Mixing
Diffeomorphism on the image
$begingroup$
Let $Phi(x,y)=(x^2+2ycosx+1, sinx+e^{x+y})$. Proof that exists neighborhood $U$ of the point $(0,0)$ such that $Phi_{|U}$ is diffeomorphism on the image. Calculate the derivative $(Phi_{|U})^{-1})$ in $(1,1)$.
Well, im struggling with this task. Idea was:
$1-1:$
$(Phi(x,y)=Phi(t1,t2)$ for $(t1,t2)in mathbb R^{2}$
$begin{cases}
x^2+2ycosx+1 = (t1)^2+2(t2)cos(t1)+1 \
sinx+e^{x+y} = sin(t1) + e^{t1+t2}
end{cases}$
What now? ... Or maybe it's obvious, because I have to do this in $(0,0)$ point?
$Phi(x,y)=Phi(t1,t2)=(0,0) :$
$begin{cases}
0+0+1 = 0+0+1 \
0+1 = 0+1
end{cases}$
(I think this is stupid a little bit and idk if I am doing it right)
$C^{1}:$
$DPhi(x,y)=begin{pmatrix}
2x-2ysinx & 2cosx \
cosx +e^{x+y} & e^{x+y} \
end{pmatrix}$
All functions are continuous, so it's $C^{1}$
$detDPhi(x,y)=begin{pmatrix}
2x-2ysinx & 2cosx \
cosx +e^{x+y} & e^{x+y} \
end{pmatrix} = (2x-2ysinx)e^{x+y} - 2cosx(cosx +e^{x+y}) not= 0$
for all $(x,y)$ especially in $(0,0)$
To find $(Phi)^{-1}$ I would do something like that:
$begin{cases}
u = x^2+2ycosx+1 \
v = sinx+e^{x+y}
end{cases}$
Find here $x,y$, check if $Phi(u,v)=(x(u,v),y(u,v))$ is continuous, calculate derivatives and put it into matrix, then check if determinant is $not=0$. That would show that this function is diffeomorphism. But how should i calculate in $(Phi_{|U})^{-1}$? Should I do following steps and then make something like this?
begin{pmatrix}
frac{d}{dx}phi_1(u,v) & frac{d}{dx}phi_2(u,v) \
frac{d}{dy}phi_1(u,v) & frac{d}{dy}phi_(u,v) \
end{pmatrix}
Edit: This matrix is wrong i think, i should use Jacobi's matrix here propably
Propably this is completly wrong, so please give me advice :(
analysis
asked Jan 27 at 10:37
VictorVictor
464
$endgroup$
add a comment |
$begingroup$
Let $Phi(x,y)=(x^2+2ycosx+1, sinx+e^{x+y})$. Proof that exists neighborhood $U$ of the point $(0,0)$ such that $Phi_{|U}$ is diffeomorphism on the image. Calculate the derivative $(Phi_{|U})^{-1})$ in $(1,1)$.
Well, im struggling with this task. Idea was:
$1-1:$
$(Phi(x,y)=Phi(t1,t2)$ for $(t1,t2)in mathbb R^{2}$
$begin{cases}
x^2+2ycosx+1 = (t1)^2+2(t2)cos(t1)+1 \
sinx+e^{x+y} = sin(t1) + e^{t1+t2}
end{cases}$
What now? ... Or maybe it's obvious, because I have to do this in $(0,0)$ point?
$Phi(x,y)=Phi(t1,t2)=(0,0) :$
$begin{cases}
0+0+1 = 0+0+1 \
0+1 = 0+1
end{cases}$
(I think this is stupid a little bit and idk if I am doing it right)
$C^{1}:$
$DPhi(x,y)=begin{pmatrix}
2x-2ysinx & 2cosx \
cosx +e^{x+y} & e^{x+y} \
end{pmatrix}$
All functions are continuous, so it's $C^{1}$
$detDPhi(x,y)=begin{pmatrix}
2x-2ysinx & 2cosx \
cosx +e^{x+y} & e^{x+y} \
end{pmatrix} = (2x-2ysinx)e^{x+y} - 2cosx(cosx +e^{x+y}) not= 0$
for all $(x,y)$ especially in $(0,0)$
To find $(Phi)^{-1}$ I would do something like that:
$begin{cases}
u = x^2+2ycosx+1 \
v = sinx+e^{x+y}
end{cases}$
Find here $x,y$, check if $Phi(u,v)=(x(u,v),y(u,v))$ is continuous, calculate derivatives and put it into matrix, then check if determinant is $not=0$. That would show that this function is diffeomorphism. But how should i calculate in $(Phi_{|U})^{-1}$? Should I do following steps and then make something like this?
begin{pmatrix}
frac{d}{dx}phi_1(u,v) & frac{d}{dx}phi_2(u,v) \
frac{d}{dy}phi_1(u,v) & frac{d}{dy}phi_(u,v) \
end{pmatrix}
Edit: This matrix is wrong i think, i should use Jacobi's matrix here propably
Propably this is completly wrong, so please give me advice :(
analysis
asked Jan 27 at 10:37
VictorVictor
464
$endgroup$
add a comment |
$begingroup$
Let $Phi(x,y)=(x^2+2ycosx+1, sinx+e^{x+y})$. Proof that exists neighborhood $U$ of the point $(0,0)$ such that $Phi_{|U}$ is diffeomorphism on the image. Calculate the derivative $(Phi_{|U})^{-1})$ in $(1,1)$.
Well, im struggling with this task. Idea was:
$1-1:$
$(Phi(x,y)=Phi(t1,t2)$ for $(t1,t2)in mathbb R^{2}$
$begin{cases}
x^2+2ycosx+1 = (t1)^2+2(t2)cos(t1)+1 \
sinx+e^{x+y} = sin(t1) + e^{t1+t2}
end{cases}$
What now? ... Or maybe it's obvious, because I have to do this in $(0,0)$ point?
$Phi(x,y)=Phi(t1,t2)=(0,0) :$
$begin{cases}
0+0+1 = 0+0+1 \
0+1 = 0+1
end{cases}$
(I think this is stupid a little bit and idk if I am doing it right)
$C^{1}:$
$DPhi(x,y)=begin{pmatrix}
2x-2ysinx & 2cosx \
cosx +e^{x+y} & e^{x+y} \
end{pmatrix}$
All functions are continuous, so it's $C^{1}$
$detDPhi(x,y)=begin{pmatrix}
2x-2ysinx & 2cosx \
cosx +e^{x+y} & e^{x+y} \
end{pmatrix} = (2x-2ysinx)e^{x+y} - 2cosx(cosx +e^{x+y}) not= 0$
for all $(x,y)$ especially in $(0,0)$
To find $(Phi)^{-1}$ I would do something like that:
$begin{cases}
u = x^2+2ycosx+1 \
v = sinx+e^{x+y}
end{cases}$
Find here $x,y$, check if $Phi(u,v)=(x(u,v),y(u,v))$ is continuous, calculate derivatives and put it into matrix, then check if determinant is $not=0$. That would show that this function is diffeomorphism. But how should i calculate in $(Phi_{|U})^{-1}$? Should I do following steps and then make something like this?
begin{pmatrix}
frac{d}{dx}phi_1(u,v) & frac{d}{dx}phi_2(u,v) \
frac{d}{dy}phi_1(u,v) & frac{d}{dy}phi_(u,v) \
end{pmatrix}
Edit: This matrix is wrong i think, i should use Jacobi's matrix here propably
Propably this is completly wrong, so please give me advice :(
analysis
asked Jan 27 at 10:37
VictorVictor
464
$endgroup$
Let $Phi(x,y)=(x^2+2ycosx+1, sinx+e^{x+y})$. Proof that exists neighborhood $U$ of the point $(0,0)$ such that $Phi_{|U}$ is diffeomorphism on the image. Calculate the derivative $(Phi_{|U})^{-1})$ in $(1,1)$.
Well, im struggling with this task. Idea was:
$1-1:$
$(Phi(x,y)=Phi(t1,t2)$ for $(t1,t2)in mathbb R^{2}$
$begin{cases}
x^2+2ycosx+1 = (t1)^2+2(t2)cos(t1)+1 \
sinx+e^{x+y} = sin(t1) + e^{t1+t2}
end{cases}$
What now? ... Or maybe it's obvious, because I have to do this in $(0,0)$ point?
$Phi(x,y)=Phi(t1,t2)=(0,0) :$
$begin{cases}
0+0+1 = 0+0+1 \
0+1 = 0+1
end{cases}$
(I think this is stupid a little bit and idk if I am doing it right)
$C^{1}:$
$DPhi(x,y)=begin{pmatrix}
2x-2ysinx & 2cosx \
cosx +e^{x+y} & e^{x+y} \
end{pmatrix}$
All functions are continuous, so it's $C^{1}$
$detDPhi(x,y)=begin{pmatrix}
2x-2ysinx & 2cosx \
cosx +e^{x+y} & e^{x+y} \
end{pmatrix} = (2x-2ysinx)e^{x+y} - 2cosx(cosx +e^{x+y}) not= 0$
for all $(x,y)$ especially in $(0,0)$
To find $(Phi)^{-1}$ I would do something like that:
$begin{cases}
u = x^2+2ycosx+1 \
v = sinx+e^{x+y}
end{cases}$
Find here $x,y$, check if $Phi(u,v)=(x(u,v),y(u,v))$ is continuous, calculate derivatives and put it into matrix, then check if determinant is $not=0$. That would show that this function is diffeomorphism. But how should i calculate in $(Phi_{|U})^{-1}$? Should I do following steps and then make something like this?
begin{pmatrix}
frac{d}{dx}phi_1(u,v) & frac{d}{dx}phi_2(u,v) \
frac{d}{dy}phi_1(u,v) & frac{d}{dy}phi_(u,v) \
end{pmatrix}
Edit: This matrix is wrong i think, i should use Jacobi's matrix here propably
Propably this is completly wrong, so please give me advice :(
analysis
analysis
asked Jan 27 at 10:37
VictorVictor
464
asked Jan 27 at 10:37
VictorVictor
464
edited Jan 27 at 10:42
Victor
asked Jan 27 at 10:37
VictorVictor
464
asked Jan 27 at 10:37
VictorVictor
464
asked Jan 27 at 10:37
VictorVictor
464
464
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Somehow, you think to complicated. Why are you using $t_1$ and $t_2$?
If you checked that $Phi$ is differentiable, you can compute the Jacobian matrix $J_{Phi}(x,y)$ of $Phi$ at $(x,y)$ and you get
$$
J_{Phi}(x,y)=begin{pmatrix}
2x-2ysin(x) & 2cos(x)\
cos(x)+e^{x+y} & e^{x+y}
end{pmatrix}
$$
and further
$$
J_{Phi}(0,0)=begin{pmatrix}0 & 2\2&1end{pmatrix}.
$$
Since $detleft(J_{Phi}(0,0)right)=-4neq 0$, there exists a neighbourhood $U$ of $(0,0)$ such that
$$
Phimid_U:Uto V:=Phi(U)
$$
is bijective and $det(J_{Phi}(x,y))neq 0$ for all $(x,y)in U$. Hence $Phi^{-1}$ is differentiable on $V$ with
$$
J_{Phi^{-1}}(Phi(x,y))=left(J_{Phi}(x,y)right)^{-1}.
$$
(This should be a theorem in your lecture)
Now, you can compute
$$
J_{Phi^{-1}}(1,1)=J_{Phi^{-1}}(Phi(0,0))=ldots
$$
answered Jan 27 at 10:58
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
"If you checked that Φ is differentiable". How to do it? I have to show that this function is 1-1? This system of equations looks bad... Yeah I've made Jacobian matrix, but in my lecture we call it "D", not "J".
$endgroup$
– Victor
Jan 27 at 11:04
$begingroup$
Now, to find $JPhi^{−1}(1,1)$, I have to find $x,y$ from $u=x^{2}+2ycos(x)+1, v=sin(x)+e^{x+y}$ equation? I am confused here. Or if u said, that $JPhi^{−1}(1,1)$ = $JPhi^{−1}(0,0)$, so the correct answer is just to find inverse matrix, so $JPhi^{−1}(1,1) = -frac{1}{4}begin{pmatrix} 1 & -2 \ -2 & 0 \ end{pmatrix}$$ ??
$endgroup$
– Victor
Jan 27 at 11:10
$begingroup$
$Phi$ is differentiable as composition of differentiable functions, so it is sufficient to mention it. And the bijectivity of $Phimid_U$ comes from a theorem, you should have in your lecture. Check it out. The theorem claims the existence of $Phi^{-1}$ but that doesn't mean, that you can get it with a nice explicit formula. But in your case, you don't need it. As the formula for the Jacobian of $Phi^{-1}$ says, you just have to compute the matrix inverse of the Jacobian of $Phi$ at the correct point.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:16
$begingroup$
"$Phi$ is differentiable as composition of differentiable function". Of course it is!...Theorem on the inverse function... Yea I got it now. Well, is my matrix inverse correct?
$endgroup$
– Victor
Jan 27 at 11:27
$begingroup$
Yes, it is correct.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:27
|
show 1 more comment
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$begingroup$
Somehow, you think to complicated. Why are you using $t_1$ and $t_2$?
If you checked that $Phi$ is differentiable, you can compute the Jacobian matrix $J_{Phi}(x,y)$ of $Phi$ at $(x,y)$ and you get
$$
J_{Phi}(x,y)=begin{pmatrix}
2x-2ysin(x) & 2cos(x)\
cos(x)+e^{x+y} & e^{x+y}
end{pmatrix}
$$
and further
$$
J_{Phi}(0,0)=begin{pmatrix}0 & 2\2&1end{pmatrix}.
$$
Since $detleft(J_{Phi}(0,0)right)=-4neq 0$, there exists a neighbourhood $U$ of $(0,0)$ such that
$$
Phimid_U:Uto V:=Phi(U)
$$
is bijective and $det(J_{Phi}(x,y))neq 0$ for all $(x,y)in U$. Hence $Phi^{-1}$ is differentiable on $V$ with
$$
J_{Phi^{-1}}(Phi(x,y))=left(J_{Phi}(x,y)right)^{-1}.
$$
(This should be a theorem in your lecture)
Now, you can compute
$$
J_{Phi^{-1}}(1,1)=J_{Phi^{-1}}(Phi(0,0))=ldots
$$
answered Jan 27 at 10:58
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
"If you checked that Φ is differentiable". How to do it? I have to show that this function is 1-1? This system of equations looks bad... Yeah I've made Jacobian matrix, but in my lecture we call it "D", not "J".
$endgroup$
– Victor
Jan 27 at 11:04
$begingroup$
Now, to find $JPhi^{−1}(1,1)$, I have to find $x,y$ from $u=x^{2}+2ycos(x)+1, v=sin(x)+e^{x+y}$ equation? I am confused here. Or if u said, that $JPhi^{−1}(1,1)$ = $JPhi^{−1}(0,0)$, so the correct answer is just to find inverse matrix, so $JPhi^{−1}(1,1) = -frac{1}{4}begin{pmatrix} 1 & -2 \ -2 & 0 \ end{pmatrix}$$ ??
$endgroup$
– Victor
Jan 27 at 11:10
$begingroup$
$Phi$ is differentiable as composition of differentiable functions, so it is sufficient to mention it. And the bijectivity of $Phimid_U$ comes from a theorem, you should have in your lecture. Check it out. The theorem claims the existence of $Phi^{-1}$ but that doesn't mean, that you can get it with a nice explicit formula. But in your case, you don't need it. As the formula for the Jacobian of $Phi^{-1}$ says, you just have to compute the matrix inverse of the Jacobian of $Phi$ at the correct point.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:16
$begingroup$
"$Phi$ is differentiable as composition of differentiable function". Of course it is!...Theorem on the inverse function... Yea I got it now. Well, is my matrix inverse correct?
$endgroup$
– Victor
Jan 27 at 11:27
$begingroup$
Yes, it is correct.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:27
|
show 1 more comment
$begingroup$
Somehow, you think to complicated. Why are you using $t_1$ and $t_2$?
If you checked that $Phi$ is differentiable, you can compute the Jacobian matrix $J_{Phi}(x,y)$ of $Phi$ at $(x,y)$ and you get
$$
J_{Phi}(x,y)=begin{pmatrix}
2x-2ysin(x) & 2cos(x)\
cos(x)+e^{x+y} & e^{x+y}
end{pmatrix}
$$
and further
$$
J_{Phi}(0,0)=begin{pmatrix}0 & 2\2&1end{pmatrix}.
$$
Since $detleft(J_{Phi}(0,0)right)=-4neq 0$, there exists a neighbourhood $U$ of $(0,0)$ such that
$$
Phimid_U:Uto V:=Phi(U)
$$
is bijective and $det(J_{Phi}(x,y))neq 0$ for all $(x,y)in U$. Hence $Phi^{-1}$ is differentiable on $V$ with
$$
J_{Phi^{-1}}(Phi(x,y))=left(J_{Phi}(x,y)right)^{-1}.
$$
(This should be a theorem in your lecture)
Now, you can compute
$$
J_{Phi^{-1}}(1,1)=J_{Phi^{-1}}(Phi(0,0))=ldots
$$
answered Jan 27 at 10:58
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
"If you checked that Φ is differentiable". How to do it? I have to show that this function is 1-1? This system of equations looks bad... Yeah I've made Jacobian matrix, but in my lecture we call it "D", not "J".
$endgroup$
– Victor
Jan 27 at 11:04
$begingroup$
Now, to find $JPhi^{−1}(1,1)$, I have to find $x,y$ from $u=x^{2}+2ycos(x)+1, v=sin(x)+e^{x+y}$ equation? I am confused here. Or if u said, that $JPhi^{−1}(1,1)$ = $JPhi^{−1}(0,0)$, so the correct answer is just to find inverse matrix, so $JPhi^{−1}(1,1) = -frac{1}{4}begin{pmatrix} 1 & -2 \ -2 & 0 \ end{pmatrix}$$ ??
$endgroup$
– Victor
Jan 27 at 11:10
$begingroup$
$Phi$ is differentiable as composition of differentiable functions, so it is sufficient to mention it. And the bijectivity of $Phimid_U$ comes from a theorem, you should have in your lecture. Check it out. The theorem claims the existence of $Phi^{-1}$ but that doesn't mean, that you can get it with a nice explicit formula. But in your case, you don't need it. As the formula for the Jacobian of $Phi^{-1}$ says, you just have to compute the matrix inverse of the Jacobian of $Phi$ at the correct point.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:16
$begingroup$
"$Phi$ is differentiable as composition of differentiable function". Of course it is!...Theorem on the inverse function... Yea I got it now. Well, is my matrix inverse correct?
$endgroup$
– Victor
Jan 27 at 11:27
$begingroup$
Yes, it is correct.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:27
|
show 1 more comment
$begingroup$
Somehow, you think to complicated. Why are you using $t_1$ and $t_2$?
If you checked that $Phi$ is differentiable, you can compute the Jacobian matrix $J_{Phi}(x,y)$ of $Phi$ at $(x,y)$ and you get
$$
J_{Phi}(x,y)=begin{pmatrix}
2x-2ysin(x) & 2cos(x)\
cos(x)+e^{x+y} & e^{x+y}
end{pmatrix}
$$
and further
$$
J_{Phi}(0,0)=begin{pmatrix}0 & 2\2&1end{pmatrix}.
$$
Since $detleft(J_{Phi}(0,0)right)=-4neq 0$, there exists a neighbourhood $U$ of $(0,0)$ such that
$$
Phimid_U:Uto V:=Phi(U)
$$
is bijective and $det(J_{Phi}(x,y))neq 0$ for all $(x,y)in U$. Hence $Phi^{-1}$ is differentiable on $V$ with
$$
J_{Phi^{-1}}(Phi(x,y))=left(J_{Phi}(x,y)right)^{-1}.
$$
(This should be a theorem in your lecture)
Now, you can compute
$$
J_{Phi^{-1}}(1,1)=J_{Phi^{-1}}(Phi(0,0))=ldots
$$
answered Jan 27 at 10:58
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
Somehow, you think to complicated. Why are you using $t_1$ and $t_2$?
If you checked that $Phi$ is differentiable, you can compute the Jacobian matrix $J_{Phi}(x,y)$ of $Phi$ at $(x,y)$ and you get
$$
J_{Phi}(x,y)=begin{pmatrix}
2x-2ysin(x) & 2cos(x)\
cos(x)+e^{x+y} & e^{x+y}
end{pmatrix}
$$
and further
$$
J_{Phi}(0,0)=begin{pmatrix}0 & 2\2&1end{pmatrix}.
$$
Since $detleft(J_{Phi}(0,0)right)=-4neq 0$, there exists a neighbourhood $U$ of $(0,0)$ such that
$$
Phimid_U:Uto V:=Phi(U)
$$
is bijective and $det(J_{Phi}(x,y))neq 0$ for all $(x,y)in U$. Hence $Phi^{-1}$ is differentiable on $V$ with
$$
J_{Phi^{-1}}(Phi(x,y))=left(J_{Phi}(x,y)right)^{-1}.
$$
(This should be a theorem in your lecture)
Now, you can compute
$$
J_{Phi^{-1}}(1,1)=J_{Phi^{-1}}(Phi(0,0))=ldots
$$
answered Jan 27 at 10:58
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 27 at 10:58
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 27 at 10:58
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 27 at 10:58
Mundron SchmidtMundron Schmidt
7,5042729
7,5042729
$begingroup$
"If you checked that Φ is differentiable". How to do it? I have to show that this function is 1-1? This system of equations looks bad... Yeah I've made Jacobian matrix, but in my lecture we call it "D", not "J".
$endgroup$
– Victor
Jan 27 at 11:04
$begingroup$
Now, to find $JPhi^{−1}(1,1)$, I have to find $x,y$ from $u=x^{2}+2ycos(x)+1, v=sin(x)+e^{x+y}$ equation? I am confused here. Or if u said, that $JPhi^{−1}(1,1)$ = $JPhi^{−1}(0,0)$, so the correct answer is just to find inverse matrix, so $JPhi^{−1}(1,1) = -frac{1}{4}begin{pmatrix} 1 & -2 \ -2 & 0 \ end{pmatrix}$$ ??
$endgroup$
– Victor
Jan 27 at 11:10
$begingroup$
$Phi$ is differentiable as composition of differentiable functions, so it is sufficient to mention it. And the bijectivity of $Phimid_U$ comes from a theorem, you should have in your lecture. Check it out. The theorem claims the existence of $Phi^{-1}$ but that doesn't mean, that you can get it with a nice explicit formula. But in your case, you don't need it. As the formula for the Jacobian of $Phi^{-1}$ says, you just have to compute the matrix inverse of the Jacobian of $Phi$ at the correct point.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:16
$begingroup$
"$Phi$ is differentiable as composition of differentiable function". Of course it is!...Theorem on the inverse function... Yea I got it now. Well, is my matrix inverse correct?
$endgroup$
– Victor
Jan 27 at 11:27
$begingroup$
Yes, it is correct.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:27
|
show 1 more comment
$begingroup$
"If you checked that Φ is differentiable". How to do it? I have to show that this function is 1-1? This system of equations looks bad... Yeah I've made Jacobian matrix, but in my lecture we call it "D", not "J".
$endgroup$
– Victor
Jan 27 at 11:04
$begingroup$
Now, to find $JPhi^{−1}(1,1)$, I have to find $x,y$ from $u=x^{2}+2ycos(x)+1, v=sin(x)+e^{x+y}$ equation? I am confused here. Or if u said, that $JPhi^{−1}(1,1)$ = $JPhi^{−1}(0,0)$, so the correct answer is just to find inverse matrix, so $JPhi^{−1}(1,1) = -frac{1}{4}begin{pmatrix} 1 & -2 \ -2 & 0 \ end{pmatrix}$$ ??
$endgroup$
– Victor
Jan 27 at 11:10
$begingroup$
$Phi$ is differentiable as composition of differentiable functions, so it is sufficient to mention it. And the bijectivity of $Phimid_U$ comes from a theorem, you should have in your lecture. Check it out. The theorem claims the existence of $Phi^{-1}$ but that doesn't mean, that you can get it with a nice explicit formula. But in your case, you don't need it. As the formula for the Jacobian of $Phi^{-1}$ says, you just have to compute the matrix inverse of the Jacobian of $Phi$ at the correct point.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:16
$begingroup$
"$Phi$ is differentiable as composition of differentiable function". Of course it is!...Theorem on the inverse function... Yea I got it now. Well, is my matrix inverse correct?
$endgroup$
– Victor
Jan 27 at 11:27
$begingroup$
Yes, it is correct.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:27
$begingroup$
"If you checked that Φ is differentiable". How to do it? I have to show that this function is 1-1? This system of equations looks bad... Yeah I've made Jacobian matrix, but in my lecture we call it "D", not "J".
$endgroup$
– Victor
Jan 27 at 11:04
$begingroup$
"If you checked that Φ is differentiable". How to do it? I have to show that this function is 1-1? This system of equations looks bad... Yeah I've made Jacobian matrix, but in my lecture we call it "D", not "J".
$endgroup$
– Victor
Jan 27 at 11:04
$begingroup$
Now, to find $JPhi^{−1}(1,1)$, I have to find $x,y$ from $u=x^{2}+2ycos(x)+1, v=sin(x)+e^{x+y}$ equation? I am confused here. Or if u said, that $JPhi^{−1}(1,1)$ = $JPhi^{−1}(0,0)$, so the correct answer is just to find inverse matrix, so $JPhi^{−1}(1,1) = -frac{1}{4}begin{pmatrix} 1 & -2 \ -2 & 0 \ end{pmatrix}$$ ??
$endgroup$
– Victor
Jan 27 at 11:10
$begingroup$
Now, to find $JPhi^{−1}(1,1)$, I have to find $x,y$ from $u=x^{2}+2ycos(x)+1, v=sin(x)+e^{x+y}$ equation? I am confused here. Or if u said, that $JPhi^{−1}(1,1)$ = $JPhi^{−1}(0,0)$, so the correct answer is just to find inverse matrix, so $JPhi^{−1}(1,1) = -frac{1}{4}begin{pmatrix} 1 & -2 \ -2 & 0 \ end{pmatrix}$$ ??
$endgroup$
– Victor
Jan 27 at 11:10
$begingroup$
$Phi$ is differentiable as composition of differentiable functions, so it is sufficient to mention it. And the bijectivity of $Phimid_U$ comes from a theorem, you should have in your lecture. Check it out. The theorem claims the existence of $Phi^{-1}$ but that doesn't mean, that you can get it with a nice explicit formula. But in your case, you don't need it. As the formula for the Jacobian of $Phi^{-1}$ says, you just have to compute the matrix inverse of the Jacobian of $Phi$ at the correct point.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:16
$begingroup$
$Phi$ is differentiable as composition of differentiable functions, so it is sufficient to mention it. And the bijectivity of $Phimid_U$ comes from a theorem, you should have in your lecture. Check it out. The theorem claims the existence of $Phi^{-1}$ but that doesn't mean, that you can get it with a nice explicit formula. But in your case, you don't need it. As the formula for the Jacobian of $Phi^{-1}$ says, you just have to compute the matrix inverse of the Jacobian of $Phi$ at the correct point.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:16
$begingroup$
"$Phi$ is differentiable as composition of differentiable function". Of course it is!...Theorem on the inverse function... Yea I got it now. Well, is my matrix inverse correct?
$endgroup$
– Victor
Jan 27 at 11:27
$begingroup$
"$Phi$ is differentiable as composition of differentiable function". Of course it is!...Theorem on the inverse function... Yea I got it now. Well, is my matrix inverse correct?
$endgroup$
– Victor
Jan 27 at 11:27
$begingroup$
Yes, it is correct.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:27
$begingroup$
Yes, it is correct.
$endgroup$
– Mundron Schmidt
Jan 27 at 11:27
|
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