设变换为u=x-2y、v=x+ay,可把方程d²z/dx²+d²z/(dxdy)-d²z/dy²=0化简为d²z/(dudv)=0且d²z/(dudv)=d²z/(dvdu)求常数a可把方程6d²z/dx²+d²z/(dxdy)-d²z/dy²=0化简为d²z/(du

Question

设变换为u=x-2y、v=x+ay,可把方程d²z/dx²+d²z/(dxdy)-d²z/dy²=0化简为d²z/(dudv)=0且d²z/(dudv)=d²z/(dvdu)
求常数a
可把方程6d²z/dx²+d²z/(dxdy)-d²z/dy²=0化简为d²z/(dudv)=0且d²z/(dudv)=d²z/(dvdu)

Answer 1

dz/dx=dz/du*1+dz/dv*1=dz/du+dz/dv;
dz/dy=dz/du*(-2)+dz/dv*(a)=adz/dv-2dz/du;于是
d^2z/dx^2=d(dz/du+dz/dv)/dx
=d^2z/du^2+d^2z/dudv+d^2z/dvdu+d^2z/dv^2
=d^2z/du^2+2d^2z/dudv+d^2z/dv^2；
d^2z/dxdy=d(dz/du+dz/dv)/dy
=d^2z/du^2(-2)+d^2z/dudv*a+d^2z/dvdu*(-2)+d^2z/dv^2*(a)
=-2d^2z/du^2+(a-2)d^2z/duvd+ad^2z/dv^2；
d^2z/dy^2=d(adz/dv-2dz/du)/dy
=ad^2z/dvdu(-2)+ad^2z/dv^2(a)-2d^2z/du^2(-2)-2d^2z/dudv(a)
=a^2d^2z/dv^2-4ad^2z/dudv+4d^2z/du^2.
代入得
-5d^2z/du^2-3ad^2z/dudv+(1+a-a^2)d^2z/dv^2=0.
怀疑你题目写得不准确.