Solutions to Systems — Question 2

Question 2 – Solution

Strategy. Distinguish solution columns from a complete basis, and normalize by the matrix at the initial time.

Step 1: Verify the columns and independence. Differentiating each exponential column gives $\Phi'=A\Phi$. Its determinant is $-2$, never zero. Every initial vector can therefore be represented by $\Phi(0)c$; uniqueness of the linear IVP proves that the resulting constant combination supplies every solution.

Step 2: Normalize at the initial time. Since $\Phi(0)^{-1}=\tfrac 12\begin{pmatrix}1&1\\1&-1\end{pmatrix}$, $$\boxed{Y(t)=\Phi(t)\Phi(0)^{-1} =\begin{pmatrix}\cosh t&\sinh t\\\sinh t&\cosh t\end{pmatrix}.}$$ Here $\cosh t=(e^t+e^{-t})/2$, $\sinh t=(e^t-e^{-t})/2$. Thus $Y(0)=I$ and $Y'=AY$ by differentiation or constant right multiplication.

Step 3: Recover arbitrary data. Solving $\Phi(0)c=(p,q)^T$ gives $$\boxed{c_1=(p+q)/2,\quad c_2=(p-q)/2, \quad X=(p\cosh t+q\sinh t,\ p\sinh t+q\cosh t)^T.}$$ This equals $Y(t)(p,q)^T$. The constants in a nonnormalized basis are not generally the initial state coordinates themselves.

Step 4: Detect a collapsed family. Constant combinations remain solutions, but $\det\Psi_a=-2(1-a^2)$, so the family is fundamental exactly when $\boxed{a\ne\pm 1}$. At $a=1$, both columns are $U+V$, whose value at $0$ is $(2,0)^T$; exactly the initial states $(p,0)^T$ are representable. At $a=-1$, the columns are $U-V$ and its negative, with initial value $(0,2)^T$; exactly $(0,q)^T$ are representable. Here $U,V$ denote the columns of $\Phi$. Neither singular case supplies all initial states, despite having two displayed solution columns.