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HW 4 Mathematics 503, Mathematical Modeling, CSUF , June 9, 2007

Nasser M. Abbasi

June 15, 2014

Contents

1 Problem 1 (section 3.3, 2(b), page 175)
2 Problem 1 (section 3.3,#10, page 176)

1 Problem 1 (section 3.3, 2(b), page 175)

problem: Find extrermals for the following functional:

(b) ∫by2+ (y′)2+ 2yex dx
 a

Solution:

Assume first that y(x)  has normal conditions on the boundaries. I.e. y(a)= ya,y(b)= yb

     ′      2   ′ 2     x
L (y,y ,x) = y + (y) + 2ye

We have the functional

       ∫ b
J (y) =    L(y,y′,x) dx
        a

and we seek to find a function y(x)  which minimizes this functional.

Let the vector space from which we can pick y(x)  from be

      2
V = C  [a,b]

And let the set of admissible functions (within V ) (Is this set a subspace?) be defined as

A = {y(x) s.t. y(x)∈ V and y(a)= y0,y(b)= y1}

And let the set of admissible directions v(x)  be

Ad = {v(x)∈ V s.t. v(a)= 0,v(b)= 0}

Use the variational method since the Lagrangian contains a quadratic terms.

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rearrange terms

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Hence if we can find ~y(x)  which will make the last term above zero, then J (y)  will have been minimized by this~y(x)

Therefor the problem now becomes of solving for y(x)  the following integral equation

∫
  byv+ y′v′+ vex dx = 0
 a
(1)

We need to try to convert the above into something like ∫
 ba f(y,y′,ex)v(x) dx= 0  so that we can say that f (y,y′,ex)= 0  , so this means in (1) we need to do integration by parts on the term y′v′ . Hence (1) can be written as

∫ b      ∫ b ′′    ∫ b  x
 a yvdx+  a yvdx +  a ve dx = 0

Now since ∫ u dz = [uz]b− ∫bz du
           a   a , now let u = y′ → du = y′′, and let dz= v′dx→  z= v , hence we have

∫         ◜◞0◟|◝  ∫
  by′v′dx= [y′v]b−   bvy′′ dx
 a            a   a

Hence (1) can be written as

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Now we apply the standard argument and say that since v(x)  is arbitrary function, and the integral above is always zero, then it must be that

     ′′  x
(y− y + e )= 0

or

|------------|
| y′′− y= ex  |
--------------

This is a linear second order ODE with constant coefficients with a forcing function. The homogeneous ODE will have 2 independent solutions, say y1(t)  and y2(t)  , so the total solution is

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To solve the homogeneous ODE

y′′− y = 0

Assume the solution is y= Aemx  , hence the characteristic equation is m2 − 1= 0 → m = ±1, hence the solution is y (x)= ex,y (x) = e−x
 1         2  , so

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Or the solution can be written in hyperbolic sin and cosine

yh(t)= c1cosh(x)+ c2sinh (x)

Now to find particular solution, use variation of parameters. Assume

yp = − y1u1 +y2u2

where

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Where

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Hence

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and

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Hence the solution is

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Hence

|------------------------------|
|                              |
| y(x)= c1ex+ c2e−x+ 12xex− 14ex  |
-------------------------------

2 Problem 1 (section 3.3,#10, page 176)

problem: Show that the minimal area of a surface of revolution in a catenoid, that is, the surface found by revolving a catenary

          (      )
            x+ c2
y = c1cosh  -c---
              1

about the x axis

solution:

First we assume that y′(x)>  0  over the integration range. And that the lower end of the integration x = a is smaller than the upper limit x = b

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If we view a small ds at y(x)  we see that it has a length of

  2    2    2
ds = dy + dx

Hence

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or we can also write

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So there are 2 ways to solve this depending if we use (1) or (2). Let us leave the choice open for a little longer.

Now a size of a differential area dA of a strip of width ds and a length given by the circumference of the circle generated by rotation is

dA = 2πy(x)ds

Hence the total surface area is the integral of the above over the range which y(x)  is defined at. Let this be from x = a, to x= b is given by

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Since we have y(x)  already in the Lagrangian, let then pick expression (2) from the above.

      ∫        ∘ (---)2----
A = 2π  x=by(x)    dx-  + 1dy
        x=a         dy

Now I need to change the limits. Let y(a)= y
       1   , and let y(b)=  y
        2   hence

                 ∘ ----------
       ∫ y=y(b)      ( dx)2
A = 2π       y(x)    ---  +1dy
        y=y(a)         dy

If we write the above in the more standard format, we have

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Remember now that y is the independent variable, and x is the dependent variable. This is different from the normal way.

Hence the Lagrangian L is

           ∘  --------
L (y,x,x′)=  y  1+ (x′)2
(3)

If I had picked expression (1) instead, I would have obtained the Lagrangian as

           ∘  --------
L (x,y,y′)=  y  1+ (y′)2
(4)

Both will give the same answer but with (3) we have ∂L   d( ∂L )
∂x − dy-∂x′ = 0  and the first term ∂L
∂x = 0  since L does not depend on x , and now we can just say that ∂L-=
∂x′  constant,. While with (4) we have       (   )
∂L−  d- ∂L  = 0
∂y   dx  ∂y′  now ∂∂Ly  is not zero.

Now we continue, and we will use (3) as our lagrangian.

We start the solution of the problem. We seek a function x~(y)  which minimizes J(x)= ∫ y=y(b)L(y,x,x′)dy
       y=y(a) .

Where x~(y)∈ V s.t. C2[a,b ]  , and let the set of admissible functions

A(x(y))= {x(y)|x∈ V and x(a)= x1,x(b)= x2} ,

and let the set of admissible directions v(y)=  {v(y) |v∈ V and v(a) = 0 and v(b)= 0}

Now that we have written down all the formal definitions, we can just solve this by applying Euler-Lagrange equation since the Lagrangian above meets the conditions of using Euler-Lagrange equations (L is a function of    ′
x,x,y and x is defined at the boundary conditions with a dirichlet type boundary conditions).

The Euler Lagrangian equation is

       (    )
∂L-− d-- ∂-L  = 0
∂x   dy  ∂ x′

Since L does NOT depend on x then ∂L∂x = 0  , and the above reduces to

   (    )
-d-  ∂L-  = 0
dy   ∂x′

Since the derivative is zero, then we can write that

∂L-= c
∂x′   1

Where c1   is some constant. So the above becomes

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Hence we have

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Hence the final ODE is

|----------------|
|                |
| x′(y)=  ∘-c21-2- |
|         (y−c1)  |
------------------

This is a linear ODE Its solution is found by integration both side as follows

pict

Let -y= u
c1 hence dy = cdu
      1 and the above becomes (do not need to worry about limits of integrations, as I will flip back the earlier variable in a minute)

        ∫
x(y)= c1  √--du---
            u2− 1

Which from table is given by

          (    √------)
x(y)= c1ln u +  u2 − 1 + c2

Where c2   is constant of integration. Hence going back to our variables, we have

          (    ∘ --------)
            y    ( y)2
x(y)= c1ln  k +    k  − 1  + c2
(3)

From tables I found that cosh−1(z)= ln(z+ √z2−-1)

Hence (3) can now be written as

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Or

y        (x − c)
--=  cosh  ----2
c1          c1

Then

|---------------------|
|              (x−c2)  |
| y(x) = c1cosh  -c2   |
-----------------------

So the above curve will minimize the surface area.

Problem 1 (section 3.3,#14, page 177)

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answer:

y(0)= Y,y(T) = 0

    ∫ T
E =    e−βtU (r(t))dt
     0

but

            ′
r (t)= αy − y

Hence

    ∫ T
E =    e−βtU (αy (t)− y′(t))dt
     0

Hence the Lagrangian is

|-------------------------------|
|        ′   −βt           ′    |
| L (t,y,y)= e   U (αy(t)− y(t)) |
--------------------------------

since y(t)  is defined at boundaries of the interval, we can use Euler-Lagrange equations

              (           )
d--     ′   d-  -d-      ′
dyL(t,y,y )− dt  dy′L(t,y,y)  = 0

Now the first term above is

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and the second term is

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Hence our E-L equations now looks like

   −βt  ′ -d ( −βt ′)
αe   U + dt  e  U   = 0
(1)

Since U is a function of r(t)  , then

-d (e−βtU′)= − βe−βtU′+ e−βtU′′r′(t)
dt

And (1) now becomes

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This is separable ODE, hence

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Integrate both sides

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where k is constant on integration

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where c= ek  is another constant

But U (r)= 2√r- hence dU=  1√-
dr    r  and the above becomes

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Where since c− 2   is constant, I call it c2

Now, Since

y′(t)= αy (t)− r(t)

Then

|------------------------|
| y′(t)− αy (t)= c2e2(α−β)t |
--------------------------

The solution is

y= y + y
    h   p

Assume       mt
yh = Ae  , hence    mt      mt
Ame  − αAe   = 0→  m = α

So the solution is

yh = c1eαt

For the particular solution, guess a solution. Since the forcing function is of the form cet  , guess

yp = Aekt

so y′p = Atekt  and we substitute this solution in the ODE above, we obtain

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so by comparing exponents, we see that k = 2(α − β)  and A (k− α) = c2   hence      c      c       c
A = k2−α = 2(α−2β)−α-= α−22β

Therefore

yp =---c2--e2(α−β)t
    α − 2β

Hence, since

y= yh+ yp

Then

|-------------------------|
|                         |
| y(t)= c1eαt+ αc−22β-e2(α−β)t |
---------------------------

We now find c1   and c2   from I.C. At t = 0,y = Y , hence

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at t = T,y = 0  , hence

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so from (2)

        ----(α-−-2β)YeαT------
c1 = Y − (α − 2β )(eαT − e2(α−β)T)

Hence

|--------------------------------------------------------|
|       (                   )                            |
|            ----(α−-2β)YeαT---   αt  (1−α+2βY)e(−α+2β)T 2(α−β)t  |
| y(t)=   Y − (α−2β)(eαT−e2(α−β)T) e  +      α−2β     e       |
---------------------------------------------------------

and

|------------------------------------|
|                                    |
| r(t)= (1− α + 2βY )e(−α+2β)T e2(α−β)t  |
-------------------------------------

Analysis on results:

These are 3 plots showing y(t)  and r(t)  . The first is for α = 0.03

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This one is for α =  0.1

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We notice that the higher the interest rate α is the more capital will accumulate, which means to achieve the goal of zero capital at death the rate r(t)  is more steep near the end. If the money hardly accumulate during life time, i.e. when the interest rate is very low, then we should expect a straight line for y(t)  , which is verified by this plot below when I set α = 0.001

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