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mathmari
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Hey!
Let $L$ be a differential operator.
We suppose that we have $n$ equations, that means $\phi: \displaystyle{\bigwedge_{j=1}^n L_j x=f_j}$ and we assume that $\phi$ can be written as $Lx=f \land \psi$, where $\psi$ doesn't contain any $x$.
We prove this by induction on the number of equations, $n$.
Is this correct so far? (Wondering)
How could we continue? (Wondering)
Let $L$ be a differential operator.
We suppose that we have $n$ equations, that means $\phi: \displaystyle{\bigwedge_{j=1}^n L_j x=f_j}$ and we assume that $\phi$ can be written as $Lx=f \land \psi$, where $\psi$ doesn't contain any $x$.
We prove this by induction on the number of equations, $n$.
- Base case: For $n=1$ we have one equation, so it is of the form $Lx=f$.
- Inductive hypothesis: We suppose that it holds for $n=k$, i.e., if $\phi$ contains $k$ equations, then we can reduce it into the form $$Lx=f \land \psi \ \ \text{ where } \psi \text{ doesn't contain any } x.$$
- Inductive step: We will show that it holds for $n=k+1$, i.e., if we have $k+1$ equations we can reduce this system into the form $$Lx=f \land \psi \ \ \text{ where } \psi \text{ doesn't contain any } x.$$
From the inductive hypothesis we know that we can reduce the first $k$ equations into the above form. So we have two equations that contain $x$ and its derivatives.
Is this correct so far? (Wondering)
How could we continue? (Wondering)