \Lambda_1$. (This already gives $\lambda_1 \le \mu_1$.) With a little bit (but not much) more work, you also show the reverse inclusion:
$$
M_1 \supseteq \Lambda_1 \qquad (\ast)
$$
Once this is done, you have $\Lambda_1 = M_1$, so $\lambda_1 = \mu_1$ follows. If you like, try to deduce $(\ast)$ yourself.

(The essential property you will use, is that both the numerator and the denominator of the original function
$$
u \mapsto \frac{(Lu,u)}{(Mu,u)}
$$
are homogeneous of degree two, so any scaling of $u$ does not change the function value.)

Ok, so I'm thinking if the denominator was $(u,u)$ (a less general case) then all we would have to do is let $v = u/||u||$. But with the $M$ operator present, I'm unsure how to proceed...I'll think more on this, but feel free to give the spoiler.

On a separate note, I'm interested in this entire process of finding the lowest eigenvalues because I am trying to solve $Lu=\lambda Mu$. If we instead look at a simpler problem $Lu=\lambda u$ I know the lowest eigenvalue is $$\lambda_1=\inf \frac{(Lu,u)}{(u,u)}.$$ Letting $$u=\sum_{i=1}^N a_i\psi_i$$ where $\psi_i$ are orthonormal basis vectors that satisfy boundary conditions, we deduce
$$(Lu,u) = \sum_{i,k=1}^NF_{ik}a_ia_k:F_{ik}\equiv (L\psi_i,\psi_k)
$$
To find the infimum from above, we can reduce the minimizing problem to minimizing $(Lu,u)$ subject to the constraint $(u,u)=1$. This evokes Lagrange multipliers. Letting $\Lambda$ be the Lagrange multiplier, take $$\frac{\partial}{\partial a_j}\left( \sum_{i,k=1}^NF_{ik}a_ia_k + \Lambda \sum_{i,k=1}^Na_ia_k(\psi_i,\psi_k)\right)=0\implies\\
\frac{\partial}{\partial a_j}\left( \sum_{i,k=1}^NF_{ik}a_ia_k + \Lambda \sum_{i}^Na_i^2\right)=0\implies\\
\sum_{k=1}^NF_{ki}a_k + \Lambda a_i=0\implies\\
\det(F_{ki}-\Lambda\delta_{ki})=0.
$$
Similarly, when solving the problem $Lu=\lambda M u$, if we let $(Mu,u) = \sum_{i,k=1}^ND_{ik}a_ia_k:D_{ik}\equiv (M\psi_i,\psi_k)$ then the solution should be $$\det(F_{ki}-\Lambda D_{ki})=0$$ where $\lambda_1$ is the lowest root $\Lambda$. Does this look correct to you?

I'm under the impression this technique is called the Rayleigh-Ritz variational approach.

 

[S.G. Janssens]

Ok, so I'm thinking if the denominator was $(u,u)$ (a less general case) then all we would have to do is let $v = u/||u||$. But with the $M$ operator present, I'm unsure how to proceed...I'll think more on this, but feel free to give the spoiler.

The symmetry and - in particular - the positivity of the operator $M$ imply that $(u_1,u_2) \mapsto (Mu_1,u_2)$ defines a bilinear form on $H \times H$ that has all the properties of an inner product, with the exception that $(Mu,u) = 0$ may not imply $u = 0$. In any case, for the problem at hand you can just let
$$
v =\frac{1}{\sqrt{(Mu,u)}}u
$$
(In other words, you scale by $(Mu,u)^{-\frac{1}{2}}$ instead of $(u,u)^{-\frac{1}{2}}$.)

 

[member 428835]

Thanks!

 

[WWGD]

Id say, if $f,g$ are differentiable, you can use the constraint $f'g-fg'=g^2$ when $g \neq 0 ; g(x) \neq 0$ or not defined , since extremes are reached at critical points.