System of Distributional Differential Equations

Question

Here are the problem and my attempt to the solution. Is it correct?

Answer 1

Let $$ A = \begin{pmatrix}2 & -1 \\ 3 & -2\end{pmatrix}, \quad y = \begin{pmatrix}y_1 \\ y_2\end{pmatrix}, \quad b = \begin{pmatrix}0 \\ 2\end{pmatrix} \delta(x). $$

The equation can then be written $y' = Ay + b.$ This can be rewritten as $y' - Ay = b.$ After multiplication from the left with $e^{-Ax}$ we get $e^{-Ax}y' - e^{-Ax}Ay = e^{-Ax}b.$

The left hand side can now be written as $(e^{-A}y)'$ giving $(e^{-Ax}y)' = e^{-Ax}b,$ i.e. $e^{-Ax}y = \int e^{-Ax}b + c,$ where $c$ is a constant vector and $\int$ denotes taking antiderivative. Thus the solution is given by $$y = e^{Ax} \int e^{-Ax}b + e^{Ax} c.$$

Now $A^2 = I$ (check it!) so $$ e^{Ax} = \sum_{n=0}^{\infty} \frac{1}{n!} (Ax)^n = \sum_{n\text{ even}} \frac{1}{n!} x^n I + \sum_{n\text{ odd}} \frac{1}{n!} x^n A = (\cosh x) I + (\sinh x) A $$ and $$ e^{-Ax} = (\cosh x) I - (\sinh x) A . $$

This makes $$\begin{align} e^{-Ax} b &= \left( (\cosh x) I - (\sinh x) A \right) b \\ &= \begin{pmatrix} \cosh x - 2 \sinh x & \sinh x \\ -3 \sinh x & \cosh x + 2 \sinh x \end{pmatrix} \begin{pmatrix} 0 \\ 2 \end{pmatrix} \delta(x) \\ &= \begin{pmatrix} 2 \sinh x \\ 2 \cosh x + 4 \sinh x \end{pmatrix} \delta(x) \\ &= \begin{pmatrix} 0 \\ 2 \end{pmatrix} \delta(x), \end{align}$$ where the last identity holds since $f(x) \, \delta(x) = f(0) \, \delta(x).$

Now, $$ \int e^{-Ax} b = \begin{pmatrix}0 \\ 2\end{pmatrix} H(x), $$ where $H(x)$ is the Heaviside function.

This gives $$ e^{Ax} \int e^{-Ax} b = \left( (\cosh x) I + (\sinh x) A \right) \begin{pmatrix}0 \\ 2\end{pmatrix} H(x) = \begin{pmatrix}-2 \sinh x \\ 2 \cosh x - 4 \sinh x\end{pmatrix} H(x). $$

Thus the general solution is $$ y(x) = \begin{pmatrix} -2 \sinh x\\ 2 \cosh x - 4 \sinh x \end{pmatrix} H(x) + \begin{pmatrix} c_1 \cosh x + 2 c_1 \sinh x - c_2 \sinh x \\ c_2 \cosh x + 3 c_1 \sinh x - 2 c_2 \sinh x \end{pmatrix}, $$ where $c_1$ and $c_2$ are constants.

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Solving the equation using Fourier transform, just like you I get $$ \begin{pmatrix}\hat{y}_1 \\ \hat{y}_2\end{pmatrix} = \frac{-1}{1+\xi^2} \begin{pmatrix}i\xi+2 & -1 \\ 3 & i\xi-2\end{pmatrix} \begin{pmatrix}0 \\ 2\end{pmatrix} $$ but then you make some mistakes. First you make a sign error. I get $$ \begin{pmatrix}\hat{y}_1 \\ \hat{y}_2\end{pmatrix} = \frac{2}{1+\xi^2} \begin{pmatrix}1 \\ 2-i\xi\end{pmatrix}. $$

The right hand side is the Fourier transform of the convolution $$ e^{-|x|} * \begin{pmatrix}\delta \\ 2\delta-\delta' \end{pmatrix} = \begin{pmatrix}e^{-|x|} * \delta \\ e^{-|x|}*(2\delta-\delta') \end{pmatrix} = \begin{pmatrix}e^{-|x|} \\ 2 e^{-|x|} - (e^{-|x|})' \end{pmatrix} \\ = \begin{pmatrix}e^{-|x|} \\ 2 e^{-|x|} + \operatorname{sign}(x) e^{-|x|} \end{pmatrix} = e^{-|x|} \begin{pmatrix}1 \\ 2 + \operatorname{sign}(x) \end{pmatrix}. $$

Thus, $$ \begin{pmatrix}{y}_1 \\ {y}_2\end{pmatrix} = e^{-|x|} \begin{pmatrix}1 \\ 2 + \operatorname{sign}(x) \end{pmatrix} + \text{solutions to the homogeneous equation}. $$

I should add something about the solutions to the homogeneous equation and also show that the two solutions are equal.