MHB Finding Restrictions for the Implicit Function Theorem

[evinda]

Hello! (Wave)

Which relation do the constants $a,b$ have to satisfy so that the implicit function theorem implies that the system of two equations

$$axu^2v+byv^2=-a \ \ \ \ bxyu-auv^2=-a$$

can be solved as for u and v as functions $u=u(x,y)$ and $v=v(x,y)$ with continuous partial derivatives of first order in some region of $(1,0)$ and with u(1,0)=1, v(1,0)=-1.

I have thought the following:$$\Delta=\begin{pmatrix}
au^2v & bv^2\\
byu & -bxu
\end{pmatrix}=\begin{pmatrix}
-a & b\\
0 & -b
\end{pmatrix}$$It should hold that $det(\Delta)\neq0 \Rightarrow ab \neq 0$.

So the condition is $a^2+b^2 \neq 0$. Am I right?

 

[I like Serena]

Hello! (Wave)

Which relation do the constants $a,b$ have to satisfy so that the implicit function theorem implies that the system of two equations

$$axu^2v+byv^2=-a \ \ \ \ bxyu-auv^2=-a$$

can be solved as for u and v as functions $u=u(x,y)$ and $v=v(x,y)$ with continuous partial derivatives of first order in some region of $(1,0)$ and with u(1,0)=1, v(1,0)=-1.

I have thought the following:$$\Delta=\begin{pmatrix}
au^2v & bv^2\\
byu & -bxu
\end{pmatrix}=\begin{pmatrix}
-a & b\\
0 & -b
\end{pmatrix}$$It should hold that $det(\Delta)\neq0 \Rightarrow ab \neq 0$.

So the condition is $a^2+b^2 \neq 0$. Am I right?

Hey evinda! (Smile)

Suppose we pick $u(x,y)=v(x,y)=a=0$, wouldn't that satisfy all conditions? (Wondering)

 

[evinda]

Hey evinda! (Smile)

Suppose we pick $u(x,y)=v(x,y)=a=0$, wouldn't that satisfy all conditions? (Wondering)

Yes, that's right... Could you give me a hint how can we can find the desired condition? (Thinking)

 

[I like Serena]

Yes, that's right... Could you give me a hint how can we can find the desired condition? (Thinking)

Shouldn't we have partial derivative like:
$$
\pd{}x(axu^2v+byv^2)=au^2v + 2axuvu_x + axu^2v_x + 2buvv_x = 0
$$
and 3 more such equations?

We should be able to solve for $(u_x,u_y)$ and $(v_x,v_y)$, after which we need to check if each of them is a well defined , continuous vector function around $(x,y)=(1,0)$.
That means in particular that if we substitute $(x,y)=(1,0)$, $u(1,0)=1$, and $v(1,0)=-1$, that we should find that $(u_x,u_y)$ and $(v_x,v_y)$ are defined (not infinite).

We can probably make the substitution immediately, before solving for $(u_x,u_y)$ and $(v_x,v_y)$.
That should give us the required restrictions for a and b.
That is:
$$
\pd{}x(axu^2v+byv^2)\Big|_{(1,0)}=a\cdot -1 + 2a\cdot -1 \cdot u_x + av_x + 2b\cdot -1 \cdot v_x
= -a -2au_x+(a-2b)v_x = 0
$$
and the same for the other 3 equations.
Then solve for $(u_x,u_y)$ and $(v_x,v_y)$. (Thinking)