Solve the Euler-type (equidimensional) system: 3tdot…
Solve the Euler-type (equidimensional) system: $$ 3t\dot x=2x+y-z,\qquad 2t\dot y=x+3y+z,\qquad 6t\dot z=-x+7y+5z $$
Explanation of Method
After dividing through so that every equation has the form $t\dot u_i=(\ldots)$, the system is a constant-coefficient Euler system $t\dot{\mathbf u}=A\mathbf u$; substituting $t=e^{s}$ converts it into an ordinary constant-coefficient system in $s$, so each eigenvalue $\lambda$ of $A$ contributes a power-law mode $t^{\lambda}$ rather than an exponential $e^{\lambda t}$.
- Divide each equation by its $t$-coefficient to reach the standard Euler form $t\dot{\mathbf u}=A\mathbf u$.
- Substitute $t=e^{s}$ so that $t\dot u=u'$ ($'=d/ds$), reducing to a constant-coefficient system.
- Find the eigenvalues and eigenvectors of $A$.
- Translate each mode $e^{\lambda s}$ back into $t^{\lambda}$ and assemble the general solution.
Step 1 · Standard Euler form
Dividing by the $t$-coefficients gives $t\dot{\mathbf u}=A\mathbf u$ with
$$ A=\frac{1}{6}\begin{pmatrix}4&2&-2\\3&9&3\\-1&7&5\end{pmatrix}. $$
Step 2 · Substitution $t=e^{s}$
With $s=\ln t$, the relation $t\dot x=x'$ (and similarly for $y,z$) converts the Euler system into the ordinary constant-coefficient system $\mathbf u'=A\mathbf u$.
Step 3 · Characteristic equation
Working with $M=6A=\begin{pmatrix}4&2&-2\\3&9&3\\-1&7&5\end{pmatrix}$, expanding $\det(M-\mu I)$ gives
$$ -\mu(\mu-6)(\mu-12)=0\ \Longrightarrow\ \mu_1=0,\ \mu_2=6,\ \mu_3=12, $$
so the eigenvalues of $A=M/6$ are $\lambda_1=0,\ \lambda_2=1,\ \lambda_3=2$.
Step 4 · Eigenvectors
$$ \mathbf v_1=\begin{pmatrix}-4\\3\\-5\end{pmatrix}\ (\lambda=0),\qquad \mathbf v_2=\begin{pmatrix}-1\\0\\1\end{pmatrix}\ (\lambda=1),\qquad \mathbf v_3=\begin{pmatrix}0\\1\\1\end{pmatrix}\ (\lambda=2). $$
Each is verified directly against $M\mathbf v=\mu\mathbf v$, and the three vectors are independent ($\det=12\neq0$). ✓
Step 5 · Translate back to $t$ and verify
Since $e^{\lambda s}=t^{\lambda}$, the modes $\lambda=0,1,2$ become $1,\,t,\,t^2$. Substituting the resulting general solution into all three original equations and matching the constant, $t$, and $t^2$ coefficients confirms every equation holds identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues of $A$ | $\lambda_1=0,\ \lambda_2=1,\ \lambda_3=2$ |
| Corresponding power-law modes | $1,\ t,\ t^2$ |
| Validity | $t>0$ |
Additional Notes
- Euler systems always convert power-law behaviour $t^{\lambda}$ into exponential behaviour $e^{\lambda s}$ under $t=e^s$, exactly mirroring the constant-coefficient theory.
- Verified by direct substitution into all three original equations, matching the constant, linear, and quadratic terms in $t$ separately.
Result
The general solution of the system is: