Calculus of Variation (5) - with Auxiliary Conditions

6. with Auxiliary Conditions

 

$ g=\sum_{i}x_{i}^{2}-\rho ^{2}=0 \rightarrow r=\rho =constant $: Equations of Constraint

(구면 위를 따라간다)

$ f=f(y_{i},y_{i}';x)=f(y,y',z,z';x) $

 

$ g(y_{i};x)=g(y,z;x)=0 $

$ \Rightarrow \frac{\partial y}{\partial \alpha }, \frac{\partial z}{\partial \alpha } $ are not independent (하나의 variable이 바뀌면, 다른 variable도 바뀐다)

$ \therefore $ for $ \alpha =0, \frac{\partial y}{\partial \alpha }\neq \frac{\partial z}{\partial \alpha }\neq 0 $

$ dg=(\frac{\partial g}{\partial y}\frac{\partial y}{\partial \alpha }+\frac{\partial g}{\partial z}\frac{\partial z}{\partial \alpha })d\alpha =0 (\frac{\partial x}{\partial \alpha }=0) $

$ \frac{\partial g}{\partial y}\frac{\partial y}{\partial \alpha }+\frac{\partial g}{\partial z}\frac{\partial z}{\partial \alpha }=\frac{\partial g}{\partial y}\eta _{1}(x)+\frac{\partial g}{\partial z}\eta _{2}(x)=0 $

($ y(\alpha ,x)=y(x)+\alpha \eta _{1}(x) \rightarrow \frac{\partial y}{\partial \alpha }=\eta _{1}(x) $

 $ z(\alpha ,x)=z(x)+\alpha \eta _{2}(x) \rightarrow \frac{\partial z}{\partial \alpha }=\eta _{2}(x) $)

$ \Rightarrow \frac{\partial g}{\partial y}\eta _{1}(x)=-\frac{\partial g}{\partial z}\eta _{2}(x) $

$ \Rightarrow \frac{\eta _{2}(x)}{\eta _{1}(x)}=-\frac{\partial g/\partial y}{\partial g/\partial z} $

 

for two variables,

$ \frac{\partial J}{\partial \alpha }=\int_{x_{1}}^{x_{2}}[(\frac{\partial f}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial y'})\frac{\partial y}{\partial \alpha }+(\frac{\partial f}{\partial z}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial z'})\frac{\partial z}{\partial \alpha }]dx $

   $ =\int_{x_{1}}^{x_{2}}[(\frac{\partial f}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial y'})\eta _{1}(x)+(\frac{\partial f}{\partial z}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial z'})\eta _{2}(x)]dx $

   $ =\int_{x_{1}}^{x_{2}}[(\frac{\partial f}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial y'})+(\frac{\partial f}{\partial z}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial z'})\frac{\eta _{2}(x)}{\eta _{1}(x)}]\eta _{1}(x)dx $

   $ =\int_{x_{1}}^{x_{2}}[(\frac{\partial f}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial y'})-(\frac{\partial f}{\partial z}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial z'})\frac{\partial g/\partial y}{\partial g/\partial z}]\eta _{1}(x)dx $

 

$ (\frac{\partial f}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial y'})(\frac{\partial g}{\partial y})^{-1}=(\frac{\partial f}{\partial z}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial z'})(\frac{\partial g}{\partial z})^{-1}\equiv -\lambda (x) $

$ \Rightarrow \frac{\partial f}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial y'}+\lambda (x)\frac{\partial g}{\partial y}=0 $

   $ \frac{\partial f}{\partial z}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial z'}+\lambda (x)\frac{\partial g}{\partial z}=0 $

($ \lambda (x) $: Lagrange Undetermined Multiplier)

 

in general

$ \frac{\partial f}{\partial y_{i}}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial y'_{i}}+\sum_{j}\lambda _{j}(x)\frac{\partial g_{j}}{\partial y_{i}}=0 $ ($ i=1,2,...,m, j=1,2,...,n $)

: (m equations) + (n equations of constraints) = (m+n unknowns)

$ g_{j}(y_{i};x)=0 \Rightarrow \sum_{i}\frac{\partial g_{i}}{\partial y_{i}}dy_{i}=0 $

 

integral constraint - isoperimetric problem

finding y=y(x) s.t. $ J[y]=\int_{a}^{b}f(y,y';x)dx $ has an extremum

B.C.: y(a)=A, y(b)=B

$ K[y]=\int_{a}^{b}g(y,y';x)dx $: a fixed value for length(l) (integral constraint)

$ \Rightarrow \int_{a}^{b}(f+\lambda g)dx $

   $ \frac{\partial f}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial y'}+\lambda (\frac{\partial g}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial g}{\partial y'})=0 $