Initial value problems

Section 1.8 Initial value problems

Subsection 1.8.1 Types of solutions

Definition 1.8.1. Explicit versus implicit solutions.

An explicit solution to (1.1.1) on an interval \(I\) is a function of \(t\) such that when substituted into (1.1.1) results in an identity. We will see later that all linear ODEs will give us an explicit solution, but for higher-order linear ODEs, some explicit solutions can be expressed in multiple different ways. An implicit solution to (1.1.1) on a region \(R\) in the plane is a relation \(\phi(t,y)=c\) that when differentiated implicitly results in the ODE (1.1.1).

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Example 1.8.2.

Determine if the following functions represent an explicit solution to the differential equation \(ty'+y=4t\) on the interval \(t>0\text{.}\)

\(\displaystyle f(t)=t^2\)

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\(\displaystyle g(t)=\dfrac{1}{t}+2t\)

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Example 1.8.3.

Determine if \(e^y=ty\) represents an implicit solution to \(\dfrac{\mathrm{d}y}{\mathrm{d}t}=\dfrac{y}{(y-1)t}\) on \(R=\{(t,y) \colon t>0, y>1 \}\text{.}\)

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Subsection 1.8.2 Initial value problems

Solutions to (1.1.1) can also be described by the number of parameters (arbitrary constants) that show up in its solution. For a linear ODE, a family of solutions represents all possible solutions.

Table 1.8.4. Some first-order ODEs and their explicit solutions

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ODE	Solution
\(y'+2ty^2=0\)	\(y(t)=\dfrac{1}{x^2+c}\)
\(\dfrac{\mathrm{d}B}{\mathrm{d}t}=k(B-70)\)	\(B(t)=70+ce^{kt}\)
\((t+1)y'-ty=0\)	\(y(t)=\dfrac{ce^t}{t+1}\)

Each of the solutions in Table Table 1.8.4 depend on a constant \(c\text{,}\) which may change while keeping the resulting function a solution of the differential equation. Unless we have more information to work with, we cannot determine a value for \(c\text{.}\) That is precisely the role of an initial value problem, which is a differential equation equipped with some values of the solution function (and possibly its derivatives) at an initial point \(t_0\text{.}\) We express a first-order initial value problem based on (1.1.1) as

\begin{equation} y'=f(t,y), \quad y(t_0)=y_0\tag{1.8.1} \end{equation}

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Consider the differential equation \(y'=y^2\) with the one-parameter family of solutions given by \(y(t)=\dfrac{1}{C-t}\text{.}\) Notice that \(y(t)=0\) is also a solution, often called the trivial solution. However, we cannot obtain the trivial solution by a choice of \(C\) in the solution family.

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Example 1.8.5.

Consider the differential equation \(y'=4y\) with the one-parameter family of solutions given by \(y(t)=ce^{4t}\text{.}\) Suppose we want a particular solution curve in the family of solutions that goes through the point \((3,17)\text{,}\) or in terms of the solution \(y\text{,}\) it obeys the initial condition \(y(3)=17\text{.}\) The general solution of this differential equation is \(y(t)=ce^{4t}\text{.}\) Find the value of \(c\) that plots the solution obeying the initial condition and alter the code in the box below to plot the correct solution.In the code box below, plot the solution corresponding to the initial condition given that the general solution

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Example 1.8.6.

Consider \(y'=3y^{\frac{2}{3}}\) with the one-parameter family of solutions given by \(y(t)=(t+c)^3\text{.}\) Find all solutions that satisfy the initial condition \(y(0)=0\text{.}\)

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The previous example raises two fundamental questions that we can ask about any initial value problem:

Existence - is there a solution matching my initial data?

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Uniqueness - is it the only such solution?

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Theorem 1.8.7. Existence and uniqueness for first-order linear IVPs.

Consider the linear IVP we obtain from (1.1.4) and (1.8.1) as guides:

\begin{equation} a_1(t)y'+a_0(t)y=g(t), \quad y(t_0)=y_0.\tag{1.8.2} \end{equation}

Rewriting it into standard form

\begin{equation*} y'+P(t)y=Q(t), \quad y(t_0)=y_0, \end{equation*}

where \(P(t)=\dfrac{a_0(t)}{a_1(t)}\) and \(Q(t)=\dfrac{g(t)}{a_1(t)}\) implies that (1.8.2) is guaranteed to have a unique solution if \(P\) and \(Q\) are both continuous on an open interval \(I\) containing \(t_0\text{.}\) The largest such interval \(I\) is called the interval of definition for the IVP (1.8.2).

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Example 1.8.8.

Find the largest interval where the initial value problem is guaranteed a unique solution.

\(\displaystyle (t-2)y'+(t-5)y=4\ln(|t|), \quad y(1)=2\)

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\(\displaystyle \sin(t)y'+7y=t^2, \quad y(2)=-1\)

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When solving a linear IVP, the interval where our solutions are defined always matters! Our solution may also not be defined the same way over the entire interval.

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Example 1.8.9.

Consider the IVP

\begin{equation*} y'+y=\mathcal{U}(t-2), \quad y(0)=1, \end{equation*}

where \(\mathcal{U}(t-2)=\left\{ \begin{align*} 0, \amp \quad 0 \le t \lt 2 \\ 1, \amp \quad t \ge 2 \end{align*}\right.\text{.}\) Its solution is of the form \(y(t)=\left\{ \begin{align*} e^{-t}, \amp \quad 0 \le t \lt 2 \\ 1+e^{-(t-2)}+e^{-t}, \amp \quad t \ge 2 \end{align*}\right.\text{.}\)

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Theorem 1.8.10. Existence and uniqueness for first-order (nonlinear) IVPs.

The IVP (1.8.1) is guaranteed to have a unique solution on a region \(R\) in the plane containing the point \((t_0,y_0)\) provided that both \(f\) and \(\dfrac{\partial f}{\partial y}\) are continuous on \(R\text{.}\)

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Proof.

later!

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Example 1.8.11.

Consider the differential equation \(\dfrac{\mathrm{d}y}{\mathrm{d}t}=\sqrt{16-t^2-y^2}\text{.}\) Determine a region \(R\) in the plane for the differential equation to have a unique solution through each point \((t_0,y_0)\text{.}\) Plot the region.

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