Question

solve; [D^3 − DD′^2 − D^2 + DD′] z = 0

Answer 1

Answer on Question 66041 - Math - Differential Equations

Question: Solve

$(D^{3}-DD^{\prime 2}-D^{2}+DD^{\prime})z=0.$ (1)

Solution: Recall that $D=\frac{\partial}{\partial x}$, $D^{\prime}=\frac{\partial}{\partial y}$. We look for the general solution $z=z(x,y)$ of the PDE with constant coefficients

$\frac{\partial^{3}z}{\partial x^{3}}-\frac{\partial^{3}z}{\partial x\partial y^{2}}-\frac{\partial^{2}z}{\partial x^{2}}-\frac{\partial^{2}z}{\partial x\partial y}=0.$

This operator is reducible

$D^{3}-DD^{\prime 2}-D^{2}+DD^{\prime}=D(D-D^{\prime})(D+D^{\prime}-1).$

Therefore the general solution of (1) has the form [1]

$z(x,y)=u(x,y)+v(x,y)+w(x,y),$

where $u$, $v$ and $w$ are general solutions of the first order PDEs

$u_{x}=0,\qquad v_{x}-v_{y}=0,\qquad w_{x}+w_{y}-w=0$

respectively. All these PDEs can be directly solved by the Lagrange method:

$u_{x}=0\quad\Rightarrow\quad u(x,y)=f(y);$

$v_{x}-v_{y}=0\quad\Rightarrow\quad dx=-dy\quad\Rightarrow\quad x+y=c\quad\Rightarrow\quad v(x,y)=g(x+y);$

$w_{x}+w_{y}=w\quad\Rightarrow\quad dx=dy=\frac{dw}{w}\quad\Rightarrow\quad x-y=c_{1},\;we^{-x}=c_{2}$

$\quad\Rightarrow\quad we^{-x}=h(x-y)\quad\Rightarrow\quad w=e^{x}h(x-y),$

where $f$, $g$ and $h$ are arbitrary $C^{1}$-functions. Finally,

$z(x,y)=f(y)+g(x+y)+e^{x}h(x-y).$

Answer: $z(x,y)=f(y)+g(x+y)+e^{x}h(x-y)$, where $f$, $g$ and $h$ are arbitrary $C^{1}$-functions.

References

- [1] Sneddon, Ian N., and J. C. Polkinghorne. Elements of partial differential equations. Physics Today 10.5 (1957): 96–109.