HW 9 Mathematics 503, Mathematical Modeling, CSUF , July 16, 2007

Nasser M. Abbasi

June 15, 2014

problem:

Consider the problem of minimizing the functional $∫ J(u)= L (x,u,∇u )dx Ω$ over all $u∈ C2 (Ω)$ with $u (x) = f(x)$ at boundary $Γ$ where $f$ is a given function. $Ω$ is bounded and well behaved in $2 ℝ .$

(a) Show that the ﬁrst variation is (Where $L$ below is meant to be $L (x,u,∇u )$ ) where $x$ is the vector $⌊ ⌋ || x1|| |⌈ |⌉ x2$

Where $L∇u$ is the vector $⌊ ⌋ ⌊ ⌋ |L( ∂u)| | ∂(∂ ∂Lu)| || ∂x1 ||= || ∂x1 || ⌈L( )⌉ ⌈ -(∂L)⌉ ∂∂xu2 ∂ ∂∂ux2$ and $h ∈ C2(Ω)$ with $h(x)= 0$ at the boundary $Γ$

(b)Show that the necessary condition for $u$ to minimize $J$ is that $u$ must satisfy the Euler equation $Lu− ∇⋅L∇u = 0$ , $x∈ Ω$

(c)If $u$ is not ﬁxed on the boundary $Γ$ ﬁnd the natural boundary conditions.

Answer

(a)

Hence

But $δ J(u,h)= limt→0 dJ(ud+tth)$ , hence at $t = 0$ the above becomes

∫ ∂L ( ∂ L ) δJ (u,h )= ---h+ ------⋅∇h dA Ω ∂u ∂(∇u )

(1)

But $⌊ ⌋ |-∂u-| |∂x1| ∇u = |⌈ |⌉ ∂∂ux2$ , hence $⌊ ⌋ |-(∂L)| -∂L- |∂ ∂∂ux1 | ∂(∇u) = |⌈ |⌉ ∂(∂ ∂Lu) ∂x2$ , and $⌊ ⌋ |-∂h| |∂x1| ∇h = |⌈ |⌉ ∂∂xh2$ , therefore

Hence (1) becomes

∫ ( ) δJ (u,h )= Luh+ Luxhx1+ Luxhx2 dA Ω 1 2

(2)

Now

-∂-( ) ∂-Luxi ∂xi Luxih = ∂xi h+ Luxihxi

Hence

∂ ( ) ∂Luxi Luxihxi = ∂x- Luxih − ∂-x-h i i

Hence substitute the above in (2) for $i = 1,2$ we obtain

Now using Green theorem, where

∫ ( ∂Q ∂ P) ∫ ∂-x − ∂x- dx1dx2 = Pdx1+ Qdx2 Ω 1 2 Γ

Let $Q ≡ Luxh,P ≡ − Luxh 1 2$ , hence Green theorem becomes

∫ ( ) ∫ ( ) -∂-L + -∂-L h dxdx = − L dx + L dx h ∂x1 ux1 ∂x2 ux2 1 2 Γ ux2 1 ux1 2 Ω

Substitute the above into second term in (3) we obtain (noting that $dA = dx1dx2$ since we are in $ℝ2$ )

∫ ( ∂ Lux1 ∂ Lux2) ∫ ( ) δ J(u,h)= Lu− -∂x--− -∂x-- h dA + Lux1dx2− Lux2dx1 h Ω 1 2 Γ

(4)

But the second integral above can be rewritten as (by dividing and multiplying by $ds$ )

∫ ( ) ∫ ( dx2 dx1) Lux1dx2− Lux2dx1 h≡ Lux1 ---− Lux2--- h ds Γ Γ ds ds

Hence (4) becomes

∫ ( ) ∫ ( ) ∂Lux1 ∂Lux2 dx2 dx1 δJ (u,h )= Lu− ∂x1 − ∂x2 h dA + Γ Lux1ds − Lux2ds h ds Ω

(5)

Now Tangent vector at the boundary at point $(x1,x2)$ is given by vector $(dx1,dx2)T ds ds$ , hence the normal is $(dx2 dx1)T n = ds ,− ds$ (since if we take dot product of these 2 vectors we get zero). Now we can rewrite the integrand in the second integral in (5) in terms of this normal vector since

Substitute the above into the second term of (5) we obtain

|----------------------------------------------------| | ∫ ( ) | | δJ (u,h) = L − ∂Lux1-− ∂Lux2- h dA+ ∫ h (L ⋅n ) ds | | u ∂x1 ∂x2 Γ ∇u | | Ω | -----------------------------------------------------

Final note on the sign before the second integral above. The book shows it as $”− ”$ . I think this is because the normal should be pointing outside? Hence if we make out normal the negative of the normal used here (which I think points inwards), we obtain the result we are asked to show for part (a). (notice, the book has a mistake/typo, it says $∫ h (L ∇u ⋅n ) dA Γ$ instead of $∫ h(L∇u⋅n) ds Γ$ , i.e. the integration is over a line segment, not over a diﬀerential area (since obviously this is contour integration).

part (b)

Necessary condition for minimum is that $δJ(u,h)= 0$ ,. ie.

∫ ( ) ∫ ∂Lux1- ∂Lux2 Lu − ∂x1 − ∂x2 h dA − Γ h(L∇u⋅n) ds = 0 Ω

Now consider the second integral in the above. Since $h= 0$ on $Γ$ , hence we are left to show that

( ) ∫ ∂Lux1- ∂Lux2 Lu − ∂x1 − ∂x2 h dA = 0 Ω

But $h$ is arbitrary function, hence by lemma 3.13 again, we argue that for the above to be zero, then

Which is Euler-Lagrange equation.

Part (c)

Here we have free boundary conditions. Hence we can not take $h= 0$ everywhere on $Γ$ . Starting with the ﬁrst variation

∫ ( ∂Lux1 ∂Lux2) ∫ δJ(u,h) = Lu − ∂-x--− -∂x-- h dA − h(L∇u⋅n) ds = 0 Ω 1 2 Γ

Since $h ⁄= 0$ on $Γ$ then by lemma 3.13 we can argue that $L∇u⋅n= 0$ on $Γ$

Hence on $ℝ2$ , this means $⌊ ⌋T ⌊ ⌋ |L | | dx2 | || ux1|| || ds || = 0 ⌈ ⌉ ⌈ dx ⌉ Lux2 −-d1s$ , i.e.

dx2 dx1 Lux1 ds − Lux2 ds = 0

Now we need to know the shape of the boundary to evaluate the above at each point. For example, for a circle, $⌊ ⌋ | x| n = || 1|| ⌈ ⌉ x2$ and the above become

Lux1x1− Lux2x2 = 0

And the above equation needs to be satisﬁed at each point on the boundary after discretization.

HW 9 Mathematics 503, Mathematical Modeling, CSUF , July 16, 2007

Contents

1 Problem 8 page 362 section 6.3