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Question

Find the type, transform to normal form, and solve. Show your work in detail.

uxx6uxy+9uyy=0u_{xx}-6u_{xy}+9u_{yy}=0

Solution

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Answered 2 years ago
Answered 2 years ago
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The coefficients of the equation

uxx6yxy+9uyy=0u_{xx} -6y_{xy}+ 9u_{yy} =0

are

A=1,B=3,C=9.A=1, \, B=-3, \, C=9 .

Since

ACB2=99=0AC - B^2 = 9-9=0

we see that this is parabolic\textbf{parabolic} type equation. The charateristic equation we solve is

(y)2+6y+9=0.(y^\prime)^2 +6y^\prime + 9 =0 .

which can be rewritten as

(y+3)2=0.(y^\prime +3)^2 =0 .

It is clear that its solutions are

y=3y^\prime =-3

which leads to the differential equation

dy=3dx.dy =-3 dx.

Integrating we get

y=3x.y=-3x.

So the characteristic(s)\textbf{characteristic(s)} is (are)

Φ(x,y)=Ψ(x,y)=y+3x\boxed{ \Phi (x,y) = \Psi (x,y) = y+3x}

So we introduce new variables v=vv= v and w=y+3xw= y+3x. We have that

ux=uvvx+uwwx=uv+3uw,uxx=xuv+3xuw==(uvvvx+uvwwx)+3(uwvvx+uwwwx)==(uvv+3uvw)+3(uvw+3uww)==uvv+6uvw+9uwwuy=uvvy+uwwy=uw,uyy=yuw=uww,uxy=yux=y(uv+3uw)=uvw+3uww\begin{aligned}u_x & = u_v \cdot \frac{\partial v}{\partial x} + u_w \cdot \frac{\partial w}{\partial x} = u_v +3u_w, \\ u_{xx}& =\frac{\partial}{\partial x} u_v +3 \frac{\partial }{\partial x} u_w =\\ & = \left( u_{vv} \frac{\partial v}{\partial x} +u_{vw} \frac{\partial w}{\partial x}\right) +3 \left( u_{wv} \frac{\partial v}{\partial x} +u_{ww} \frac{\partial w}{\partial x}\right) = \\ & = ( u_{vv} +3 u_{vw}) +3 ( u_{vw} +3 u_{ww} ) = \\ & = u_{vv} +6 u_{vw} +9 u_{ww} \\ u_{y} & = u_v \cdot \frac{\partial v}{\partial y} + u_w \cdot \frac{\partial w}{\partial y} =u_w, \\ u_{yy} &= \frac{\partial }{\partial y} u_w = u_{ww}, \\ u_{xy} & = \frac{\partial }{\partial y} u_x = \frac{\partial }{\partial y} (u_v +3 u_w) = u_{vw} +3 u_{ww} \end{aligned}

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