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

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

Nasser M. Abbasi

June 15, 2014

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.

rearrange terms

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

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

|------------| | 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

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

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

Where

Hence

and

Hence the solution is

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 $x = b$

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

2 2 2 ds = dy + dx

Hence

or we can also write

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

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

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

Hence we have

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

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

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)

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

and the second term is

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

This is separable ODE, hence

Integrate both sides

where $k$ is constant on integration

where $c= ek$ is another constant

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

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

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

at $t = T,y = 0$ , hence

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 $α = 0.03$

This one is for $α = 0.1$

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 $α = 0.001$