Transforming to a Normal Form (PDE)

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SUMMARY

The discussion focuses on solving the partial differential equation (PDE) for u(x,y) in normal form, specifically addressing the equation u_{ww}=0. Participants confirm that the solution takes the form u(w,v)=A(v)w+B(v), where A(v) is a function of v and B(v) is an arbitrary function of v. The key takeaway is that the indefinite integral of zero with respect to w results in an arbitrary function of v, rather than a constant.

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Homework Statement
Find the type, transform to normal form and solve. Show your work in detail: Uxx+2Uxy+Uyy=0
Relevant Equations
Hyperbolic- AC-B^2<0
Parabolic- AC-B^2=0
Elliptic- AC-B^2>0
I don't know how to solve for u(x,y) from where I left of after 5.

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Shouldn't you be getting: ##u_{ww}=0##, i.e. ##u(w,v)=Aw+B(v)## where ##B(v)## is a function of ##v## and ##A## is a constant?
 
MathematicalPhysicist said:
Shouldn't you be getting: ##u_{ww}=0##, i.e. ##u(w,v)=Aw+B(v)## where ##B(v)## is a function of ##v## and ##A## is a constant?

The indefinite integral of 0 with respect to w is an arbitrary function of v, not an arbitrary ocnstant.
 
pasmith said:
The indefinite integral of 0 with respect to w is an arbitrary function of v, not an arbitrary ocnstant.
yes correct then A should be A(v).
 

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