$\text{Question: }$

Consider the following system for $u(x, y)$

$$ \begin{array}{l} \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial x \partial y}-2 \frac{\partial^{2} u}{\partial y^{2}}+1=0, \quad 0<x<1, y>0 \\ u(x, 0)=x, \quad \frac{\partial u}{\partial y}(x, 0)=x \end{array} $$

(a) Determine if the equation is elliptic, parabolic, or hyperbolic.

So, we have the case: $A(x, y) u_{x x}+2 B(x, y) u_{x y}+C(x, y) u_{y y}+D=0.$

With $A(x, y) \equiv 1 ;\space B(x, y) \equiv \frac{1}{2} \space; C(x, y) \equiv-2 \quad \& \quad D \equiv 1$.

$\therefore A C-B^{2}=-2-\left(\frac{1}{2}\right)^{2}=-2-\frac{1}{4}=-\frac{9}{4}<0. \quad \therefore P D E\space$ is hyperbolic.

(b) Find a solution to the above system. As for this question I've tried various things such as the following $ \left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial x \partial y}-2 \frac{\partial^{2}}{\partial y^{2}}\right) u=-1 \Leftrightarrow\left(\frac{\partial}{\partial x}-\sqrt{2} \frac{\partial}{\partial y}\right)\left(\frac{\partial}{\partial y}+\sqrt{2} \frac{\partial}{\partial y}\right) u+\frac{\partial^{2} u}{\partial x \partial y}=-1 $ I've tried to do this so I can define a $v$ e.g. $v:=\left(\frac{\partial}{\partial y}+\sqrt{2} \frac{\partial}{\partial y}\right)u$ and then solve an easier system with such function. But after a while now I have had no inspiration so it would be great if I could get some help :)

Any help or corrections to the above workings would be great.


2 Answers 2


$ \begin{aligned} &\left(\partial_{x x}+\partial_{x y}-2 \partial_{y y}\right) u+1=0 \Rightarrow\left(\partial_{x}+2 \partial_{y}\right)\left(\partial_{x}-\partial_{y}\right) u+1=0 \\ \Rightarrow &\left(\frac{1}{2} \partial_{x}+\partial_{y}\right)\left(\partial_{x}+\partial_{y}\right) u+1=0 \\ &\frac{d x}{d y}=\frac{1}{2} \Rightarrow x=\frac{y}{2}+\xi \Rightarrow \xi=x-\frac{y}{2} ; \frac{d x}{d y}=-1 \Rightarrow x=-y+\eta \Rightarrow \eta=x+y \end{aligned} $

$ \text{So, we have: } \frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial x}+\frac{\partial u}{\partial \eta} \frac{\partial \eta}{\partial x} ; \frac{\partial u}{\partial y}=\frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial y}+\frac{\partial u}{\partial \eta} \frac{\partial \eta}{\partial y} $

$ \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial \xi^{2}}\left(\frac{\partial \xi}{\partial x}\right)^{2}+2 \frac{\partial^{2} u}{\partial \xi \partial \eta} \frac{\partial \xi}{\partial x} \frac{\partial \eta}{\partial x}+\frac{\partial^{2} u}{\partial \eta^{2}}\left(\frac{\partial \eta}{\partial x}\right)^{2}+\frac{\partial u}{\partial \xi}\left(\frac{\partial^{2} \xi}{\partial x^{2}}\right)+\frac{\partial u}{\partial \eta}\left(\frac{\partial^{2} \eta}{\partial x^{2}}\right) $

$ \frac{\partial^{2} u}{\partial y^{2}}=\frac{\partial^{2} u}{\partial \xi^{2}}\left(\frac{\partial \xi}{\partial y}\right)^{2}+2 \frac{\partial^{2} u}{\partial \xi \partial \eta} \frac{\partial \xi}{\partial y} \frac{\partial \eta}{\partial y}+\frac{\partial^{2} u}{\partial \eta^{2}}\left(\frac{\partial \eta}{\partial y}\right)^{2}+\frac{\partial u}{\partial \xi}\left(\frac{\partial^{2} \xi}{\partial y^{2}}\right)+\frac{\partial u}{\partial \eta}\left(\frac{\partial^{2} \eta}{\partial y^{2}}\right) $

$ \text{And, also: } \frac{\partial^{2} u}{\partial x \partial y}=\frac{\partial^{2} u}{\partial \xi^{2}}\left(\frac{\partial \xi}{\partial x}\right)^{2}+2 \frac{\partial^{2} u}{\partial \xi \partial \eta} \frac{\partial \xi}{\partial x} \frac{\partial \eta}{\partial y}+\frac{\partial^{2} u}{\partial \eta^{2}}\left(\frac{\partial \eta}{\partial y}\right)^{2}+\frac{\partial u}{\partial \xi}\left(\frac{\partial \xi}{\partial x}\right)+\frac{\partial u}{\partial \eta}\left(\frac{\partial^{2} \eta}{\partial y^{2}}\right) $

$ \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial \xi^{2}}+2 \frac{\partial^{2} u}{\partial \xi \partial \eta}+\frac{\partial^{2} u}{\partial \eta^{2}} ; \frac{\partial^{2} u}{\partial y^{2}}=\frac{1}{4} \frac{\partial^{2} u}{\partial \xi^{2}}-\frac{\partial^{2} u}{\partial \xi \partial \eta}+\frac{\partial^{2} u}{\partial \eta^{2}} ; \frac{\partial^{2} u}{\partial x \partial y}=-\frac{1}{2} \frac{\partial^{2} u}{\partial \xi^{2}}+\frac{1}{2}\frac{\partial^{2} u}{\partial \xi \partial \eta}+\frac{\partial^{2} u}{\partial \eta^{2}} $

$ \therefore u_{\xi \xi}+2 u_{\xi \eta}+u_{\eta \eta}-\frac{1}{2} u_{\xi \xi}+\frac{1}{2} u_{\xi \eta}+u_{\eta \eta}-\frac{1}{2} u_{\xi \xi}+2 u_{\xi \eta}-2 u_{\eta\eta}+1=0 $

$ \begin{array}{l} \Rightarrow \frac{9}{2} u_{\xi \eta}+1=0 \Rightarrow u_{\xi\eta} =-\frac{2}{9} \Rightarrow u_{\xi}=-\frac{2}{9} \eta+F(\xi) \\ \Rightarrow u(\xi, \eta)=-\frac{2}{9} \eta \xi+f(\xi)+g(\eta). \\ \\ \text {Using initial conditions: } y=0, \xi=\eta=x \\ u(\xi=x, \eta=x)=-\frac{2}{9} x^{2}+f(x)+g(x)=x \Rightarrow f(x)+g(x)=\frac{2}{9} x^{2}+x \end{array} $

\begin{array}{l} \text { Observe that: } \\ u_{y}=-\frac{1}{2} u_{\xi}+u_{\eta}=\frac{1}{9} \eta-\frac{1}{2} f^{\prime}(\xi)+g^{\prime}(\eta) \\ u_{y}(\xi=x, \eta=x)=-\frac{1}{9} x-\frac{1}{2} f^{\prime}(x)+g^{\prime}(x)=x \Rightarrow g^{\prime}(x)-\frac{1}{2} f^{\prime}(x)=\frac{10}{9} x \end{array}

$ \Rightarrow g(x)-\frac{1}{2} f(x)=\frac{5}{9} x^{2}+c \Rightarrow f(x)=-\frac{2}{9} x^{2}+\frac{2}{3} x+k $

$ \begin{array}{l} \Rightarrow g(x)=\frac{4}{9} x^{2}+\frac{1}{3} x-k \\ \therefore u(\xi, \eta)=-\frac{2}{9} \eta \xi-\frac{2}{9} \xi^{2}+\frac{2}{3} \xi+\frac{4}{9} \eta^{2}+\frac{1}{3} \eta \\ \therefore u(x, y)=-\frac{2}{9}(x+y)\left(x-\frac{y}{2}\right)-\frac{2}{9}\left(x-\frac{y}{2}\right)^{2}+\frac{2}{3}\left(x-\frac{y}{2}\right)+\frac{4}{9}(x+y)^{2}+\frac{1}{3}(x+y) \end{array} $

\begin{array}{l} =\frac{-2}{9}\left(x^{2}+\frac{x y}{2}-\frac{y^{2}}{2}\right)-\frac{2}{9}\left(x^{2}-x y+\frac{y^{2}}{4}\right)+\frac{2}{3} x-\frac{1}{3} y+\frac{4}{9}\left(x^{2}+2 x y+y^{2}\right)+\frac{1}{2} x+\frac{1}{3} y \\ =-\frac{4}{9} x^{2}+\frac{1}{9} x y+\frac{1}{18} y^{2}+x+\frac{4}{9} x^{2}+\frac{8}{9} x y+\frac{4}{9} y^{2} \\ =x y+x+\frac{y^{2}}{2} \end{array}



The PDE is linear then the solution can be build as a sum of the homogeneous plus a particular solution, A particular solution can be obtained by making $u_p(x,y)=ax^2+b xy +c y^2$ and after substitution we obtain

$$ u_p(x,y)=ax^2+b xy +c y^2,\ \ 2a+b-4c+1=0 $$

The homogeneous is

$$ u_h(x,y) = \phi\left(y-2x\right)+\psi\left(y+x\right) $$

hence $u(x,y) = u_h(x,y)+u_p(x,y)$



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