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Let $f(t,x)$ be a function whereat $x\in\mathbb{R}^n$ and $t\in\mathbb{R}$ fixed. Furthermore both $\frac{\partial}{\partial t}f(t,x)$ and $\frac{\partial}{\partial x}f(t,x)$ exists and are continious. For fixed $t$, $\frac{\partial}{\partial x}f(t,x)=D_x f(t,x)$ then is the Jacobi-matrix of $x\mapsto f(t,x)$. Is then $$ \frac{\partial}{\partial t}\text{det}D_x f(t,x)=\text{det}\frac{\partial}{\partial t}D_x f(t,x)? $$ And is $\text{det}D_x f(t,x)$ differentiable in $t=0$?

For the first question I do not know how to show this.

For the second question I have to show that $$ \lim_{h\to 0}\frac{\text{det}D_x f(h,x)-\text{det}D_x\Phi(0,x)}{h} $$ exisits and I think one can use then that (if it can be proved) $$ \lim_{h\to 0}\frac{\text{det}D_x f(h,x)-\text{det}D_x\Phi(0,x)}{h}=\text{det}\frac{\partial}{\partial t}D_x f(t,x)? $$

mathfemi
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2 Answers2

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Let $A_i$ be the $i$th column of the matrix $A = D_x f(t,x)$, so that $A$ can be written in the form $A = (A_1, A_2, \ldots, A_n)$. Since $\det A$ is a multilinear function of $A_1, \ldots, A_n$, we have $$\frac{\partial (\det A)}{\partial t} = \det\left(\frac{\partial A_1}{\partial t}, A_2, \ldots, A_n\right) + \det\left(A_1, \frac{\partial A_2}{\partial t}, \ldots, A_n\right) + \cdots + \det\left(A_1, A_2, \ldots, \frac{\partial A_n}{\partial t}\right). $$ An analogous formula can be given using the rows of $A$ instead of the columns.

In the absence of other information, in order to deduce differentiability of $\det A$ with respect to $t$, one should at least assume the existence of the mixed partial derivatives $\frac{\partial^2 f_i}{\partial t\,\partial x_j}$ for $i,j=1,\ldots,n$.

ivanpenev
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  • Ok, and what does this say to me? – mathfemi May 30 '14 at 14:45
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    The interchangeability of '$\det$' and '$\partial/\partial t$' would imply that $$ \frac{\partial (\det A)}{\partial t} = \det\left(\frac{\partial A_1}{\partial t}, \frac{\partial A_1}{\partial t}, \ldots, \frac{\partial A_n}{\partial t}\right),$$ which is not (in general) the case. – ivanpenev May 30 '14 at 14:50
  • Under waht conditions is that right? – mathfemi May 30 '14 at 14:51
  • If $f(t,x)$ does not depend on $t$ for instance. But I doubt that this would be an interesting case – ivanpenev May 30 '14 at 14:55
  • Or if the mixed partial derivatives are $o$? – mathfemi May 30 '14 at 15:02
  • Yes, this would be the case if $f(t,x) = g(t) + h(x)$. But again, this is a rather exceptional case. – ivanpenev May 30 '14 at 15:07
  • If $f(t,x)$ does not depend on $t$, why is is true then? – mathfemi May 30 '14 at 15:25
  • If $f(t,x)$ does not depend on $t$, then it's partial derivatives with respect to the variables $x_j$ do not either, so both $\partial(\det A)/\partial t$ and $\det(\partial A/\partial t)$ vanish. – ivanpenev May 30 '14 at 16:03
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I think that a counterexample is: $$ f(x,t) = A(t) x $$

Then $D_x f(x,t) = A(t)$ and you are asking if $$ \det A'(t) = (\det A(t))'. $$

If you take as $A(t)$ the rotation matrix: $((\cos t, \sin t),(-\sin t, \cos t))$ you have $\det A(t) = 1$ hence $(\det A(t))'=0$. While $\det A'(t) = 1$.

Also the second question has negative answer. If $n=1$ you are asking if $D_x f$ is differentiable with respect to $t$. This is not true if you only know that $f$ is differentiable, take for example $f(x,t) = |tx|x$

Emanuele Paolini
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