Problem set 1: Solutions Math 207B, Winter 2016

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1 Problem set 1: Solutions Mth 27B, Winter Define f : R 2 R by f(,) = nd f(x,y) = xy3 x 2 +y 6 if (x,y) (,). ()Show tht thedirectionl derivtives of f t (,)exist inevery direction. Wht is its Gâteux derivtive t (,)? (b) Show tht f is not Fréchet differentible t (,). (Hint. A Fréchet differentible function must be continuous.) () The directionl derivtive of f t (,) in the direction (h,k) (,) is df(,;h,k) = d dǫ ǫ= f(ǫh,ǫk) ( ) f(ǫh,ǫk) f(,) = lim ǫ ǫ ( ) ǫhk 3 = lim ǫ h 2 +ǫ 4 k 6 =. So ll of the directionl derivtives exist nd df(,;h,k) =. (b) If f : R n R is Fréchet differentible t x R n, then it follows directly from the definition tht f( x + h) f( x) s h, so f is continuous t x. On the curve x = t 3, y = t, we hve f(t 3,t) = 1/2 for t, so f(t 3,t) s t. It follows tht f is not continuous t (,) nd therefore f is not Fréchet differentible t (, ).

2 2. Define f,g : R 2 R by f(x,y) = x 2 +y 2, g(x,y) = (y 1) 3 x 2. Find the minimum vlue of f(x,y) subject to the constrint g(x,y) =. Show tht there does not exist ny constnt λ such tht f = λ g t some point (x,y) R 2. Why does the method of Lgrnge multipliers fil in this exmple? On the curve g(x,y) =, we hve y = 1+x 2/3 1, so f(x,y) 1. On the other hnd, f(,1) = 1 nd g(,1) =, so the minimum vlue of f(x,y) is 1, ttined t (x,y) = (,1). The Lgrnge-multiplier-equtions f = λ g re 2x = 2λx, 2y = 3λ(y 1) 2. The first eqution is stisfied if either x = nd λ is rbitrry, or λ = 1. If x =, then the constrint g(x,y) = implies tht y = 1, in which cse the second eqution does not hold for ny vlue of λ. On the other hnd, if λ = 1, then 3y 2 4y + 3 =, which implies tht y = (2± 5)/3, so there re no rel-vlued solutions for y. The Lgrnge-multiplier method fils becuse g = t the point (x,y) = (,1) where f ttins its minimum on g =. As result, the curve g(x,y) = is not smooth with well-defined norml vector t tht point (see figure) g(x,y) = y f(x,y)= x

3 3. Derive the Euler-Lgrnge eqution for functionl of the form J(u) = F(x,u,u,u )dx. Wht re the nturl boundry conditions for this functionl? Computing the directionl derivtive of J t u in the direction φ, nd using integrtion by prts, we get tht dj(u;φ) = d F(x,u+ǫφ,u +ǫφ,u +ǫφ )dx dǫ = = + ǫ= {F u (x,u,u,u )φ+f u (x,u,u )φ +F u (x,u,u,u )φ } dx { F u (x,u,u ) d } dx F u (x,u,u )+ d2 dx 2F u (x,u,u,u ) φdx [ F u (x,u,u,u )φ d ] b dx F u (x,u,u,u ) φ+f u (x,u,u,u )φ If u is smooth extreml of J, then dj(u;φ) = for every φ Cc ([,b]). In prticulr, if φ nd its derivtives re zero t x =,b, then the boundry terms in the integrtion by prts vnish nd { F u (x,u,u ) d } dx F u (x,u,u )+ d2 dx 2F u (x,u,u,u ) φdx = for ll φ Cc (,b). The fundmentl lemm of the clculus of vritions implies tht u stisfies the Euler-Lgrnge eqution d 2 dx 2F u (x,u,u,u ) d dx F u (x,u,u )+F u (x,u,u ) = < x < b. It follows tht if φ C c ([,b]) is non-zero t x =,b, then [ F u (x,u,u,u )φ d ] b dx F u (x,u,u,u ) φ+f u (x,u,u,u )φ =..

4 Choosing functions φ such tht only one of φ(), φ(b), φ (), or φ (b) is nonzero, we conclude tht the nturl boundry conditions for u t x =,b re d dx F u (x,u,u,u )+F u (x,u,u,u ) =, F u (x,u,u,u ) =. This gives four nturl boundry conditions for the Euler-Lgrnge eqution, which is fourth-order ODE (provided tht F u u ).

5 4. A curve y = u(x) with x b, u(x) >, nd u() = u, u(b) = u 1 is rotted bout the x-xis. Find the curve tht minimizes the re of the surfce of revolution, J(u) = u 1+(u ) 2 dx. Since the Lgrngin F(u,u ) = u 1+(u ) 2 is independent of x, the Euler-Lgrnge eqution for J(u) hs the first integrl u F u +F u = c 1 where c 1 is constnt of integrtion, which gives u 1+(u ) 2 = c 1. The solution for u is u = u 2 c 2 1 1, nd seprtion of vribles gives du u2 /c = dx. Mking the substitution u = c 1 cosht, where (cosht) = sinht nd cosh 2 t sinh 2 t = 1, we get tht c 1 t = x+c 2, so ( ) x u(x) = c 1 cosh +c 3. c 1 We choose the constnts c 1, c 3 so tht ( u = c 1 cosh +c 3 c 1 ), u 1 = c 1 cosh ( ) b +c 3. c 1 These lgebric equtions might not hve solution for (c 1,c 3 ), in which cse there is no smooth curve tht gives miniml surfce of revolution with rdius u t x = nd rdius u 1 t x = b.

6 For exmple, consider the cse when = b, with b >, nd u = u 1. Then c 3 = nd ( ) b u = ccosh. c Writing y = u /c, t = b/c, nd m = u /b, we see tht this eqution hs solution for c if y = cosht, y = mt. The line y = mt is tngent to the curve y = cosht t t = t when m = m (see figure), where m t = cosht, m = sinht. If < m < m, mening tht u < m b, then the line y = mt does not intersect the curve y = cosht, nd there re no solutions; if m > m, mening tht u > m b, then the line y = mt intersects y = cosht in two points, nd there re two solutions. The criticl vlue of m is given by m = sinht where t > is the solutionof t tnht = 1. The numericl solution of this trnscendentl eqution is t , which gives m y t

7 5. Let X be the spce of smooth functions u : [,1] R such tht u() =, u(1) =. Define functionls J,K : X R by J(u) = (u ) 2 dx, K(u) = u 2 dx. () Introduce Lgrnge multiplier nd write down the Euler-Lgrnge eqution for extremls in X of the functionl J(u) subject to the constrint K(u) = 1. (b) Solve the eigenvlue problem in () nd find ll of the extremls. Which one minimizes J(u)? () We hve δj δu = δk u, δu = u, so the Lgrnge-multiplier eqution δj/δu = λδk/δu is u = λu, u() =, u(1) =. If λ = k 2 <, then the generl solution of the ODE is u(x) = c 1 coshx+c 2 sinhx. The BC u() = implies tht c 1 =, ndthen the BC u(1) = implies tht c 2 =, so u = nd it does not stisfy the constrint K(u) = 1. If λ =, then u = c 1 +c 2 x, nd the BCs gin imply tht u =. If λ = k 2 >, then the generl solution of the ODE is u(x) = c 1 cosx+c 2 sinx. The BC u() = implies tht c 1 =, nd then the BC u(1) = is stisfied for c 2 if sink =. Without loss of generlity, we cn tke k = nπ with n = 1,2,3,..., when u(x) = csin(nπx) nd λ = n 2 π 2.

8 The constrint K(u) = 1 is stisfied if 1 2 c2 1 sin 2 (nπx)dx = 1, or c 2 /4 = 1, so the constrined extremls re u = ±u n where u n (x) = 2sin(nπx), for n = 1,2,3,... We hve J(u n ) = n 2 π 2 cos 2 (nπx)dx = n 2 π 2. The minimum of J is π 2, ttined t u 1. For n 2, the extremls u n re sddle points of J.

9 6. () Mke chnge of vrible x = φ(t), v(t) = u(φ(t)), where φ (t) >, in the functionl J(u) = F(x,u,u )dx. Show tht J(u) = K(v) where K(v) hs the form K(v) = d c G(t,v,v )dt nd express G in terms of F nd φ. (b) Show tht the Euler-Lgrnge eqution for K(v) is the sme s wht you get by chnging vribles in the Euler-Lgrnge eqution for J(u). By the chin rule v (t) = d dt u(φ(t)) = φ (t)u (φ(t)). Mking the chnge of vribles x = φ(t) in the integrl for J(u), we get J(u) = d where c = φ(), d = φ(b). It follows tht c F (φ(t),v(t),v (t)/φ (t)) φ (t)dt G(t,v,v ) = φ (t)f (φ(t),v,v /φ (t)). The Euler-Lgrnge eqution for K(v) is d dt G v +G v =. From the previous expression for G, we hve G v = F u nd G v = φ F u, so the Euler-Lgrnge eqution becomes d dt F u +φ (t)f u =. Since d dt = dx d dt dx = φ (t) d dx,

10 it follows tht d dx F u +F u =, which shows tht the Euler-Lgrnge eqution for K(v) is equivlent to the one for J(u). Remrk. The Euler-Lgrnge equtions re lso invrint under more generl trnsformtions of the independent nd dependent vribles. It is often convenient to obtin equtions tht re invrint under some group of trnsformtions by deriving them from n invrint Lgrngin. For exmple, in reltivistic clssicl field theory, Lorentz-invrint Lgrngins led to Lorentz-invrint field equtions.