Mixing
Change of variables and the partial derivative
$begingroup$
From time to time, I suddenly get confused with a change of variables in a partial derivative.
Here, I am trying to perform a change of variables $(x,t) mapsto (xi, eta)$ where
$$xi = t qquad qquad text{and} qquad qquad eta = x+t$$
The question is, how to compute $$frac{partial u}{partial t}$$ in the new coordinate system?
Intuitively, since $xi = t$ (or rather $t=xi$), we should have
$$frac{partial u}{partial t} = frac{partial u}{partial xi}$$
However, applying the chain rule for partial derivatives, we instead get
$$frac{partial u}{partial t} = frac{partial u}{partial xi}frac{partial xi}{partial t} + frac{partial u}{partial eta}frac{partial eta}{partial t} = frac{partial u}{partial xi}(1) + frac{partial u}{partial eta}(1) = frac{partial u}{partial xi} + frac{partial u}{partial eta}$$
So which one is correct?
multivariable-calculus partial-derivative change-of-variable
edited Jan 15 at 18:36
Git Gud
28.9k1050100
asked Jan 15 at 13:54
glowstonetreesglowstonetrees
2,368418
$endgroup$
add a comment |
$begingroup$
From time to time, I suddenly get confused with a change of variables in a partial derivative.
Here, I am trying to perform a change of variables $(x,t) mapsto (xi, eta)$ where
$$xi = t qquad qquad text{and} qquad qquad eta = x+t$$
The question is, how to compute $$frac{partial u}{partial t}$$ in the new coordinate system?
Intuitively, since $xi = t$ (or rather $t=xi$), we should have
$$frac{partial u}{partial t} = frac{partial u}{partial xi}$$
However, applying the chain rule for partial derivatives, we instead get
$$frac{partial u}{partial t} = frac{partial u}{partial xi}frac{partial xi}{partial t} + frac{partial u}{partial eta}frac{partial eta}{partial t} = frac{partial u}{partial xi}(1) + frac{partial u}{partial eta}(1) = frac{partial u}{partial xi} + frac{partial u}{partial eta}$$
So which one is correct?
multivariable-calculus partial-derivative change-of-variable
edited Jan 15 at 18:36
Git Gud
28.9k1050100
asked Jan 15 at 13:54
glowstonetreesglowstonetrees
2,368418
$endgroup$
add a comment |
$begingroup$
From time to time, I suddenly get confused with a change of variables in a partial derivative.
Here, I am trying to perform a change of variables $(x,t) mapsto (xi, eta)$ where
$$xi = t qquad qquad text{and} qquad qquad eta = x+t$$
The question is, how to compute $$frac{partial u}{partial t}$$ in the new coordinate system?
Intuitively, since $xi = t$ (or rather $t=xi$), we should have
$$frac{partial u}{partial t} = frac{partial u}{partial xi}$$
However, applying the chain rule for partial derivatives, we instead get
$$frac{partial u}{partial t} = frac{partial u}{partial xi}frac{partial xi}{partial t} + frac{partial u}{partial eta}frac{partial eta}{partial t} = frac{partial u}{partial xi}(1) + frac{partial u}{partial eta}(1) = frac{partial u}{partial xi} + frac{partial u}{partial eta}$$
So which one is correct?
multivariable-calculus partial-derivative change-of-variable
edited Jan 15 at 18:36
Git Gud
28.9k1050100
asked Jan 15 at 13:54
glowstonetreesglowstonetrees
2,368418
$endgroup$
From time to time, I suddenly get confused with a change of variables in a partial derivative.
Here, I am trying to perform a change of variables $(x,t) mapsto (xi, eta)$ where
$$xi = t qquad qquad text{and} qquad qquad eta = x+t$$
The question is, how to compute $$frac{partial u}{partial t}$$ in the new coordinate system?
Intuitively, since $xi = t$ (or rather $t=xi$), we should have
$$frac{partial u}{partial t} = frac{partial u}{partial xi}$$
However, applying the chain rule for partial derivatives, we instead get
$$frac{partial u}{partial t} = frac{partial u}{partial xi}frac{partial xi}{partial t} + frac{partial u}{partial eta}frac{partial eta}{partial t} = frac{partial u}{partial xi}(1) + frac{partial u}{partial eta}(1) = frac{partial u}{partial xi} + frac{partial u}{partial eta}$$
So which one is correct?
multivariable-calculus partial-derivative change-of-variable
multivariable-calculus partial-derivative change-of-variable
edited Jan 15 at 18:36
Git Gud
28.9k1050100
asked Jan 15 at 13:54
glowstonetreesglowstonetrees
2,368418
edited Jan 15 at 18:36
Git Gud
28.9k1050100
asked Jan 15 at 13:54
glowstonetreesglowstonetrees
2,368418
edited Jan 15 at 18:36
Git Gud
28.9k1050100
edited Jan 15 at 18:36
Git Gud
28.9k1050100
edited Jan 15 at 18:36
Git Gud
28.9k1050100
28.9k1050100
asked Jan 15 at 13:54
glowstonetreesglowstonetrees
2,368418
asked Jan 15 at 13:54
glowstonetreesglowstonetrees
2,368418
asked Jan 15 at 13:54
glowstonetreesglowstonetrees
2,368418
2,368418
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The second one is correct. Why is the other one wrong? Well when you write $$frac{partial u}{partial xi}$$what you really are saying is $$frac{partial u}{partial xi}{huge|}_eta$$, i.e. taking the derivative with respect to $xi$, but keeping $eta$ fixed. If we do this, we are restricting ourselves to the path $x=-t$, where $t$ varies. But we wanted to represent $$frac{partial u}{partial t}{huge|}_x$$with a fixed $x$. So this expression fails to express the same quantity. Using the chain rule works to ensure this error does not happen.
answered Jan 15 at 14:07
John DoeJohn Doe
11.1k11238
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add a comment |
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Consider $u(x,t)=x^2+t^2$ and $xi=t, eta=x+t$.
Then: $u(xi(x,t),eta(x,t))=(eta-xi)^2+xi^2$. We get:
$$u_t=2(eta-xi)cdot (eta_t-xi_t)+2xicdot xi_t=\
2(x+t-t)cdot (1-1)+2tcdot 1=\
2t,$$
which is true:
$$u_t=(x^2+t^2)_t=2t.$$
answered Jan 15 at 16:33
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
$begingroup$
What follows is not only the correct way to handle this issue (most of the time people resort to what are to me confusing notational abuses and artifacts from a time when it was hard to be a bit more rigorous), it's a sketch of the answer that I'd like to be given.
To simplify, in what's below I will only consider the whole spaces as domains, but it can be generalized to any open subset of these spaces.
Theorem. Let $m, n$ and $p$ be natural numbers, and $Fcolon mathbb R^nto mathbb R^m$ and $Gcolonmathbb R^mto mathbb R^p$ differentiable functions. If $H=Gcirc F$, then $H$ is differentiable and for all $X$ in $mathbb R^n$ it holds that $$(DH)_X=(DG)_{F(X)}(DF)_X tag 1$$
Considering that $F, G$ and $H$ can be written as
$$
F=(f_1, ldots, f_m),\
G=(g_1, ldots, g_p),\
H=(h_1, ldots , h_p)
$$
where:
$f_1, ldots, f_m$ are real functions whose domain is $mathbb R^n$ and whose partial derivatives all exist,
$g_1, ldots, g_p$ are real functions whose domain is $mathbb R^m$ and whose partial derivatives all exist,
$h_1, ldots, h_p$ are real functions whose domain is $mathbb R^n$ and whose partial derivatives all exist,
$(1)$ can be rewritten as
$$
begin{bmatrix}
partial _1h_1 & ldots & partial_nh_1\
vdots & ddots & vdots\
partial_1h_p & ldots & partial_nh_p
end{bmatrix}_X
=
begin{bmatrix}
partial _1g_1 & ldots & partial_mg_1\
vdots & ddots & vdots\
partial_1g_p & ldots & partial_mg_p
end{bmatrix}_{F(X)}
begin{bmatrix}
partial _1f_1 & ldots & partial_nf_1\
vdots & ddots & vdots\
partial_1f_m & ldots & partial_nf_m
end{bmatrix}_X
tag{2}
$$
Armed with this formulation of this standard theorem, all that's left is to look at the problem through this lens.
As I understand it you have a differentiable function $ucolon mathbb R^2to mathbb R$, a function $varphi colon mathbb R^2tomathbb R^2, (x,t)mapsto (t, x+t)$ (which is also differentiable) and you're asked to find $partial _2(ucirc varphi)$.
Now it's easy. In the notation above we have $m=2=n$, $p=1$, $G=u$, $F=varphi$, $f_1colon mathbb R^2to mathbb R, (x,t)mapsto t$ and $f_2colon mathbb R^2to mathbb R, (x,t)mapsto x+t$. Set $H=ucirc varphi$ and conclude that for all $(x,t)$ in $mathbb R^2$ it holds that
$$
begin{align}
begin{bmatrix}
partial _1h_1 & partial_nh_1
end{bmatrix}_{(x,t)}
&=
begin{bmatrix}
partial _1u & partial_2u
end{bmatrix}_{F(x,t)}
begin{bmatrix}
partial _1f_1 & partial_2f_1\
partial_1f_2 & partial_2f_2
end{bmatrix}_{(x,t)}\
&=
begin{bmatrix}
left(partial _1uright)(t,x+t) & left(partial_2uright)(t,x+t)
end{bmatrix}
begin{bmatrix}
0& 1\
1 & 1
end{bmatrix}\
&=
begin{bmatrix}
left(partial _2uright)(t,x+t) & left(partial _1uright)(t,x+t)+left(partial_2uright)(t,x+t)
end{bmatrix}_.
end{align}
$$
answered Jan 15 at 18:35
Git GudGit Gud
28.9k1050100
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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active
oldest
votes
$begingroup$
The second one is correct. Why is the other one wrong? Well when you write $$frac{partial u}{partial xi}$$what you really are saying is $$frac{partial u}{partial xi}{huge|}_eta$$, i.e. taking the derivative with respect to $xi$, but keeping $eta$ fixed. If we do this, we are restricting ourselves to the path $x=-t$, where $t$ varies. But we wanted to represent $$frac{partial u}{partial t}{huge|}_x$$with a fixed $x$. So this expression fails to express the same quantity. Using the chain rule works to ensure this error does not happen.
answered Jan 15 at 14:07
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
The second one is correct. Why is the other one wrong? Well when you write $$frac{partial u}{partial xi}$$what you really are saying is $$frac{partial u}{partial xi}{huge|}_eta$$, i.e. taking the derivative with respect to $xi$, but keeping $eta$ fixed. If we do this, we are restricting ourselves to the path $x=-t$, where $t$ varies. But we wanted to represent $$frac{partial u}{partial t}{huge|}_x$$with a fixed $x$. So this expression fails to express the same quantity. Using the chain rule works to ensure this error does not happen.
answered Jan 15 at 14:07
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
The second one is correct. Why is the other one wrong? Well when you write $$frac{partial u}{partial xi}$$what you really are saying is $$frac{partial u}{partial xi}{huge|}_eta$$, i.e. taking the derivative with respect to $xi$, but keeping $eta$ fixed. If we do this, we are restricting ourselves to the path $x=-t$, where $t$ varies. But we wanted to represent $$frac{partial u}{partial t}{huge|}_x$$with a fixed $x$. So this expression fails to express the same quantity. Using the chain rule works to ensure this error does not happen.
answered Jan 15 at 14:07
John DoeJohn Doe
11.1k11238
$endgroup$
The second one is correct. Why is the other one wrong? Well when you write $$frac{partial u}{partial xi}$$what you really are saying is $$frac{partial u}{partial xi}{huge|}_eta$$, i.e. taking the derivative with respect to $xi$, but keeping $eta$ fixed. If we do this, we are restricting ourselves to the path $x=-t$, where $t$ varies. But we wanted to represent $$frac{partial u}{partial t}{huge|}_x$$with a fixed $x$. So this expression fails to express the same quantity. Using the chain rule works to ensure this error does not happen.
answered Jan 15 at 14:07
John DoeJohn Doe
11.1k11238
answered Jan 15 at 14:07
John DoeJohn Doe
11.1k11238
answered Jan 15 at 14:07
John DoeJohn Doe
11.1k11238
answered Jan 15 at 14:07
John DoeJohn Doe
11.1k11238
11.1k11238
add a comment |
add a comment |
$begingroup$
Consider $u(x,t)=x^2+t^2$ and $xi=t, eta=x+t$.
Then: $u(xi(x,t),eta(x,t))=(eta-xi)^2+xi^2$. We get:
$$u_t=2(eta-xi)cdot (eta_t-xi_t)+2xicdot xi_t=\
2(x+t-t)cdot (1-1)+2tcdot 1=\
2t,$$
which is true:
$$u_t=(x^2+t^2)_t=2t.$$
answered Jan 15 at 16:33
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
$begingroup$
Consider $u(x,t)=x^2+t^2$ and $xi=t, eta=x+t$.
Then: $u(xi(x,t),eta(x,t))=(eta-xi)^2+xi^2$. We get:
$$u_t=2(eta-xi)cdot (eta_t-xi_t)+2xicdot xi_t=\
2(x+t-t)cdot (1-1)+2tcdot 1=\
2t,$$
which is true:
$$u_t=(x^2+t^2)_t=2t.$$
answered Jan 15 at 16:33
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
$begingroup$
Consider $u(x,t)=x^2+t^2$ and $xi=t, eta=x+t$.
Then: $u(xi(x,t),eta(x,t))=(eta-xi)^2+xi^2$. We get:
$$u_t=2(eta-xi)cdot (eta_t-xi_t)+2xicdot xi_t=\
2(x+t-t)cdot (1-1)+2tcdot 1=\
2t,$$
which is true:
$$u_t=(x^2+t^2)_t=2t.$$
answered Jan 15 at 16:33
farruhotafarruhota
20.4k2739
$endgroup$
Consider $u(x,t)=x^2+t^2$ and $xi=t, eta=x+t$.
Then: $u(xi(x,t),eta(x,t))=(eta-xi)^2+xi^2$. We get:
$$u_t=2(eta-xi)cdot (eta_t-xi_t)+2xicdot xi_t=\
2(x+t-t)cdot (1-1)+2tcdot 1=\
2t,$$
which is true:
$$u_t=(x^2+t^2)_t=2t.$$
answered Jan 15 at 16:33
farruhotafarruhota
20.4k2739
answered Jan 15 at 16:33
farruhotafarruhota
20.4k2739
answered Jan 15 at 16:33
farruhotafarruhota
20.4k2739
answered Jan 15 at 16:33
farruhotafarruhota
20.4k2739
20.4k2739
add a comment |
add a comment |
$begingroup$
What follows is not only the correct way to handle this issue (most of the time people resort to what are to me confusing notational abuses and artifacts from a time when it was hard to be a bit more rigorous), it's a sketch of the answer that I'd like to be given.
To simplify, in what's below I will only consider the whole spaces as domains, but it can be generalized to any open subset of these spaces.
Theorem. Let $m, n$ and $p$ be natural numbers, and $Fcolon mathbb R^nto mathbb R^m$ and $Gcolonmathbb R^mto mathbb R^p$ differentiable functions. If $H=Gcirc F$, then $H$ is differentiable and for all $X$ in $mathbb R^n$ it holds that $$(DH)_X=(DG)_{F(X)}(DF)_X tag 1$$
Considering that $F, G$ and $H$ can be written as
$$
F=(f_1, ldots, f_m),\
G=(g_1, ldots, g_p),\
H=(h_1, ldots , h_p)
$$
where:
$f_1, ldots, f_m$ are real functions whose domain is $mathbb R^n$ and whose partial derivatives all exist,
$g_1, ldots, g_p$ are real functions whose domain is $mathbb R^m$ and whose partial derivatives all exist,
$h_1, ldots, h_p$ are real functions whose domain is $mathbb R^n$ and whose partial derivatives all exist,
$(1)$ can be rewritten as
$$
begin{bmatrix}
partial _1h_1 & ldots & partial_nh_1\
vdots & ddots & vdots\
partial_1h_p & ldots & partial_nh_p
end{bmatrix}_X
=
begin{bmatrix}
partial _1g_1 & ldots & partial_mg_1\
vdots & ddots & vdots\
partial_1g_p & ldots & partial_mg_p
end{bmatrix}_{F(X)}
begin{bmatrix}
partial _1f_1 & ldots & partial_nf_1\
vdots & ddots & vdots\
partial_1f_m & ldots & partial_nf_m
end{bmatrix}_X
tag{2}
$$
Armed with this formulation of this standard theorem, all that's left is to look at the problem through this lens.
As I understand it you have a differentiable function $ucolon mathbb R^2to mathbb R$, a function $varphi colon mathbb R^2tomathbb R^2, (x,t)mapsto (t, x+t)$ (which is also differentiable) and you're asked to find $partial _2(ucirc varphi)$.
Now it's easy. In the notation above we have $m=2=n$, $p=1$, $G=u$, $F=varphi$, $f_1colon mathbb R^2to mathbb R, (x,t)mapsto t$ and $f_2colon mathbb R^2to mathbb R, (x,t)mapsto x+t$. Set $H=ucirc varphi$ and conclude that for all $(x,t)$ in $mathbb R^2$ it holds that
$$
begin{align}
begin{bmatrix}
partial _1h_1 & partial_nh_1
end{bmatrix}_{(x,t)}
&=
begin{bmatrix}
partial _1u & partial_2u
end{bmatrix}_{F(x,t)}
begin{bmatrix}
partial _1f_1 & partial_2f_1\
partial_1f_2 & partial_2f_2
end{bmatrix}_{(x,t)}\
&=
begin{bmatrix}
left(partial _1uright)(t,x+t) & left(partial_2uright)(t,x+t)
end{bmatrix}
begin{bmatrix}
0& 1\
1 & 1
end{bmatrix}\
&=
begin{bmatrix}
left(partial _2uright)(t,x+t) & left(partial _1uright)(t,x+t)+left(partial_2uright)(t,x+t)
end{bmatrix}_.
end{align}
$$
answered Jan 15 at 18:35
Git GudGit Gud
28.9k1050100
$endgroup$
add a comment |
$begingroup$
What follows is not only the correct way to handle this issue (most of the time people resort to what are to me confusing notational abuses and artifacts from a time when it was hard to be a bit more rigorous), it's a sketch of the answer that I'd like to be given.
To simplify, in what's below I will only consider the whole spaces as domains, but it can be generalized to any open subset of these spaces.
Theorem. Let $m, n$ and $p$ be natural numbers, and $Fcolon mathbb R^nto mathbb R^m$ and $Gcolonmathbb R^mto mathbb R^p$ differentiable functions. If $H=Gcirc F$, then $H$ is differentiable and for all $X$ in $mathbb R^n$ it holds that $$(DH)_X=(DG)_{F(X)}(DF)_X tag 1$$
Considering that $F, G$ and $H$ can be written as
$$
F=(f_1, ldots, f_m),\
G=(g_1, ldots, g_p),\
H=(h_1, ldots , h_p)
$$
where:
$f_1, ldots, f_m$ are real functions whose domain is $mathbb R^n$ and whose partial derivatives all exist,
$g_1, ldots, g_p$ are real functions whose domain is $mathbb R^m$ and whose partial derivatives all exist,
$h_1, ldots, h_p$ are real functions whose domain is $mathbb R^n$ and whose partial derivatives all exist,
$(1)$ can be rewritten as
$$
begin{bmatrix}
partial _1h_1 & ldots & partial_nh_1\
vdots & ddots & vdots\
partial_1h_p & ldots & partial_nh_p
end{bmatrix}_X
=
begin{bmatrix}
partial _1g_1 & ldots & partial_mg_1\
vdots & ddots & vdots\
partial_1g_p & ldots & partial_mg_p
end{bmatrix}_{F(X)}
begin{bmatrix}
partial _1f_1 & ldots & partial_nf_1\
vdots & ddots & vdots\
partial_1f_m & ldots & partial_nf_m
end{bmatrix}_X
tag{2}
$$
Armed with this formulation of this standard theorem, all that's left is to look at the problem through this lens.
As I understand it you have a differentiable function $ucolon mathbb R^2to mathbb R$, a function $varphi colon mathbb R^2tomathbb R^2, (x,t)mapsto (t, x+t)$ (which is also differentiable) and you're asked to find $partial _2(ucirc varphi)$.
Now it's easy. In the notation above we have $m=2=n$, $p=1$, $G=u$, $F=varphi$, $f_1colon mathbb R^2to mathbb R, (x,t)mapsto t$ and $f_2colon mathbb R^2to mathbb R, (x,t)mapsto x+t$. Set $H=ucirc varphi$ and conclude that for all $(x,t)$ in $mathbb R^2$ it holds that
$$
begin{align}
begin{bmatrix}
partial _1h_1 & partial_nh_1
end{bmatrix}_{(x,t)}
&=
begin{bmatrix}
partial _1u & partial_2u
end{bmatrix}_{F(x,t)}
begin{bmatrix}
partial _1f_1 & partial_2f_1\
partial_1f_2 & partial_2f_2
end{bmatrix}_{(x,t)}\
&=
begin{bmatrix}
left(partial _1uright)(t,x+t) & left(partial_2uright)(t,x+t)
end{bmatrix}
begin{bmatrix}
0& 1\
1 & 1
end{bmatrix}\
&=
begin{bmatrix}
left(partial _2uright)(t,x+t) & left(partial _1uright)(t,x+t)+left(partial_2uright)(t,x+t)
end{bmatrix}_.
end{align}
$$
answered Jan 15 at 18:35
Git GudGit Gud
28.9k1050100
$endgroup$
add a comment |
$begingroup$
What follows is not only the correct way to handle this issue (most of the time people resort to what are to me confusing notational abuses and artifacts from a time when it was hard to be a bit more rigorous), it's a sketch of the answer that I'd like to be given.
To simplify, in what's below I will only consider the whole spaces as domains, but it can be generalized to any open subset of these spaces.
Theorem. Let $m, n$ and $p$ be natural numbers, and $Fcolon mathbb R^nto mathbb R^m$ and $Gcolonmathbb R^mto mathbb R^p$ differentiable functions. If $H=Gcirc F$, then $H$ is differentiable and for all $X$ in $mathbb R^n$ it holds that $$(DH)_X=(DG)_{F(X)}(DF)_X tag 1$$
Considering that $F, G$ and $H$ can be written as
$$
F=(f_1, ldots, f_m),\
G=(g_1, ldots, g_p),\
H=(h_1, ldots , h_p)
$$
where:
$f_1, ldots, f_m$ are real functions whose domain is $mathbb R^n$ and whose partial derivatives all exist,
$g_1, ldots, g_p$ are real functions whose domain is $mathbb R^m$ and whose partial derivatives all exist,
$h_1, ldots, h_p$ are real functions whose domain is $mathbb R^n$ and whose partial derivatives all exist,
$(1)$ can be rewritten as
$$
begin{bmatrix}
partial _1h_1 & ldots & partial_nh_1\
vdots & ddots & vdots\
partial_1h_p & ldots & partial_nh_p
end{bmatrix}_X
=
begin{bmatrix}
partial _1g_1 & ldots & partial_mg_1\
vdots & ddots & vdots\
partial_1g_p & ldots & partial_mg_p
end{bmatrix}_{F(X)}
begin{bmatrix}
partial _1f_1 & ldots & partial_nf_1\
vdots & ddots & vdots\
partial_1f_m & ldots & partial_nf_m
end{bmatrix}_X
tag{2}
$$
Armed with this formulation of this standard theorem, all that's left is to look at the problem through this lens.
As I understand it you have a differentiable function $ucolon mathbb R^2to mathbb R$, a function $varphi colon mathbb R^2tomathbb R^2, (x,t)mapsto (t, x+t)$ (which is also differentiable) and you're asked to find $partial _2(ucirc varphi)$.
Now it's easy. In the notation above we have $m=2=n$, $p=1$, $G=u$, $F=varphi$, $f_1colon mathbb R^2to mathbb R, (x,t)mapsto t$ and $f_2colon mathbb R^2to mathbb R, (x,t)mapsto x+t$. Set $H=ucirc varphi$ and conclude that for all $(x,t)$ in $mathbb R^2$ it holds that
$$
begin{align}
begin{bmatrix}
partial _1h_1 & partial_nh_1
end{bmatrix}_{(x,t)}
&=
begin{bmatrix}
partial _1u & partial_2u
end{bmatrix}_{F(x,t)}
begin{bmatrix}
partial _1f_1 & partial_2f_1\
partial_1f_2 & partial_2f_2
end{bmatrix}_{(x,t)}\
&=
begin{bmatrix}
left(partial _1uright)(t,x+t) & left(partial_2uright)(t,x+t)
end{bmatrix}
begin{bmatrix}
0& 1\
1 & 1
end{bmatrix}\
&=
begin{bmatrix}
left(partial _2uright)(t,x+t) & left(partial _1uright)(t,x+t)+left(partial_2uright)(t,x+t)
end{bmatrix}_.
end{align}
$$
answered Jan 15 at 18:35
Git GudGit Gud
28.9k1050100
$endgroup$
What follows is not only the correct way to handle this issue (most of the time people resort to what are to me confusing notational abuses and artifacts from a time when it was hard to be a bit more rigorous), it's a sketch of the answer that I'd like to be given.
To simplify, in what's below I will only consider the whole spaces as domains, but it can be generalized to any open subset of these spaces.
Theorem. Let $m, n$ and $p$ be natural numbers, and $Fcolon mathbb R^nto mathbb R^m$ and $Gcolonmathbb R^mto mathbb R^p$ differentiable functions. If $H=Gcirc F$, then $H$ is differentiable and for all $X$ in $mathbb R^n$ it holds that $$(DH)_X=(DG)_{F(X)}(DF)_X tag 1$$
Considering that $F, G$ and $H$ can be written as
$$
F=(f_1, ldots, f_m),\
G=(g_1, ldots, g_p),\
H=(h_1, ldots , h_p)
$$
where:
$f_1, ldots, f_m$ are real functions whose domain is $mathbb R^n$ and whose partial derivatives all exist,
$g_1, ldots, g_p$ are real functions whose domain is $mathbb R^m$ and whose partial derivatives all exist,
$h_1, ldots, h_p$ are real functions whose domain is $mathbb R^n$ and whose partial derivatives all exist,
$(1)$ can be rewritten as
$$
begin{bmatrix}
partial _1h_1 & ldots & partial_nh_1\
vdots & ddots & vdots\
partial_1h_p & ldots & partial_nh_p
end{bmatrix}_X
=
begin{bmatrix}
partial _1g_1 & ldots & partial_mg_1\
vdots & ddots & vdots\
partial_1g_p & ldots & partial_mg_p
end{bmatrix}_{F(X)}
begin{bmatrix}
partial _1f_1 & ldots & partial_nf_1\
vdots & ddots & vdots\
partial_1f_m & ldots & partial_nf_m
end{bmatrix}_X
tag{2}
$$
Armed with this formulation of this standard theorem, all that's left is to look at the problem through this lens.
As I understand it you have a differentiable function $ucolon mathbb R^2to mathbb R$, a function $varphi colon mathbb R^2tomathbb R^2, (x,t)mapsto (t, x+t)$ (which is also differentiable) and you're asked to find $partial _2(ucirc varphi)$.
Now it's easy. In the notation above we have $m=2=n$, $p=1$, $G=u$, $F=varphi$, $f_1colon mathbb R^2to mathbb R, (x,t)mapsto t$ and $f_2colon mathbb R^2to mathbb R, (x,t)mapsto x+t$. Set $H=ucirc varphi$ and conclude that for all $(x,t)$ in $mathbb R^2$ it holds that
$$
begin{align}
begin{bmatrix}
partial _1h_1 & partial_nh_1
end{bmatrix}_{(x,t)}
&=
begin{bmatrix}
partial _1u & partial_2u
end{bmatrix}_{F(x,t)}
begin{bmatrix}
partial _1f_1 & partial_2f_1\
partial_1f_2 & partial_2f_2
end{bmatrix}_{(x,t)}\
&=
begin{bmatrix}
left(partial _1uright)(t,x+t) & left(partial_2uright)(t,x+t)
end{bmatrix}
begin{bmatrix}
0& 1\
1 & 1
end{bmatrix}\
&=
begin{bmatrix}
left(partial _2uright)(t,x+t) & left(partial _1uright)(t,x+t)+left(partial_2uright)(t,x+t)
end{bmatrix}_.
end{align}
$$
answered Jan 15 at 18:35
Git GudGit Gud
28.9k1050100
edited Jan 15 at 22:34
answered Jan 15 at 18:35
Git GudGit Gud
28.9k1050100
answered Jan 15 at 18:35
Git GudGit Gud
28.9k1050100
answered Jan 15 at 18:35
Git GudGit Gud
28.9k1050100
28.9k1050100
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