3.2: Existence and Uniqueness for Linear DEs

First lets recall what we know about 1st order LINEAR differential equation:

Homogeneous:
\[\notag
y^{(n)}+p_1(t) y^{(n-1)}+\ldots p_{n-1}(t) y^{\prime}+p_n(t) y=0
\]

Non-homogeneous: \(g(t) \neq 0\)
\[\notag
y^{(n)}+p_1(t) y^{(n-1)}+\ldots p_{n-1}(t) y^{\prime}+p_n(t) y=g(t)
\]

Theorem \(\PageIndex{1}\)

If \(p:(a, b) \rightarrow R\) and \(g:(a, b) \rightarrow R\) are continuous and \(a<t_0<b\), then there exists a unique function \(y=\phi(t), \phi:(a, b) \rightarrow R\) that satisfies the
IVP: \(y^{\prime}+p(t) y=g(t), \quad y\left(t_0\right)=y_0\)

Theorem \(\PageIndex{2}\)

If \(y=\phi_1(t)\) is a solution to homogeneous equation, \(y^{\prime}+p(t) y=0\), then \(y=c \phi_1(t)\) is the general solution to this equation.

If in addition \(y=\psi(t)\) is a solution to non-homogeneous equation, \(y^{\prime}+p(t) y=g(t)\), then \(y=c \phi_1(t)+\psi(t)\) is the general solution to this equation.

Partial proof

\(y=\phi_1(t)\) is a solution to \(y^{\prime}+p(t) y=0\) implies

Thus \(y=c \phi_1(t)\) is a solution to \(y^{\prime}+p(t) y=0\) since
\(y=\psi(t)\) is a solution to \(y^{\prime}+p(t) y=g(t)\) implies

Thus \(y=c \phi_1(t)+\psi(t)\) is a solution to \(y^{\prime}+p(t) y=g(t)\) since

Now we move onto 2nd order Linear differential equation:

Theorem \(\PageIndex{3}\)

If \(p:(a, b) \rightarrow R, q:(a, b) \rightarrow R\), and \(g:\) \((a, b) \rightarrow R\) are continuous and \(a<t_0<b\), then there exists a unique function \(y=\phi(t), \phi:(a, b) \rightarrow R\) that satisfies the initial value problem
\[\notag
\begin{array}{c}
y^{\prime \prime}+p(t) y^{\prime}+q(t) y=g(t), \\
y\left(t_0\right)=y_0, \\
y^{\prime}\left(t_0\right)=y_0^{\prime}
\end{array}
\]

Theorem \(\PageIndex{4}\)

If \(\phi_1\) and \(\phi_2\) are two solutions to a homogeneous linear differential equation, then \(c_1 \phi_1+c_2 \phi_2\) is also a solution to this linear differential equation.

Proof

Since \(y(t)=\phi_i(t)\) is a solution to the linear homogeneous differential equation \(y^{\prime \prime}+p y^{\prime}+q y=0\) where \(p\) and \(q\) are functions of \(t\) (note this includes the case with constant coefficients), then

Claim: \(y(t)=c_1 \phi_1(t)+c_2 \phi_2(t)\) is also a solution to \(y^{\prime \prime}+\) \(p y^{\prime}+q y=0\)

Proof of claim: Solve: \(y^{\prime \prime}+y=0, y(0)=-1, y^{\prime}(0)=-3\) \(r^2+1=0\) implies \(r^2=-1\). Thus \(r= \pm i\). Since \(r=0 \pm 1 i, y=k_1 \cos (t)+k_2 \sin (t)\). Then \(y^{\prime}=-k_1 \sin (t)+k_2 \cos (t)\) \(y(0)=-1:-1=k_1 \cos (0)+k_2 \sin (0)\) implies \(-1=k_1\) \(y^{\prime}(0)=-3:-3=-k_1 \sin (0)+k_2 \cos (0)\) implies \(-3=k_2.\) Thus Initial value problem has solution: \(y=-\cos (t)-3 \sin (t).\)

Question

When does the following initial value problem have a unique solution:
\[\notag
\text { IVP: } a y^{\prime \prime}+b y^{\prime}+c y=0, y\left(t_0\right)=y_0, y^{\prime}\left(t_0\right)=y_1 \text {. }
\]

Solution

Suppose \(y=c_1 \phi_1(t)+c_2 \phi_2(t)\) is a solution to \(a y^{\prime \prime}+b y^{\prime}+c y=0.\) Then \( y^{\prime}=c_1 \phi_1^{\prime}(t)+c_2 \phi_2^{\prime}(t)\) and
\[\notag
\begin{array}{ll}
y\left(t_0\right)=y_0: & y_0=c_1 \phi_1\left(t_0\right)+c_2 \phi_2\left(t_0\right) \\
y^{\prime}\left(t_0\right)=y_1: & y_1=c_1 \phi_1^{\prime}\left(t_0\right)+c_2 \phi_2^{\prime}\left(t_0\right)
\end{array}
\]

To find the solution, we need to solve above system of two equations for the unknowns, namely, \(c_1\) and \(c_2\).

Note: the IVP has a unique solution if and only if the above system of two equations has a unique solution for \(c_1\) and \(c_2\)

Next, observe that in these equations \(c_1\) and \(c_2\) are the unknowns and \(y_0, \phi_1\left(t_0\right), \phi_2\left(t_0\right), y_1, \phi_1^{\prime}\left(t_0\right), \phi_2^{\prime}\left(t_0\right)\) are the constants. We can translate this linear system of equations into matrix form:

\[\notag
\begin{array}{l}
c_1 \phi_1\left(t_0\right)+c_2 \phi_2\left(t_0\right)=y_0 \\
c_1 \phi_1^{\prime}\left(t_0\right)+c_2 \phi_2^{\prime}\left(t_0\right)=y_1
\end{array} \Rightarrow\left[\begin{array}{ll}
\phi_1\left(t_0\right) & \phi_2\left(t_0\right) \\
\phi_1^{\prime}\left(t_0\right) & \phi_2^{\prime}\left(t_0\right)
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]=\left[\begin{array}{l}
y_0 \\
y_1
\end{array}\right]
\]

Note this equation has a unique solution if and only if
\[\notag
\operatorname{det}\left[\begin{array}{ll}
\phi_1\left(t_0\right) & \phi_2\left(t_0\right) \\
\phi_1^{\prime}\left(t_0\right) & \phi_2^{\prime}\left(t_0\right)
\end{array}\right]=\left|\begin{array}{ll}
\phi_1 & \phi_2 \\
\phi_1^{\prime} & \phi_2^{\prime}
\end{array}\right|=\phi_1 \phi_2^{\prime}-\phi_1^{\prime} \phi_2 \neq 0
\]

Definition: The Wronskian

The Wronskian of two differential functions , \(\phi_1\) and \(\phi_2\) (denoted as \(W\left(\phi_1, \phi_2\right)\)) is
\[ \notag
W\left(\phi_1, \phi_2\right)=\phi_1 \phi_2^{\prime}-\phi_1^{\prime} \phi_2=\left|\begin{array}{ll}
\phi_1 & \phi_2 \\
\phi_1^{\prime} & \phi_2^{\prime}
\end{array}\right|
\]

Example \(\PageIndex{1}\)

Find the Wronskian of \(\cos (t)\) and  \(\sin (t)\)

Solution

\[\notag
\begin{aligned}
\mathrm{W}(\cos (t), \sin (t)) & =\left|\begin{array}{cc}
\cos (t) & \sin (t) \\
-\sin (t) & \cos (t)
\end{array}\right| \\
& =\cos ^2(t)+\sin ^2(t)=1>0
\end{aligned}
\]

Example \(\PageIndex{1}\)

Find the Wronskian of \(e^{dt}\cos (nt)\) and  \(e^{dt}\sin (nt)\)

Solution

\[ \notag
\mathrm{W}\left(e^{d t} \cos (n t), e^{d t} \sin (n t)\right)=
\left|\begin{array}{cc}
e^{d t} \cos (n t) & e^{d t} \sin (n t) \\
d e^{d t} \cos (n t)-n e^{d t} \sin (n t) & d e^{d t} \sin (n t)+n e^{d t} \cos (n t)
\end{array}\right|
\]

\[ \notag
\begin{array}{l}
=e^{d t} \cos (n t)\left(d e^{d t} \sin (n t)+n e^{d t} \cos (n t)\right)-e^{d t} \sin (n t)\left(d e^{d t} \cos (n t)-n e^{d t} \sin (nt)\right). \\
=e^{2 d t}[\cos (n t)(d \sin (n t)+n \cos (n t))-\sin (n t)(d \cos (n t)-n \sin (n t))] \\
\left.=e^{2 d t}\left[d \cos (n t) \sin (n t)+n \cos ^2(n t)-d \sin (n t) \cos (n t)+n \sin ^2(n t)\right]\right) \\
=e^{2 d t}\left[n \cos ^2(n t)+n \sin ^2(n t)\right] \\
=n e^{2 d t}\left[\cos ^2(n t)+\sin ^2(n t)\right]=n e^{2 d t}>0 \text { for all } t
\end{array}
\]

 

Theorem \(\PageIndex{5}\)

Suppose that \(\phi_1\) and \(\phi_2\) are two solutions to \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\).
There is a unique choice of constants \(c_1\) and \(c_2\) such that \(c_1 \phi_1+c_2 \phi_2\) satisfies this homog linear differential equation and initial conditions, \(y\left(t_0\right)=y_0, y^{\prime}\left(t_0\right)=y_0^{\prime}\).
iff
\[\notag
W\left(\phi_1, \phi_2\right)\left(t_0\right)=\phi_1\left(t_0\right) \phi_2^{\prime}\left(t_0\right)-\phi_1^{\prime}\left(t_0\right) \phi_2\left(t_0\right) \neq 0
\]

Theorem \(\PageIndex{6}\)

Given the hypothesis of thm 3.2.1, suppose that \(\phi_1\) and \(\phi_2\) are two solutions to
\[ \notag
y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0 .
\]

 

If \(W\left(\phi_1, \phi_2\right)\left(t_0\right) \neq 0\), for some \(t_0 \in(a, b)\), then any solution to this homogeneous linear differential equation can be written as \(y=c_1 \phi_1+c_2 \phi_2\) for some \(c_1\) and \(c_2\).

Definition: Term

If \(\phi_1\) and \(\phi_2\) satisfy the conditions in thm 3.2.4, then \(\phi_1\) and \(\phi_2\) form a fundamental set of solutions to \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\).

Thm 3.2.5: Given any second order homogeneous linear differential equation, there exist a pair of functions which form a fundamental set of solutions.

FYI: Linear Independence and the Wronskian
Defn: \(\phi_1\) and \(\phi_2\) are linearly dependent if there exists constants \(c_1, c_2\) such that \(c_1 \neq 0\) or \(c_2 \neq 0\) and
\[\notag
c_1 \phi_1(t)+c_2 \phi_2(t)=0 \text { for all } t \in(a, b)
\]

Thm 3.3.1: If \(\phi_1:(a, b) \rightarrow R\) and \(\phi_2(a, b) \rightarrow R\) are differentiable functions on \((a, b)\) and if \(W\left(\phi_1, \phi_2\right)\left(t_0\right) \neq 0\) for some \(t_0 \in(a, b)\), then \(\phi_1\) and \(\phi_2\) are linearly independent on \((a, b)\). Moreover, if \(\phi_1\) and \(\phi_2\) are linearly dependent on \((a, b)\), then \(W\left(\phi_1, \phi_2\right)(t)=0\) for all \(t \in(a, b)\)

Proof idea:
If \(c_1 \phi_1(t)+c_2 \phi_2(t)=0\) for all \(t \in(a, b)\), then \(c_1 \phi_1^{\prime}(t)+c_2 \phi_2^{\prime}(t)=0\) for all \(t \in(a, b)\)

Solve the following linear system of equations for \(c_1, c_2\)
\[\notag
\begin{array}{l}
c_1 \phi_1\left(t_0\right)+c_2 \phi_2\left(t_0\right)=0 \\
c_1 \phi_1^{\prime}\left(t_0\right)+c_2 \phi_2^{\prime}\left(t_0\right)=0 \\
{\left[\begin{array}{ll}
\phi_1\left(t_0\right) & \phi_2\left(t_0\right) \\
\phi_1^{\prime}\left(t_0\right) & \phi_2^{\prime}\left(t_0\right)
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]=\left[\begin{array}{l}
0 \\
0
\end{array}\right]}
\end{array}
\]

In other words the fundamental set of solutions \(\left\{\phi_1, \phi_2\right\}\) to \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\) form a basis for the set of all solutions to this linear homogeneous DE.