Internal problem ID [10231]
Internal file name [OUTPUT/9178_Monday_June_06_2022_01_34_36_PM_15570617/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 8, system of first order odes
Problem number: 1909.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "system of linear ODEs" Unable to solve or complete the solution.
Solve \begin {align*} x^{\prime }\left (t \right )&=a x \left (t \right )+g y \left (t \right )+\beta z \left (t \right )\\ y^{\prime }\left (t \right )&=g x \left (t \right )+b y \left (t \right )+\alpha z \left (t \right )\\ z^{\prime }\left (t \right )&=\beta x \left (t \right )+\alpha y \left (t \right )+c z \left (t \right ) \end {align*}
In this method, we will assume we have found the matrix exponential \(e^{A t}\) allready. There are different methods to determine this but will not be shown here. This is a system of linear ODE’s given as Warning. Unable to find the matrix exponential.
This is a system of linear ODE’s given as \begin {align*} \vec {x}'(t) &= A\, \vec {x}(t) \end {align*}
Or \begin {align*} \left [\begin {array}{c} x^{\prime }\left (t \right ) \\ y^{\prime }\left (t \right ) \\ z^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{ccc} a & g & \beta \\ g & b & \alpha \\ \beta & \alpha & c \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \left (t \right ) \\ z \left (t \right ) \end {array}\right ] \end {align*}
The first step is find the homogeneous solution. We start by finding the eigenvalues of \(A\). This is done by solving the following equation for the eigenvalues \(\lambda \) \begin {align*} \operatorname {det} \left ( A- \lambda I \right ) &= 0 \end {align*}
Expanding gives \begin {align*} \operatorname {det} \left (\left [\begin {array}{ccc} a & g & \beta \\ g & b & \alpha \\ \beta & \alpha & c \end {array}\right ]-\lambda \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right ) &= 0 \end {align*}
Therefore \begin {align*} \operatorname {det} \left (\left [\begin {array}{ccc} a -\lambda & g & \beta \\ g & b -\lambda & \alpha \\ \beta & \alpha & c -\lambda \end {array}\right ]\right ) &= 0 \end {align*}
Which gives the characteristic equation \begin {align*} \lambda ^{3}-\left (a +b +c \right ) \lambda ^{2}-\left (-a b -c a +\alpha ^{2}-b c +\beta ^{2}+g^{2}\right ) \lambda +a \,\alpha ^{2}-a b c -2 \alpha \beta g +b \,\beta ^{2}+c \,g^{2}&=0 \end {align*}
The roots of the above are the eigenvalues. \begin {align*} \lambda _1 &= \text {Expression too large to display}\\ \lambda _2 &= \text {Expression too large to display}\\ \lambda _3 &= \text {Expression too large to display} \end {align*}
This table summarises the above result
| eigenvalue | algebraic multiplicity | type of eigenvalue |
| \(\text {Expression too large to display}\) | \(1\) | real eigenvalue |
| \(\text {Expression too large to display}\) | \(1\) | complex eigenvalue |
| \(\text {Expression too large to display}\) | \(1\) | complex eigenvalue |
Now the eigenvector for each eigenvalue are found.
Considering the eigenvalue \(\lambda _{1} = \text {Expression too large to display}\)
We need to solve \(A \vec {v} = \lambda \vec {v}\) or \((A-\lambda I) \vec {v} = \vec {0}\) which becomes \begin {align*} \left (\left [\begin {array}{ccc} a & g & \beta \\ g & b & \alpha \\ \beta & \alpha & c \end {array}\right ] - \left (\text {Expression too large to display}\right ) \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\\ \text {Expression too large to display} \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \text {Expression too large to display}&g&\beta& 0\\ g&\text {Expression too large to display}&\alpha& 0\\ \beta& \alpha& \text {Expression too large to display}&0 \end {array} \right ] \] \begin {align*} \text {Expression too large to display} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \text {Expression too large to display}&g&\beta &0\\ 0&\text {Expression too large to display}&\text {Expression too large to display}&0\\ \beta &\alpha &\text {Expression too large to display}&0 \end {array} \right ] \end {align*}
\begin {align*} \text {Expression too large to display} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \text {Expression too large to display}&g&\beta &0\\ 0&\text {Expression too large to display}&\text {Expression too large to display}&0\\ 0&\text {Expression too large to display}&\text {Expression too large to display}&0 \end {array} \right ] \end {align*}
\begin {align*} \text {Expression too large to display} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \text {Expression too large to display}&g&\beta &0\\ 0&\text {Expression too large to display}&\text {Expression too large to display}&0\\ 0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \text {Expression too large to display} \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{3}\}\) and the leading variables are \(\{v_{1}, v_{2}\}\). Let \(v_{3} = t\). Now we start back substitution. Failed to find eigenvector. Terminating
✓ Solution by Maple
Time used: 14.265 (sec). Leaf size: 32449
dsolve([diff(x(t),t)=a*x(t)+g*y(t)+beta*z(t),diff(y(t),t)=g*x(t)+b*y(t)+alpha*z(t),diff(z(t),t)=beta*x(t)+alpha*y(t)+c*z(t)],singsol=all)
\begin{align*} \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.034 (sec). Leaf size: 1639
DSolve[{x'[t]==a*x[t]+g*y[t]+\[Beta]*z[t],y'[t]==g*x[t]+b*y[t]+\[Alpha]*z[t],z'[t]==\[Beta]*x[t]+\[Alpha]*y[t]+c*z[t]},{x[t],y[t],z[t]},t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to -c_3 \text {RootSum}\left [-\text {$\#$1}^3+\text {$\#$1}^2 a+\text {$\#$1}^2 b+\text {$\#$1}^2 c+\text {$\#$1} \alpha ^2-\text {$\#$1} a b-\text {$\#$1} a c+\text {$\#$1} \beta ^2-\text {$\#$1} b c+\text {$\#$1} g^2-a \alpha ^2+a b c-b \beta ^2-c g^2+2 \alpha \beta g\&,\frac {-b \beta e^{\text {$\#$1} t}+\alpha g e^{\text {$\#$1} t}+\text {$\#$1} \beta e^{\text {$\#$1} t}}{-3 \text {$\#$1}^2+2 \text {$\#$1} a+2 \text {$\#$1} b+2 \text {$\#$1} c+\alpha ^2-a b-a c+\beta ^2-b c+g^2}\&\right ]+c_2 \text {RootSum}\left [-\text {$\#$1}^3+\text {$\#$1}^2 a+\text {$\#$1}^2 b+\text {$\#$1}^2 c+\text {$\#$1} \alpha ^2-\text {$\#$1} a b-\text {$\#$1} a c+\text {$\#$1} \beta ^2-\text {$\#$1} b c+\text {$\#$1} g^2-a \alpha ^2+a b c-b \beta ^2-c g^2+2 \alpha \beta g\&,\frac {-c g e^{\text {$\#$1} t}+\text {$\#$1} g e^{\text {$\#$1} t}+\alpha \beta e^{\text {$\#$1} t}}{3 \text {$\#$1}^2-2 \text {$\#$1} a-2 \text {$\#$1} b-2 \text {$\#$1} c-\alpha ^2+a b+a c-\beta ^2+b c-g^2}\&\right ]+c_1 \text {RootSum}\left [-\text {$\#$1}^3+\text {$\#$1}^2 a+\text {$\#$1}^2 b+\text {$\#$1}^2 c+\text {$\#$1} \alpha ^2-\text {$\#$1} a b-\text {$\#$1} a c+\text {$\#$1} \beta ^2-\text {$\#$1} b c+\text {$\#$1} g^2-a \alpha ^2+a b c-b \beta ^2-c g^2+2 \alpha \beta g\&,\frac {\text {$\#$1}^2 e^{\text {$\#$1} t}+b c e^{\text {$\#$1} t}-\text {$\#$1} b e^{\text {$\#$1} t}-\text {$\#$1} c e^{\text {$\#$1} t}+\alpha ^2 \left (-e^{\text {$\#$1} t}\right )}{3 \text {$\#$1}^2-2 \text {$\#$1} a-2 \text {$\#$1} b-2 \text {$\#$1} c-\alpha ^2+a b+a c-\beta ^2+b c-g^2}\&\right ] \\ y(t)\to c_1 \text {RootSum}\left [-\text {$\#$1}^3+\text {$\#$1}^2 a+\text {$\#$1}^2 b+\text {$\#$1}^2 c+\text {$\#$1} \alpha ^2-\text {$\#$1} a b-\text {$\#$1} a c+\text {$\#$1} \beta ^2-\text {$\#$1} b c+\text {$\#$1} g^2-a \alpha ^2+a b c-b \beta ^2-c g^2+2 \alpha \beta g\&,\frac {-c g e^{\text {$\#$1} t}+\text {$\#$1} g e^{\text {$\#$1} t}+\alpha \beta e^{\text {$\#$1} t}}{3 \text {$\#$1}^2-2 \text {$\#$1} a-2 \text {$\#$1} b-2 \text {$\#$1} c-\alpha ^2+a b+a c-\beta ^2+b c-g^2}\&\right ]+c_3 \text {RootSum}\left [-\text {$\#$1}^3+\text {$\#$1}^2 a+\text {$\#$1}^2 b+\text {$\#$1}^2 c+\text {$\#$1} \alpha ^2-\text {$\#$1} a b-\text {$\#$1} a c+\text {$\#$1} \beta ^2-\text {$\#$1} b c+\text {$\#$1} g^2-a \alpha ^2+a b c-b \beta ^2-c g^2+2 \alpha \beta g\&,\frac {-a \alpha e^{\text {$\#$1} t}+\beta g e^{\text {$\#$1} t}+\text {$\#$1} \alpha e^{\text {$\#$1} t}}{3 \text {$\#$1}^2-2 \text {$\#$1} a-2 \text {$\#$1} b-2 \text {$\#$1} c-\alpha ^2+a b+a c-\beta ^2+b c-g^2}\&\right ]+c_2 \text {RootSum}\left [-\text {$\#$1}^3+\text {$\#$1}^2 a+\text {$\#$1}^2 b+\text {$\#$1}^2 c+\text {$\#$1} \alpha ^2-\text {$\#$1} a b-\text {$\#$1} a c+\text {$\#$1} \beta ^2-\text {$\#$1} b c+\text {$\#$1} g^2-a \alpha ^2+a b c-b \beta ^2-c g^2+2 \alpha \beta g\&,\frac {\text {$\#$1}^2 e^{\text {$\#$1} t}+a c e^{\text {$\#$1} t}-\text {$\#$1} a e^{\text {$\#$1} t}-\text {$\#$1} c e^{\text {$\#$1} t}+\beta ^2 \left (-e^{\text {$\#$1} t}\right )}{3 \text {$\#$1}^2-2 \text {$\#$1} a-2 \text {$\#$1} b-2 \text {$\#$1} c-\alpha ^2+a b+a c-\beta ^2+b c-g^2}\&\right ] \\ z(t)\to -c_1 \text {RootSum}\left [-\text {$\#$1}^3+\text {$\#$1}^2 a+\text {$\#$1}^2 b+\text {$\#$1}^2 c+\text {$\#$1} \alpha ^2-\text {$\#$1} a b-\text {$\#$1} a c+\text {$\#$1} \beta ^2-\text {$\#$1} b c+\text {$\#$1} g^2-a \alpha ^2+a b c-b \beta ^2-c g^2+2 \alpha \beta g\&,\frac {-b \beta e^{\text {$\#$1} t}+\alpha g e^{\text {$\#$1} t}+\text {$\#$1} \beta e^{\text {$\#$1} t}}{-3 \text {$\#$1}^2+2 \text {$\#$1} a+2 \text {$\#$1} b+2 \text {$\#$1} c+\alpha ^2-a b-a c+\beta ^2-b c+g^2}\&\right ]+c_2 \text {RootSum}\left [-\text {$\#$1}^3+\text {$\#$1}^2 a+\text {$\#$1}^2 b+\text {$\#$1}^2 c+\text {$\#$1} \alpha ^2-\text {$\#$1} a b-\text {$\#$1} a c+\text {$\#$1} \beta ^2-\text {$\#$1} b c+\text {$\#$1} g^2-a \alpha ^2+a b c-b \beta ^2-c g^2+2 \alpha \beta g\&,\frac {-a \alpha e^{\text {$\#$1} t}+\beta g e^{\text {$\#$1} t}+\text {$\#$1} \alpha e^{\text {$\#$1} t}}{3 \text {$\#$1}^2-2 \text {$\#$1} a-2 \text {$\#$1} b-2 \text {$\#$1} c-\alpha ^2+a b+a c-\beta ^2+b c-g^2}\&\right ]+c_3 \text {RootSum}\left [-\text {$\#$1}^3+\text {$\#$1}^2 a+\text {$\#$1}^2 b+\text {$\#$1}^2 c+\text {$\#$1} \alpha ^2-\text {$\#$1} a b-\text {$\#$1} a c+\text {$\#$1} \beta ^2-\text {$\#$1} b c+\text {$\#$1} g^2-a \alpha ^2+a b c-b \beta ^2-c g^2+2 \alpha \beta g\&,\frac {\text {$\#$1}^2 e^{\text {$\#$1} t}+a b e^{\text {$\#$1} t}-\text {$\#$1} a e^{\text {$\#$1} t}-\text {$\#$1} b e^{\text {$\#$1} t}+g^2 \left (-e^{\text {$\#$1} t}\right )}{3 \text {$\#$1}^2-2 \text {$\#$1} a-2 \text {$\#$1} b-2 \text {$\#$1} c-\alpha ^2+a b+a c-\beta ^2+b c-g^2}\&\right ] \\ \end{align*}