HW 5 Mathematics 503, Mathematical Modeling, CSUF , June 18, 2007

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HW 5 Mathematics 503, Mathematical Modeling, CSUF , June 18, 2007

Nasser M. Abbasi

June 15, 2014

1 Problem 1 (section 3.5, 5(b), page 185)
2 Problem 2 (section 3.5,#6, page 185)

1 Problem 1 (section 3.5, 5(b), page 185)

problem: Find extrermals for the following functional:

(b) $J(y)= ∫3e2x(y′2− y2)dx 0$

$y(0)= 1,y(3)=$ free

Solution:

′ 2x( ′2 2) L(x,y,y)= e y − y

Starting from ﬁrst principles. First the preliminary standard setup:

Let $J :(A⊂ V )→ ℝ$ , where $A$ is the set of admissible functions, and $V :C2 [a,b]$ , hence $A = {y ∈V : y(a)= 0,y(b)= free}$

Let $v(x)$ be the set $Ad(y)$ of the permissible directions deﬁned as $Ad(y)= {v ∈V :y +tv ∈A }$ for some real scalar $− ξ < t < ξ$

And $Ly(x,y,y′)≡ ∂LL (x,y,y′) ∂y$ and $Ly′(x,y,y′)≡ ∂L′L (x,y,y′) ∂y$

Now we write

Therefor a necessary condition for $y(x)∈ A$ to be a local minimum for the functional $J (y)$ is that $δ J(y,v)= 0$ for all $v ∈ Ad$ , which means

∫ b L (x,y,y′)v+ L ′(x,y,y′)v′ dx = 0 a y y

Integrating by parts the second term above results in the general expression for the necessary condition for $y(x)$ to be a local minimum for $J(y)$ , which is

∫ b{ d } Ly(x,y,y′)− --Ly′(x,y,y′) v dx + [Ly′(x,y,y′)v]ba = 0 a dx

(see 3.15 in text)

Since $v(a) = 0$ , the second term above simpliﬁes, and the above equation becomes

∫ { } b ′ -d- ′ ′ a Ly(x,y,y)− dx Ly′(x,y,y ) v dx+ Ly′(b,y(b),y (b))v (b) = 0

(1)

Now we apply the following argument: Out of all functions $v ∈ Ad$ , we can ﬁnd a set which has the property such that $v(b) = 0$ . For these $v′s$ only (1) becomes

{ } ∫ b ′ d-- ′ a Ly(x,y,y )− dxLy′(x,y,y) v dx = 0

Where now we apply the other standard argument: Since the above is true for every arbitrary $v$ (but remember now $v$ is such that $v(b)= 0,$ but since there are so many such $′ vs$ still, then the argument still holds) , then it must mean that

′ -d- ′ Ly(x,y,y)− dx Ly′(x,y,y )= 0

(2)

This will generate a second order ODE, which we will solve, with the boundary conditions $y(0) = 1$

But we need another boundary condition. Then we hold oﬀ solving this for one moment. Let us now consider those functions $v∈ Ad$ which have the property that $v(b)⁄= 0.$ For these $v$ 's, and for the second term in (1) to become zero, we now must have

′ Ly′(b,y(b),y (b)) = 0

(3)

Now from (3) we have $∂L′ = ∂′e2x(y′2− y2)= 2e2xy′ ∂y ∂y$ , which means

Hence

|-----------| | y′(b)= 0 | | | ------------

This gives us the second boundary condition we needed to solve (2).

Hence to summarize the problem becomes that of solving for $y$ given

d Ly(x,y,y′)− ---Ly′(x,y,y′)= 0 dx

with the boundary conditions $y(0)= 1$ and $′ y (3) = 0$

Now (2) can be written as

Hence

|----------------------------------------------| | y′′+ 2y′+ y= 0 y(0) = 1, y′(3) = 0 | | | -----------------------------------------------

Assume $mx y= Ae$ , hence the characteristic equation is $2 −b±√b2−4ac −2±√4−4- m + 2m + 1= 0 → m = 2a = 2 =$ $− 1$

Since we have repeated root, then the solution is

|---------------------| | y (x) = ce−x+ c xe−x | ----------1-----2------

When $x= 0, y= 1$ , hence $c1 = 1$

′ −x −x − x y(x)= − e + c2(e − xe )

when $x = 3,y′ = 0,$ hence $e−3 0= − e−3+ c2(e−3− 3e−3) → c2 = − e−3(2) →$ $1 c2 = − 2$

Hence the solution is

y(x)= e− x− 1xe−x 2

|--------------------| | y(x)= e−x(1 − 1x) | | 2 | ---------------------

2 Problem 2 (section 3.5,#6, page 185)

problem: determine the natural boundary condition at $x= b$ for the variational problem deﬁned by $∫b ′ J (y) = a L(x,y,y )dx+ G (y(b))$ where $2 y ∈ C [a,b],y(a)= y0$ and $G$ is a given diﬀerentiable function on $ℝ$

Solution:

Starting from ﬁrst principles, ﬁrst the preliminary standard setup.

Let $J :(A⊂ V )→ ℝ$ , where $A$ is the set of admissible functions, and $V :C2 [a,b]$ Hence $A = {y ∈V : y(a)= 0,y(b)= free} .$ Let $v(x)$ be a set $Ad(y)$ of permissible directions deﬁned as $Ad (y) = {v∈ V :y+ tv∈ A}$ for some real scalar $− ξ < t < ξ$ , and Let $′ ∂L ′ Ly(x,y,y )≡ ∂yL (x,y,y)$ , and $Ly′(x,y,y′) ≡ ∂∂Ly′L(x,y,y′)$

Now we write

Therefor a necessary condition for $y(x)∈ A$ to be a local minimum for $J(y)$ is that $δJ(y,v) = 0$ for all $v ∈ Ad$ , which means

( ∫ b ) Ly(x,y,y′)v+ Ly′(x,y,y′) v′ dx + v(b)G ′(y (b)) = 0 a

Integrating the second term in the integral above by parts results in the general expression for the necessary condition for $y(x)$ to be a local minimum for $J(y)$ , which is

∫ b{ } Ly(x,y,y′)− d-Ly′(x,y,y′) v dx + [Ly′(x,y,y′)v]b+ v(b)G′(y(b))= 0 a dx a

Hence

∫ b{ d }
Ly(x,y,y′)− ---Ly′(x,y,y′) v dx+
a dx
Ly′(b,y(b),y′(b))v(b)− Ly′(a,y (a) ,y′(a))v(a)+ v(b)G′(y(b))= 0

Since $y(a) = y0$ , we must have $v(a)= 0$ , then the above simpliﬁes to

∫ { } b L (x,y,y′)− -d-L′(x,y,y′) v dx+ {L ′(b,y(b),y′(b)) + G′(y(b))}v (b) = 0 a y dx y y

(1)

Let us now consider those functions $v∈ Ad$ which have the property that $v(b)⁄= 0.$ For these $v$ 's, for the second term in (1) to become zero, we now must have

Ly′(b,y(b),y′(b)) +G ′(y (b)) = 0

Hence

|----|-------------------| | -∂L | = − G ′(y(x))| | | ∂y′|x=b x=b | -------------------------|

Hence the natural boundary condition on $y (x)$ at $x = b$ must satisfy the above. (I do not see how can one go further without being given what $L$ and $G$ are.)