The Calculus of Variations All You Need to Know in One Easy Lesson Richard Taylor [email protected] TRU Math Seminar Putnam 2006: Problem B5 For each continuous function f : [0, 1] → IR let 1 I(f) = x2f(x) − xf(x)2 dx. Z 0 Find the maximum value of I(f) over all such functions f. Putnam 2006: Problem B5 For each continuous function f : [0, 1] → IR let 1 I(f) = x2f(x) − xf(x)2 dx. Z 0 Find the maximum value of I(f) over all such functions f. f(x) = sin(πx) 0.0 0.4 0.8 0.0 0.2 0.4 0.6 0.8 1.0 x 4π2−π3−16 ≈ − I(f) = 4π3 0.061 Putnam 2006: Problem B5 For each continuous function f : [0, 1] → IR let 1 I(f) = x2f(x) − xf(x)2 dx. Z 0 Find the maximum value of I(f) over all such functions f. f(x) = sin(πx) g(x)= x2 0.0 0.4 0.8 0.0 0.4 0.8 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x x 4π2−π3−16 ≈ − 1 ≈ I(f) = 4π3 0.061 I(g) = 30 0.033 Putnam 2006: Problem B5 Could try maximizing I(f) over a family of functions, e.g. f(x) = cx (c is a parameter). Putnam 2006: Problem B5 Could try maximizing I(f) over a family of functions, e.g. f(x) = cx (c is a parameter). 1 Then I(f) = x2(cx) − x(cx)2 dx Z 0 Putnam 2006: Problem B5 Could try maximizing I(f) over a family of functions, e.g. f(x) = cx (c is a parameter). 1 Then I(f) = x2(cx) − x(cx)2 dx = 1 (c − c2). Z 4 0 I(f) So, restricted to this family, I(f) is 1 maximized with c = 2 . 0.00 0.02 0.04 0.06 0.0 0.2 0.4c 0.6 0.8 1.0 A Solution: Claim: The maximum of I(f) is achieved for f(x) = x/2. A Solution: Claim: The maximum of I(f) is achieved for f(x) = x/2. x Proof. Any f can be written f(x) = 2 + g(x). A Solution: Claim: The maximum of I(f) is achieved for f(x) = x/2. x Proof. Any f can be written f(x) = 2 + g(x). Then 1 I(f) = x2[ x + g(x)] − x[ x + g(x)]2 dx. Z 2 2 0 A Solution: Claim: The maximum of I(f) is achieved for f(x) = x/2. x Proof. Any f can be written f(x) = 2 + g(x). Then 1 2 x − x 2 I(f) = x [ 2 + g(x)] x[ 2 + g(x)] dx. Z0 1 1 1 3 − 2 = 4 x dx xg(x) dx Z0 Z0 A Solution: Claim: The maximum of I(f) is achieved for f(x) = x/2. x Proof. Any f can be written f(x) = 2 + g(x). Then 1 I(f) = x2[ x + g(x)] − x[ x + g(x)]2 dx. Z 2 2 0 1 1 = − xg(x)2 dx 16 Z0 ≥0 | {z } A Solution: Claim: The maximum of I(f) is achieved for f(x) = x/2. x Proof. Any f can be written f(x) = 2 + g(x). Then 1 I(f) = x2[ x + g(x)] − x[ x + g(x)]2 dx. Z 2 2 0 1 1 = − xg(x)2 dx 16 Z0 ≥0 | {z } 1 So the maximum of I(f), achieved with g(x) = 0, is 16 . Some Multivariable Calculus That approach works only if you “guess” the right family of functions (i.e., containing the optimal f(x)) . how to tell? Some Multivariable Calculus That approach works only if you “guess” the right family of functions (i.e., containing the optimal f(x)) . how to tell? A more reliable method uses ideas from multivariable calculus: Definition. Given a function f : IRn → IR, the directional derivative at x, in the direction of a unit vector u, is x u − x f( + h ) f( ) d u x x u D f( ) = lim = dh f( + h ) h→0 h h=0 Some Multivariable Calculus That approach works only if you “guess” the right family of functions (i.e., containing the optimal f(x)) . how to tell? A more reliable method uses ideas from multivariable calculus: Definition. Given a function f : IRn → IR, the directional derivative at x, in the direction of a unit vector u, is x u − x f( + h ) f( ) d u x x u D f( ) = lim = dh f( + h ) h→0 h h=0 Duf(x) gives the rate of change of f(x) as we move in the direction u at unit speed. (e.g. rate of change of temperature along a given direction). Some Multivariable Calculus 2 For f : IR → IR, the graph of y = f(x) is a surface, and Duf(x) is the slope of this surface along the direction u: z = f(x, y) x y x Some Multivariable Calculus 2 For f : IR → IR, the graph of y = f(x) is a surface, and Duf(x) is the slope of this surface along the direction u: z = f(x, y) x u y x Some Multivariable Calculus 2 For f : IR → IR, the graph of y = f(x) is a surface, and Duf(x) is the slope of this surface along the direction u: z = f(x, y) x u y x Some Multivariable Calculus 2 For f : IR → IR, the graph of y = f(x) is a surface, and Duf(x) is the slope of this surface along the direction u: z = f(x, y) slope = Duf(x) x u y x Some Multivariable Calculus The directional derivative helps find extreme values of f(x): Some Multivariable Calculus The directional derivative helps find extreme values of f(x): n Theorem. If f : IR → IR has a local extremum at x, then Duf(x) = 0 for every direction u. z = f(x, y) y x Some Infinite-Dimensional Calculus 1 Problem: Find f to maximize I(f) = x2f(x) − xf(x)2 dx. Z 0 Some Infinite-Dimensional Calculus 1 Problem: Find f to maximize I(f) = x2f(x) − xf(x)2 dx. Z 0 y y = I(f) I(f) is a real-valued function on the space C of continuous C functions on the interval [0, 1]. f Some Infinite-Dimensional Calculus 1 Problem: Find f to maximize I(f) = x2f(x) − xf(x)2 dx. Z 0 y y = I(f) I(f) is a real-valued function on the space C of continuous C functions on the interval [0, 1]. f C is an infinite-dimensional vector space (i.e. equipped with addition and scalar multiplication, hence a notion of “direction”). Some Infinite-Dimensional Calculus 1 Problem: Find f to maximize I(f) = x2f(x) − xf(x)2 dx. Z 0 y y = I(f) I(f) is a real-valued function on the space C of continuous C functions on the interval [0, 1]. f C is an infinite-dimensional vector space (i.e. equipped with addition and scalar multiplication, hence a notion of “direction”). I(f) depends continuously (even differentiably) on f ∈ C. Some Infinite-Dimensional Calculus Directional derivative of I(f) on the function space C: Pick a function f ∈ C C f Some Infinite-Dimensional Calculus Directional derivative of I(f) on the function space C: Pick a function f ∈ C and a “direction” — say, the direction of the function g ∈ C. C f + λg f λg Some Infinite-Dimensional Calculus Directional derivative of I(f) on the function space C: Pick a function f ∈ C and a “direction” — say, the direction of the function g ∈ C. How quickly does I(f) change as we move f in the direction of g? C f + λg f λg Some Infinite-Dimensional Calculus Directional derivative of I(f) on the function space C: Pick a function f ∈ C and a “direction” — say, the direction of the function g ∈ C. How quickly does I(f) change as we move f in the direction of g? C f + λg f λg The rate of change of I(f) in the g-direction is the directional or Gâteaux derivative: − I(f + λg) I(f) d DgI(f) = lim = dλ I(f + λg) λ→0 λ λ=0 Application to the Putnam Problem: 1 For I(f) = x2f(x) − xf(x)2 dx this is an easy calculation: Z 0 Application to the Putnam Problem: 1 For I(f) = x2f(x) − xf(x)2 dx this is an easy calculation: Z 0 d DgI(f) = dλ I(f + λg) λ=0 1 d 2 = x2 f(x) + λg(x) − x f(x) + λg(x) dx dλ Z 0 λ=0 Application to the Putnam Problem: 1 For I(f) = x2f(x) − xf(x)2 dx this is an easy calculation: Z 0 d DgI(f) = dλ I(f + λg) λ=0 1 d 2 = x2 f(x) + λg(x) − x f(x) + λg(x) dx dλ Z 0 λ=0 d 1 1 1 = x2 f(x) − xf(x)2 dx + λ x2g(x) − 2xf(x)g(x) dx − λ2 xg(x)2 dx dλ » 0 0 0 – R ` ´ R ` ´ R λ=0 Application to the Putnam Problem: 1 For I(f) = x2f(x) − xf(x)2 dx this is an easy calculation: Z 0 d DgI(f) = dλ I(f + λg) λ=0 1 d 2 = x2 f(x) + λg(x) − x f(x) + λg(x) dx dλ Z 0 λ=0 d 1 1 1 = x2 f(x) − xf(x)2 dx + λ x2g(x) − 2xf(x)g(x) dx − λ2 xg(x)2 dx dλ » 0 0 0 – R ` ´ R ` ´ R λ=0 1 1 = x2g(x) − 2xf(x)g(x) − 2λ xg(x)2 dx Z Z 0 0 λ=0 Application to the Putnam Problem: 1 For I(f) = x2f(x) − xf(x)2 dx this is an easy calculation: Z 0 d DgI(f) = dλ I(f + λg) λ=0 1 d 2 = x2 f(x) + λg(x) − x f(x) + λg(x) dx dλ Z 0 λ=0 d 1 1 1 = x2 f(x) − xf(x)2 dx + λ x2g(x) − 2xf(x)g(x) dx − λ2 xg(x)2 dx dλ » 0 0 0 – R ` ´ R ` ´ R λ=0 1 1 = x2g(x) − 2xf(x)g(x) − 2λ xg(x)2 dx Z0 Z0 λ=0 1 = x2 − 2xf(x) g(x) dx.