# Finding function given its Jacobian and the initial condition

Consider continuously differentiable function $f:\mathbb{R}^k\mapsto \mathbb{R}^k$. We know that $f(x_0)=y_0$ and the Jacobian matrix is given for all $x$. I'd like to know the explicit for of the function $f$?

If $k=1$, then I could know the explicit for of $f$ is able to be determined using the following method:

Note that $f(x)=\int_{x_0}^xf^\prime(t)dt+c$ for some $c$. Then we can find $c=y_0$ by substituting $x$ with $x_0$. So, the explicit for of $f$ is $f(x)=\int_{x_0}^xf^\prime(t)dt+y_0$.

However, I found when $k>1$. Any suggestion? Thanks for any help.

• No, unfortunately the fundamental theorem of calculus doesn't extend verbatim to higher dimensions. To understand why this happens you might want to study de Rham cohomology. – J.R. Mar 4 '14 at 11:29
• Thanks @TooOldForMath. I'll go to the website – Jlamprong Mar 4 '14 at 11:44
• By the way, so in my problem above. What should I do to takle it? – Jlamprong Mar 4 '14 at 11:57

When you are given a matrix-valued function ${\bf x}\to A({\bf x})$ and know for sure that it is the Jacobian of some vector-valued function $${\bf f}:\quad{\mathbb R}^n\to{\mathbb R}^m,\qquad {\bf x}\to{\bf f}({\bf x})$$ then it is easy to recover ${\bf f}$ from $A$. Indeed, the columns of $A$ are nothing else but the partial derivatives $${\bf f}_{.k}({\bf x})=\left({\partial f_1\over\partial x_k},{\partial f_2\over\partial x_k},\ldots,{\partial f_m\over\partial x_k}\right)\qquad(1\leq k\leq n)\ ,$$

and assuming ${\bf x}_0={\bf y}_0={\bf 0}$ it is pretty obvious that we have $${\bf f}({\bf x})=\int_0^{x_1}{\bf f}_{.1}(t,0,\ldots,0)\ dt+\int_0^{x_2}{\bf f}_{.2}(x_1,t,0,\ldots,0)\ dt+\ldots+\int_0^{x_n}{\bf f}_{.n}(x_1,\ldots, x_{n-1},t)\ dt\ .$$ This comes from integrating the partial derivatives of ${\bf f}$ along a multi-L-shaped path with legs parallel to the coordinate axes from ${\bf 0}$ to ${\bf x}$. A coordinate-free way to write ${\bf f}$ would be $${\bf f}({\bf x})=\int_0^1 A(t{\bf x}).{\bf x}\ dt\ .$$

The deep problem here is somewhere else: You cannot give an arbitrary matrix-valued function ${\bf x}\to A({\bf x})$ and hope that there is a function ${\bf x}\to{\bf f}({\bf x})$ with $d{\bf f}({\bf x})=A({\bf x})$ for each ${\bf x}$. There are so-called integrability conditions as soon as $n\geq2$, corresponding to the condition ${\rm curl}\,{\bf F}={\bf 0}$ for a vector field ${\bf F}$ in ${\mathbb R}^3$ to be a gradient field $\nabla f$. To put it simply: These conditions encode that the mixed second partials of the prospective ${\bf f}$ should be equal.

Given a matrix-valued function ${\bf x}\to A({\bf x})=\bigl[a_{ik}({\bf x})\bigr]_{1\leq i\leq m, \ 1\leq k\leq n}$ on a simply connected domain $\Omega\subset{\mathbb R}^n$ the integrability conditions read as follows: $${\partial a_{ik}({\bf x})\over\partial x_l}\equiv {\partial a_{il}({\bf x})\over\partial x_k}\quad({\bf x}\in\Omega, \ 1\leq i\leq m, \ 1\leq k<l\leq n)\ .$$ So there are $m\cdot{n\choose 2}$ of them.

• Thanks @Chistian Blatter. So, in what condition I could recover my function $f$ if its Jacobian and a value of $f$ is given? Any reference? – Jlamprong Mar 4 '14 at 12:45

The situations of differential equations in one variable and several variable are deeply different.

In several variables, you have $f(x_1,\dots,x_k)=(f_1(x_1,\dots,x_k),\dots,f_m(x_1,\dots,x_k))$ and the question is about knowing the $f_i$ once one knows the partial derivatives of them (the Jacobian). Well, the world of PDEs (Partial Differential Equations) is more complicated than that of ODE (Ordinary Differential Equations). The theorem of existence and uniqueness fails in general.