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  1. MHB Master
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    #1
    Hello!!!

    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?

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    #2
    Quote Originally Posted by evinda View Post
    Hello!!!

    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!!

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

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    #3 Thread Author
    Quote Originally Posted by I like Serena View Post
    Hey evinda!!

    Suppose we pick $u(x,y)=v(x,y)=a=0$, wouldn't that satisfy all conditions?
    Yes, that's right... Could you give me a hint how can we can find the desired condition?

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    #4
    Quote Originally Posted by evinda View Post
    Yes, that's right... Could you give me a hint how can we can find the desired condition?
    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 $(v_x,v_y)$.

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