Systems of Differential Equations — Question 10

Question 10 – Solution

Strategy. Subtract the velocity at the origin to isolate the matrix columns, then test the model assumption with new data.

Step 1: Recover the affine coefficients. At the origin, $b=F(0,0)=(1,-1)^T$. The differences $F(1,0)-b=(-2,3)^T$ and $F(0,1)-b=(1,-2)^T$ are the first and second columns of $A$. Therefore $$\boxed{A=\begin{pmatrix}-2&1\\3&-2\end{pmatrix},\qquad b=(1,-1)^T.}$$ These columns and $b$ are forced, proving uniqueness within the affine class. The measurements prescribe the field at three different states; they are not three independently assignable derivatives at one initial state.

Step 2: Translate the unique equilibrium. The equation $AX_*+b=0$ reads $-2x+y+1=0$ and $3x-2y-1=0$. It gives $X_*=(1,1)^T$; uniqueness also follows from $\det A=1$. For $Z=X-X_*$, the translation is constant, and $$Z'=A(Z+X_*)+b=AZ,\qquad \boxed{Z'=AZ.}$$ The forcing vanishes because the translation is by an equilibrium, not because every translation removes every forcing term.

Step 3: Check the fourth report. The recovered model predicts $F(1,1)=(0,0)^T$, while the report is $(0,1)^T$. The reported-minus-predicted discrepancy is $(0,1)^T$. No affine system fits all four measurements, since the first three already fix its coefficients. This conclusion tests the affine model assumption.

Step 4: Exhibit a nonlinear alternative. Define $$\boxed{\widetilde F(x,y)=(-2x+y+1,\,3x-2y-1+xy)^T.}$$ The extra term $xy$ vanishes at the first three states and equals $1$ at $(1,1)$, so all four measurements are met. This polynomial field is smooth and nonlinear. In it, $(1,1)$ is not an equilibrium. More generally, fields obtained by adding $(0,cxy)^T$ all fit the first three data for any real $c$. Finite measurements uniquely identify this affine class, but leave unrestricted nonlinear models undetermined.