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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.

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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

$$S^\prime(t)=-\left(D_S(t,S(t)\right)^{-1}.D_t(t,S(t))=-\left(A(t)\right)^{-1}.\left(A^\prime(t).S(t)-B^\prime(t)\right)$$

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  • $\begingroup$ 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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