Introduction : In this question, We will apply Duhamel principle to solve system of nonhomogeneous differential equations u’=v+2e^(-4t),v’=u-e^t , u(0)=-1/2,v(0)=2.
Question : Apply Duhamel principle to solve system of non homogeneous differential equations
\[\left(\begin{array}{c}u'(t) \\v'(t) \end{array}\right)=\left(\begin{array}{cc}0 & 1 \\1 & 0 \end{array}\right)\left(\begin{array}{c}u(t) \\v(t) \end{array}\right)+\left( {\begin{array}{*{20}{c}}{2{e^{ – 4t}}}\\{ – {e^t}}\end{array}} \right)\]
such that \(u(0)=-1/2,v(0)=2\)
Solution :
Matrix form of the system is
\[\left(\begin{array}{c}u'(t) \\v'(t) \end{array}\right)=\left(\begin{array}{cc}0 & 1 \\1 & 0 \end{array}\right)\left(\begin{array}{c}u(t) \\v(t) \end{array}\right)+\left( {\begin{array}{*{20}{c}}{2{e^{ – 4t}}}\\{ – {e^t}}\end{array}} \right)\]
Let \[A =\left(\begin{array}{cc}0 & 1 \\1 & 0 \end{array}\right), B(t)=\left( {\begin{array}{*{20}{c}}{2{e^{ – 4t}}}\\{ – {e^t}}\end{array}} \right)\]
Since eigen values of matrix A are -1 and 1 (distinct) hence A is a diagonalizable matrix and it’s diagonalization is
\[A = \left(\begin{array}{cc}-1 & 1 \\1 & 1 \end{array}\right)\left(\begin{array}{cc}-1 & 0 \\0 & 1 \end{array}\right)\left(\begin{array}{cc}-1 & 1 \\1 & 1 \end{array}\right)^{-1}\]
\[= \left(\begin{array}{cc}-1 & 1 \\1 & 1 \end{array}\right)\left(\begin{array}{cc}-1 & 0 \\0 & 1 \end{array}\right)\left(\begin{array}{cc}-\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{array}\right)\]
Hence
\[e^{tA} =\left(\begin{array}{cc}-1 & 1 \\1 & 1 \end{array}\right)\left(\begin{array}{cc} e^{-t}&0 \\0&e^t \end{array}\right)\left(\begin{array}{cc}-\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{array}\right)\]
Now by Duhamel principle, solution of the system is given as
\[\left( {\begin{array}{*{20}{c}}{u(t)}\\{v(t)}\end{array}} \right) =e^{tA} \left( {\begin{array}{*{20}{c}}{u(0)}\\{v(0)}\end{array}}\right)+e^{tA}\int_{0}^{t}e^{-sA}B\left(s\right)ds\]
\[=\left(\begin{array}{cc} e^{-t}&0 \\0&e^t \end{array}\right)\left( {\begin{array}{*{20}{c}}{\frac{-1}{2}}\\{2}\end{array}}\right)+\left(\begin{array}{cc} e^{-t}&0 \\0&e^t \end{array}\right)\int_{0}^{t}\left(\begin{array}{cc} e^{-s}&0 \\0&e^s\end{array}\right)\left( {\begin{array}{*{20}{c}}{2{e^{ – 4s}}}\\{ – {e^s}}\end{array}} \right)ds\]
\[=\left( {\begin{array}{*{20}{c}}{\frac{-e^t}{2}}\\{2e^t}\end{array}}\right)+\left(\begin{array}{cc} e^{-t}&0 \\0&e^t \end{array}\right)\int_{0}^{t}\left( {\begin{array}{*{20}{c}}{2{e^{ – 5s}}}\\{ – {e^{2s}}}\end{array}} \right)ds\]
\[=\left( {\begin{array}{*{20}{c}}{\frac{-e^t}{2}}\\{2e^t}\end{array}}\right)+\left(\begin{array}{cc} e^{-t}&0 \\0&e^t \end{array}\right)\left( {\begin{array}{*{20}{c}}{\frac{-2}{5}({e^{ – 5t}-1})}\\{ – {\frac{1}{2}(e^{2t}-1))}}\end{array}} \right)\]
\[=\left( {\begin{array}{*{20}{c}}{\frac{-e^t}{2}}\\{2e^t}\end{array}}\right)+\left( {\begin{array}{*{20}{c}}{\frac{-2}{5}({e^{ – 6t}-e^{-t}})}\\{ – {\frac{1}{2}}}(e^{3t}-e^t)\end{array}} \right)\]
\[=\left( \begin{array}{l}\frac{{{-e^t}}}{2} + \frac{2}{5}({e^{ – t}} – {e^{ – 6t}})\\{\frac{5}{2}e^t – {\frac{1}{2}}}(e^{2t})\end{array} \right)\]