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This equation is bugging me for a while now. It seems pretty simple but what is the general solution?

There are ways to find solutions like $u(x,y)=Ae^{Bx+\frac{B}{1-B}y}+C(x-y)+D$ But I wonder if there is a way to represent all the solutions.

Any thoughts will be appreciated.

MOMO
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Let $u(x,y)=v(\xi,\eta)$, where $\xi=x+y$ and $\eta=x-y$; then $$ u_{xy}=u_x+u_y \implies v_{\xi\xi}-v_{\eta\eta}=2v_{\xi}. \tag{1} $$ Next, substitute $v=e^{\xi}w$ in $(1)$ and obtain $$ w_{\eta\eta}-w_{\xi\xi}+w=0. \tag{2} $$ Equation $(2)$ is an instance of the Klein-Gordon equation. Its general solution can be obtained using the Fourier transform, and is given by $$ w(\xi,\eta)=\int_{-\infty}^{\infty}\left[A(k)e^{ik\xi+i\omega_k\eta}+B(k)e^{ik\xi-i\omega_k\eta}\right]dk, \tag{3} $$ where $\omega_k=\sqrt{k^2+1}$.

Gonçalo
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    With $w_{\eta\eta}-w_{\xi\xi}+w=0$ separation of variables can be used. – Cesareo Mar 19 '24 at 13:53
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    With an unbounded domain, you will have an uncountable set of feasible separation constants which is equivalent to the Fourier transform. – whpowell96 Mar 19 '24 at 14:54
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    If $(1)$ is correct, then $v_{\xi}=(e^{\xi}w){\xi}=e^{\xi}(w{\xi}+w)$, $v_{\xi\xi}=e^{\xi}(w_{\xi\xi}+2w_{\xi}+w)$, and $v_{\eta\eta}=e^{\xi}w_{\eta\eta}$, hence $$(1)\implies e^{\xi}(w_{\xi\xi}+2w_{\xi}+w-w_{\eta\eta})=2e^{\xi}(w_{\xi}+w)\implies w_{\xi\xi}-w_{\eta\eta}=w \implies (2).$$ – Gonçalo Mar 19 '24 at 21:13
  • Yes you are right – MOMO Mar 20 '24 at 05:57