Matrix exponential

In mathematics , the matrix exponential , also referred to as the matrix exponential function , is a function on the set of square matrices , which is defined analogously to the usual (scalar) exponential function . The matrix exponential establishes the connection between Lie algebra and the associated Lie group .

definition

Be a real or complex - matrix . The exponential of , which is denoted by or , is the matrix, which is defined by the following power series ( Taylor expansion ): ${\displaystyle X}$ ${\displaystyle n\times n}$ ${\displaystyle X}$ ${\displaystyle e^{X}}$ ${\displaystyle \exp(X)}$ ${\displaystyle n\times n}$

${\displaystyle e^{X}=\sum _{k=0}^{\infty }{\frac {X^{k}}{k!}}=E+X+{\frac {X^{2}}{2}}+\dots }$.

This series, like that of the ordinary exponential function , always converges . Hence the exponential of is well defined. When is a matrix, the matrix exponential is equal to the ordinary exponential function . A generalization that is also useful for infinite matrices is the exponential function on arbitrary Banach algebras . ${\displaystyle X}$ ${\displaystyle X}$ ${\displaystyle 1\times 1}$ ${\displaystyle X}$

properties

The matrix exponential shares a number of the properties of the ordinary exponential function. For example, the exponential of the - zero matrix is equal to the - unit matrix : ${\displaystyle (n\times n)}$ ${\displaystyle 0}$ ${\displaystyle (n\times n)}$ ${\displaystyle E}$

${\displaystyle e^{0}=E}$.

For any complex matrices and any complex numbers and holds ${\displaystyle n\times n}$ ${\displaystyle X}$ ${\displaystyle a}$ ${\displaystyle b}$

${\displaystyle e^{aX}\cdot e^{bX}=e^{(a+b)X}}$.

It follows

${\displaystyle e^{X}\cdot e^{-X}=e^{(1-1)X}=e^{0}=E}$,

this means

${\displaystyle \left(e^{X}\right)^{-1}=e^{-X}}$.

The matrix that is too inverse denotes . ${\displaystyle \left(e^{X}\right)^{-1}}$ ${\displaystyle e^{X}}$

The exponential function satisfies for all numbers and . The same applies to commuting matrices and , that is, from ${\displaystyle e^{x+y}=e^{x}\,e^{y}}$ ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle X}$ ${\displaystyle Y}$

${\displaystyle X\cdot Y=Y\cdot X}$

follows

${\displaystyle e^{X+Y}=e^{X}\cdot e^{Y}}$.

This equation is generally incorrect for non-commutating matrices. In this case one can calculate with the help of the Baker-Campbell-Hausdorff formula . ${\displaystyle e^{X+Y}}$

The exponential of the matrix to be transposed is equal to the transposition of the exponential of : ${\displaystyle X}$ ${\displaystyle X}$

${\displaystyle \exp \left(X^{\mathrm {T} }\right)=\left(\exp X\right)^{\mathrm {T} }}$

It follows that the matrix exponential function maps symmetric matrices to symmetric matrices and skew-symmetric matrices to orthogonal matrices . The relationship between adjunction and exponentiation applies analogously

${\displaystyle \exp \left(X^{*}\right)=\left(\exp X\right)^{*}}$,

so that the matrix exponential function maps Hermitian matrices to Hermitian matrices and skew Hermitian matrices to unitary matrices .

The following also applies:

• If is invertible , then is . ${\displaystyle Y}$ ${\displaystyle e^{YXY^{-1}}=Ye^{X}Y^{-1}}$
• ${\displaystyle \det(e^{X})=e^{{\mbox{tr}}(X)}}$, Here denotes the trace of the square matrix . ${\displaystyle {\mbox{tr}}(X)}$ ${\displaystyle X}$
• ${\displaystyle e^{\operatorname {diag} (x_{1},\ldots ,x_{n})}=\operatorname {diag} \left(e^{x_{1}},\ldots ,e^{x_{n}}\right)}$.

The exponential map

The exponential of a matrix is ​​always an invertible matrix . The inverse of is given by. The (complex) matrix exponential thus provides a mapping ${\displaystyle e^{X}}$ ${\displaystyle e^{-X}}$

${\displaystyle \exp \colon M_{n}(\mathbb {C} )\to {\mbox{GL}}(n,\mathbb {C} )}$

from the vector space of all (complex) matrices into the general linear group , the group of all (complex) invertible matrices. This mapping is surjective , which means that every (real or complex) invertible matrix can be written as the exponential matrix of a complex matrix. Archetypes (or local sections ) can be calculated using matrix logarithms . ${\displaystyle (n\times n)}$

For every two matrices and applies ${\displaystyle X}$ ${\displaystyle Y}$

${\displaystyle \|e^{X+Y}-e^{X}\|\leq \|Y\|e^{\|X\|}e^{\|Y\|}}$,

where denotes any matrix norm . From this it follows that the exponential mapping is continuous and even lipschitz continuous on compact subsets of . But there is a more precise limit for the norm of the matrix exponential itself ${\displaystyle \|\cdot \|}$ ${\displaystyle M_{n}(\mathbb {C} )}$

${\displaystyle \|e^{X}\|\leq e^{\mu (X)}}$

with the logarithmic matrix norm and the numeric range of values . ${\displaystyle \mu }$

The assignment

${\displaystyle t\mapsto e^{tX},\qquad t\in \mathbb {R} ,}$

defines a smooth curve in the general linear group that yields the identity matrix. This provides a one-parameter subset of the general linear group, da ${\displaystyle t=0}$

${\displaystyle e^{tX}e^{sX}=e^{(t+s)X}}$

applies. The derivation of this function in the point is through ${\displaystyle t}$

${\displaystyle {\frac {d}{dt}}e^{tX}=Xe^{tX}\qquad (1)}$

given. The derivation for is just the matrix , that is, creates this one-parameter subgroup. ${\displaystyle t=0}$ ${\displaystyle X}$ ${\displaystyle X}$

More generally applies:

${\displaystyle {\frac {d}{dt}}e^{X(t)}=\int _{0}^{1}e^{(1-\alpha )X(t)}{\frac {dX(t)}{dt}}e^{\alpha X(t)}d\alpha }$

Examples of Lie algebras and associated Lie groups

Lie group	example
General linear group   ${\displaystyle \mathrm {GL} (n,K)}$	${\displaystyle {\mathfrak {gl}}(n)=\{X\in M_{n}(\mathbb {R} )\}}$ ${\displaystyle \mathrm {GL} (n)=\{X\in M_{n}(\mathbb {R} )\vert \det(X)\neq 0\}}$.
Orthogonal group   ${\displaystyle \mathrm {O} (n,K)}$	${\displaystyle {\mathfrak {o}}(n)=\{X\in M_{n}(\mathbb {R} )\vert X^{T}=-X\}}$ ${\displaystyle \mathrm {O} (n)=\{X\in M_{n}(\mathbb {R} )\vert X^{T}=X^{-1}\}}$
Unitary group   ${\displaystyle \mathrm {U} (n)}$	${\displaystyle {\mathfrak {u}}(n)=\{X\in M_{n}(\mathbb {C} )\vert X^{*}=-X\}}$ ${\displaystyle \mathrm {U} (n)=\{X\in M_{n}(\mathbb {C} )\vert X^{*}=X^{-1}\}}$
Special unitary group   ${\displaystyle \mathrm {SU} (n)}$	${\displaystyle {\mathfrak {su}}(2)=\{X\in M_{2}(\mathbb {C} )\vert X^{*}=-X,\operatorname {tr} (X)=0\}}$is mapped from surjective to . ${\displaystyle \exp }$ ${\displaystyle \mathrm {SU} (2)=\{A\in M_{2}(\mathbb {C} )\vert A^{*}=A^{-1},\det(A)=1\}}$
Special orthogonal group   ${\displaystyle \mathrm {SO} (n,K)}$	${\displaystyle {\mathfrak {so}}(n,\mathbb {R} )={\mathfrak {o}}(n,\mathbb {R} )=\{X\in M_{n}(\mathbb {R} )\vert X^{T}=-X\}}$( skew-symmetrical matrices ) is mapped from surjective to . ${\displaystyle \exp }$ ${\displaystyle \mathrm {SO} (n,\mathbb {R} )=\{A\in M_{n}(\mathbb {R} )\vert A^{T}=A^{-1},\det(A)=1\}}$
Special linear group   ${\displaystyle \mathrm {SL} (n,K)}$	${\displaystyle {\mathfrak {sl}}(2,\mathbb {C} )}$is mapped from non- surjective to . Notorious counterexample with is not in the image of . ${\displaystyle \exp }$ ${\displaystyle \mathrm {SL} (2,\mathbb {C} )}$ ${\displaystyle {\begin{pmatrix}-1&a\\0&-1\end{pmatrix}}\in \mathrm {SL} (2,\mathbb {C} )}$ ${\displaystyle a\neq 0}$ ${\displaystyle {\mathfrak {sl}}(2,\mathbb {C} )}$

The last example shows that the exponential mapping for generating Lie groups (depending on the Lie algebra) is generally not surjective.

Linear differential equations

One of the advantages of matrix exponential is that it can be used to solve systems of linear ordinary differential equations . For example, from equation (1) above it follows that the solution of the initial value problem

${\displaystyle {\frac {d}{dt}}y(t)=Ay(t),\quad y(t_{0})=y_{0}}$

with a square matrix by ${\displaystyle A}$

${\displaystyle y(t)=e^{(t-t_{0})A}y_{0}}$

given is.

The matrix exponential can also be used to solve the inhomogeneous equation

${\displaystyle {\frac {d}{dt}}y(t)=Ay(t)+z(t),\quad y(t_{0})=y_{0}}$,

be used. Examples can be found below in the Applications chapter .

For differential equations of the form

${\displaystyle {\frac {d}{dt}}y(t)=A(t)\,y(t),\quad y(t_{0})=y_{0}}$

with non-constant there are generally no closed solutions. The Magnus series , however, provides a general solution in matrix notation via the matrix exponential function even in the case of non-constant coefficients (as an infinite series of the exponent). ${\displaystyle A}$

Calculation of the matrix exponential

Taylor series

The exponential function of the matrix and can in principle be calculated using its Taylor expansion: ${\displaystyle A\in \mathbb {C} ^{n\times n}}$ ${\displaystyle t\in \mathbb {R} }$

${\displaystyle \exp(tA)=\sum _{k=0}^{\infty }{\frac {t^{k}A^{k}}{k!}}=E+tA+{\frac {t^{2}A^{2}}{2}}+{\frac {t^{3}A^{3}}{6}}+{\frac {t^{4}A^{4}}{24}}+\cdots }$

Here refers to the faculty of . If the accuracy is sufficient (the series is absolutely convergent), the series should terminate at a finite number of calculation steps. The larger the entries in the matrix, the more terms in the series have to be calculated (e.g. for solving the linear differential equation for a large time step). In order to improve the solution algorithm in this regard, the entries in the matrix can be elegantly scaled using the calculation rule ("Scaling & Squaring" method). If the (natural) matrix norm is not too large, the series can also be calculated using the Padé approximation . The scaling & squaring method has an expense of the order of magnitude (essentially matrix multiplications ). The factor of depends on the scaling parameters and, in particular, on the matrix standard. ${\displaystyle k!}$ ${\displaystyle k}$ ${\displaystyle e^{tA}=(e^{tA/m})^{m}}$ ${\displaystyle \|tA\|}$ ${\displaystyle {\mathcal {O}}(n^{3})}$ ${\displaystyle n^{3}}$

Nile Potent Fall

A matrix is nilpotent if holds for a suitable natural number . In this case the series expansion of breaks off after a finite number of terms and the matrix exponential can be used as ${\displaystyle N}$ ${\displaystyle N^{q}=0}$ ${\displaystyle q}$ ${\displaystyle e^{N}}$

${\displaystyle \exp(N)=E+N+{\frac {1}{2}}N^{2}+{\frac {1}{6}}N^{3}+\cdots +{\frac {1}{(q-1)!}}N^{q-1}}$

be calculated.

Diagonalization of the matrix

Is the matrix a diagonal matrix ${\displaystyle D}$

${\displaystyle D={\begin{pmatrix}a_{1}&0&\ldots &0\\0&a_{2}&\ldots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\ldots &a_{n}\end{pmatrix}}}$,

then one can find its exponential by applying the usual exponential function to each entry of the main diagonal :

${\displaystyle e^{D}={\begin{pmatrix}e^{a_{1}}&0&\ldots &0\\0&e^{a_{2}}&\ldots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\ldots &e^{a_{n}}\end{pmatrix}}}$.

With this one can also calculate the exponential of a diagonalizable matrix . For diagonalization ${\displaystyle A}$

${\displaystyle A=VDV^{-1}}$

by a diagonal matrix , the associated be eigenbasis and - eigenvalues of the matrix determined. For the matrix exponential function it follows from this ${\displaystyle D}$ ${\displaystyle V}$ ${\displaystyle n}$ ${\displaystyle \lambda }$ ${\displaystyle A}$

${\displaystyle \exp(tA)=Ve^{tD}\,V^{-1}=V\operatorname {diag} (e^{\lambda _{1}t},e^{\lambda _{2}t},\dotsc ,e^{\lambda _{n}t})\,V^{-1}}$

with the ordinary exponential function . The proof follows directly from the Taylor expansion of the exponential function. ${\displaystyle e^{\lambda _{i}t}}$

The diagonalization of the matrix, like the QR algorithm or the Jordan normal form , belongs to the matrix decomposition methods for calculating the exponential function. The diagonalization and the QR algorithm each have an effort of the order of magnitude and, in comparison to methods based on the Taylor expansion, are independent of . The main calculation effort (here: determination of the eigenvalues ​​and eigenvectors ) is also independent of the variables . To solve, for example, linear differential equations for several time steps , this amount of work only has to be done once. With the diagonalization method, the calculation of the further time steps is carried out by simple matrix multiplication and with the QR algorithm, the effort is only in the order of magnitude . ${\displaystyle {\mathcal {O}}(n^{3})}$ ${\displaystyle \|tA\|}$ ${\displaystyle t}$ ${\displaystyle t}$ ${\displaystyle {\mathcal {O}}(n^{2})}$

example 1

The following matrix exponential function is to be calculated:

${\displaystyle \exp(tA)=\exp \left[t{\begin{pmatrix}0&1\\a_{2,1}&a_{2,2}\end{pmatrix}}\right]}$.

For this purpose, the matrix is first diagonalized using the eigenvalues ​​and the eigenvectors. With the diagonal matrix and the eigen basis it follows: ${\displaystyle (2\times 2)}$ ${\displaystyle A}$ ${\displaystyle D}$ ${\displaystyle V}$

${\displaystyle \exp(tA)=Ve^{tD}\,V^{-1}=V{\begin{pmatrix}e^{\lambda _{1}t}&0\\0&e^{\lambda _{2}t}\end{pmatrix}}V^{-1}}$.

The eigenvalues of the characteristic polynomial determined to

${\displaystyle \lambda _{1,2}={\frac {a_{2,2}}{2}}\pm {\sqrt {\left({\frac {a_{2,2}}{2}}\right)^{2}+a_{2,1}}}}$.

For the two eigenvectors or the eigen basis applies:

${\displaystyle V={\begin{pmatrix}1&1\\\lambda _{1}&\lambda _{2}\end{pmatrix}}~~~}$such as ${\displaystyle ~~V^{-1}={\frac {1}{\lambda _{2}-\lambda _{1}}}{\begin{pmatrix}\lambda _{2}&-1\\-\lambda _{1}&1\end{pmatrix}}}$

Substituting for the matrix exponential function finally yields

${\displaystyle \exp(tA)={\frac {1}{\lambda _{2}-\lambda _{1}}}{\begin{pmatrix}\lambda _{2}e^{\lambda _{1}t}-\lambda _{1}e^{\lambda _{2}t}&e^{\lambda _{2}t}-e^{\lambda _{1}t}\\\lambda _{1}\lambda _{2}(e^{\lambda _{1}t}-e^{\lambda _{2}t})&\lambda _{2}e^{\lambda _{2}t}-\lambda _{1}e^{\lambda _{1}t}\end{pmatrix}}}$

as a closed analytical solution.

Example 2

The matrix exponential function

${\displaystyle \exp(tA)=\exp \left[t{\begin{pmatrix}3&-4\\4&-5\end{pmatrix}}\right]={\begin{pmatrix}(1+4t)e^{-t}&-4te^{-t}\\4te^{-t}&(1-4t)e^{-t}\end{pmatrix}}}$

has a solution, but the matrix itself cannot be diagonalized. The matrix has the two eigenvalues . Although the eigenvalue has the algebraic multiplicity 2, there is only one linearly independent eigenvector. The basis from the eigenvectors ${\displaystyle A}$ ${\displaystyle \lambda _{1,2}=-1}$

${\displaystyle V={\begin{pmatrix}a_{2,1}&a_{2,1}\\\lambda _{1}-a_{1,1}&\lambda _{2}-a_{1,1}\end{pmatrix}}={\begin{pmatrix}-4&-4\\-4&-4\end{pmatrix}}}$

is not invertible . The discriminant of the characteristic polynomial

${\displaystyle D_{2}=\left({\frac {a_{1,1}-a_{2,2}}{2}}\right)^{2}+a_{1,2}\,a_{2,1}=\left({\frac {3-(-5)}{2}}\right)^{2}-16}$

always becomes zero. In this case, i.e. when the same eigenvalues ​​or eigenvectors occur, the Jordan normal form can be used formally for the transformation.

Splitting method

If the minimal polynomial (or the characteristic polynomial ) of the matrix breaks down into linear factors (this is always the case above ), then it can be uniquely into a sum ${\displaystyle X}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle X}$

${\displaystyle X=A+N}$

be disassembled, with

• ${\displaystyle A}$ is diagonalizable,
• ${\displaystyle N}$ is nilpotent and
• ${\displaystyle A}$commutates with (i.e. ). ${\displaystyle N}$ ${\displaystyle AN=NA}$

This allows you the exponential of charge, by reducing it to the aforementioned cases: . In the last step you need the commutativity of and . ${\displaystyle X}$ ${\displaystyle e^{X}=e^{A+N}=e^{A}e^{N}}$ ${\displaystyle A}$ ${\displaystyle N}$

Use of the Jordan normal form

Another method is to use the Jordan normal form of , whereby the splitting method is also used. Let be the Jordan normal form of with the base change matrix , then ${\displaystyle X}$ ${\displaystyle J}$ ${\displaystyle X}$ ${\displaystyle P}$

${\displaystyle e^{X}=Pe^{J}P^{-1}.}$

Because of

${\displaystyle J=J_{a_{1}}(\lambda _{1})\oplus J_{a_{2}}(\lambda _{2})\oplus \cdots \oplus J_{a_{n}}(\lambda _{n})}$

applies

${\displaystyle {\begin{aligned}e^{J}&=\exp {\big (}J_{a_{1}}(\lambda _{1})\oplus J_{a_{2}}(\lambda _{2})\oplus \cdots \oplus J_{a_{n}}(\lambda _{n}){\big )}\\&=\exp {\big (}J_{a_{1}}(\lambda _{1}){\big )}\oplus \exp {\big (}J_{a_{2}}(\lambda _{2}){\big )}\oplus \cdots \oplus \exp {\big (}J_{a_{k}}(\lambda _{k}){\big )}.\end{aligned}}}$

Therefore one only needs to know the exponential of a Jordan block. Now every Jordan block is of the shape

${\displaystyle J_{a}(\lambda )=\lambda \,E+N,}$

where is a special nilpotent matrix. So the exponential of the Jordan block is ${\displaystyle N}$

${\displaystyle e^{\lambda E+N}=e^{\lambda }e^{N}.}$

example

Look at the matrix

${\displaystyle B={\begin{pmatrix}21&17&6\\-5&-1&-6\\4&4&16\end{pmatrix}}}$,

which is the Jordan normal form

${\displaystyle J=P^{-1}BP={\begin{pmatrix}4&0&0\\0&16&1\\0&0&16\end{pmatrix}}}$

with the transition matrix

${\displaystyle P={\begin{pmatrix}-1&1&{\tfrac {5}{8}}\\1&-1&-{\tfrac {1}{8}}\\0&2&0\end{pmatrix}}}$

Has. Then applies

${\displaystyle J=J_{1}(4)\oplus J_{2}(16)}$

and

${\displaystyle e^{B}=Pe^{J}P^{-1}=P(e^{J_{2}(16)}\oplus e^{J_{1}(4)})P^{-1}}$.

So is

${\displaystyle \exp \left[{\begin{pmatrix}16&0\\0&16\end{pmatrix}}+{\begin{pmatrix}0&1\\0&0\end{pmatrix}}\right]=e^{16}E\cdot \left[{\begin{pmatrix}1&0\\0&1\end{pmatrix}}+{\begin{pmatrix}0&1\\0&0\end{pmatrix}}+{1 \over 2!}{\begin{pmatrix}0&0\\0&0\end{pmatrix}}+\cdots \right]={\begin{pmatrix}e^{16}&e^{16}\\0&e^{16}\end{pmatrix}}}$.

The exponential of a 1 × 1 matrix is ​​trivial. With follows ${\displaystyle e^{J_{1}(4)}=e^{4}}$

${\displaystyle {\begin{aligned}\exp(B)&=P\exp(J)P^{-1}=P{\begin{pmatrix}e^{4}&0&0\\0&e^{16}&e^{16}\\0&0&e^{16}\end{pmatrix}}P^{-1}\\[6pt]&={1 \over 4}{\begin{pmatrix}13e^{16}-e^{4}&13e^{16}-5e^{4}&2e^{16}-2e^{4}\\-9e^{16}+e^{4}&-9e^{16}+5e^{4}&-2e^{16}+2e^{4}\\16e^{16}&16e^{16}&4e^{16}\end{pmatrix}}.\end{aligned}}}$

To calculate the Jordan normal form and from it the exponential is very tedious on this way. Usually it is sufficient to calculate the effect of the exponential matrix on a few vectors.

Numerical calculation

The Jordan normal form decomposition is numerically unstable because, due to the floating point arithmetic, rounding errors are introduced into the eigenvalues, which make it impossible to group the eigenvalues ​​into groups of identical eigenvalues. Therefore, other techniques are used in numerics to calculate the matrix exponential. The most effective algorithms available include the Padé approximation with scaling and squaring (see calculation using the Taylor series ) or the matrix decomposition methods such as diagonalization of the matrix . In the case of large matrices, the computational effort can also be reduced by using Krylow spaces whose basis vectors have been orthogonalized with the Arnoldi method .

Cleaner algorithm

→ Main article : Putzer's algorithm

Another way to calculate the matrix expontial is the Putzer algorithm. With a given matrix and recursively, continuously differentiable functions and matrices are defined , so that: ${\displaystyle A\in \mathbb {C} ^{n\times n}}$ ${\displaystyle t\in \mathbb {R} }$ ${\displaystyle p_{k}}$ ${\displaystyle M_{k}}$

${\displaystyle \exp(tA)=\sum _{k=1}^{n}p_{k}(t)M_{k-1}}$

The solution of the matrix exponential of a matrix is ​​obtained as a polynomial. The computation involves an effort of the order of magnitude (computation of the eigenvalues ​​and especially matrix multiplications) and is therefore only suitable for small matrices. ${\displaystyle (n\times n)}$ ${\displaystyle {\mathcal {O}}(n^{4})}$ ${\displaystyle n}$

Applications

Homogeneous linear differential equations

The matrix exponential can be used to solve a system of linear differential equations . A differential equation of the form

${\displaystyle y'=Cy}$

has the solution . If you have the vector ${\displaystyle e^{Ct}}$

${\displaystyle \mathbf {y} (t)={\begin{pmatrix}y_{1}(t)\\\vdots \\y_{n}(t)\end{pmatrix}}}$

considered, then a system of coupled linear differential equations can be considered as

${\displaystyle \mathbf {y} '(t)=A\mathbf {y} (t)+\mathbf {b} }$.

If you apply the integration factor and multiply on both sides, you get ${\displaystyle e^{-tA}}$

${\displaystyle e^{-tA}\mathbf {y} '(t)-e^{-tA}A\mathbf {y} =e^{-tA}\mathbf {b} }$,

${\displaystyle D(e^{-tA}\mathbf {y} )=e^{-tA}\mathbf {b} }$.

If one calculates, one obtains a solution of the differential equation system. ${\displaystyle e^{tA}}$

Example (homogeneous)

The following differential equation system is given

${\displaystyle {\begin{aligned}y_{1}'&=3y_{1}-y_{2},\\y_{2}'&=y_{1}+y_{2}.\end{aligned}}}$

It can be written as using the coefficient matrix ${\displaystyle y'(t)=Ay(t)}$

${\displaystyle A={\begin{pmatrix}3&-1\\1&1\end{pmatrix}}}$.

This gives the associated matrix exponential to

${\displaystyle e^{tA}=e^{2t}{\begin{pmatrix}1+t&-t\\t&1-t\end{pmatrix}}}$.

The general solution of the differential equation system is thus obtained

${\displaystyle {\begin{pmatrix}y_{1}(t)\\y_{2}(t)\end{pmatrix}}=C_{1}e^{2t}{\begin{pmatrix}1+t\\t\end{pmatrix}}+C_{2}e^{2t}{\begin{pmatrix}-t\\1-t\end{pmatrix}}}$.

Inhomogeneous case - variation of the constants

For the inhomogeneous case, a method similar to the variation of the constants can be used. A solution of the form is sought: ${\displaystyle \mathbf {y} _{p}(t)=e^{tA}\mathbf {z} (t)}$

${\displaystyle \mathbf {y} _{p}'=(e^{tA})'\mathbf {z} (t)+e^{tA}\mathbf {z} '(t)=Ae^{tA}\mathbf {z} (t)+e^{tA}\mathbf {z} '(t)=A\mathbf {y} _{p}(t)+e^{tA}\mathbf {z} '(t)}$

To find the solution , you bet ${\displaystyle \mathbf {y} _{p}}$

${\displaystyle e^{tA}\mathbf {z} '(t)=\mathbf {b} (t)}$

${\displaystyle \mathbf {z} '(t)=(e^{tA})^{-1}\mathbf {b} (t)}$

${\displaystyle \mathbf {z} (t)=\int _{0}^{t}e^{-uA}\mathbf {b} (u)\,du+\mathbf {c} }$

This results in

${\displaystyle \mathbf {y} _{p}=e^{tA}\int _{0}^{t}e^{-uA}\mathbf {b} (u)\,du+e^{tA}\mathbf {c} =\int _{0}^{t}e^{(t-u)A}\mathbf {b} (u)\,du+e^{tA}\mathbf {c} }$,

where is determined by the initial conditions. ${\displaystyle c}$

Example (inhomogeneous)

The differential equation system is given

${\displaystyle {\begin{aligned}y_{1}'&=3y_{1}-y_{2}-2e^{t},\\y_{2}'&=y_{1}+y_{2}-e^{t}.\end{aligned}}}$

The system writes itself with the matrix above ${\displaystyle A}$

${\displaystyle y'(t)=Ay(t)+b(t)}$

With

${\displaystyle b(t)=e^{t}{\begin{pmatrix}-2\\-1\end{pmatrix}}}$.

The general solution of the homogeneous equation has already been calculated above. The sum of homogeneous and special solutions results in the solution for the inhomogeneous problem. Now you just have to find a special solution (by varying the constants). From the equation above we get: ${\displaystyle y_{p}(t)}$

${\displaystyle y_{p}(t)=e^{tA}\int _{0}^{t}e^{-uA}b(u)\,du}$,

so

${\displaystyle y_{p}(t)=e^{2t}{\begin{pmatrix}1+t&-t\\t&1-t\end{pmatrix}}\int _{0}^{t}e^{-2u}{\begin{pmatrix}1-u&u\\-u&1+u\end{pmatrix}}{\begin{pmatrix}-2e^{u}\\-e^{u}\end{pmatrix}}\,du={\begin{pmatrix}-e^{2t}(1+t)+e^{t}\\-te^{2t}\end{pmatrix}}}$.

See also

• Matrices
• Matrix logarithm
• Exponential function
• Exponential mapping

literature

• Roger A. Horn, Charles R. Johnson: Topics in Matrix Analysis . Cambridge University Press, 1991, ISBN 0-521-46713-6 (English).
• Cleve Moler, Charles F. Van Loan: Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later . In: SIAM Review . tape 45 , no. 1 , 2003, ISSN  1095-7200 , p. 1-49 , doi : 10.1137 / S00361445024180 ( cornell.edu [PDF]).
• VI Arnolʹd: Ordinary differential equations . Springer-Verlag, Berlin / New York 1980, ISBN 3-540-09216-1 .

Individual evidence

1. ^ S. Blanes, F. Casas, JA Oteo, J. Ros: The Magnus expansion and some of its applications. (= Physics Reports. Volume 470). Cornell University Library, 2009, OCLC 635162561 .
2. ↑ T. Möller: Symbolic mathematics-based simulation of cylinder spaces for regenerative gas cycles. In: Int J Energy Environ Eng. Springer Berlin / Heidelberg, Feb. 2015. http://link.springer.com/article/10.1007/s40095-015-0163-3