Math 115 ( ) YumTong Siu 1. Lagrange Multipliers and Variational Problems with Constraints. F (x,y,y )dx


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1 Mth ) YumTong Siu Lgrnge Multipliers nd Vritionl Problems with Constrints Integrl Constrints. Consider the vritionl problem of finding the extremls for the functionl J[y] = F x,y,y )dx with y) = A nd yb) = B subject to the dditionl integrl constrint tht K[y] = G x,y,y )dx = l, where l is given constnt. Suppose we hve n extreml y = yx) for this vritionl problem. To derive necessry condition for the extreml, we embed it in 2prmeter fmily y = yx,s,t) so tht the given extreml y = yx) corresponds to s,t) = 0, 0). The reson why we need 2prmeter fmily insted of prmeter fmily is tht the fmily y = ys,t) hs to stisfy the integrl constrint ) K[y] = G x,yx,s,t), x yx,s,t) dx = l for ll s,t). Consider now J[y] s function of two vribles s,t) subject to the condition K[y] = l. Since J[y] hs criticl point t s,t) = 0, 0) subject to the condition tht the point s,t) lies on the curve K[y] = l in the spce of the two rel vribles s,t), it follows tht the grdient of J[y] with respect to s,t) is proportionl to the grdient of K[y] with respect to s,t) t the point s,t) = 0, 0). Thus there exists some rel number λ such tht s J[y] = λ s K[y], t J[y] = λ t K[y] t the point s,t) = 0, 0). In other words, we hve F y ) ) b s y dx = λ G y ) ) s y dx, F y ) ) b t y dx = λ G y ) ) t y dx,
2 Mth ) YumTong Siu 2 The lefthnd side F y ) ) s y dx of the first eqution is the derivtive of J with respect to the vector y in s the spce of functions. The lefthnd side F y ) ) t y dx of the second eqution is the derivtive of J with respect to the vector y t in the spce of functions. The righthnd side G y ) ) s y dx of the first eqution is the derivtive of K with respect to the vector y in s the spce of functions. The righthnd side G y ) ) t y dx of the second eqution is the derivtive of K with respect to the vector t y in the spce of functions. The constnt λ is lredy determined by the first eqution which sys tht the component of the grdient of J in the direction of the vector y in s the spce of functions is equl to λ times the component of the grdient of K in the direction of the vector y in the spce of functions. The second s eqution sys tht s result the component of the grdient of J in the direction of ny other vector y in the spce of functions is equl to the t sme constnt λ times the component of the grdient of K in the direction of the vector y in the spce of functions. t As consequence we cn sy tht the full grdient of J in the spce of functions is equl to λ times the full grdient of K in the spce of functions. In other words, F y ) λ G y )) δy)dx = 0
3 Mth ) YumTong Siu 3 for ll δy with δy) ) = δy) b) = 0. Therefore we get the EulerLgrnge eqution ) F y = λ G y ) for some constnt λ which is known s the Lgrnge multiplier. Another wy to write it is F λg) y dx F λg) y = 0. Besides the two initil conditions y) = A nd yb) = B to determine the two constnt of integrtions for the solution of the secondorder differentil eqution ), we hve n extr unknown λ which will be determined by the integrl constrint G x,y,y )dx = l. Exmple. Given > 0 nd l > 0. Find n extreml for the vritionl problem subject to y ) = y) = 0 nd K[y] = J[y] = ydx + y 2 dx = l. Solution. The EulerLgrnge eqution is ) y λ + y y 2 )) y λ + y dx y 2 = 0 for some Lgrnge multiplier λ. We cn rewrite it s Integrting it once, we get y + λ d dx = 0. + y 2 y x + λ = C. + y 2 Squring to remove the squre root, we get x C ) 2 = y 2 + y 2 = λ ) + y 2.
4 Mth ) YumTong Siu 4 As result, y 2 = x C ) 2 = + y 2, + y 2 = x C ) 2, + y 2 =, x C ) 2 = x C ) 2 y = ± x C λ x C ) 2 x C ) 2 x C ) 2, dx. Let x C λ = cos θ. Then cos θ sin θ)λdθ y = ± = λ sin θ + C 2. cos2 θ Eliminting θ, we get x C ) 2 + y C 2 ) 2 =. The constnts C, C 2, nd λ re to be determined by y ) = y) = 0 nd K[y] = + y 2 dx = l. Pointwise Constrints. Consider the vritionl problem of finding the extremls for the functionl J[y,z] = F x,y,z,y,z )dx with y) = A nd yb) = B subject to the dditionl pointwise constrint tht gx,y,z) = 0. Suppose we hve n extreml y = yx),z = zx) for this vritionl problem. To derive necessry condition for the extreml, we embed it in prmeter fmily y = yx,t),z = zx,t) which stisfy the pointwise constrint gx,yx,t),zx,t)) 0 for ll x,t so tht the given
5 Mth ) YumTong Siu 5 extreml y = yx),z = zx) corresponds to t = 0. Tking the first vrition of J, we get ) δj = F y ) δy)dx + F z ) dx F z δz) dx, where δj = dj nd δy = y nd δz = z. Differentiting t=0 t=0 t=0 dt gx, yx, t), zx, t)) 0 with respect to t yields for ll x. Solving for δz, we get Putting it into ) gives us 0 = δj = t g y δy) + δz) 0 δz = g y δy). F y g y t F z )) dx F z δy) dx. By the rbitrriness of δy, we conclude tht the integrnd vnishes nd get F y g y F z ) dx F z = 0 which cn be rewritten s F y g y = F z dx F z. Let λx) be either of the two equl sides of the bove eqution. We get F λg) y dx F λg) y = 0, F λg) z dx F λg) z = 0. Heuristiclly, we cn lso interpret the pointwise condition s prmeter fmily of integrl constrint δ ξ ) x)gx,y,z)dx = 0 with prmeter ξ [,b], where δ ξ x) is the Dirc delt t ξ so tht we end up with one constnt multiplier λξ) for ech of such n integrl constrint.
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