Answer to Question #93399 in Differential Equations for Subhasis Padhy

Question

Solve :
(D²-2DD´+D´²)z = 12xy

Answer 1

Consider a quadratic form $x^2-2xy+y^2$ (for a differential equation $A_{11}\frac{\partial^2 z}{\partial x^2}+A_{12}\frac{\partial^2 z}{\partial x\partial y}+A_{22}\frac{\partial^2 z}{\partial y^2}+\ldots=0$ we consider a quadratic form $A_{11}x^2+A_{12}xy+A_{22}y^2$ ). Its matrix is $\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$ with eigenvectors $(1,1)$ and $(1,-1)$ .

So we have change of the variables: $u=1\cdot x+1\cdot y = x+y$ and $v=1\cdot x+(-1)\cdot y = x-y$ .

Then $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial f}{\partial u}+\frac{\partial f}{\partial v}$ and $\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=\frac{\partial f}{\partial u}-\frac{\partial f}{\partial v}$ . Move from the variables $x,y$ to the variables $u,v$ .

In case $f=z$ we have $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}$ and $\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}-\frac{\partial z}{\partial v}$ .

In case $f=\frac{\partial z}{\partial x}$ we have $\frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}\bigl(\frac{\partial z}{\partial x}\bigr)=\frac{\partial}{\partial u}\bigl(\frac{\partial z}{\partial x}\bigr)+\frac{\partial}{\partial v}\bigl(\frac{\partial z}{\partial x}\bigr)=\frac{\partial^2 z}{\partial u^2}+2\frac{\partial^2 z}{\partial u\partial v}+\frac{\partial^2 z}{\partial v^2}$ and $\frac{\partial^2 z}{\partial x\partial y}=\frac{\partial}{\partial y}\bigl(\frac{\partial z}{\partial x}\bigr)=\frac{\partial}{\partial u}\bigl(\frac{\partial z}{\partial x}\bigr)-\frac{\partial}{\partial v}\bigl(\frac{\partial z}{\partial x}\bigr)=\frac{\partial^2 z}{\partial u^2}-\frac{\partial^2 z}{\partial v^2}$

In case $f=\frac{\partial z}{\partial y}$ we have $\frac{\partial^2 z}{\partial y^2}=\frac{\partial}{\partial y}\bigl(\frac{\partial z}{\partial y}\bigr)=\frac{\partial}{\partial u}\bigl(\frac{\partial z}{\partial y}\bigr)-\frac{\partial}{\partial v}\bigl(\frac{\partial z}{\partial y}\bigr)=\frac{\partial^2 z}{\partial u^2}-2\frac{\partial^2 z}{\partial u\partial v}+\frac{\partial^2 z}{\partial v^2}$ .

In addition $x=\frac{u+v}{2}$ and $y=\frac{u-v}{2}$ .

Rewrite the original equation: $\bigl(\frac{\partial^2 z}{\partial u^2}+2\frac{\partial^2 z}{\partial u\partial v}+\frac{\partial^2 z}{\partial v^2}\bigr)-2\bigl(\frac{\partial^2 z}{\partial u^2}-\frac{\partial^2 z}{\partial v^2}\bigr)+$

$+\bigl(\frac{\partial^2 z}{\partial u^2}-2\frac{\partial^2 z}{\partial u\partial v}+\frac{\partial^2 z}{\partial v^2}\bigr)=12\frac{u+v}{2}\frac{u-v}{2}$ . That is $\frac{\partial^2 z}{\partial v^2}=\frac{3}{4}u^2-\frac{3}{4}v^2$

Integrate it with respect to $v$ : $\frac{\partial z}{\partial v}=\frac{3}{4}u^2v-\frac{1}{4}v^3+R(u)$ and again $z=\frac{3}{8}u^2v^2-\frac{1}{16}v^4+R(u)v+S(u)$,

where $R(u), S(u)$ are arbitrary twice differentiable functions.

Return to the variables $x$ and $y$ : $z=\frac{3}{8}(x+y)^2(x-y)^2-\frac{1}{16}(x-y)^4+(x-y)\cdot R(x+y)+$

$+S(x+y)$