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Consider the differential equation $$y'=Ay+b$$ , where $A$ is a $n\times n$-matrix, y and b are vectors of functions $(y_1(x),...y_n(x))^T$ and $(b_1(x),...,b_n(x))^T$

Suppose, we have found the general solution of the homogenous equation $$y'=Ay$$

The standard approach is to continue with the "variation of the constants". But in a script, I found that the matrix $e^{Ax}$ has something to do with the special solution of the inhomogenous equation.

What is the meaning of $e^{Ax}$ ?

For example, let $$A=\pmatrix{2&-1\\-1&2}$$ , $$b=\pmatrix{-e^{3x}\\e^{3x}}$$

What is $e^{Ax}$ and a special solution of $$y'=Ay+b$$ ?

Peter
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  • Don't you mean $e^Ax$? – Hagen von Eitzen Feb 11 '15 at 21:23
  • I am pretty sure, that the script means $e^{Ax}$ and not $e^Ax=exp(A)x$ – Peter Feb 11 '15 at 21:24
  • I assumed that for each entry $a_{ij}$, we get $\large e^{a_{ij}x}$, but this does not work. – Peter Feb 11 '15 at 21:26
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    Do you mean http://en.wikipedia.org/wiki/Matrix_exponential? @Peter, also see: http://people.math.gatech.edu/~xchen/teach/ode/NonhomoSys.pdf – Amzoti Feb 11 '15 at 21:27
  • I cannot see any link between $exp(Ax)$ and a solution of the differential equation, so probably this is not meant. – Peter Feb 11 '15 at 21:28
  • @Peter The link is completely explained here. – Git Gud Feb 11 '15 at 21:33
  • @Peter, $y = e^{Ax}b$ is the unique solution of $\frac{dy}{dx} = Ay, y(0) = b.$ – abel Feb 11 '15 at 21:34
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    Actually, on a second reading of your question, I think my answer answers you completely. I'm voting to close as a duplicate. – Git Gud Feb 11 '15 at 21:35
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    @GitGud, i think question is much more specific than the you answered. i see no reason for this question to be closed. – abel Feb 11 '15 at 21:44
  • @abel You're right, I completely overlooked the term special. Retracting my vote. Edit: But a special matrix is easy to find from my answer. – Git Gud Feb 11 '15 at 21:45
  • If $A$ is invertible take $z = y + A^{-1}b$ to get $z' = Az$. Edit: This only works if $b$ is indepentent of $x$.! – Winther Feb 11 '15 at 21:47
  • Sometimes we use notations $x$ and $y$ in parallel ways. But here $y$ is a vector function, and $x$ is a scalar variable. So they are not parallel objects. – GEdgar Feb 11 '15 at 22:44

3 Answers3

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It is the matrix exponential: $$ e^{Ax}=\sum_{n=0}^\infty\frac{x^n}{n!}\,A^n. $$ It is a fundamental solution of the homogeneous equation $y'=A\,y$, and is the unique one equal to the identity matrix when $x=0$. The unique solution such that $y(0)=y_0$ is $y(x)=e^{Ax}y_0$.

The method of variation of constants looks for a particular solution of the complete equation of the form $e^{Ax}C(c)$, where $C(x)=(C_1(x),\dots,C_n(x))$.

Julián Aguirre
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in your specific problem, it is easy to find the eigenvalues and eigenvectors of the coefficient matrix $\pmatrix{2&-1\\-1&2}.$ the eigenvalues are $3$ and $1$ the corresponding orthonormal eigenvectors are first and second columns of the matrix $V = \pmatrix{1/\sqrt 2& 1/\sqrt 2\\-1/\sqrt 2& 1/\sqrt 2}.$ we have $$A = V \pmatrix{3&0\\0&1}V^T, \, e^{Ax} = V \pmatrix{e^{3x}&0\\0&e^x}V^T = \frac{1}{2}\pmatrix{e^{x} + e^{3x}& e^x - e^{3x}\\ e^x - e^{3x}&e^{x} + e^{3x}}$$

a particular solution of the nonhomogeneous problem $\frac{dy}{dx} = Ay + b$ is $$y_p = \int_0^x e^{A(x-\xi)}e^{3\xi}\pmatrix{-1\\1}\, d\xi=\int_0^x e^{3\xi} \pmatrix{0\\e^{3(x-\xi)}}\, d\xi = e^{3x}\pmatrix{0\\x}.$$

the general solution is $$y = e^{Ax}y(0) + y_p. $$

abel
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There are several ways to solve this problem. One way consists in applying the Laplace transform to the differential equation so that: \begin{equation} L\{y'(x)\}=A L\{y(x)\}+L\{b(x)\} \end{equation} By writing $L\{y(x)\}=Y(s)$ and $L\{b(x)\}=B(s)$ the equation above reduces to: \begin{equation} s Y(s)-y(0)= A Y(s) + B(s) \end{equation} Then \begin{equation} Y(s) (I s -A) = I y(0) + B(s) \end{equation} And \begin{equation} Y(s) = y(0) (s I-A)^{-1} + B(s) (s I-A)^{-1} \end{equation} And finally we go back in the x domain by applying the inverse Laplace transform \begin{equation} y(x) = y(0) L^{-1}\{(s I-A)^{-1}\} + b(x) \star L^{-1}\{ (s I-A)^{-1}\} \end{equation} So that when comparing the above relation to the solution of the differential equation \begin{equation} y(x) = y(0) e^{A x}+\int_0^x e^{A(x-\psi)} b(\psi) d\psi \end{equation} we can conclude that
\begin{equation} e^{A x} = L^{-1}\{(s I-A)^{-1}\} \end{equation}

Upax
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