Terminology: Question 9-3.7 & Solution

Question 7

For a real second-order linear PDE in two spatial variables, use the convention $A u_{xx}+2B u_{xy}+C u_{yy}+\text{lower-order terms}=0, \qquad D=B^2-AC.$ At points where the second-order part is nonzero, call the equation hyperbolic if $D>0$ , elliptic if $D<0$ , and parabolic if $D=0$ . Consider $u_{xx}+2y u_{xy}+x u_{yy}+3u_x-u=0.$

Tasks

Identify $A,B,C$ and classify the equation at every point of the plane, including the separating curve. Explain why the coefficient of $u_{xy}$ is not $B$ itself.
Classify the points $(0,1),(2,1),(1,1)$ , and sketch the type regions with correct axes and boundary curve.
At a fixed point, rewrite the principal quadratic form $Q(\xi,\eta)=\xi^2+2y\xi\eta+x\eta^2$ as a completed square. Explain its signs and rank in all three regions.
Replace $3u_x-u$ by $e^{x+y}u_x+7u$ and then multiply the entire PDE by a smooth nowhere-zero function $m(x,y)$ . Decide which changes can alter the type and justify your answer.

Original worksheet page 1: question and worked solution for 9-3-007

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Question 7 – Solution

Strategy. Classify the principal second-order part point by point, keeping the mixed-derivative convention fixed.

Step 1: Read the discriminant correctly. Here $A=1$ , $B=y$ and $C=x$ , since the mixed coefficient is $2B=2y$ . Thus $D=y^2-x$ , giving $\boxed{\begin{array}{ll} x<y^2:&\text{hyperbolic},\\ x=y^2:&\text{parabolic},\\ x>y^2:&\text{elliptic}. \end{array}}$ Since $A=1$ , the principal part never vanishes completely. The curve is therefore parabolic under the stated convention, not an order-zero exception.

Step 2: Check the points and the geometry. The three discriminants are respectively $1,-1,0$ , so $(0,1)$ is hyperbolic, $(2,1)$ elliptic and $(1,1)$ parabolic. With $x$ horizontal and $y$ vertical, $x=y^2$ is a right-opening parabola; ellipticity lies to its right.

Step 3: Interpret the principal form. Completing the square gives $Q=(\xi+y\eta)^2+(x-y^2)\eta^2$ . For $x>y^2$ , it is positive definite, with rank two. For $x<y^2$ , the choices $(\xi,\eta)=(1,0)$ and $(-y,1)$ give opposite signs; it is indefinite with rank two. On $x=y^2$ , it is a nonnegative square of rank one, vanishing on the nonzero direction $(-y,1)$ . This explains the type boundary algebraically.

Step 4: Identify changes that preserve type. Changing lower-order terms leaves $A,B,C$ unchanged. Multiplying by $m$ gives $\widetilde D=(mB)^2-(mA)(mC)=m^2D$ , with the same sign because $m\ne 0$ . Neither operation changes type. Allowing $m=0$ could erase the principal part at those points, which is why the nowhere-zero restriction matters.

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 9-3-007

Terminology — Question 7

Question 7

Question 7 – Solution