Consider the system \[ x'=y+tz, \qquad y'=z+t^2x, \qquad z'=x+e^t y. \] Prove that there exists a solution on \([0,1]\) satisfying \[ \begin{pmatrix} 1&2&3\\ 4&5&6\\ 7&8&9 \end{pmatrix} \begin{pmatrix} x(0)\\y(0)\\z(0) \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}, \] and also \[ \int_0^1\bigl(x(t)^2+y(t)^2+z(t)^2\bigr)\,dt=1. \]
Solution
Write the system as \[ v'(t)=A(t)v(t), \qquad A(t)= \begin{pmatrix} 0&1&t\\ t^2&0&1\\ 1&e^t&0 \end{pmatrix}, \] where \[ v(t)= \begin{pmatrix} x(t)\\y(t)\\z(t) \end{pmatrix}. \] Also put \[ B= \begin{pmatrix} 1&2&3\\ 4&5&6\\ 7&8&9 \end{pmatrix}. \]
1. The vector\[ v_0= \begin{pmatrix} 1\\-2\\1 \end{pmatrix} \]is nonzero and satisfies\[ Bv_0=0. \]
Direct multiplication gives \[ Bv_0 = \begin{pmatrix} 1-4+3\\ 4-10+6\\ 7-16+9 \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}. \] Its first entry is \(1\), so \(v_0\ne0\).
2. There is a unique solution\[ u:[0,1]\longrightarrow\RR^3 \]of\[ u'(t)=A(t)u(t), \qquad u(0)=v_0. \]
Every entry of \(A(t)\) is continuous on \([0,1]\). The standard existence-and-uniqueness theorem for linear systems therefore gives a unique solution on the whole interval for the prescribed initial value.
3. The number\[ I\coloneqq\int_0^1\|u(t)\|^2\,dt \]is finite and strictly positive.
The solution \(u\) is continuous, so \(\|u(t)\|^2\) is continuous on the compact interval \([0,1]\) and therefore integrable with finite integral. Moreover, \[ \|u(0)\|^2=\|v_0\|^2=6>0. \] By continuity, \(\|u(t)\|^2\) remains positive on some interval \([0,\delta]\), so its integral over \([0,1]\) is strictly positive.
4. Define\[ v(t)\coloneqq I^{-1/2}u(t). \]Then \(v\) is a solution of the same differential system and satisfies thedisplayed initial condition.
The system is homogeneous and linear, so multiplication of a solution by the constant \(I^{-1/2}\) gives another solution. By steps <1>1–<1>2, \[ Bv(0) = I^{-1/2}Bu(0) = I^{-1/2}Bv_0 = 0. \] Thus the required matrix equation at \(t=0\) holds.
5. The solution \(v\) from step 4 satisfies\[ \boxed{ \int_0^1\bigl(x(t)^2+y(t)^2+z(t)^2\bigr)\,dt=1. } \]
Since \(\|v(t)\|^2=I^{-1}\|u(t)\|^2\), the definition of \(I\) gives \[ \int_0^1\|v(t)\|^2\,dt = I^{-1}\int_0^1\|u(t)\|^2\,dt = 1. \] In coordinates, \(\|v(t)\|^2=x(t)^2+y(t)^2+z(t)^2\).
6. Q.E.D.
Step 4 gives a solution satisfying the displayed initial condition, and step 5 gives the required normalization.