# Systems of Linear Differential Equations

(Difference between revisions)
 Revision as of 16:39, 19 July 2013 (edit)← Previous diff Revision as of 16:43, 19 July 2013 (edit) (undo)Next diff → Line 187: Line 187: In physics, springs are modeled by an equation known as Hooke's law. In an ideal environment (no air resistance, no friction, no heat generated by movement, etc.), an object of mass ''m'', when set into motion attached to a spring, should oscillate forever. Hooke's law states that the mass's displacement ''x'' from the equilibrium and the force ''F'' the spring is exerting on the mass at time ''t'' is linearly related by a constant ''k'', the spring constant. This constant quantifies the stiffness of the spring. Furthermore, this linear relationship is negative; the force is acting in a direction opposite of the displacement. We can write Hooke's law as In physics, springs are modeled by an equation known as Hooke's law. In an ideal environment (no air resistance, no friction, no heat generated by movement, etc.), an object of mass ''m'', when set into motion attached to a spring, should oscillate forever. Hooke's law states that the mass's displacement ''x'' from the equilibrium and the force ''F'' the spring is exerting on the mass at time ''t'' is linearly related by a constant ''k'', the spring constant. This constant quantifies the stiffness of the spring. Furthermore, this linear relationship is negative; the force is acting in a direction opposite of the displacement. We can write Hooke's law as - :$F = -kx$ + :$F = -kx$. We combine Hooke's law with Newton's second law of motion, ''F'' = ''ma'', to get We combine Hooke's law with Newton's second law of motion, ''F'' = ''ma'', to get Line 200: Line 200: :$x' = y \quad\quad\quad x(0) = x_0$ :$x' = y \quad\quad\quad x(0) = x_0$ - :$y' = -x \quad\quad\, y(0) = 0$ + :$y' = -x \quad\quad\, y(0) = 0$. From here, solving this is the same as solving any other system of first order linear differential equations. We write this as the matrix equation From here, solving this is the same as solving any other system of first order linear differential equations. We write this as the matrix equation - :$\vec{u}' = \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = A \vec{u}$ + :$\vec{u}' = \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = A \vec{u}$. Solve for the eigenvalues of ''A'': Solve for the eigenvalues of ''A'': Line 238: Line 238: &= \frac{x_0}{2}(\cos t + i\sin t + \cos(-t) + i\sin(-t)) \\ &= \frac{x_0}{2}(\cos t + i\sin t + \cos(-t) + i\sin(-t)) \\ &= \frac{x_0}{2}(\cos t + i\sin t + \cos t - i\sin t) \text{ (Recall that } \cos(-x) = \cos x \text{ and } \sin(-x) = -\sin x \text{)} \\ &= \frac{x_0}{2}(\cos t + i\sin t + \cos t - i\sin t) \text{ (Recall that } \cos(-x) = \cos x \text{ and } \sin(-x) = -\sin x \text{)} \\ - &= \frac{x_0}{2}(2 \cos t) = x_0 \cos t + &= \frac{x_0}{2}(2 \cos t) = x_0 \cos t \text{ .} - \end{align} [/itex]. + \end{align} [/itex] ==Stability of the Two Dimensional System== ==Stability of the Two Dimensional System==

## Revision as of 16:43, 19 July 2013

Difference equation image, just a placeholder for now

Differential equations have always been a popular research topic due to their various applications. A system of linear differential equations is no exception; it can model basic arms races, simple predator prey models, and more.

# Basic Description

In 1945, the United States dropped atomic bombs on the Japanese cities Hiroshima and Nagasaki, ending World War II and establishing itself as a new superpower. The Soviet Union, while an ally of the United States during WWII, feared the bomb and spent the next few years developing their own atomic bomb, finally detonating their first bomb in 1949. This was just the start of a long and expensive arms race between the two powers. The competition for nuclear might, along with the countries' different ideologies (communism vs. capitalism), caused a political and psychological war during the second half of the 20th century, now known as the Cold War. During this period, both powers invested tremendous resources into their technology and weaponry, worried that the other was pulling ahead.

We will attempt to build a simple model to see the effects of the nuclear arms race, but what type of model would fit an arms race? Consider these two images.

Figure 1
Figure 2

From these images, we see the following: Starting 1947, the Soviet Union rapidly increased its defense spending (Figure 2). In response, the United States rapidly increased its defense spending in 1948 (Figure 1). From that point on, the spending of the two powers pushed and pulled, never changing much (the two higher points of United States defense spending around 1953 and 1968 are because of the Korean and Vietnam Wars). These fluctuating changes are mostly caused by each power's analysis of its own defense spending and the other power's defense spending. For instance, if the United States is already spending a tremendous amount on defense, then it's less willing to increase its defense spending for the following year. In contrast, if the Soviet Union is spending a tremendous amount on defense, then the United States will increase its defense spending to not be left behind. In other words, we can analyze the defense spending of each country by analyzing the change in defense spending from year to year. Since derivatives are the mathematical tool to analyze change, we will build our model around derivatives.

Let x(t) and y(t) be the defense spending of the United States and the Soviet Union respectively at time t (in years). Then x′(t) is the derivative of x over t and it represents how much the U.S. spending changes over t years. Likewise, y′(t) is the derivative of y over t and it represents how much the Soviet spending changes over t years. As mentioned above, the rate of U.S. defense spending depends negatively on its current defense spending and positively on the Soviet Union's current defense spending. Consequently, the rate of Soviet defense spending depends negatively on its current defense spending and positively on the United States's current defense spending. Furthermore, the model must have a starting point: the year in which the arms race started, when t = 0. Let x0 and y0 be the initial(t = 0) defense spending of the United States and the Soviet Union respectively The following equations fit these conditions.

$x'(t) = ax(t) + by(t)\quad\quad\quad x(0) = x_0$ , where a is negative and b is positive.
$y'(t) = cx(t) + dy(t)\quad\quad\quad\, y(0) = y_0$ , where c is positive and d is negative.

This is a linked system of first order linear differential equations. In words, equation x′(t) states that the United States lowers its budget by a dollars for every dollar it spent the previous year, and raises it by b dollars for every dollar in the Soviet Union's budget. Equation y′(t) states that the Soviet Union lowers its budget by d for every dollar it spent the previous year, and raises it by c dollars for every dollar in the United States's budget.

Note: We must use our model with a degree of skepticism because there are far more factors affecting the arms race than just the two countries' defense spending, such as available money/resources and the spending requirements of other important programs. Nevertheless, our model is useful for analyzing the general outline of the Cold War arms race and predicting its outcomes.

### Introduction to Systems of Linear Differential Equations

We clarify a few terms:

• A differential equation is an equation that involves both function(s) and their derivatives.
• A nth order differential equation only involves up to the nth derivative of any function.
• A linked system of differential equations has at least one equation that involves multiple different derivatives or functions.
• The expressions of linear equations are linear combinations of functions.

Each equation in our system is a linear combination of functions x(t) and y(t) and both have one derivative of either x or y. Hence we have a linked system of first order linear differential equations.

We can use eigentheory to solve for any system of first order linear differential equations. Ultimately, the outcome of the nuclear arms race depends only on the values of a, b, m, and n (read the More Mathematical Explanation for details).

# A More Mathematical Explanation

## Solving the Two Dimensional System

We will first solve the general system of two linear different [...]

## Solving the Two Dimensional System

We will first solve the general system of two linear differential equations, then work through an example that follows the rules of the model stated in the Basic Description.

The general equations are

$x'(t) = ax(t) + by(t) \quad\quad\quad x(0) = x_0$
$y'(t) = cx(t) + dy(t) \quad\quad\quad\, y(0) = y_0$.

Since we are solving the general system, a, b, c, and d are elements of ALL numbers (including imaginary numbers). Naturally, the systems generated from our arms race model is a subset of this general system, so knowing how to solve the general system will lead to knowing how to solve the arms race model.

We can write this system with matrices in the form

$\vec{u}'=A\vec{u}$,    where     $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,     $\vec{u} = \begin{bmatrix} x \\ y \end{bmatrix}$,    and     $\vec{u}' = \begin{bmatrix} x' \\ y' \end{bmatrix}$.

In order to gain an intuition on how to solve this equation, we consider the one dimensional case.

### The One Dimensional Case

If the two dimensional case involved the equation :$\vec{u}' = A\vec{u}$ with A a 2x2 matrix, then the A in the equation of the one dimensional case must be a 1x1 matrix, or just a constant. Consequently, $\vec{u}$ will actually be u instead; it is no longer a vector. Hence the one dimensional equation is

$u' = ku$, where k is a real number.

Solving this requires basic knowledge of calculus. We can write this as

$u' = \frac{du}{dt} = ku$ (remember that u is a function of t)
$\frac{1}{u} du = kdt$ (rearranged variables).

Integrate both sides:

$\int \frac{1}{u} du = \int kdt$
$ln(u) = kt + c$
$e^{ln(u)} = e^{kt+c}$
$u = e^{kt}e^{c}$
$u = Ce^{kt}$. (So C = ec.)

Now we can apply the initial condition. Note that

$u(0) = C(1) = C$, so u(0) = C.

Thus the solution to the one dimensional case is

$u(t) = u_0e^{kt}$.

### Back to the Two Dimensional System

The solution to the one dimensional case suggests that the solution to our system is $\vec{u}(t) = e^{At}\vec{u}_0$. We will explore the concept of raising e to a matrix later. Since the solution to the one dimensional case is u = Cekt, a good guess of the solution to the two dimensional case is $\vec{u}(t) = e^{\lambda t} \vec{v}$. We plug our guess into the original equations to find λ and $\vec{v}$ that work. We write $\vec{v} = \begin{bmatrix} p \\ q \end{bmatrix}$.

We verify our guess by plugging it into our original system:

$p \lambda e^{\lambda t} = x'(t) = ap e^{\lambda t} + bq e^{\lambda t}$
$q \lambda e^{\lambda t} = y'(t) = cp e^{\lambda t} + dq e^{\lambda t}$

After we divide by eλt (which is never 0), we can rewrite our system as $\lambda \vec{v} = A \vec{v}$. Thus, our guess is correct if and only if λ is an eigenvalue and $\vec{v}$ is an eigenvector of A.

Now we must account for the fact that a 2x2 matrix can have two eigenvalues λ1 and λ2. Since differentiation is a linear operator, we know that the solution to our system is linear. We write the general solution as a linear combination

$\vec{u} = e^{\lambda_1 t} \vec{v}_1 + e^{\lambda_2 t} \vec{v}_2$.

We now apply our initial condition. When t = 0, our general solution becomes

$\vec{u}(0) = \vec{v}_1 + \vec{v}_2$.

It is rare that $\vec{v}_1$ and $\vec{v}_2$ add up to the initial condition, so how do we solve this? Ideally, we multiply the two vectors by real number constants c1 and c2, but we must check that this still works for our system.

By multiplying constants to each term of our solution expression, our new proposed solution is

$\vec{u} = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2$.

We verify this by plugging into our matrix equation $\vec{u} = A\vec{u}$.

\begin{align} \vec{u}' &= \lambda_1 c_1 e^{\lambda_1 t} \vec{v}_1 + \lambda_2 c_2 e^{\lambda_2 t} \vec{v}_2 \\ &= A\vec{u} = A(c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2) \\ &= A(c_1 e^{\lambda_1 t} \vec{v}_1) + A(c_2 e^{\lambda_2 t} \vec{v}_2) \text{ (Matrix multiplication is distributive.)} \\ &= c_1 e^{\lambda_1 t} (A \vec{v}_1) + c_2 e^{\lambda_2 t} (A \vec{v}_2) \text{ (Scalars pass through.)} \\ &= c_1 e^{\lambda_1 t} (\lambda_1 \vec{v}_1) + c_2 e^{\lambda_2 t} (\lambda_2 \vec{v}_2) \\ &= \lambda_1 c_1 e^{\lambda_1 t} \vec{v}_1 + \lambda_2 c_2 e^{\lambda_2 t} \vec{v}_2 \quad \blacksquare \end{align}

In fact, multiplying constants to the terms is equivalent to finding a linear combination of our basis vectors $e^{\lambda_1 t} \vec{v}_1$ and $e^{\lambda_2 t} \vec{v}_2$. Remember that differentiation is a linear transformation, so the linear combinations of existing solutions are naturally solutions as well.

Now that we have a more refined general solution, we need to know how to solve for the constants. Our initial condition now becomes

$\vec{u}(0) = \begin{bmatrix} x_0 \\ y_0 \end{bmatrix} = c_1\begin{bmatrix} p_1 \\ q_1 \end{bmatrix} + c_2\begin{bmatrix} p_2 \\ q_2 \end{bmatrix}$.

We can rewrite this as

$\begin{bmatrix} p_1 & p_2 \\ q_1 & p_2 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} x_0 \\ y_0 \end{bmatrix}$, which can be solved using Gaussian elimination.

Thus we have our particular solution for any initial condition. With the constants solvable, our general solution is

$\vec{u} = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2$.

### Arms Race Example

We will solve a system that follows the rules of our arms race model. Since the real Cold War arms race is hard to model, we will make assumptions on the actions of the United States and the Soviet Union. Assume that the United States lowers its budget by 3 dollars for every dollar it spent the previous year, and raises it by 6 dollars for every dollar in the Soviet Union's budget. Further assume that the Soviet Union lowers its budget by d for every dollar it spent the previous year, and raises it by c dollars for every dollar in the United States's budget. Finally assume that the United States defense spending is 130 billion dollars and the Soviet Union defense spending is 150 billion dollars at t = 0.

$x'(t) = -3x(t) + 6y(t) \quad\quad\quad x(0)=13$
$y'(t) = 5x(t) - 2y(t) \quad\quad\quad\;\;\; y(0)=15$.

All of the constants are made up. The values of x and y are in tens of billions of dollars.

We can rewrite this as

$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} -3 & 6 \\ 5 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$.

To find the eigenvalues of A, remember that λ is an eigenvalue if and only if the determinant of (A - λI) is 0:

\begin{align} |A-\lambda I| &= \begin{vmatrix} -3-\lambda & 6 \\ 5 & -2-\lambda \end{vmatrix} \\ &= (-3-\lambda)(-2-\lambda) - 30 \\ &= \lambda^2 + 5\lambda - 24 \\ &= (\lambda + 8)(\lambda - 3) = 0 \end{align}

So our eigenvalues are λ = -8, 3. Using the eigenvalues, find the eigenvectors:

For λ1 = -8: $A-\lambda I = \begin{bmatrix} 5&6 \\ 5&6 \end{bmatrix}$, so $\vec{v}_1 = \begin{bmatrix} -6 \\ 5 \end{bmatrix}$.
For λ2 = 3: $A-\lambda I = \begin{bmatrix} -6&6 \\ 5&-5 \end{bmatrix}$, so $\vec{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.

Now we will find the constants for our initial condition. Remember that we can use Gaussian elimination to find them.

Start with $\left[\begin{array}{rr|r} -6 & 1 & 13 \\ 5 & 1 & 15 \end{array}\right]$, after Gaussian elimination --> $\left[\begin{array}{rr|r} 1 & 0 & \frac{2}{11} \\[6pt] 0 & 1 & \frac{155}{11} \end{array}\right]$, so c1 = 211 and c2 = 15511.

Finally, we just plug all of our values back into our general solution to get our particular solution.

$\vec{u} = \begin{bmatrix} x \\ y \end{bmatrix} = \frac{2}{11} e^{-8t} \begin{bmatrix} -6 \\ 5 \end{bmatrix} + \frac{155}{11} e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Thus we have the equations

$x(t) = \frac{-12}{11} e^{-8t} + \frac{155}{11} e^{3t}$
$y(t) = \frac{10}{11} e^{-8t} + \frac{155}{11} e^{3t}$.

These equations satisfy both the system of differential equations and the initial condition (check if doubtful).

We can now use these equations to help analyze the results of our hypothetical arms race. The results will happen many years later, so we can get a good estimate by taking the limit of x(t) and y(t) as t goes to ∞.

$\lim_{t \to \infty}x(t) = \frac{-12}{11} e^{-8(\infty)} + \frac{155}{11} e^{3(\infty)} = \frac{-12}{11} (1) + \frac{155}{11} (\infty) = \infty$
$\lim_{t \to \infty}y(t) = \frac{10}{11} e^{-8(\infty)} + \frac{155}{11} e^{3(\infty)} = \frac{10}{11} (1) + \frac{155}{11} (\infty) = \infty$

The results of this hypothetical arms race are disastrous; they indicate that tensions between the United States and Soviet will cause the arms race to spiral out of control, with both countries investing more and more of their resources into defense. The outcome of that is the economic collapse of at least one of the two powers. Thus, both countries should lessen the military and nuclear competition to avoid national turmoil.

We can represent our arms race example visually:

Figure 3

This is a vector field of all the paths in our made up model. The horizontal axis represent U.S. defense spending, and the vertical axis represent Soviet defense spending. In order to find the solution path for our initial condition, find the point (13, 15) on the plot and follow the arrows. While the vector field of all four quadrants is displayed, we are mainly interested in the first quadrant; it makes no sense for either country to have negative defense spending. Note that regardless of where the initial condition is in the first quadrant, the arrows lead to positive infinity . This is an unstable system, or a system that will reach infinity. Mathematically, the only factor that determines the stability of a system is its eigenvalues (more about this later).

Figure 3 also shows the vital role of the eigenvectors. Using Geometer's Sketchpad, we calculated the slopes of the two predominant lines. Line AC corresponds to $\vec{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$; they both have a slope of 1. Similarly, line AB corresponds to $\vec{v}_2 = \begin{bmatrix} -6 \\ 5 \end{bmatrix}$; they both have a slope of 5-6, or -0.833. These eigenvectors dictate the possible paths our model can go through given different initial conditions.

### The Complex Eigenvalue Case

So far, we have explored the systems with real eigenvalues. Now we will investigate a system that will contain complex eigenvalues.

In physics, springs are modeled by an equation known as Hooke's law. In an ideal environment (no air resistance, no friction, no heat generated by movement, etc.), an object of mass m, when set into motion attached to a spring, should oscillate forever. Hooke's law states that the mass's displacement x from the equilibrium and the force F the spring is exerting on the mass at time t is linearly related by a constant k, the spring constant. This constant quantifies the stiffness of the spring. Furthermore, this linear relationship is negative; the force is acting in a direction opposite of the displacement. We can write Hooke's law as

$F = -kx$.

We combine Hooke's law with Newton's second law of motion, F = ma, to get

$F = ma = -kx$, so $a = -\frac{k}{m} x$.

Remember that acceleration a is the second derivative of displacement x. Thus we observe that Hooke's law is actually the differential equation

$x'' = -\frac{k}{m} x$.

For the sake of simplificity, we assume k = m = 1. The only function in our differential equation is the displacement x, and it's a function of time t. The goal is to find the equation for x in terms of t. Note that our equation is actually a second order differential equation. So how do we solve this? We transform the equation into two first order differential equations by introducing a dummy variable y. Set $y = x'$, so $x'' = y' = -x$. In physics, y is the velocity of the mass at time t (derivative of displacement is velocity). At time t = 0, the displacement is x0 and the velocity is always 0 (the velocity of the mass the instant it is released is 0). Now we have the system of first order linear differential equations

$x' = y \quad\quad\quad x(0) = x_0$
$y' = -x \quad\quad\, y(0) = 0$.

From here, solving this is the same as solving any other system of first order linear differential equations. We write this as the matrix equation

$\vec{u}' = \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = A \vec{u}$.

Solve for the eigenvalues of A:

$|A-\lambda I| = \begin{vmatrix} 0-\lambda & 1 \\ -1 & 0-\lambda \end{vmatrix} = \lambda^2 + 1 = 0$

So our eigenvalues are λ = i, -i. Using these eigenvalues, we find the eigenvectors.

For λ1 = i: $A-\lambda I = \begin{bmatrix} -i & 1 \\ -1 & -i \end{bmatrix}$, so $\vec{v}_1 = \begin{bmatrix} 1 \\ i \end{bmatrix}$.
For λ2 = -i: $A-\lambda I = \begin{bmatrix} i & 1 \\ -1 & i \end{bmatrix}$, so $\vec{v}_2 = \begin{bmatrix} 1 \\ -i \end{bmatrix}$.

Now we will find the constants for our initial condition. Since we do not have a specific starting value for x, we will solve the system of equations instead of use Gaussian Elimination. When t = 0, we have

$x(0) = x_0 = c_1 + c_2$ and $y(0) = 0 = ic_1 - ic_2$.

Looking at the second equation, we see that since ic1 = ic2, c1 = c2. Since the constants are equal, we will just call both constants c. We plug this fact into the first equation and we get x0 = 2c, so c = x02. Our general solution is then

$\vec{u} = \begin{bmatrix} x \\ y \end{bmatrix} = \frac{x_0}{2} \left(e^{it} \begin{bmatrix} 1 \\ i \end{bmatrix} + e^{-it} \begin{bmatrix} 1 \\ -i \end{bmatrix} \right)$.

Remember that we are only interested in x, the displacement of the mass. The general equation for x is

$x(t) = \frac{x_0}{2}(e^{it}+e^{-it})$.

Our solution still has imaginary numbers. We use Euler's formula to understand what raising e by imaginary numbers mean. Euler's formula states

$e^{ix} = \cos x + i\sin x$ for any x.

We now use Euler's formula in equation x:

\begin{align} x(t) &= \frac{x_0}{2}(e^{it}+e^{-it}) \\ &= \frac{x_0}{2}(e^{it}+e^{i(-t)}) \\ &= \frac{x_0}{2}(\cos t + i\sin t + \cos(-t) + i\sin(-t)) \\ &= \frac{x_0}{2}(\cos t + i\sin t + \cos t - i\sin t) \text{ (Recall that } \cos(-x) = \cos x \text{ and } \sin(-x) = -\sin x \text{)} \\ &= \frac{x_0}{2}(2 \cos t) = x_0 \cos t \text{ .} \end{align}

## Stability of the Two Dimensional System

We judge the stability of a system by what happens when time goes to infinity. Each system also has an equilibrium. In our arms race example, the equilibrium is just the origin (in most cases it's the origin). An unstable system will go to infinity as time goes to infinity, and a stable system will not. There are two types of being stable: asymptotically stable and neutrally stable. Neutrally stable systems either spirals inward towards the equilibrium or circles in a closed path around the equilibrium. Asymptotically stable systems approach the equilibrium in open and non-spiral paths. The arms race example that we worked through is a case of an unstable system with a saddle point equilibrium. As mentioned above, the eigenvalues of a system are the only things required to determine the system's stability. We use our arms race example to help explain this. Since we judge the stability of a system by what happens when time goes to infinity, we can take the limit of our solutions as time goes to infinity. Note that the only factor that affected the outcome of the solution (whether it converges or diverges) is the exponents on the e′s, and the exponents of the e′s are based on the eigenvalues. Thus, we can explore the relationship between the eigenvalues of our system and its stability.

## The More Rigorous Proof

We have mentioned above that since the one-dimensional case is $u=u_0 e^{at}$ for initial condition u0, the solution to our system is $\vec{u}(t) = e^{At}\vec{u}_0$. Using Taylor Series, we define

$e^A = \sum_{n=0}^{\infty}\frac{A^n}{n!}$ for a square kxk matrix A. Note: A0 = I.

Now we have to show that $\vec{u} = e^{tA}\vec{u}_0$ solves $\vec{u}'=A\vec{u}$:

\begin{align} \vec{u}' = \frac{d}{dt} (e^{tA} \vec{u}_0) & = \frac{d}{dt} \left( \left(\sum_{n=0}^{\infty}\frac{tA^n}{n!} \right) \vec{u}_0 \right) \\ & = \left( \frac{d}{dt}\sum_{n=0}^{\infty}\frac{t^nA^n}{n!} \right)\vec{u}_0 \text{ (}\vec{u}_0 \text{ is a constant)} \\ & = \left( \sum_{n=0}^{\infty} \frac{d}{dt} \left( \frac{A^n}{n!} \right) \right) \vec{u}_0 \\ & = \left( \sum_{n=1}^{\infty}\frac{n t^{n-1} A^n}{n!} \right) \vec{u}_0 \text{ (Note the index change)} \\ & = \left( A \sum_{n=1}^{\infty} \frac{t^{n-1}A^{n-1}}{(n-1)!} \right) \vec{u}_0 \\ & = \left( A \sum_{j=0}^{\infty} \frac{(tA)^j}{j!} \right) \vec{u}_0 \\ & = Ae^{tA} \vec{u}_0 = A \vec{u} \quad \blacksquare \end{align}.

So we know that $\vec{u} = e^{tA}\vec{u}_0$ solves $\vec{u}'=A\vec{u}$. But what does $e^{tA}\vec{u}_0$ mean? We use diagonalization. If the sum of the dimensions of a square matrix A's eigenspaces is equal to dim(A), then we can write A as $A=PLP^{-1}$, where P has the eigenvectors of A as its columns and L is a diagonal matrix with the eigenvalues of A as its diagonal entries. The eigenvalues of A in L must match up with the eigenvectors in P, meaning that if we choose a random eigenvector/column (say this is the kth eigenvector/column) of P, then the eigenvalue in the kth column of L must correspond to that eigenvector. Let us first consider etA.

\begin{align} e^{tA} = e^{t(PLP^{-1})} = e^{P(tL)P^{-1}} &= \sum_n \frac{(PtLP^{-1})^n}{n!} \\ &= \sum_n \frac{P(tL)^nP^{-1}}{n!} \text{ (} (PLP^{-1})^n = P(L)^nP^{-1} \text{ is a property of diagonalization.)} \\ &= P \left(\sum_n \frac{(tL)^n}{n!} \right) P^{-1} \\ &= P \left(\sum_n \begin{bmatrix} (\lambda_1 t)^n/n! & & & \\ & (\lambda_2 t)^n/n! & & \\ & & \ddots & \\ & & & (\lambda_k t)^n/n! \end{bmatrix} \right) P^{-1} \\ &= P \begin{bmatrix} \displaystyle\sum\limits_{n} (\lambda_1 t)^n/n! & & & \\ & \displaystyle\sum\limits_{n} (\lambda_2 t)^n/n! & & \\ & & \ddots & \\ & & & \displaystyle\sum\limits_{n} (\lambda_k t)^n/n! \end{bmatrix} P^{-1} \\ &= P \begin{bmatrix} e^{\lambda_1 t} & & & \\ & e^{\lambda_2 t} & & \\ & & \ddots & \\ & & & e^{\lambda_k t} \end{bmatrix} P^{-1} \text{ (Remember Taylor Series.)} \end{align}

Before we continue, consider $P^{-1}\vec{u}_0$. Remember in solving the two dimensional case, we made up constants c1 and c2 to satisfy the initial condition. For this more rigorous proof, we define

$\vec{c} = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_k \end{bmatrix}$.

We solve for the constants by knowing that $P \vec{c} = \vec{u}_0$, so $\vec{c} = P^{-1} \vec{u}_0$. Thus

\begin{align} e^{tA}\vec{u}_0 &= P \begin{bmatrix} e^{\lambda_1 t} & & & \\ & e^{\lambda_2 t} & & \\ & & \ddots & \\ & & & e^{\lambda_k t} \end{bmatrix} P^{-1}\vec{u}_0 \\ &= P \begin{bmatrix} e^{\lambda_1 t} & & & \\ & e^{\lambda_2 t} & & \\ & & \ddots & \\ & & & e^{\lambda_k t} \end{bmatrix} \vec{c} \\ &= P \begin{bmatrix} c_1 e^{\lambda_1 t} \\ c_2 e^{\lambda_2 t} \\ \vdots \\ c_k e^{\lambda_k t} \end{bmatrix} = \sum_{i=1}^k c_i e^{\lambda_i t} \vec{v}_i \end{align}.

When k = 2, note that this final expression expands into our two dimensional general solution. Thus the solution to any system of differential equations that can be written in the form $\vec{u}' = A\vec{u}$ is a linear combination of exponential functions with the eigenvectors and eigenvalues of A.

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