Mathematics Asked by Lucas G on January 7, 2022

I was working through a robotics book, and came across the following equation, solution, and proof where $A$ is a $n times n$ constant matrix, and $x_0 = x(0)$

$$dot x (t)=Ax(t)$$

whose solution is :

$$x(t)=e^{At}x_0$$

Proof:

begin{align}

dot{x}(t)&=left(frac{d}{dt}e^{At}right)x_0

\&=frac{d}{dt}left(I+At+frac{A^2t^2}{2!}+frac{A^3t^3}{3!}+dotsright)x_0

\&=left(A+A^2t+frac{A^3t^2}{2!}+dotsright)x_0

\&=Ae^{At}x_0

\&=Ax(t)

end{align}

I’m not super familiar with linear differential equations and operations, and am kind of confused as to what happens around the third line of the proof and onward. If anyone understands this well, an explanation would be appreciated

I'm going to go ahead and put in my two cents, which does *not* answer your actual question (but perhaps can shed some light for you). A linear differential equation is of the form:

$$ A_nx^{(n)} + A_{n - 1}x^{(n - 1)} + A_{n - 2}x^{(n - 2)} + dots A_2ddot{x} + A_1dot{x} + A_0x = 0 $$

where $x^{(n)}$ is the $text{n}^text{th}$ derivative of the function $x$. The general solution to this type of differential equation, called a *homogeneous* differential equation--because the RHS is equal to 0--is an exponential, of the form: $x(t) = x_0e^{lambda t}$ (note that $x_0 neq lambda$ or maybe it is--point is that there is no relation between the two).

So in your case, we have a first-order, homogeneous, linear differential equation (first order because it only involves a first derivative):

$$ dot{x} = Ax $$

The typical way to solve is that $x(t) = x_0x^{lambda t}$ and the procedure is to find $x_0$ and $lambda$:

begin{align*} &x(t) = x_0e^{lambda t} \ &dot{x}(t) = x_0lambda e^{lambda t} \ &dot{x} = Ax leadsto x_0lambda e^{at} = Ax_0e^{lambda t} end{align*}

To solve we set this equal to zero and factor out the common $x_0e^{lambda t}$:

$$ x_0e^{lambda t} left(lambda - Aright) = 0 $$

Regardless of the value of $e^{lambda t}$, this equation is satisfied so long as $lambda = A$ (since then $lambda - A = A - lambda = 0$). This is similar to when solving a quadratic, we factor and set each factor to zero to solve, e.g. $3x^2 + 5x + 2= (3x + 2)(x + 1) = 0 leadsto x = -frac{2}{3}, -1$.

So we now have the general solution: $x(t) = x_0e^{At}$. Now we must find $x_0$. Unfortunately, regardless of $lambda$, $x(0) = x_0$--usually there is an equation to be solved to find the initial conditions.

So there it is--this is the first week of a differential equations class.

Answered by Jared on January 7, 2022

The proof is

begin{align} dot{x}(t)&=left(frac{d}{dt}e^{At}right)x_0tag{1} \&=frac{d}{dt}left(I+At+frac{A^2t^2}{2!}+frac{A^3t^3}{3!}+dotsright)x_0tag{2} \&=left(A+A^2t+frac{A^3t^2}{2!}+dotsright)x_0tag{3} \&=Ae^{At}x_0tag{4} \&=Ax(t)tag{5}. end{align}

Explanation.

$text{(1)}$: Follows from the definition of $x(t)$. We have $$dot{x}(t)=frac{d}{dt}left(x(t)right)=frac{d}{dt}left(e^{At}x_0right)=frac{d}{dt}left(e^{At}right)x_0.$$ $text{(2)}$: The Taylor series of $e^x$ is $$e^x=1+frac{x}{1!}+frac{x^2}{2!}+frac{x^3}{3!}+ldots=sum_{n=0}^{infty}frac{x^n}{n!}.$$ If $A$ is a $n times n$ matrix with real entries, define $$e^{At}=I+At+frac{A^2t^2}{2!}+frac{A^3t^3}{3!}+dots=sum_{n=0}^{infty}frac{t^nA^n}{n!},$$ where $e^{0 ,cdot A}=I$.

$text{(3)}$: Differentiate the series term-by-term. A more direct way is begin{align}frac{d}{dt}(e^{At})&=sum_{n=0}^{infty}nfrac{t^{n-1}A^n}{n!}\&=sum_{n=1}^{infty}frac{t^{n-1}A^n}{(n-1)!}\&=Asum_{n=1}^{infty}frac{t^{n-1}A^{n-1}}{(n-1)!}\&=Asum_{m=0}^{infty}frac{t^mA^m}{m!}\&=Ae^{At}. end{align} $text{(4)}$: This follows from $$left(A+A^2t+frac{A^3t^2}{2!}+dotsright)x_0=Aleft(I+At+frac{A^2t^2}{2!}+dotsright)x_0=Ae^{At}x_0,$$ because $$e^{At}=left(I+At+frac{A^2t^2}{2!}+dotsright).$$ $text{(5)}$: Directly from the definition $x(t), ~x(t)=e^{At}x_0$.

Answered by Axion004 on January 7, 2022

The proof you've attached shows why the given answer is the correct one. It assumes you are aware of the Taylor expansion of $e^x$, which is given in the proof.

With some experience in calculus, you might be aware of the fact that $e^x$ is its own derivative, and see the similarity to that statement with the first equation. If you know how to differentiate $e^{Ax}$, you can easily confirm the solution without the added machinery of Taylor expansion.

In general, solving differential equations like these come from practice and experience. It's about trying different functions with certain analytical properties which might work in the given context. You will only get better at them as you keep exploring differential equations like these.

Answered by Subhasish Mukherjee on January 7, 2022

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