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I am trying to solve the following linear system of 2 PDE's in 3 variables: $$u_t-u_x+v_y=0$$ $$v_t-u_y-v_x=0$$ With initial condition $u(x,y,0)=a(x,y),v(x,y,0)=b(x,y)$. In matrix form, I can write this system as:

$-\begin{pmatrix}u_x\\v_x\end{pmatrix}\\ $$+ \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}\begin{pmatrix}u_y\\v_y\end{pmatrix}\\ $$+ \begin{pmatrix}u_t\\v_t\end{pmatrix}\\ $$= \begin{pmatrix}0\\0\end{pmatrix}\\$

I can diagonalize the matrix above and get: $$\begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}=\begin{pmatrix}1 & 1 \\-i & i\end{pmatrix}\begin{pmatrix}-i & 0 \\0 & i\end{pmatrix}\begin{pmatrix}1 & 1 \\-i & i\end{pmatrix}^{-1}$$ Thus, I can define $\begin{pmatrix}f\\g\end{pmatrix}=\begin{pmatrix}1 & 1 \\-i & i\end{pmatrix}^{-1}\begin{pmatrix}u\\v\end{pmatrix}$, multiply both sides by $\begin{pmatrix}1 & 1 \\-i & i\end{pmatrix}^{-1}$ and get the separated system: $$-f_x-if_y+f_t=0$$ $$-g_x+ig_y+g_t=0$$ With initial condition: $$f(x,y,0)=\frac{1}{2}(a(x,y)+ib(x,y))$$ $$g(x,y,0)=\frac{1}{2}(a(x,y)-ib(x,y))$$ So far, I am basically following the method from this post. I can now solve this system via the standard method of characteristics and then go back to the original functions $u,v$. The only problem is, that since the new PDE I got has complex coefficients (due to the fact that the eigenvalues are complex), I get the following solution for $u$: $$u(x,y,t)=\frac{1}{2}[a(x+t,y+it)+ib(x+t,y+it)+a(x+t,y-it)-ib(x+t,y-it)]$$ And the solution for $v$ is similar. This is no good, since $a,b$ are real functions, but in the solution I plug in a complex number (which is not in the domain of $a,b$).

Does anyone understand what the problem is? Will this happen for any system with complex eigenvalues? Does that mean that there is no solution in this case? Is there something I can do to solve this?

Thanks in advance.

GSofer
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  • You haven't included the coefficient matrix for $(u_{x}, v_{x})$ in your post above. It would be modified by the the matrix of eigenvectors from the diagonalisation of the coefficient matrix of $(u_{y}, v_{y})$ as well. – Matthew Cassell Jun 07 '21 at 18:05
  • @mattos The coefficient matrices are minus the identity for $x$ and the identity for $y$, so I have not written them explicitly (and they are not affected by the diagonalization). – GSofer Jun 07 '21 at 18:09
  • I don't understand why sorry. You have $$\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \vec{u}{t} + \begin{pmatrix} -1 & 0 \ 0 & 1 \end{pmatrix} \vec{u}{x} + \begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix} \vec{u}{y} = \vec{0}$$ The only identity matrix is for the time derivative. The coefficient matrix for $\vec{u}{x} = (u_{x}, v_{x})$ is not the identity. – Matthew Cassell Jun 07 '21 at 18:14
  • @mattos You are right, I did not write this correctly in the first line. Fixed it. – GSofer Jun 07 '21 at 18:23

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