Some formatting updates

src/7_differential_equations_1.md

@@ -44,24 +44,25 @@ initial conditions,
 $$x^{(n-1)}(t_{0}) = x^{(n-1)}_{0}, \cdots, x(t_0)=x_0. $$
 
 This is because to fully specify the solution of an $n$*-th* order differential 
-equation, $n-1$ initial conditions are necessary. To understand why we need 
+equation, $n$ initial conditions are necessary (we need to specify the value of $n-1$ derivatives
+of $x(t)$ and as well the value of the function $x(t)$ for some  $t_0$). To understand why we need 
 initial conditions, look at the following example.
 
 !!! check "Example: Initial conditions"
 
     Consider the following calculus problem,
     
-    $$\dot{f}(x)=x. $$ 
+    $$\dot{x}(t)=t. $$ 
 
     By integrating, one finds that the solution to this equation is 
 
-    $$\frac{1}{2}x^2 + c,$$
+    $$\frac{1}{2}t^2 + c,$$
 
     where $c$ is an integration constant. In order to specify the integration 
     constant, an initial condition is needed. For instance, if we know that when 
-    $x=2$ then $f(2)=4$, we can plug this into the equation to get 
+    $t=2$ then $x(2)=4$, we can plug this into the equation to get 
 
-    $$\frac{1}{2}*4 + c = 4, $$
+    $$\frac{1}{2}\cdot 4 + c = 4, $$
 
     which implies that $c=2$. 
     
@@ -82,7 +83,7 @@ In this course, we will be focusing on *Linear Differential Equations*, meaning
 that we consider differential equations $x^{(n)}(t) = f(x^{(n-1)}(t), \cdots, x(t), t)$
 where the function $f$ is a linear polynomial function of the unknown function
 $x(t)$. A simple way to spot a non-linear differential equation is to look for 
-non-linear terms, such as $x(t)*\dot{x}(t)$ or $x^{(n)}(t)*x^{(2)}(t)$. 
+non-linear terms, such as $x(t) \cdot \dot{x}(t)$ or $x^{(n)}(t) \cdot x^{(2)}(t)$. 
 
 Often, we will be dealing with several coupled differential equations. In this 
 situation, we can write the entire system of differential equations as a vector 
@@ -96,11 +97,17 @@ x_{m}(t) \\
 
 A system of first order linear equations is then written as 
 
-$$\dot{\vec{x}(t)} = \vec{f}(\vec{x}(t),t) $$
+$$\dot{\vec{x}}(t) = \vec{f}(\vec{x}(t),t) $$
 
 with the initial condition $\vec{x}(t_0) = \vec{x}_0$.
 
-###7.1.2. Basic examples and strategies
+###7.1.2. Basic examples and strategies for a (single) dfirst-order differential equation
+
+Before focusing on systems of first order equations, we will first consider 
+examplary cases of single first-order equations with only one unknown function $x(t)$.
+In this case, we can distinguish important cases.
+
+#### Type 1: $\dot{x}(t) = f(t)$
 
 The simplest type of differential equation is the type usually learned about in the 
 integration portion of a calculus course. Such equations have the form,
@@ -112,7 +119,22 @@ to this type of equation are
 
 $$x(t) = F(t) + c. $$
 
-!!! check "Example: First order linear differential equation with constant coefficients"
+
+!!! info "What is the antiderivative?"
+    
+    You may know the antiderivative $F(t)$ of a function $f(t)$ under a different name -
+    it is the same as the indefinite integral: $F(t) = \int f(t) dt$. Remember that taking
+    an integral is essentially the opposite of differentiation, and indeed taking an integral
+    means finding a function $F(t)$ such that $\dot{F}(t) = \frac{dF}{dt} = \frac{d}{dt}
+    \int f(t) dt = f(t)$. In the context of differential equations we prefer to call this the
+    antiderivative as solving the differential equation means essentially undoing the derivative.
+
+    Note that the antiderivative is only defined up to a constant (as is the indefinite integral).
+    In practice, you will thus find some particular expression for $F(t)$ through integration. To
+    capture all possible solutions, don't forget the integration constant $c$ in the expression
+    above!
+
+!!! check "Example"
 
     Given the equation
     
@@ -120,16 +142,19 @@ $$x(t) = F(t) + c. $$
     
     one finds by integrating that the solution is $\frac{1}{2}t^2 + c$. 
     
-For first order linear differential equations, it is possible to use the 
-concept of an anti-derivative from calculus to write a general solution in 
-terms of the independent variable:
+
+#### Type 2: $\dot{x}(t) = f(x(t))$
+
+The previous example was easy, as the function $x(t)$ did not enter in the right-hand side.
+A second important case that we can solve explicitly is when the right-hand side is some
+function of $x(t)$:
     
 $$\dot{x}(t)=f(x(t)).$$ 
     
 This implies that $\frac{\dot{x}(t)}{f(x)} = 1$. Let $F(x)$ be the 
 anti-derivative of $\frac{1}{f(x)}$. Then, by making use of the chain rule: 
     
-$$\frac{\dot{x}(t)}{f(x(t))} = \frac{dx}{dt}\,\frac{dF}{dx} = \frac{d}{dt} F(x(t)) = 1$$
+$$\frac{d}{dt} F(x(t)) = \frac{dx}{dt}\,\frac{dF}{dx} = \frac{\dot{x}(t)}{f(x(t))} = 1$$
     
 $$\Leftrightarrow F(x(t)) = t + c.$$
     
@@ -137,7 +162,7 @@ From this, we notice that if we can solve for $x(t)$, then we have the
 solution! Having a specific form for the function $f(x)$ can often make it 
 possible to solve either implicitly or explicitly for the function $x(t)$.
 
-!!! check "Example: Autonomous first order linear differential equation with constant coefficients"
+!!! check "Example"
 
     Given the equation
     
@@ -151,20 +176,24 @@ possible to solve either implicitly or explicitly for the function $x(t)$.
     and $F(x)$ be the anti-derivative of the $\frac{1}{f(x)}$. Integrating 
     allows us to find the form of this anti-derivative. 
     
-    $$F(x):= \int \frac{dx}{\lambda x} = \frac{1}{\lambda}log{\lambda x} $$
+    $$F(x):= \int \frac{dx}{\lambda x} = \frac{1}{\lambda}\log{\lambda x} $$
     
     Now, making use of the general solution we also have that $F(x(t)) =t+c$. 
     These two equations can be combined to form an equation for $x(t)$,
     
-    $$Log(\lambda x)  = \lambda t + c$$
+    $$\log(\lambda x)  = \lambda t + c$$
     $$x(t) = \frac{1}{\lambda} e^c e^{\lambda t} $$
     $$x(t) = c_0 e^{\lambda t}$$
     
     where in the last line we defined a new constant $c_0 =\frac{1}{\lambda}e^c$.
     Given an initial condition, we could immediately determine this constant $c_0$.
     
-So far we have considered only DE's with constant coefficients, but it is very 
-common to encounter equations such as the following,
+
+#### Type 3: $\dot{x}(t) = g(t) f(x(t))$
+
+So far we have considered onle DE's where the right-hand side is either a function of $t$
+*or* of $x(t)$. We can still solve a more generic case, if we can separate the two dependencies
+as:
 
 $$\dot{x}(t)=g(t)f(x(t)).$$
 
@@ -186,7 +215,7 @@ $$\Rightarrow F(x(t)) = G(t) + c $$
 Given this form of a general solution, the knowledge of specific functions $f, g$ would
 make it possible to solve for $x(t)$. 
 
-!!! check "Example: First order linear differential equation with coefficient t"
+!!! check "Example"
 
     Let us apply the above strategy to the following equation,
     
@@ -367,43 +396,42 @@ $$ \dot{\vec{x}}(t) = A(t) \vec{x}(t) + \vec{b}(t).$$
     Now we need a strategy for finding the solution of the inhomogeneous equation. 
     Begin by making an ansatz that $\vec{x}(t)$ can be written as a linear combination 
     of the basis functions for the homogeneous system, with coefficients that are 
-    functions of the independent variable. Ansatz:
-
-    $$\vec{x}(t) = c_1(t) \vec{\phi}_1 (t)+ c_2(t) \vec{\phi}_2(t) + \cdots + c_n(t) \vec{\phi}_n (t) $$
+    functions of the independent variable.
+    
+    1. Ansatz:
+        $$\vec{x}(t) = c_1(t) \vec{\phi}_1 (t)+ c_2(t) \vec{\phi}_2(t) + \cdots + c_n(t) \vec{\phi}_n (t) $$
 
-    Define the vector $\vec{c}(t)$ and matrix $\vec{\Phi}(t)$ as
+    2. Define the vector $\vec{c}(t)$ and matrix $\vec{\Phi}(t)$ as
 
-    $$\vec{c}(t) = \begin{bmatrix}
-    c_1(t) \\
-    \vdots \\
-    c_n(t) \\
-    \end{bmatrix} $$
-    $$\vec{\Phi}(t) = \big{(} \vec{\phi}_1 (t) | \cdots | \vec{\phi}_n (t) \big{)} $$
+        $$\vec{c}(t) = \begin{bmatrix}
+        c_1(t) \\
+        \vdots \\
+        c_n(t) \\
+        \end{bmatrix} $$
+        $$\vec{\Phi}(t) = \big{(} \vec{\phi}_1 (t) | \cdots | \vec{\phi}_n (t) \big{)} $$
 
-    With these definitions, it is possible to re-write the ansatz for $\vec{x}(t)$,
+    3. With these definitions, it is possible to re-write the ansatz for $\vec{x}(t)$,
 
-    $$ \vec{x}(t) = \vec{\Phi}(t) \vec{c}(t).$$
+        $$ \vec{x}(t) = \vec{\Phi}(t) \vec{c}(t).$$
 
-    Using the Leibniz rule, we then have the following expanded equation,
+    4. Using the Leibniz rule, we then have the following expanded equation,
 
-    $$\dot{\vec{x}}(t) = \dot{\vec{\Phi}}(t) \vec{c}(t) + \vec{\Phi}(t) \dot{\vec{c}}(t).$$
+        $$\dot{\vec{x}}(t) = \dot{\vec{\Phi}}(t) \vec{c}(t) + \vec{\Phi}(t) \dot{\vec{c}}(t).$$
 
-    Substituting the new expression into the differential equation gives,
+    5. Substituting the new expression into the differential equation gives,
 
-    $$\dot{\vec{\Phi}}(t) \vec{c}(t) + \vec{\Phi}(t) \dot{\vec{c}}(t) = A(t) \vec{\Phi}(t) \vec{c}(t) + \vec{b}(t) $$
-    $$\vec{\Phi}(t) \dot{\vec{c}}(t) = \vec{b}(t). $$
+        $$\dot{\vec{\Phi}}(t) \vec{c}(t) + \vec{\Phi}(t) \dot{\vec{c}}(t) = A(t) \vec{\Phi}(t) \vec{c}(t) + \vec{b}(t) $$
+        $$\vec{\Phi}(t) \dot{\vec{c}}(t) = \vec{b}(t). $$
 
-    In order to cancel terms in the previous line, we made use of the fact that 
+        In order to cancel terms in the previous line, we made use of the fact that 
     $\vec{\Phi}(t)$ solves the homogeneous equation $\dot{\vec{\Phi}} = A \vec{\Phi}$.
-    By way of inverting and integrating, we can write the equation for the coefficient
-    vector $\vec{c}(t)$
+    6. By way of inverting and integrating, we can write the equation for the coefficient vector $\vec{c}(t)$
 
-    $$\vec{c}(t) = \int \vec{\Phi}^{-1}(t) \vec{b}(t) dt.$$
+        $$\vec{c}(t) = \int \vec{\Phi}^{-1}(t) \vec{b}(t) dt.$$
 
-    With access to a concrete form of the coefficient vector, we can then write down
-    the particular solution,
+    7. With access to a concrete form of the coefficient vector, we can then write down the particular solution,
 
-    $$\vec{\psi}(t)= \vec{\Phi}(t) \cdot \int \vec{\Phi}^{-1}(t) \vec{b}(t) dt .$$
+        $$\vec{\psi}(t)= \vec{\Phi}(t) \cdot \int \vec{\Phi}^{-1}(t) \vec{b}(t) dt .$$
 
 !!! check "Example: Inhomogeneous first order linear differential equation"
 
@@ -493,39 +521,41 @@ $$A \vec{v}_i = \lambda_i \vec{v}_i, \qquad \forall i \epsilon \{1, \cdots, n \}
 Here, we give consideration to the case of distinct eigenvectors, in which case 
 the $n$ eigenvectors form a basis for $\mathbb{R}^{n}$. 
 
-To solve the equation $\dot{\vec{x}}(t) = A \vec{x}(t)$, define a set of scalar functions $\{u_{1}(t), \cdots u_{n}(t) \}$ and make the following ansatz:
+!!! info "Strategy for finding solution when $A$ is diagonizable"
 
-$$\vec{\phi}_{i}(t) = u_{i}(t) \vec{v}_{i}.$$
+    1.  To solve the equation $\dot{\vec{x}}(t) = A \vec{x}(t)$, define a set of scalar functions $\{u_{1}(t), \cdots u_{n}(t) \}$ and make the following ansatz:
+        $$\vec{\phi}_{i}(t) = u_{i}(t) \vec{v}_{i}.$$
+    2.  Then, by differentiating,
+        $$\dot{\vec{\phi}_i}(t) = \dot{u_i}(t) \vec{v}_{i}.$$
 
-Then, by differentiating,
+    3.  The above equation can be combined with the differential equation for 
+        $\vec{\phi}_{i}(t)$, 
+        $$\dot{\vec{\phi}_{i}}(t)=A \vec{\phi}_{i}(t) \, ,$$
+        to derive the following equations,
 
-$$\dot{\vec{\phi}_i}(t) = \dot{u_i}(t) \vec{v}_{i}.$$
+        $$\dot{u_i}(t) \vec{v}_{i} = A u_{i}(t) \vec{v}_{i}$$
+        $$\dot{u_i}(t) \vec{v}_{i} = u_{i}(t) \lambda_{i} \vec{v}_{i} $$
+        $$\vec{v}_{i} (\dot{u_i}(t) - \lambda_i u_{i}(t)) = 0, $$
 
-The above equation can be combined with the differential equation for 
-$\vec{\phi}_{i}(t)$, 
-$$\dot{\vec{\phi}_{i}}(t)=A \vec{\phi}_{i}(t) \, ,$$
-to derive the following equations,
+        where in the second last line, we make use of the fact that $\vec{v}_i$ is an eigenvector of $A$. 
+    4.  The obtained relation implies that 
 
-$$\dot{u_i}(t) \vec{v}_{i} = A u_{i}(t) \vec{v}_{i}$$
-$$\dot{u_i}(t) \vec{v}_{i} = u_{i}(t) \lambda_{i} \vec{v}_{i} $$
-$$\vec{v}_{i} (\dot{u_i}(t) - \lambda_i u_{i}(t)) = 0, $$
+        $$\dot{u_i}(t) = \lambda_i u_{i}(t).$$
 
-where in the second last line, we make use of the fact that $\vec{v}_i$ is an eigenvector of $A$. The obtained relation implies that 
+        This is a simple differential equation, of the type dealt with in the third example. 
+    5.  The solution is found to be
 
-$$\dot{u_i}(t) = \lambda_i u_{i}(t).$$
+        $$u_{i}(t) = c_i e^{\lambda_i t},$$
 
-This is a simple differential equation, of the type dealt with in the third example. The solution is found to be
+        with $c_i$ being a constant. 
+    6.  The general solution is found by adding all $n$ of the
+        solutions $\vec{\phi}_{i}(t)$,
 
-$$u_{i}(t) = c_i e^{\lambda_i t},$$
+        $$\vec{x}(t) = c_{1} e^{\lambda_1 t} \vec{v}_{1} + c_{2} e^{\lambda_2 t} \vec{v}_{2} + \cdots + c_{n} e^{\lambda_n t} \vec{v}_{n}.$$
 
-with $c_i$ being a constant. The general solution is found by adding all $n$ of the
-solutions $\vec{\phi}_{i}(t)$,
-
-$$\vec{x}(t) = c_{1} e^{\lambda_1 t} \vec{v}_{1} + c_{2} e^{\lambda_2 t} \vec{v}_{2} + \cdots + c_{n} e^{\lambda_n t} \vec{v}_{n}.$$
-
-and the vectors $\{e^{\lambda_1 t} \vec{v}_{1}, \cdots, e^{\lambda_n t} \vec{v}_{n} \}$
-form a basis for the solution space since $\det(\vec{v}_1 | \cdots | \vec{v}_n) \neq 0$
-(the $n$ eigenvectors are linearly independent). 
+        and the vectors $\{e^{\lambda_1 t} \vec{v}_{1}, \cdots, e^{\lambda_n t} \vec{v}_{n} \}$
+        form a basis for the solution space since $\det(\vec{v}_1 | \cdots | \vec{v}_n) \neq 0$
+        (the $n$ eigenvectors are linearly independent). 
 
 !!! check "Example: Homogeneous first order linear system with diagonalizable constant coefficient matrix"
 
@@ -671,35 +701,33 @@ has a root $\lambda$ with multiplicity 2, but only one eigenvector $\vec{v}_1$.
 What is the problem in this case? Since there are $n$ equations to be solved and an $n \times n$ linear operator $A$, the solution space for the equation requires a basis of $n$ solutions. In this case however, there are $n-1$ eigenvectors, so we cannot use only these eigenvectors in forming a basis for 
 the solution space. 
 
-Suppose that we have a system of $2$ coupled equations, so that $A$ is a $2 \times 2$ matrix, which has eigenvalue $\lambda_1$ with multiplicity $2$. As in the previous section, we can form one solution using the single eigenvector $\vec{v}_1$,
+!!! info "Strategy for finding a solution when $A (2 $ by $ 2)$ is defective"
 
-$$\vec{\phi}_1(t) = e^{\lambda_1 t} \vec{v}_1.$$
+    1. Suppose that we have a system of $2$ coupled equations, so that $A$ is a $2 \times 2$ matrix, which has eigenvalue $\lambda_1$ with multiplicity $2$. As in the previous section, we can form one solution using the single eigenvector $\vec{v}_1$,
 
-To determine a second, linearly independent solution, make the following ansatz:
+        $$\vec{\phi}_1(t) = e^{\lambda_1 t} \vec{v}_1.$$
 
-$$\vec{\phi}_2(t) = t e^{\lambda_1 t} \vec{v}_1 + e^{\lambda_1 t} \vec{v}_2.$$
+    2. To determine the second linearly independent solution, make the following ansatz:
 
-With this ansatz, it is then necessary to determine an appropriate vector $\vec{v}_2$
-such that $\vec{\phi}_2(t)$ is really a solution of this problem. To achieve that, take
-the derivative of $\vec{\phi}_2(t)$,
+        $$\vec{\phi}_2(t) = t e^{\lambda_1 t} \vec{v}_1 + e^{\lambda_1 t} \vec{v}_2.$$
 
-$$\dot{\vec{\phi}_2}(t) = e^{\lambda_1 t} \vec{v}_1 + \lambda_1 t e^{\lambda_1 t} \vec{v}_1 + \lambda_1 e^{\lambda_1 t} \vec{v}_2 $$
+    3. With this ansatz, it is then necessary to determine an appropriate vector $\vec{v}_2$ such that $\vec{\phi}_2(t)$ is really a solution of this problem. To achieve that, take the derivative of $\vec{\phi}_2(t)$,
 
-Also, write the matrix equation for $\vec{\phi}_2(t)$,
+        $$\dot{\vec{\phi}_2}(t) = e^{\lambda_1 t} \vec{v}_1 + \lambda_1 t e^{\lambda_1 t} \vec{v}_1 + \lambda_1 e^{\lambda_1 t} \vec{v}_2 $$
 
-$$A \vec{\phi}_2(t) = A t e^{\lambda_1 t} \vec{v}_1 + A e^{\lambda_1 t} \vec{v}_2 $$
-$$A \vec{\phi}_2(t) = \lambda_1 t e^{\lambda_1 t} \vec{v}_1 + A e^{\lambda_1 t}\vec{v}_2$$
+    4. Also, write the matrix equation for $\vec{\phi}_2(t)$,
 
-Since $\vec{\phi}_2(t)$ must solve the equation 
-$\dot{\vec{\phi}_2(t)} = A \vec{\phi}_2(t)$, we can combine and simplify the 
-previous equations to write 
+        $$A \vec{\phi}_2(t) = A t e^{\lambda_1 t} \vec{v}_1 + A e^{\lambda_1 t} \vec{v}_2 $$
+        $$A \vec{\phi}_2(t) = \lambda_1 t e^{\lambda_1 t} \vec{v}_1 + A e^{\lambda_1 t}\vec{v}_2$$
 
-$$A \vec{v}_2 - \lambda_1  \vec{v}_2 = \vec{v}_1$$
-$$(A- \lambda_1 I) \vec{v}_2 = \vec{v}_1 $$
+    5. Since $\vec{\phi}_2(t)$ must solve the equation $\dot{\vec{\phi}_2(t)} = A \vec{\phi}_2(t)$, we can combine and simplify the previous equations to write 
 
-With this condition, it is possible to write the general solution as
+        $$A \vec{v}_2 - \lambda_1  \vec{v}_2 = \vec{v}_1$$
+        $$(A- \lambda_1 I) \vec{v}_2 = \vec{v}_1 $$
 
-$$\vec{x}(t) = c_1  e^{\lambda_1 t} \vec{v}_1 + c_2(t e^{\lambda_1 t} \vec{v}_1 + e^{\lambda_1 t} \vec{v}_2).$$
+    6. With this condition, it is possible to write the general solution as
+
+        $$\vec{x}(t) = c_1  e^{\lambda_1 t} \vec{v}_1 + c_2(t e^{\lambda_1 t} \vec{v}_1 + e^{\lambda_1 t} \vec{v}_2).$$
 
 !!! check "Example: Continuation of the example with $A$ defective (Part 2)"
     
@@ -766,14 +794,15 @@ Then, for comparison, multiply $\vec{\phi}_k(t)$ by $A$
 
 $$\begin{align} 
 A \vec{\phi}_k (t) &= e^{\lambda t} \big{(} \frac{t^{k-1}}{(k-1)!}\lambda \vec{v}_1 + \frac{t^{k-2}}{(k-2)!} A \vec{v}_2 + \cdots + A \vec{v}_{k-1} + A \vec{v}_k \big{)}\\
-A \vec{\phi}_k (t) &= \lambda \vec{\phi}_k (t) + e^{\lambda t} \big{(} \frac{t^{k-2}}{(k-2)!}(A- \lambda I)\vec{v}_2 + \cdots + t (A- \lambda I)\vec{v}_{k-1} + (A- \lambda I)\vec{v}_{k}  \big{)}\\
-A \vec{\phi}_k (t) &=  \lambda \vec{\phi}_k (t) + e^{\lambda t} \big{(} \frac{t^{k-2}}{(k-2)!} \vec{v}_1 + \cdots + t \vec{v}_{k-2} + \vec{v}_{k-1} \big{)}\\
-A \vec{\phi}_k (t) &= \dot{\vec{\phi}}_{k}(t).
+&= \lambda \vec{\phi}_k (t) + e^{\lambda t} \big{(} \frac{t^{k-2}}{(k-2)!}(A- \lambda I)\vec{v}_2 + \cdots + t (A- \lambda I)\vec{v}_{k-1} + (A- \lambda I)\vec{v}_{k}  \big{)}\\
+&=  \lambda \vec{\phi}_k (t) + e^{\lambda t} \big{(} \frac{t^{k-2}}{(k-2)!} \vec{v}_1 + \cdots + t \vec{v}_{k-2} + \vec{v}_{k-1} \big{)}\\
+&= \dot{\vec{\phi}}_{k}(t).
 \end{align}$$
 
 Notice that in the second last line we made use of the relations 
-$(A- \lambda I)\vec{v}_{i} = \vec{v}_{i-1}$. This completes the proof 
-since we have demonstrated that $\vec{\phi}_{k}(t)$ is a solution of the DE.
+$(A- \lambda I)\vec{v}_{i} = \vec{v}_{i-1}$. 
+
+This completes the proof since we have demonstrated that $\vec{\phi}_{k}(t)$ is a solution of the DE.
 
 *** 
 ##7.4. Problems