7.7

Example \(\PageIndex{1}\)

Solve \(\mathbf{x}^{\prime}(t)=\left[\begin{array}{ll}1 & 3 \\ 4 & 5\end{array}\right] \mathbf{x}(\mathbf{t})\)

Solution

Step 1: Find eigenvalues:
\[ \notag
\begin{aligned}
& A-\lambda I=\left|\begin{array}{cc}
1-\lambda & 3 \\
4 & 5-\lambda
\end{array}\right|=(1-\lambda)(5-\lambda)-12 \\
& \quad=\lambda^2-6 \lambda+5-12=\lambda^2-6 \lambda-7=(\lambda-7)(\lambda+1)=0
\end{aligned}
\]

Thus \(\lambda=7,-1\)

Step 2: Find eigenvectors:
\[ \notag
\lambda=7: \quad A-7 I=\left[\begin{array}{cc}
1-7 & 3 \\
4 & 5-7
\end{array}\right]=\left[\begin{array}{cc}
-6 & 3 \\
4 & -2
\end{array}\right]
\]

We then see that \(\left[\begin{array}{cc}-6 & 3 \\ 4 & -2\end{array}\right]\left[\begin{array}{l}1 \\ 2\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right].\) So the dimension of the nullspace of \(\left[\begin{array}{cc}-6 & 3 \\ 4 & -2\end{array}\right]\) is 1 .
Or in other words, solution space for \(\left[\begin{array}{cc}-6 & 3 \\ 4 & -2\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right]\) is 1-dimensional.

Thus a basis for the eigenspace for \(\lambda=7\) is \(\left\{\left[\begin{array}{l}1 \\ 2\end{array}\right]\right\}\)

\[ \notag
\lambda=-1 \quad A-(-1) I=\left[\begin{array}{cc}
1+1 & 3 \\
4 & 5+1
\end{array}\right]=\left[\begin{array}{ll}
2 & 3 \\
4 & 6
\end{array}\right]
\]

Similar to above \(\left[\begin{array}{ll}2 & 3 \\ 4 & 6\end{array}\right]\left[\begin{array}{c}3 \\ -2\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right].\) Thus a basis for the eigenspace for \(\lambda=-1\) is \(\left\{\left[\begin{array}{c}3 \\ -2\end{array}\right]\right\}\)

Thus a basis for the solution space to \(\mathbf{x}^{\prime}=\left[\begin{array}{ll}1 & 3 \\ 4 & 5\end{array}\right] \mathbf{x}\) is
\[ \notag
\left\{\left[\begin{array}{l}
1 \\
2
\end{array}\right] e^{7 t},\left[\begin{array}{c}
3 \\
-2
\end{array}\right] e^{-t}\right\}
\]

Hence the general solution is
\[ \notag
\mathbf{x}(t)=c_1\left[\begin{array}{l}
1 \\
2
\end{array}\right] e^{7 t}+c_2\left[\begin{array}{c}
3 \\
-2
\end{array}\right] e^{-t}
\]

Note we can take any basis for the solution space to create the general solution

Alternate basis: \(\left\{\left[\begin{array}{l}2 \\ 4\end{array}\right] e^{7 t},\left[\begin{array}{c}-9 \\ 6\end{array}\right] e^{-t}\right\}\)
Alternate format of general solution:
\[ \notag
\mathbf{x}(t)=c_1\left[\begin{array}{l}
2 \\
4
\end{array}\right] e^{7 t}+c_2\left[\begin{array}{c}
-9 \\
6
\end{array}\right] e^{-t}
\]

---

Example \(\PageIndex{2}\)

Solve the initial value problem \(\mathbf{x}^{\prime}=\left[\begin{array}{ll}
1 & 3 \\
4 & 5
\end{array}\right] \mathbf{x}, \quad \mathbf{x}(0)=\left[\begin{array}{l}
e \\
f
\end{array}\right]\)

Solution

\[ \notag
\begin{aligned}
& \text { IVP: } \mathbf{x}^{\prime}=\left[\begin{array}{ll}
1 & 3 \\
4 & 5
\end{array}\right] \mathbf{x}, \quad \mathbf{x}(0)=\left[\begin{array}{l}
e \\
f
\end{array}\right] \\
& {\left[\begin{array}{l}
e \\
f
\end{array}\right]=c_1\left[\begin{array}{l}
1 \\
2
\end{array}\right]+c_2\left[\begin{array}{c}
3 \\
-2
\end{array}\right]=\left[\begin{array}{c}
c_1+3 c_2 \\
2 c_1-2 c_2
\end{array}\right]}
\end{aligned}
\]

Solve using any method you like. We will use matrix form:
\[ \notag
\left[\begin{array}{l}
e \\
f
\end{array}\right]=\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]
\]

Solution exists if Wronskian evaluated at 0 is not zero.
\[ \notag
\begin{aligned}
W\left(\left[\begin{array}{l}
1 \\
2
\end{array}\right] e^{7 t},\left[\begin{array}{c}
3 \\
-2
\end{array}\right] e^{-t}\right)= & \left|\begin{array}{cc}
e^{7 t} & 3 e^{-t} \\
2 e^{7 t} & -2 e^{-t}
\end{array}\right| \\
& =-2 e^{6 t}-6 e^{6 t}=-8 e^{6 t} \neq 0
\end{aligned}
\notag \]

Fundamental matrix: \(\quad \Psi(t)=\left[\begin{array}{cc}e^{7 t} & 3 e^{-t} \\ 2 e^{7 t} & -2 e^{-t}\end{array}\right]\)
Back to IVP: \(\left[\begin{array}{l}e \\ f\end{array}\right]=\left[\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right]\left[\begin{array}{l}c_1 \\ c_2\end{array}\right]\)
\[ \notag
\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right]^{-1}\left[\begin{array}{l}
e \\
f
\end{array}\right]=\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right]^{-1}\left[\begin{array}{cc} 
& 3 \\
2 & -2
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]
\]

Thus \(\left[\begin{array}{l}c_1 \\ c_2\end{array}\right]=\left[\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right]^{-1}\left[\begin{array}{l}e \\ f\end{array}\right]\)
But I would prefer a fundamental matrix whose inverse is easier to calculate, at least when \(t_0=0\).

Thus we will find another basis for the solution set to \(\mathbf{x}^{\prime}=A \mathbf{x}\) so that the corresponding fundamental matrix \(\Phi\) has the property that \(\Phi(0)=\left[\notag \begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\), the \(2 \mathrm{x} 2\) identity matrix via long method:

Step 1: Solve IVP: \(\quad \mathbf{x}^{\prime}=A \mathbf{x}, \quad \mathbf{x}(0)=\left[\begin{array}{l}1 \\ 0\end{array}\right]\)
\[ \notag
\begin{gathered}
{\left[\begin{array}{l}
1 \\
0
\end{array}\right]=c_1\left[\begin{array}{l}
1 \\
2
\end{array}\right] e^0+c_2\left[\begin{array}{c}
3 \\
-2
\end{array}\right] e^0=\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]} \\
{\left[\begin{array}{l}
1 \\
0
\end{array}\right]=\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right] \text { implies }\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]=\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right]^{-1}\left[\begin{array}{l}
1 \\
0
\end{array}\right]} \\
\left(-\frac{1}{8}\right)\left[\begin{array}{cc}
-2 & -3 \\
-2 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
\end{gathered}
\]

\[ \notag
\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right]^{-1}=\left[\begin{array}{cc}
\frac{1}{4} & \frac{3}{8} \\
\frac{1}{4} & -\frac{1}{8}
\end{array}\right] \&\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]=\left[\begin{array}{cc}
\frac{1}{4} & \frac{3}{8} \\
\frac{1}{4} & -\frac{1}{8}
\end{array}\right]\left[\begin{array}{l}
1 \\
0
\end{array}\right]=\left[\begin{array}{l}
\frac{1}{4} \\
\frac{1}{4}
\end{array}\right]
\]

Thus IVP solution where \(\mathbf{x}(0)=\left[\begin{array}{l}1 \\ 0\end{array}\right]\) is
\[
\mathbf{x}(t)=\frac{1}{4}\left[\begin{array}{l}
1 \\
2
\end{array}\right] e^{7 t}+\frac{1}{4}\left[\begin{array}{c}
3 \\
-2
\end{array}\right] e^{-t}=\left[\begin{array}{c}
\frac{1}{4} e^{7 t}+\frac{3}{4} e^{-t} \\
\frac{1}{2} e^{7 t}-\frac{1}{2} e^{-t}
\end{array}\right]
\]

Step 2: Solve IVP: \(\quad \mathbf{x}^{\prime}=A \mathbf{x}, \quad \mathbf{x}(0)=\left[\begin{array}{l}0 \\ 1\end{array}\right]\)
\[ \notag
\left[\begin{array}{l}
0 \\
1
\end{array}\right]=c_1\left[\begin{array}{l}
1 \\
2
\end{array}\right] e^0+c_2\left[\begin{array}{c}
3 \\
-2
\end{array}\right] e^0=\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]
\]

Thus \(\left[\begin{array}{l}c_1 \\ c_2\end{array}\right]=\left[\begin{array}{cc}\frac{1}{4} & \frac{3}{8} \\ \frac{1}{4} & -\frac{1}{8}\end{array}\right]\left[\begin{array}{l}0 \\ 1\end{array}\right]=\left[\begin{array}{c}\frac{3}{8} \\ -\frac{1}{8}\end{array}\right]\)
Thus IVP solution where \(\mathbf{x}(0)=\left[\begin{array}{l}0 \\ 1\end{array}\right]\) is
\[ \notag
\mathbf{x}(t)=\frac{3}{8}\left[\begin{array}{l}
1 \\
2
\end{array}\right] e^{7 t}-\frac{1}{8}\left[\begin{array}{c}
3 \\
-2
\end{array}\right] e^{-t}=\left[\begin{array}{l}
\frac{3}{8} e^{7 t}-\frac{3}{8} e^{-t} \\
\frac{3}{4} e^{7 t}+\frac{1}{4} e^{-t}
\end{array}\right]
\]

Thus another basis for the solution space to \(\mathbf{x}^{\prime}=\left[\begin{array}{ll}1 & 3 \\ 4 & 5\end{array}\right] \mathbf{x}\)
\[ \notag
\text { is }\left\{\left[\begin{array}{l}
\frac{1}{4} e^{7 t}+\frac{3}{4} e^{-t} \\
\frac{1}{2} e^{7 t}-\frac{1}{2} e^{-t}
\end{array}\right],\left[\begin{array}{l}
\frac{3}{8} e^{7 t}-\frac{3}{8} e^{-t} \\
\frac{3}{4} e^{7 t}+\frac{1}{4} e^{-t}
\end{array}\right]\right\}
\]

Its corresponding fundamental matrix is
\[ \notag
\Phi(t)=\left[\begin{array}{ll}
\frac{1}{4} e^{7 t}+\frac{3}{4} e^{-t} & \frac{3}{8} e^{7 t}-\frac{3}{8} e^{-t} \\
\frac{1}{2} e^{7 t}-\frac{1}{2} e^{-t} & \frac{3}{4} e^{7 t}+\frac{1}{4} e^{-t}
\end{array}\right]
\]

Thus to solve IVP where \(\mathbf{x}\left(t_0\right)=\left[\begin{array}{l}e \\ f\end{array}\right]\), we solve
\[ \notag
\left[\begin{array}{l}
e \\
f
\end{array}\right]=\left[\begin{array}{ll}
\frac{1}{4} e^{7 t_0}+\frac{3}{4} e^{-t_0} & \frac{3}{8} e^{7 t_0}-\frac{3}{8} e^{-t_0} \\
\frac{1}{2} e^{7 t_0}-\frac{1}{2} e^{-t_0} & \frac{3}{4} e^{7 t_0}+\frac{1}{4} e^{-t_0}
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]
\]

When \(t_0=0\). I.e., we have an IVP where \(\mathbf{x}(0)=\left[\begin{array}{l}e \\ f\end{array}\right]\)
\[ \notag
\left[\begin{array}{l}
e \\
f
\end{array}\right]=\left[\begin{array}{cc}
\frac{1}{4} e^0+\frac{3}{4} e^0 & \frac{3}{8} e^0-\frac{3}{8} e^0 \\
\frac{1}{2} e^0-\frac{1}{2} e^0 & \frac{3}{4} e^0+\frac{1}{4} e^0
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]
\]
which simplifies to
\[ \notag
\left[\begin{array}{l}
e \\
f
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]=\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]
\]

In other words, \(c_1=e\) and \(c_2=f\).

 

The Matrix Exponential

Definition: Matrix exponential

Let \(A\) be an \(n \times n\) matrix. Then the matrix exponential of \(A\) is
\[ \notag
\exp (A t)=e^{A t}=I+\Sigma_{n=1}^{\infty} \frac{A^k t^k}{k!}
\]
where \(I\) is the \(n \times n\) identity matrix.

Note

\(e^{A(0)}=I+\Sigma_{n=1}^{\infty} \frac{A^k 0^k}{k!}=I\)

Note

\(\left[e^{A t}\right]^{\prime}=\Sigma_{n=1}^{\infty} \frac{k A^k t^{k-1}}{k!}=\Sigma_{n=1}^{\infty} \frac{A^k t^{k-1}}{(k-1)!}=\Sigma_{n=0}^{\infty} \frac{A^{k+1} t^k}{k!}\)
\[ \notag
=A+\Sigma_{n=1}^{\infty} \frac{A^{k+1} t^k}{k!}=A\left(I+\Sigma_{n=1}^{\infty} \frac{A^k t^k}{k!}\right)=A e^{A t}
\]

Thus \(\left[e^{A t}\right]^{\prime}=A e^{A t}\) and \(e^{A(0)}=I.\) Thus \(e^{A t}\) is the solution to the IVP \( M^{\prime}=A M, M(0)=I \) where \(M\) is an \(n \times n\) matrix.

Let \(\Psi\) be a fundamental matrix for \(\mathbf{x}^{\prime}=A \mathbf{x}\)

Example \(\PageIndex{3}\)

Suppose \(\Psi\) is the \(2 \times 2\) matrix \(\left[\begin{array}{ll}\mathbf{f}_{\mathbf{1}} & \mathbf{f}_{\mathbf{2}}\end{array}\right]\).

Solution

Thus \(\mathbf{f}_{\mathbf{1}}\) and \(\mathbf{f}_{\mathbf{2}}\) are solutions to \(\mathbf{x}^{\prime}=A \mathbf{x}\).
Hence \(\mathbf{f}_{\mathbf{1}}{ }^{\prime}=A \mathbf{f}_{\mathbf{1}}\) and \(\mathbf{f}_{\mathbf{2}}{ }^{\prime}=A \mathbf{f}_{\mathbf{2}}\)
Thus \(A\left[\begin{array}{ll}\mathbf{f}_1 & \mathbf{f}_{\mathbf{2}}\end{array}\right]=\)
Since \(\left[\begin{array}{ll}\mathbf{f}_{\mathbf{1}} & \mathbf{f}_{\mathbf{2}}\end{array}\right]^{\prime}=A\left[\begin{array}{ll}\mathbf{f}_{\mathbf{1}} & \mathbf{f}_{\mathbf{2}}\end{array}\right],\Psi(t)=\left[\begin{array}{ll}
\mathbf{f}_{\mathbf{1}}(\mathbf{t}) & \mathbf{f}_{\mathbf{2}}(\mathbf{t})
\end{array}\right] \text { is the general solution to } M^{\prime}=A M\).

Let \(\Phi(t)=\Psi(t)[\Psi(0)]^{-1}\). Then \(\Phi(0)=\Psi(0)[\Psi(0)]^{-1}=I\).
Note \(\Phi(t)=\Psi(t)[\Psi(0)]^{-1}\) is also a solution to \(M^{\prime}=A M\) :
\[ \notag
\left(\Psi(t)[\Psi(0)]^{-1}\right)^{\prime}=\Psi^{\prime}(t)[\Psi(0)]^{-1}=A \Psi(t)[\Psi(0)]^{-1}
\]

Thus \(\Phi\) is the solution to the IVP
\[ \notag
M^{\prime}=A M, M(0)=I
\]

Since solution to IVP is unique (assuming entries of \(A\) are continuous functions), \(\Phi=e^{A t}\)

Theorem \(\PageIndex{1}\)

\(e^{A t}=\Phi(t)=\Psi(t)[\Psi(0)]^{-1}\) where \(\Psi\) is a fundamental matrix for \(\mathbf{x}^{\prime}=A \mathbf{x}\)

Example \(\PageIndex{4}\)

Calculate \(\exp \left(\left[\begin{array}{ll}1 & 3 \\ 4 & 5\end{array}\right] t\right):=e^{\left[\begin{array}{ll}1 & 3 \\ 4 & 5\end{array}\right] t}\)

Solution

Solve \(\mathbf{x}^{\prime}(t)=\left[\begin{array}{ll}1 & 3 \\ 4 & 5\end{array}\right] \mathbf{x}(\mathbf{t})\)
From previous work, the general solution is
\[
\mathbf{x}(t)=c_1\left[\begin{array}{l}
1 \\
2
\end{array}\right] e^{7 t}+c_2\left[\begin{array}{c}
3 \\
-2
\end{array}\right] e^{-t}
\]

Thus a fundamental matrix is \(\Psi(t)=\left[\begin{array}{cc}e^{7 t} & 3 e^{-t} \\ 2 e^{7 t} & -2 e^{-t}\end{array}\right]\)

A "better" fundamental matrix is \(\Phi(t)=\Psi(t)[\Psi(0)]^{-1}\)
\[
\begin{gathered}
\Psi(0)=\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right] . \text { Thus }[\Psi(0)]^{-1}=\left[\begin{array}{cc}
1 & 3 \\
2 & -2
\end{array}\right]^{-1}=\left(-\frac{1}{8}\right)\left[\begin{array}{cc}
-2 & -3 \\
-2 & 1
\end{array}\right] \\
\Phi(t)=\Psi(t)[\Psi(0)]^{-1}=\left(-\frac{1}{8}\right)\left[\begin{array}{cc}
e^{7 t} & 3 e^{-t} \\
2 e^{7 t} & -2 e^{-t}
\end{array}\right]\left[\begin{array}{cc}
-2 & -3 \\
-2 & 1
\end{array}\right] \\
=\left[\begin{array}{cc}
\frac{1}{4} e^{7 t}+\frac{3}{4} e^{-t} & \frac{3}{8} e^{7 t}-\frac{3}{8} e^{-t} \\
\frac{1}{2} e^{7 t}-\frac{1}{2} e^{-t} & \frac{3}{4} e^{7 t}+\frac{1}{4} e^{-t}
\end{array}\right]
\end{gathered}
\]

Thus \[ \notag \exp \left(\left[\begin{array}{ll}
1 & 3 \\
4 & 5
\end{array}\right] t\right)=e^{\left[\begin{array}{ll}
1 & 3 \\
4 & 5
\end{array}\right]^t}=\left[\begin{array}{ll}
\frac{1}{4} e^{7 t}+\frac{3}{4} e^{-t} & \frac{3}{8} e^{7 t}-\frac{3}{8} e^{-t} \\
\frac{1}{2} e^{7 t}-\frac{1}{2} e^{-t} & \frac{3}{4} e^{7 t}+\frac{1}{4} e^{-t}
\end{array}\right] \]

Example \(\PageIndex{5}\)

Solve IVP:\(\quad \mathbf{x}^{\prime}=A \mathrm{x}, \mathbf{x}(0)=\mathbf{e}\)

Solution

Observe if \(\Phi(t)=\left[\begin{array}{ll}\mathbf{f}_{\mathbf{1}} & \mathbf{f}_{\mathbf{2}}\end{array}\right]\), then general solution is
\[
\begin{aligned}
& \mathbf{x}(\mathbf{t})=c_1 \mathbf{f}_{\mathbf{1}}(t)+c_2 \mathbf{f}_2(t)=\left[\begin{array}{ll}
\mathbf{f}_{\mathbf{1}}(t) & \mathbf{f}_{\mathbf{2}}(t)
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]=\Phi(t)\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right] \\
& \mathbf{x}(\mathbf{0})=c_1 \mathbf{f}_{\mathbf{1}}(0)+c_2 \mathbf{f}_{\mathbf{2}}(0)=\left[\begin{array}{ll}
\mathbf{f}_{\mathbf{1}}(0) & \mathbf{f}_{\mathbf{2}}(0)
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right] \\
& =\Phi(0)\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right]=\left[\begin{array}{l}
c_1 \\
c_2
\end{array}\right] \\
&
\end{aligned}
\]

 

 

 

Theorem \(\PageIndex{1}\)

If \(e^{A t}=\Phi(t)=\Psi(t)[\Psi(0)]^{-1}\) where \(\Psi\) is a fundamental matrix for \(\mathbf{x}^{\prime}=A \mathbf{x}\)
or equivalently, if \(\Phi\) is the solution to IVP \(M^{\prime}=A M, M(0)=I\), Then solution to IVP \(\mathbf{x}^{\prime}=A \mathbf{x}, \mathbf{x}(0)=\mathbf{e}\) is \(\mathbf{x}=\Phi \mathbf{e}\)

 

Example \(\PageIndex{6}\)

Solve IVP: \(\mathbf{x}^{\prime}(t)=\left[\begin{array}{ll}1 & 3 \\ 4 & 5\end{array}\right] \mathbf{x}(\mathbf{t}), \quad \mathbf{x}(0)=\left[\begin{array}{l}17 \\ 92\end{array}\right]\)

Solution

That is, \(\mathbf{x}=17\left[\begin{array}{l}\frac{1}{4} e^{7 t}+\frac{3}{4} e^{-t} \\ \frac{1}{2} e^{7 t}-\frac{1}{2} e^{-t}\end{array}\right]+92\left[\begin{array}{l}\frac{3}{8} e^{7 t}-\frac{3}{8} e^{-t} \\ \frac{3}{4} e^{7 t}+\frac{1}{4} e^{-t}\end{array}\right]\)
Note: This only works if you use the fundamental matrix \(\Phi(t)\) where \(\Phi(0)=I\).