Question #197611

Reduce to canonical form

uxx - uxy + uyy + ux = 0


1
Expert's answer
2021-05-24T18:13:08-0400
A=1,B=1,C=1A=1, B=-1, C=1

B24AC=14=3<0B^2-4AC=1-4=-3<0

The given partial differential equation is elliptic.

 The characteristic equations are given by 


dydx=BB24AC2A=132(1)\dfrac{dy}{dx}=\dfrac{B-\sqrt{B^2-4AC}}{2A}=\dfrac{-1-\sqrt{-3}}{2(1)}

=1i32=\dfrac{-1-i\sqrt{3}}{2}

and



dydx=B+B24AC2A=1+32(1)\dfrac{dy}{dx}=\dfrac{B+\sqrt{B^2-4AC}}{2A}=\dfrac{-1+\sqrt{-3}}{2(1)}

=1+i32=\dfrac{-1+i\sqrt{3}}{2}


Integrating these two equations, we get 


y=x(1i32)+c1y=x(\dfrac{-1-i\sqrt{3}}{2})+c_1

and


y=x(1+i32)+c2y=x(\dfrac{-1+i\sqrt{3}}{2})+c_2

Assume that


ξ=y+x(1+i32)\xi=y+x(\dfrac{1+i\sqrt{3}}{2})

η=y+x(1i32)\eta=y+x(\dfrac{1-i\sqrt{3}}{2})

The transformation


α=12(ξ+η)\alpha=\dfrac{1}{2}(\xi+\eta)

β=12i(ξη)\beta=\dfrac{1}{2i}(\xi-\eta)


α=y+x\alpha=y+x

β=32x\beta=\dfrac{\sqrt{3}}{2}x

αx=1,αy=1\alpha_x=1, \alpha_y=1

αxx=0,αxy=0,αyy=0\alpha_{xx}=0, \alpha_{xy}=0, \alpha_{yy}=0

βx=32,βy=0\beta_x=\dfrac{\sqrt{3}}{2}, \beta_y=0

βxx=0,βxy=0,βyy=0\beta_{xx}=0, \beta_{xy}=0, \beta_{yy}=0

A=1,B=1,C=1,D=1,E=0,F=0,G=0A=1, B=-1, C=1,D=1, E=0, F=0, G=0


Aˉ=Aαx2+Bαxαy+Cαy2=11+1=1\bar{A}=A\alpha_x^2+B\alpha_x\alpha_y+C\alpha_y^2=1-1+1=1

Bˉ=2Aαxβx+B(αxβy+αyβx)+2Cαyβy\bar{B}=2A\alpha_x\beta_x+B(\alpha_x\beta_y+\alpha_y\beta_x)+2C\alpha_y\beta_y

=2(1)(32)1(0+(1)32)+2(1)(0)=32=2(1)(\dfrac{\sqrt{3}}{2})-1(0+(1)\dfrac{\sqrt{3}}{2})+2(1)(0)=\dfrac{\sqrt{3}}{2}


Cˉ=Aβx2+Bβxβy+2Cαyβy=\bar{C}=A\beta_x^2+B\beta_x\beta_y+2C\alpha_y\beta_y=

=34+0+0=34=\dfrac{3}{4}+0+0=\dfrac{3}{4}

Dˉ=Aαxx+Bαxy+Cαyy+Dαx+Eαy=1\bar{D}=A\alpha_{xx}+B\alpha_{xy}+C\alpha_{yy}+D\alpha_x+E\alpha_y=1

Eˉ=Aβxx+Bβxy+Cβyy+Dβx+Eβy=32\bar{E}=A\beta_{xx}+B\beta_{xy}+C\beta_{yy}+D\beta_x+E\beta_y=\dfrac{\sqrt{3}}{2}

Fˉ=F=0,Gˉ=G=0\bar{F}=F=0, \bar{G}=G=0


uαα+32uαβ+32uββ+uα+32uβ=0u_{\alpha\alpha}+\dfrac{\sqrt{3}}{2}u_{\alpha\beta}+\dfrac{\sqrt{3}}{2}u_{\beta\beta}+u_{\alpha}+\dfrac{\sqrt{3}}{2}u_\beta=0



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS