At the end of the section on Implicit Functions in Spivak's Calculus on Manifolds, we have Problem 2-40:

Problem 2-40. Use the implicit function theorem to re-do Problem 2-15(c).

And, Problem 2-15(c) is:

Problem 2-15(c). If $\det(a_{ij}(t)) \neq 0$ for all $t$ and $b_1,\dots,b_n : \mathbb{R} \to \mathbb{R}$ are differentiable, let $s_1,\dots,s_n : \mathbb{R} \to \mathbb{R}$ be the functions such that $s_1(t),\dots,s_n(t)$ are the solutions of the equations $$ \sum_{j=1}^n a_{ji}(t) s_j(t) = b_i(t) \qquad i = 1,\dots,n. $$ Show that $s_i$ is differentiable and find ${s_i}'(t)$.

We assume that the functions $a_{ij}$ and $b_i$ are all continuously differentiable.

I believe I have solved this problem correctly, but I am not sure that my method is what is expected. One reason is that the final expression for ${s_i}'(t)$ appears awkward. Another reason is that this expression is also quite different from the one I arrived at when solving Problem 2-15(c) earlier without using the implicit function theorem. My previous solution is given in an answer to this question: Differentiation of solution to time-dependent system of equations: Problem 2-15(c) from Spivak's Calculus on Manifolds.

It would be great if someone can go through my solution below and give me comments or suggestions. Thanks.

Soln. Let $f: \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n$ be defined by $$ f^i(t,x) = -b_i(t) + \sum_{j=1}^n a_{ji}(t) x^j, \qquad 1 \leq i \leq n. $$ Clearly $f$ is continuously differentiable. The $n \times n$ matrix $M(t,x) = \left(D_{1+j} f^i(t,x)\right)$ equals $\left(a_{ji}(t)\right)$. So, if $t_0 \in \mathbb{R}$ and $x_0 \in \mathbb{R}^n$ such that $f(t_0,x_0) = 0$, then $$ \det M(t_0,x_0) = \det \left(D_{1+j}f^i(t_0,x_0) \right) = \det \left(a_{ji}(t_0) \right) \neq 0. $$ Therefore, by the implicit function theorem, there exists a function $g : \mathbb{R} \to \mathbb{R}^n$ such that $f(t,g(t)) = 0$ for all $t \in (t_0 - \epsilon, t_0 + \epsilon)$, for some $\epsilon > 0$. But, this means that $g^1(t),\dots,g^n(t)$ is a solution to the system of equations $$ \sum_{j=1}^n a_{ji}(t) s_j(t) = b_i(t), \qquad i = 1,\dots,n. $$ Since the solution is unique for every $t \in (t_0 - \epsilon, t_0 + \epsilon)$, $g^i(t) = s_i(t)$ in this interval. But since $t_0 \in \mathbb{R}$ is arbitrary, we have that $f(t,s_1(t),\dots,s_n(t)) = 0$ for all $t \in \mathbb{R}$. Let $s : \mathbb{R} \to \mathbb{R}^n$ be defined by $s(t) = (s_1(t),\dots,s_n(t))$. By the implicit function theorem, $s$ is differentiable, and so the $s_i$ are differentiable.

To calculate ${s_i}'(t)$, we will take the derivative on both sides of the expression $f^i(t,s(t)) = 0$, for each $1 \leq i \leq n$. This gives us $$ 0 = D_1 f^i(t,s(t)) + \sum_{k = 1}^n D_{1+k} f^i(t,s(t)) \cdot D_1 s_k(t), \qquad 1 \leq i \leq n. $$ This is precisely the system of equations given by $$ M(t,s(t)) \cdot s'(t) = B(t), $$ where $B(t)$ is an $n \times 1$ matrix with the entry in the $i$th row equal to $$ {b_i}'(t) - \sum_{j=1}^n {a_{ji}}'(t)s_j(t). $$ Since $M(t,s(t)) = \left( a_{ji}(t) \right)$ and $\det \left( a_{ji}(t) \right) \neq 0$ for all $t \in \mathbb{R}$, $M(t,s(t))$ is invertible. So, $$ s'(t) = \left( a_{ji}(t) \right)^{-1} B(t). \tag*{$\blacksquare$} $$

Is the final expression the best one can do using the implicit function theorem? I haven't found the precise expression for each ${s_i}'(t)$ through this method. It feels like I was able to arrive at a more concrete result without using the implicit function theorem as in my previous attempt.

Again, any comments or suggestions are appreciated.


What you did looks good to me. On my side I would have started from the beginning using matrix/vector notations in order to have a more synthetic view of the question. Which means that you can define your map $$f(t,S)= A(t).S - B(t)$$ where capital letters are vectors / matrix « derived » from the ones whose coordinates are the same with lower letters.

Then $D_S f(t,S): H \mapsto A(t).H$ is invertible as $\det A(t)$ is supposed to be non zero. Therefore, we can apply the Implicit function theorem and


  • $\begingroup$ Thank you! I've been diligently following Spivak's notation to get a handle on how he does things. But your notation actually makes this argument easier to follow. So, this answer is really helpful. :) $\endgroup$ – Brahadeesh Aug 9 '18 at 16:00
  • $\begingroup$ Can I trouble you to also look at the question (and answer) that I've linked to, in which I solve the problem without the implicit function theorem? $\endgroup$ – Brahadeesh Aug 9 '18 at 16:01
  • $\begingroup$ I provided an alternate answer based on matrix / vector notation. $\endgroup$ – mathcounterexamples.net Aug 9 '18 at 16:25
  • $\begingroup$ When you find the derivative, where did the $b_i$'s go ? I mean they are function of $t$. $\endgroup$ – onurcanbektas Aug 10 '18 at 4:08
  • $\begingroup$ @onurcanbektas As I explained, $B$ is the vector whose coordinates are the $b_i$. $\endgroup$ – mathcounterexamples.net Aug 10 '18 at 6:10

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