I want to solve the following system $$\eqalign{ & y_t = y_x + z\text{ in }(0,\text{T}) \times (0,1) \cr & z_t = z_x + y\text{ in }(0,\text{T}) \times (0,1) \cr & y(0,x) = y_0,\,\,z(0,x) = z_0, \cr} $$ I have tried to solve explcitly the fisrt and the second, but the problem is in the coupling, if the coupling is just on one of them there will be no problems. I have tried also to compute the associated semigroup to the system but I didn't get a result. Do you know any method to deal with such a system? Thank you.

$\begingroup$ Boundary conditions ought to make an appearance somewhere ... $\endgroup$ – Michael Renardy Dec 12 '18 at 2:59
Define the vector $X=(y,z)$ and matrices $\sigma_1={{0 1}\choose{1 0}}$, $\sigma_3={{1\; 0}\choose{0\, 1}}$, then the differential equations read $$X_t = \sigma_3 X_x + \sigma_1 X.$$ No boundary conditions were specified at $x=0$ and $x=1$, depending on the boundary conditions you could now write the $x$dependence as a Fourier sine or cosine series. Let me here consider an infinite range for $x$, so no boundary condition is needed. Fourier transformation $X(k,t)=\int e^{ikx} X(x,t)\,dx$ with respect to $x$ gives $$X_t = (ik\sigma_3+ \sigma_1)X\Rightarrow X(k,t)=e^{(ik\sigma_3+ \sigma_1)t}X(k,0).$$ In components this becomes $$\begin{pmatrix}y(k,t)\\ z(k,t)\end{pmatrix}=\left( \begin{array}{cc} \cosh \left(q t\right)+(ik/q)\sinh \left(q t\right)& (1/q)\sinh \left(q t\right)\\ (1/q)\sinh \left(q t\right) & \cosh \left(q t\right)(ik/q) \sinh \left(q t\right) \\ \end{array}\right)\begin{pmatrix}y(k,0)\\ z(k,0)\end{pmatrix},$$ where I abbreviated $q=\sqrt{1k^2}$. Notice that the matrix is even in $q$, so there is no ambiguity in the sign of the square root.

$\begingroup$ Thank you sir. Can I use the method of characterestics on a bounded interval let say (0,1)? If it works, can you give me just a hint. Thank you. $\endgroup$ – Gustave Dec 11 '18 at 15:20

1$\begingroup$ no, I don't see how the method of characteristics can be applied to your system of equations. $\endgroup$ – Carlo Beenakker Dec 11 '18 at 15:24

$\begingroup$ So the only method is by using Fourier representation ? $\endgroup$ – Gustave Dec 11 '18 at 15:25

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