Calculus of Variations & Variational Methods

Calculus of Variations & Variational Methods The calculus of variations is used to obtain extrema of expressions involving unknown functions called functionals. Applications range from simple geometric problems to finite-element methods to optimization theory. References: Ewing, G. M., Calculus of Variations with Applications, Dover, 1985. Hildebrand, F. B., Methods of Applied Mathematics, 2nd Ed., Dover, 1992. Reddy & Rasmussen, Advanced Engineering Analysis, Kreiger 1990. Weinstock, R., Calculus of Variations with Applications to Physics and Engineering, Dover. 292

Calculus of Variations & Variational Methods Unconstrained Minimization In preparation for an introduction to the calculus of variations, recall maxima, minima (extrema), and inflections of functions from the differential calculus. x0 = critical (stationary) local maximum f ( x) point df f max =0 f max = f ( x0 ) dx = critical value x0

f ( x)

local minimum

inflection

f ( x)

df =0 dx

df =0 dx

f max x0

x0

293

Calculus of Variations & Variational Methods df dx

df dx

= 0, necessary condition for extremum x = x0

x = x0

⎧d2 f ⎪ 2 ⎪ dx ⎪ 2 ⎪d f =0 ⎨ 2 ⎪ dx ⎪ 2 ⎪d f ⎪ dx 2 ⎩

< 0 → local max x0

= 0 → inflection x0

> 0 → local min x0

294

Calculus of Variations & Variational Methods Given f ( x, y ) , a necessary condition for a minimum at ( x0 , y0 ) is df

( x0 , y0 )

⎛ ∂f ∂f ⎞ = ⎜ dx + dy ⎟ = dr ⋅∇f ∂y ⎠ ( x , y ) ⎝ ∂x 0 0

( x0 , y0 )

=0

Since x and y are independent variables, dx and dy are independent so, ∂f ∂f = 0 and =0 ∂x ∂y

295

Calculus of Variations & Variational Methods Constrained Minimization Now minimize f with a constraint, e.g., G ( x, y ) = 0 Lagrange Multiplier Method From (1),

(1)

∂G ∂G dx + dy = dr ⋅∇G = 0 ∂x ∂y

We introduce the modified function with no constraints FL ( x, y, λ ) ≡ f ( x, y ) + λ G ( x, y ) and set ∂F ∂F ∂F dFL = L dx + L dy + L d λ = 0 ∂x ∂y ∂λ

(2) (3) 296

Calculus of Variations & Variational Methods Example: Find the stationary (critical) point of the function f ( x, y ) = 2 x 2 + y 2 − 8 x + y + 1

with the constraint 2 x − y = 0 Sol’n: FL ( x, y, λ ) = (2 x 2 + y 2 − 8 x + y + 1) + λ (2 x − y )

⎫ ∂FL = 4 x − 8 + 2λ = 0, ⎪ ∂x ⎪ ∂FL ⎪ = 2 y + 1 − λ = 0, ⎬ → x = 0.5, ∂y ⎪ ⎪ ∂FL = 2 x − y = 0. ⎪ ∂λ ⎭

y = 1.0, λ = 3.0

297

Calculus of Variations & Variational Methods Calculus of Variations: Functionals and Euler Equations The focus of the calculus of variations is the determination of maxima and minima of expressions that involve unknown functions. Here we look at a few classical problems to introduce some of the concepts.

The Brachistochrone (the one that started it all): Weinstock gives the problem as it was originally stated by John Bernoulli in 1696: “Given two points A and B in a vertical plane, to find for the moveable particle M, the path AMB, descending along which by its own gravity, the beginning to be urged from point A, it may in the shortest time reach the point B.”

298

Calculus of Variations & Variational Methods Reddy & Rasmussen state in engineering terms: “Design a chute between two points A: (0,0) and B: (xb,yb) in a vertical plane such that a material particle, sliding without friction under its own weight, travels from point A to point B along the chute in the shortest time. xb A

yb

B

299

Calculus of Variations & Variational Methods Other Classic Problems Geodesic Problem What is the curve of minimum length that connects two points?

L=∫

b

a

2

(b, yb )

y

⎛ dy ⎞ 1 + ⎜ ⎟ dx ⎝ dx ⎠

y ( a ) = ya ,

y (b) = yb

( a , ya ) x

300

Calculus of Variations & Variational Methods Minimum Surface of Revolution What is the curve of minimum length that connects two points?

b

S = 2π ∫ y ds

(b, yb )

y

a

2

⎛ du ⎞ = 2π ∫ u 1 + ⎜ ⎟ dx a ⎝ dx ⎠ b

y ( a ) = ya ,

y (b) = yb

( a , ya )

y = u ( x) x

301

Calculus of Variations & Variational Methods The Euler Equation Each of these problems presents the problem of finding a continuously differentiable function u(x) that minimizes the integral of the form b

I (u ) = ∫ F ( x, u , u′) dx a

(1)

And that satisfies the end conditions u (a ) = ua , u (b) = ub We now suppose that u(x) is the minimizing function, then choose any continuously differentiable function η(x) and create a oneparameter ‘trial’ function y = u ( x) = u ( x) + αη ( x) 302

Calculus of Variations & Variational Methods that vanishes at the end points u = ua and u = ub. Then for any constant α, u ( x) = u ( x) + αη ( x) with η (a ) = η (b) = 0 satisfies the end conditions. u (a ) = ua , u (b) = ub u(x)

u(x) u ( x) = u ( x) + αη ( x)

η ( x) a

b

x 303

Calculus of Variations & Variational Methods Now we substitute this function into the integral to be minimized b

b

a

a

I (α ) = ∫ F ( x, u , u ′) dx = ∫ F ( x, u + αη , u ′ + αη ′) dx

where I is now a function of α once u(x) and η(x) are assigned. Since we have assumed that u(x) is the minimizing function then I(α) is minimized when α = 0. So, like the differential calculus, we now determine the stationary function (analogous to stationary points), but in this case we know in advance that α = 0. So, dI (α ) =0 dα b ⎛ ∂F du dI (α ) ∂F du′ ⎞ dx =∫ ⎜ + ⎟ a dα ⎝ ∂u dα ∂u ′ dα ⎠ b ⎛ ∂F ∂F ⎞ =∫ ⎜ η+ η ′ ⎟ dx a ∂u ∂u ′ ⎠ ⎝ 304

Calculus of Variations & Variational Methods Now when α → 0 b ⎛ ∂F dI (0) ∂F ⎞ =∫ ⎜ η+ η ′ ⎟ dx = 0. a dα ∂u ′ ⎠ ⎝ ∂u

Now we integrate the second term by parts, ⎡ ∂F d ⎛ ∂F ⎞ ⎤ ∂F dI (0) η dx + η = 0. =∫ ⎢ − ⎜ ⎟ ⎥ a ∂u′ a dα ⎣ ∂u dx ⎝ ∂u′ ⎠ ⎦ b

b

Since this must hold for all choices of η(x), then

∂F d ⎛ ∂F ⎞ − ⎜ = 0. ⎟ ∂u dx ⎝ ∂u′ ⎠

(2)

This is the Euler equation or Euler-Lagrange equation. 305

Calculus of Variations & Variational Methods So, if u(x) minimizes I(u), it must satisfy the Euler equation. Solutions of the Euler equation are called extremals of the problem and an extremal that satisfies the end conditions is called a stationary function. First Integrals Equation (2) is written with partial derivatives that treat x, u, and u′ as independent variables. The second term can be expanded to give ∂ ⎛ ∂F ⎞ ∂ ⎛ ∂F ⎞ du ∂ ⎛ ∂F ⎞ du′ ⎜ ′⎟+ ⎜ ′⎟ + ′⎜ ′⎟ ∂x ⎝ ∂u ⎠ ∂u ⎝ ∂u ⎠ dx ∂u ⎝ ∂u ⎠ dx So, Eq. (2) is equivalent to

306

Calculus of Variations & Variational Methods d 2u du + ( Fu′x − Fu ) = 0. Fu′u′ 2 + Fu′u dx dx

(3)

This is a second-order ODE, unless ∂ 2 F / ∂u′2 ≡ 0, so in general there are two arbitrary constants to satisfy the end conditions. One can show that (3) is also equivalent to 1⎡d ⎛ ∂F du ⎞ ∂F ⎤ F− − = 0. ⎜ ⎟ ⎢ ⎥ u ′ ⎣ dx ⎝ ∂u′ dx ⎠ ∂x ⎦

So, if F does not involve x explicitly, ∂F ∂F = C if ≡ 0. ∂u′ ∂x This is a first integral of Euler’s equation. F − u′

307

Calculus of Variations & Variational Methods Also, if F does not involve u explicitly, another first integral is ∂F = C if ∂u ′

∂F ≡ 0. ∂u

Examples

308

Calculus of Variations & Variational Methods Variational Notation The notation of the calculus of variations shows many similarities to the differential calculus. Beginning with the integrand of the functional I(u)

F = F ( x, u , u ′) We substituted for

y = u ( x) with the trial function

y = u ( x) = u ( x) + αη ( x) = u ( x) + δ u

where δ u = αη ( x) is defined as the first variation of u(x). 309

Calculus of Variations & Variational Methods For a fixed x, ∆F = F ( x, u + αη , u′ + αη ′) − F ( x, u , u′). Expanding via the Taylor series, ⎡ ⎤ ∂F ∂F ⎞ ⎛ ∆F = ⎢ F ( x, u , u ′) + ⎜ δ u + δ u′ + H.O.T.⎥ − F ( x, u , u ′) ⎟ ∂u ∂u ′ ⎠ ⎝ ⎣ ⎦ ∂F ∂F ′ = αη + αη + H.O.T. ∂u ∂u′ Thus, the first variation of F is ∂F ∂F ∆F = δu + δ u′ ∂u ∂u′ 310

Calculus of Variations & Variational Methods This is analogous to the total differential, ∂F ∂F ∂F dF = dx + du + du′ ∂x ∂u ∂u′ Graphically,

η ( x)

δu = u −u

du

dx

u ( x) = u ( x) + αη ( x)

x fixed

du = the first-order approximation of the change along u(x) corresponding to a change in x of dx

δu = the first-order approximation to the change u to u at a fixed x

311

Calculus of Variations & Variational Methods Variational laws are analogous to differentiation,

δ ( F1 F2 ) = F1δ F2 + F2δ F1 , ⎛ F1 ⎞ F2δ F1 − F1δ F2 δ⎜ ⎟= . 2 F2 ⎝ F2 ⎠

The variation and derivative are commutative operators, i.e., d ⎛ du ⎞ (δ u ) = δ ⎜ ⎟ dx ⎝ dx ⎠ This will be a very important result for us later.

Applying the notation to the functional b

I (u ) = ∫ F ( x, u , u ′)dx, a

312

Calculus of Variations & Variational Methods Then ∂F d ⎡ ∂F ⎤ δ I = ∫ ⎢ δu + δ u ⎥ dx = 0. a ∂u ∂u′ dx ⎦ ⎣ Integrating the second term by parts, b

∫

b

a

b d ⎛ ∂F ⎞ ∂F d ∂F δ u dx = δu − ∫ δ u dx. ⎜ ⎟ ∂u ′ dx ∂u ′ a a dx ⎝ ∂u ′ ⎠ b

Then, ⎡ ∂F d ⎛ ∂F ⎞ ⎤ ∂F δI =∫ ⎢ − ⎜ δ u dx + δu = 0 ⎟ ⎥ a ∂u dx ⎝ ∂u′ ⎠ ⎦ ∂u′ a ⎣ b

b

δu = 0 at end points

⎡ ∂F d ⎛ ∂F ⎞ ⎤ δI =∫ ⎢ − ⎜ δ u dx = 0 ⎟ ⎥ a ∂u dx ⎝ ∂u′ ⎠ ⎦ ⎣ b

313

Calculus of Variations & Variational Methods So, ∂F d ⎛ ∂F ⎞ − ⎜ = 0. ⎟ ∂u dx ⎝ ∂u′ ⎠ Thus, we have recovered the Euler-Lagrange equation using the variational notation. End Conditions We now look at the end conditions (more generally, boundary conditions). Start with b

δ I = δ ∫ F ( x, u , u ′)dx a

Applying Liebniz’s rule, 314

Calculus of Variations & Variational Methods b

0

δ I = ∫ δ F dx + [ F ]b δ b − [ F ]a δ a a

0

(fixed end points)

∂F ⎡ ∂F ⎤ ′ =∫ ⎢ δ u − δ u ⎥ dx. a ∂u ∂u ′ ⎣ ⎦ b

Integrating by parts, ⎡ ∂F d ⎛ ∂F ⎞ ⎤ ⎡ ∂F ⎤ δ I =∫ ⎢ − ⎜ δ δ u⎥ + u dx ⎟ ⎥ ⎢ a ∂u dx ⎝ ∂u′ ⎠ ⎦ ⎣ ∂u ′ ⎦ a ⎣ b

b

⎡ ∂F d ⎛ ∂F ⎞ ⎤ ⎛ ∂F ⎞ ⎛ ∂F ⎞ =∫ ⎢ − ⎜ δ u dx + ⎜ (δ u )b − ⎜ (δ u ) a ⎟ ⎟ ⎟ ⎥ a ∂u dx ⎝ ∂u ′ ⎠ ⎦ ⎝ ∂u ′ ⎠b ⎝ ∂u′ ⎠ a ⎣ b

315

Calculus of Variations & Variational Methods This then gives us two possible end point conditions: δ u = 0 at x = a, b → essential boundary conditions (u specified at x = a, b) ∂F = 0 at x = a, b → natural boundary conditions ∂u ′ (u not specified at x = a, b) b

Example: Calculate the first variation of I ( y ) = ∫ y 1 + ( y′) 2 dx a

b

{

}

δ I = ∫ δ y 1 + ( y′) 2 dx a

= 0 (fixed endpts)

Applying the variational chain rule: b ⎤ b⎡ yy ′ 2 2 δ I = ∫ ⎢ 1 + ( y ′) δ y + δ y ′⎥ dx + ⎡ y 1 + ( y ′) δ x ⎤ a ⎣ ⎦a 1 + ( y ′) 2 ⎢⎣ ⎥⎦

⇐ 316

Calculus of Variations & Variational Methods Higher Dimensions Now we extend the methods just presented to multiple variables, in particular we fist look at the function u(x,y,z) and a region in space R enclosed by the surface S. In this case, the functional I(u) has the form R

I (u ) = ∫∫∫ F ( x, y, z , u , u x , u y , u z ) dx dy dz

dτ

R

or using vector notation I (u ) = ∫∫∫ F (r, u , ∇u ) dτ .

r

S

R

317

Calculus of Variations & Variational Methods With the variational notation, the first variation of I(u) is ⎡ ∂F ⎤ ∂F ∂F ∂F δ I = ∫∫∫ ⎢ δ u + δ ux + δ uy + δ u z ⎥ dτ R ∂u ∂u x ∂u y ∂u z ⎢⎣ ⎥⎦ Note that the variation is evaluated at a fixed point (x,y,z), thus δx, δy, δz = 0. ⎡ ∂F ⎛ ∂F ∂ ∂F ∂ ∂F ∂ ⎞ ⎤ + + δ I = ∫∫∫ ⎢ δ u + ⎜⎜ ⎟⎟ δ u ⎥ dτ R ∂u ⎝ ∂u x ∂x ∂u y ∂y ∂u z ∂z ⎠ ⎥⎦ ⎣⎢

Note the operator in the parentheses looks a lot like a dot product. If we set ∂F ∂F ∂F a= eˆ x + eˆ y + eˆ z ∂u x ∂u y ∂u z 318

Calculus of Variations & Variational Methods Then ⎡ ∂F ⎤ δ I = ∫∫∫ ⎢ δ u + a ⋅∇(δ u ) ⎥ dτ . R ∂u ⎣ ⎦ For integration by parts, recall ∇ ⋅ (φ a) = φ ∇ ⋅ a + ∇φ ⋅ a a ⋅∇φ = ∇ ⋅ (φ a) − φ ∇ ⋅ a So, ⎡ ∂F ⎤ δ I = ∫∫∫ ⎢ − ∇ ⋅ a ⎥ δ u dτ + ∫∫∫ ∇ ⋅ (a δ u ) dτ . R ∂u R ⎣ ⎦ 319

Calculus of Variations & Variational Methods Now use the divergence theorem on the last integral ⎡ ∂F ⎤ δ I = ∫∫∫ ⎢ − ∇ ⋅ a ⎥ δ u dτ + w (nˆ ⋅ a) δ u dS . ∫∫ R ∂u S ⎣ ⎦ Again, the necessary condition for minimizing I(u) is that δI = 0. Thus, the integrand of the volume integral must be zero, i.e., ∂F − ∇ ⋅a = 0 ∂u or ∂F ∂ ⎛ ∂F ⎞ ∂ ⎛ ∂F ⎞ ∂ ⎛ ∂F − ⎜ ⎟− ⎜ ⎟ − ⎜⎜ ⎟ ∂u ∂x ⎝ ∂u x ⎠ ∂y ⎝ ∂u y ⎠ ∂z ⎝ ∂u z

⎞ ⎟ = 0. ⎠ 320

Calculus of Variations & Variational Methods This is the 3D Euler-Lagrange equation. The surface integral must also vanish, this gives three possible boundary conditions i.) u specified on S → δ u = 0 on S (essential) ii.) nˆ ⋅ a = iii.)

∂F ∂F ∂F nx + ny + nz = 0 on S (natural) ∂u x ∂u y ∂u z

δ u = 0 on part of S

⎫ ⎬ (mixed) nˆ ⋅ a = on remainder of S ⎭

321

Calculus of Variations & Variational Methods Note the steps used to obtain the Euler equation and the associated boundary conditions: 1. Take the first variation δI(u) 2. Use integration by parts to factor δu in the integrand 3. Use the divergence theorem to create a surface integral containing the boundary conditions

322

Calculus of Variations & Variational Methods Example (from Reddy & Rasmussen): Two-dimensional conduction and convective heat transfer y Convective heat transfer T∞

Tˆ, Specified temperature

P

CT Ch

R Q

R

Cq

Specified heat transfer x

323

Calculus of Variations & Variational Methods

T = temperature T∞ = ambient temperature Q ( x, y ) = internal heat generation I k = k1ˆˆ ii + k2 ˆˆ jj = thermal conductivity qˆ = conductive heat transfer h = convective heat transfer coefficient

324

Calculus of Variations & Variational Methods Minimize: I (T ) = ∫∫

1 2

I ˆ ds + ∫ h ( 12 T − T∞ )T ds {⎡⎣∇T ⋅ k ⋅ ∇T ⎤⎦ + QT }dx dy − ∫ qT

R

Cq

Ch

First variation:

δ I (T ) = ∫∫ R

1 2

I {⎡⎣∇(δ T ) ⋅ k ⋅ ∇T ⎤⎦ + Qδ T }dx dy − ∫ qˆδ T ds + ∫ h( 12 T − T∞ )δ T ds Cq

Ch

Note that in this problem, we already have explicit boundary integrals. As before the interior integral (over R) will also generate a boundary integral after it is integrated by parts.

325

Calculus of Variations & Variational Methods Integration by parts: First, use the vector identity I ∇ δN T ⋅k ∇

T = ∇ φ ⋅ a, ⋅ φ

a

∇ ⋅ (φ a ) = φ ∇ ⋅ a + ∇ φ ⋅ a, ∇ φ ⋅ a = ∇ ⋅ ( φ a ) − φ ∇ ⋅ a. So, I I I ∇ δ T ⋅ k ⋅ ∇ T = ∇ ⋅ (δ T k ⋅ ∇ T ) − δ T ∇ ⋅ ( k ⋅ ∇ T ). Now,

I I δ I = ∫∫ ⎡⎣ ∇ ⋅ (δ T k ⋅ ∇ T ) − δ T ∇ ⋅ ( k ⋅ ∇ T ) + Q δ T ⎤⎦ dx dy R

−

∫ qˆδ T ds + ∫ h (T − T

∞

Cq

)δ T ds.

Ch

326

Calculus of Variations & Variational Methods Now concentrate on the area integral. In 3D, the divergence theorem gives, I I ∫∫∫ ∇ ⋅ (δ T k ⋅ ∇T ) dτ = w ∫∫ (δ T k ⋅ ∇T ) ⋅ nˆ dS R

S

So, in 2D, I I ∫∫ ∇ ⋅ (δ T k ⋅ ∇T ) dS = v∫ δ T nˆ ⋅ ( k ⋅ ∇T ) dS , R

S

Then,

I δ I = ∫∫ ⎡⎣ −δ T ∇ ⋅ ( k ⋅ ∇T ) + Qδ T ⎤⎦ dx dy − ∫ qˆδ T ds + ∫ h(T − T∞ )δ T ds R

+

v∫

I δ T nˆ ⋅ ( k ⋅ ∇T ) dS ,

Cq

Ch

CT + Cq +Ch

327

Calculus of Variations & Variational Methods I δ I = ∫∫ ⎡⎣ −∇ ⋅ ( k ⋅ ∇T ) + Q ⎤⎦ δ T dx dy − R

I ∫ nˆ ⋅ ( k ⋅ ∇T )δ T ds CT

I I + ∫ ⎡⎣ nˆ ⋅ ( k ⋅ ∇T ) − qˆ ⎤⎦ δ T ds + ∫ ⎡⎣ nˆ ⋅ ( k ⋅ ∇T ) − h(T − T∞ ) ⎤⎦ δ T ds. Cq

Ch

Now that δ T is isolated we can write the Euler equation with accompanying boundary conditions I − ∇ ⋅ ( k ⋅ ∇T ) + Q = 0

Euler Equation

essential boundary condition: I set ∫ nˆ ⋅ ( k ⋅ ∇T )δ T ds = 0 by setting δ T = 0 (T = Tˆ on CT ) CT

328

Calculus of Variations & Variational Methods • Multiple Dependent Variables So far we looked at the case where the functional is a function of one dependent variable, i.e., I(u). (Note, u is a dependent variable since it depends on x.) Now we look at the case I(u,v) where the functional is a function of two dependent variables. The following development is for the 2D case, leaving the 3D case as an obvious extension. Begin with the functional to be minimized

I (u, v ) = ∫∫ F ( x, y , u, v, u x , v x , u y , v y ) dx dy R

329

Calculus of Variations & Variational Methods Then for the first variation ∂F ∂F ∂F ∂F ∂F ∂F δ F = δu + δv + δ ux + δ vx + δ uy + δ vy . ∂u ∂v ∂u x ∂v x ∂u y ∂v y essential: u, v specified on C → δ u = δ v = 0 on C. natural: u, v not specified on C. ds nˆ

R C

330

Calculus of Variations & Variational Methods As before, the necessary condition for a minimum is δI = 0 for min I(u,v) Substituting the first variation δF into the integral, ⎡⎛ ∂F

δ I = ∫∫ ⎢⎜ R

⎣⎝ ∂u

δu +

⎞ ⎛ ∂F ⎞⎤ ∂F ∂F ∂F ∂F δ ux + δ uy ⎟ + ⎜ δ v + δ vx + δ v y ⎟ ⎥ dxdy = 0. ∂u x ∂u y ∂v x ∂v y ⎠ ⎝ ∂v ⎠⎦

Now that we’ve separated the integral, it should be clear how to proceed based on our earlier developments. We next rearrange the integral to integrate by parts, ⎡⎛ ∂F

δ I = ∫∫ ⎢⎜ R

⎣⎝ ∂u

δu +

⎞ ⎛ ∂F ⎞⎤ ∂F ∂ ∂F ∂ ∂F ∂ ∂F ∂ (δ u ) + (δ u ) ⎟ + ⎜ δ v + (δ v ) + (δ v ) ⎟ ⎥ dxdy = 0. ∂u x ∂x ∂u y ∂y ∂v x ∂x ∂v y ∂y ⎠ ⎝ ∂v ⎠⎦

331

Calculus of Variations & Variational Methods Now, for integration by parts let’s focus on one of the four terms in the previous expression, ⎞ ∂ ⎛ ∂F ⎞ ∂F ∂ ∂ ⎛ ∂F (δ u ) = ⎜ δu⎟ − ⎜ δu ⎟ ∂u x ∂x ∂x ⎝ ∂u x ⎠ ∂x ⎝ ∂u x ⎠

Recall the gradient theorem,

∫∫∫ ∇φ dτ = w ∫∫ nˆφ dS. R

S

Then,

∫∫ ∇φ dx dy = v∫ nˆ φ ds R

C

⇒

⎧ ∂φ dx dy = φ n ds x ∫ v ⎪⎪ ∫∫ ∂x R C ⎨ ⎪ ∂φ dx dy = φ n y ds ∫ v y ∂ ⎪⎩ ∫∫ R C 332

Calculus of Variations & Variational Methods So, ⎡ ∂ ⎛ ∂F ⎞ ∂ ⎛ ∂F ⎞ ⎤ ∂F ∂ ∫∫R ∂ux ∂x (δ u) dx dy = ∫∫R ⎢⎣ ∂x ⎜⎝ ∂ux δ u ⎟⎠ − ∂x ⎜⎝ ∂ux ⎟⎠ δ u ⎥⎦ dx dy ∂F ∂ ⎛ ∂F ⎞ =v ∫C ∂ux nx δ u ds − ∫∫R ∂x ⎜⎝ ∂ux ⎟⎠ δ u dx dy

Just as in the previous cases, the integration by parts contributes one portion to the interior of the region and one portion to the boundary. The other three similar terms are expanded in the same fashion to give, ⎡ ∂F ∂ ⎛ ∂F ⎞ ∂ ⎛ ∂F ⎞ ⎤ ⎫⎪ ⎪⎧ ⎡ ∂F ∂ ⎛ ∂F ⎞ ∂ ⎛ ∂F ⎞ ⎤ − ⎜ − ⎜ δ I = ∫∫ ⎨ ⎢ − ⎜ ⎟⎥ δ u + ⎢ − ⎜ ⎟ ⎥ δ v ⎬ dx dy ⎟ ⎟ ⎪ ⎣ ∂u ∂x ⎝ ∂u x ⎠ ∂y ⎝ ∂u y ⎠ ⎦ R ⎩ ⎣ ∂v ∂x ⎝ ∂v x ⎠ ∂y ⎝ ∂v y ⎠ ⎦ ⎭⎪ ⎡⎛ ∂F ⎛ ∂F ∂F ⎞ ∂F ⎞ ⎤ +v ∫C ⎢⎜⎝ ∂ux nx + ∂u y n y ⎟⎠ δ u + ⎜⎝ ∂vx nx + ∂v y n y ⎟⎠ δ v ⎥ ds = 0 ⎣ ⎦ 333

Calculus of Variations & Variational Methods So, we have ⎫ ∂F ∂ ⎛ ∂F ⎞ ∂ ⎛ ∂F ⎞ − ⎜ − ⎜ ⎟ = 0⎪ ⎟ ∂u ∂x ⎝ ∂u x ⎠ ∂y ⎝ ∂u y ⎠ ⎪ ⎬ Euler Equations ∂F ∂ ⎛ ∂F ⎞ ∂ ⎛ ∂F ⎞ ⎪ 0 − ⎜ − = ⎜ ⎟ ⎪ ∂v ∂x ⎝ ∂v x ⎟⎠ ∂y ⎝ ∂v y ⎠ ⎭

δ u = δ v = 0 on C (essential) ∂F ∂F nx + ny = 0⎫ ⎪ ∂u x ∂u y ⎪ ⎬ on C (natural) ∂F ∂F nx + ny = 0 ⎪ ∂v x ∂v y ⎪⎭

or mixed boundary conditions. 334