Page 391, use pencil and paper calculations to solve exercise #60. Show all the steps to convince your reader that you understand

tb) Choose a nonzero eigenvector v1. Show that (A – ).1)2v :0 for every vector v that is a multiple of v1.

lc) Suppose that v is not a multiple of vr. Show that v and v1 form a basis ofR2.

rdl Set w : (A -.11)v. Show that there are numbers a and b such that w : avt * bv.

re) Show that (A – 1.1)w : Dw, and conclude from the fact that .1. is the only eigenvalue that b : 0.

r0 Conclude that (A – i./)2v:0. Figure 7 shows two tanks, each containing 500 gallons of a salt solution. Pure water pours into the top tank at a rate of 5 gal/s. Salt solution pours out of the bottom of the tank and into the tank below at a rate of 5 galls. There is a drain at the bottom of the second tank, out of which salt >olution flows at a rate of 5 galls. As a result, the'amount of solution in each tank remains constant at 500 gallons. Initially (time r : 0) there is 100 pounds of salt present in the first tank, and zero pounds of salt present in the tank immediately below.

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5 galls

Figure 7. Tlvo cascaded tanks. ta) Set up, in matrix-vector form, an initial value problem

that models the salt content in each tank over time. tb) Find the eigenvalues and eigenvectors of the coeffi-

cient matrix in part (a), then flnd the general solution in vector form. Find the solution that satisfies the ini- tial conditions posed in part (a).

t c) Plot each component of your solution in part (b) over a period of four time constants (see Section 4.7 or Sec- tion 2.2, Exercise 29) 10,47,.). What is the eventual salt content in each tank? W!y!9iv,e b_qth a physical

and a mathematicalreason-Tor your answer. {Jfigure 8 shows two tanks, each containing 360 liters of

9.2 Planar Systems 391

5 Limin. There are two pipes connecting tank A to tank B. The first pumps salt solution from tank B into tank A at a rate of 4 L/min. The second pumps salt solution from tank A into tank B at a rate of 9 L/min. Finally, there is a drain on tank B from which salt solution drains at a rate of 5 L/min. Thus, each tank maintains a constant. volume of 360 liters of salt solution. Initially, there are 60 kg of salt present in tank A, but tank B contains pure water.

.

C-+ 5 L/min !.1 . a .1. ;l

Fisure 8. Two interconnected tanks. (a) Set up, in matrix-vector form, an initial value problem

that models the salt content in each tank over time.

(b) Find the eigenvalues and eigenvectors of the coeffl- cient matrix in part (a), then find the general solution in vector form. Find the solution that satisfles the ini- tial conditions posed in part (a).

(c) Plot each component of your solution in part (b) over a period of four time constants (see Section 4.7 or Sec- tion 2.2, Exercise 29) [0,47,.). What is the eventual salt content in each tank? Why? Give both a physical and a mathematicakeason for your answer.

60. In Exercisr u were given the circuit in Figure 9 anTE 6-w that the voltage V across the capacitor and the current 1 across the inductor satisfled the system

V,- V *1.RCC

Suppose that the resistance is R : 112 ohm, the capaci- tance is C : 1 farad, and the inductance is L : I l2henry. If the initial voltage across the capacitor is V(0) : 10 volts and there is no initial curent across the inductor, solve the system to determine the voltage and current as a function of time. Plot the voltage and current as a function of time. Assume current flows in the directions indicateil.

Figure 9. A parallel circuit with capacitor, resistor, and inductor.

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).1)2v 1 – r salt solution. Pure water pours into tank A at a rate of

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5 L/min

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Section 9.2 Lecture Notes (Part 4) The goal in this activity is to investigate some modeling using systems of differential equations. Our first example will be an electrical circuit, so let's take a moment to recall the following laws.

The current I is the rate at which the charge Q flows. Consequently, .

Component Laws: The voltage is measured in volts, denoted by V. The current is measured in amperes, denoted by A.

Ohm's Law: We will let represent the voltage drop across a resistor. By Ohm's Law,

The resistance R is measured in ohms, denoted by Ω.

Faraday's Law: We will let represent the voltage drop across an inductor. By Faraday's Law,

The inductance L is measured in henrys, denoted by H.

Capacitance Law: We will let represent the voltage drop across an inductor. By the Capacitance Law,

The capacitance C is measured in farads, denoted by F.

Kirchoff's Voltage Law: The sum of the voltage drops around any closed loop in a circuit must equal zero.

Kirchoff's Current Law: The sum of currents flowing into or out of any junction equals zero. In the following image, note that the currents , , and are each flowing into the junction (red filled point).

Using Kirchoff's Current Law, we would write:

Another way of saying Kirchoff's Current Law is to say "the flow into the junction must equal the flow coming out of the junction. So is the flow into the junction in the picture above and there is no flow coming out of the junction, so the flow out of the junction equals zero. Let's look at another picture.

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This time the flow into the junction is and the flow out of the junction is , so we would write:

Now, let's look at an example.

Example #1: Consider the following circuit.

Write a system that models the flow of the currents and .

Solution: Note that I is flowing into the junction point and and are flowing out of the junction point, so:

If we circle clockwise around the loop on the left containing the resistor and induction, by Ohm's Law and Faraday's Law, we get:

Now, let's substitute and and simplify.

2

We have our first equation. Now we need a second equation. This time we will loop clockwise around the outer loop containing the resistor and capacitor. By Ohm's Law and the Capacitance Law, we get:

Now, the current flowing through the capacitor is . The current is equal to the rate at which the charge is

changing, so , or equivalently, . Let's differentiate our last equation and insert this result.

Now, let's insert , differentiate, and manipulate a bit.

Now, how do we get rid of the on the right side of this last equation? The answer is, substitute our first

solution .

We now have our second equation, so our system of equations for the currents is:

3

Example #2: Using the equations in Example #1, suppose that ohms, henry, and farads. Assume the

initial currents at time are given by amperes and amperes.

Solution: Enter the values ohms, henry in our first equation.

Next, enter the values ohms, henry, and farads in our second equation.

Thus, our two equations are:

We can check these with Matlab.

syms I1(t) I2(t) R=2;L=1;C=1/8; eqns=[diff(I1,t)==-R/L*I1-R/L*I2;… diff(I2,t)==R/L*I1+(R/L-1/(R*C))*I2]

eqns(t) =

Same equations. Now, we can write this system in matrix form.

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The trace is and the determinant is .

We can check this with Matlab.

syms lambda A=[-2 -2;2 -2]; p=charpoly(A,lambda)

p =

solve(p,lambda)

ans =

Next, use our shortcut for finding the corresponding eigenvectors, subtracting λ from each diagonal element of matrix A.

Looking at the first row, the following vector will zero out the first row.

Choosing the eigenvalue , we get:

Because any multiple of an eigenvector is also an eigenvector, we will multiply this last result by and use:

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We can check this with Matlab.

[V,D]=eig(sym(A))

V =

D =

Again, our eigenvector is a multiple of the Matlab's eigenvector. Let's multiply the second column of the matrix V by .

-i*V(:,2)

ans =

That's our hand-calculated eigenvector. Because any multiple of an eigenvector is also an eigenvector, our work is correct to this point. Now, we substitute our eigenvalue and eigenvector to get a complex solution and expand the result.

We can check our effort with Matlab.

syms t real lambda=-2+2i; v=[1;-i]; z=exp(lambda*t)*v; real(z)

ans =

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imag(z)

ans =

Same as our hand-calculated answers above. The real and imaginary parts of this complex solution gives us a fundamental set of real solutions, so our general solution for our system is:

Now, our initial condition is:

Substituting this in our general solution gives us:

This makes us giggle! Note that this is equivalent to:

We never get this lucky. Now we can substitute and into our general solution.

Therefore, the following equations give us the currents in the circuit at time t.

7

Let's check our answer with Matlab. Recall that the original initial value problem was:

syms I1(t) I2(t) eqns=[diff(I1,t)==-2*I1-2*I2;diff(I2,t)==2*I1-2*I2]; conds=[I1(0)==5;I2(0)==4]; [I1sol,I2sol]=dsolve(eqns,conds)

I1sol =

I2sol =

Same answer! How about visualizing our result?

fplot(I1sol,I2sol,[0,2*pi]) line(5,4,'Marker','o') text(5.2,4.2,'(5,4)') line(0,0,'Marker','o') text(0.2,0.2,'(0,0)') axis([-1,6,-1,5]) grid on xlabel('I_1-axis') ylabel('I_2-axis') title('Both currents head towards zero.')

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Note that the currents start at the initial condition . Now, there is no voltage source in our circuit, so one would expect the current to die. Sure enough, both currents wind up at zero (see the point ).

Example #3: Consider the two tanks picture below.

Each tank contains 200 gallons of a salt solution. Pure water pours into the top tank at a rate of 10 gal/min. Salt solution pours out of the top tank at a rate of 10 gal/min and enters the bottom tank at a rate of 10 gal/min. Salt solution pours out of the bottom tank at a rate of 10 gal/min. At time , the salt content in the top tank is 50 lb and the salt content in the bottom tank is 0 lb (pure water).

Part 1: Set up, in matrix-vector form, an initial value problem that models the salt content in each tank over time.

Part 2: Sketch both solutions and versus time t. What is the eventual salt content in each tank? Give both a physical and a mathematical reason for your answer.

Solution: Because pure water (no salt content) is poured in the top tank from above, the rate at which salt is coming in is:

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Because solution flows into and out of the top tank at the same rate (10 gal/min), the volume of the solution in the top tank remains a constant gallons. Because represents the salt content (in pounds) in the solution in the top tank,

Therefore, our equation for the solution change in the top tank is:

For the tank at the bottom, the rate at which the salt flows into the tank equals the rate at which it flows out of the tank above, so:

Because solution flows into and out of the bottom tank at the same rate (10 gal/min), the volume of the solution in the bottom tank remains a constant gallons. Because represents the salt content (in pounds) in the solution in the bottom tank,

Therefore, our equation for the solution change in the bottom tank is:

Because the salt content in the top tank at time is 50 pounds, . Because the salt content in the

bottom tank at time is 0 pounds (pure water), . Thus, we have the following initial value problem.

Now we can put this in matrix form.

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The trace of the matrix is and the determinant is . Therefore,

We can check this with Matlab.

syms lambda A=[-1/20, 0;1/20, -1/20]; p=charpoly(A,lambda)

p =

solve(p)

ans =

Perfect! This is a single repeated eigenvalue of algebraic multiplicity 2. Now we can use our shortcut to find the corresponding eigenvector.

Examining the second row (remember, our shortcut does not work for a row containing a zero), we see that

will zero out the second row. Hence,

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Because any multiple of an eigenvector is also an eigenvector, we will multiply this eigenvector by 20 and choose

Again, we can check this with Matlab.

[V,D]=eig(sym(A))

V =

D =

Perfect! Thus, our first solution is:

To find our second solution, we have to solve the equation:

Note that if we multiply 20 times the first column and 0 times the second column, we have a solution.

Thus,

and our second solution is:

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Therefore, our general solution is:

Now we substitute our initial condition.

Therefore, and . Substituting these into our general solution, our solution is:

Thus, we have solutions for the salt content in each tank.

We can check our work with Matlab.

syms x1(t) x2(t) eqns=[diff(x1,t)==-1/20*x1;diff(x2,t)==1/20*x1-1/20*x2]; conds=[x1(0)==50,x2(0)==0]; [x1sol,x2sol]=dsolve(eqns,conds)

x1sol =

x2sol =

Same as our hand-calculated answers!

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Physically, because pure water is pouring into the top tank, one would imagine that eventually both tanks will have nothing but pure water in them. Let's sketch the graphs of and to visualize this idea. First, let's examine the numeric value of .

[exp(-1);exp(-2);exp(-3);exp(-4)]

ans = 4×1 0.3679 0.1353 0.0498 0.0183

Note that these numbers go from about 36% down to 1%. If we consider our first equation, , if

, then will be about 36% of 50 pounds. Similarly, if , then will be about 1% of 50 pounds. Note that:

The number is called a time constant and is a interval of four time constants, a good interval over which to sketch our graph.

fplot([x1sol,x2sol],[0,80]) grid on xlabel('t-axis') ylabel('x_1 and x_2 axis') title('Salt content in two tanks.') legend('Top Tank','Bottom Tank')

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Sure enough, the salt content in both tanks appears to be heading toward zero pounds. Our image also shows that the salt content in the top tank approaches zero quicker, which makes sense. Our image also shows that the salt content in the bottom tank first increases, then decreases to zero. Also makes sense.

Now, what analytic approach can we use to verify that both salt contents are eventually zero? Let's take the limit of as .

We can check this with Matlab.

limit(x1sol,t,Inf)

ans =

Checks. Now, we can use l'Hopital's rule to help with the second solution.

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We can also check this with Matlab.

limit(x2sol,t,Inf)

ans =

Same answer! We're done! 🙂

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