MHB Induction on the number of equations

[mathmari]

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, $n$.

• 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)

 

[mathmari]

Is the inductive step as follows? (Wondering) 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. We add these two equations and we get a new differential equation $Lx=f$. So the initial system is equivalent to the new differential equation $Lx=f$.