2.6: Classification of Second Order PDEs
( \newcommand{\kernel}{\mathrm{null}\,}\)
We have studied several examples of partial differential equations, the heat equation, the wave equation, and Laplace’s equation. These equations are examples of parabolic, hyperbolic, and elliptic equations, respectively. Given a general second order linear partial differential equation, how can we tell what type it is? This is known as the classification of second order PDEs.
Let u=u(x,y). Then, the general form of a linear second order partial differential equation is given by
a(x,y)uxx+2b(x,y)uxy+c(x,y)uyy+d(x,y)ux+e(x,y)uy+f(x,y)u=g(x,y).
In this section we will show that this equation can be transformed into one of three types of second order partial differential equations.
Let x=x(ξ,η) and y=y(ξ,η) be an invertible transformation from coordinates (ξ,η) to coordinates (x,y). Furthermore, let u(x(ξ,η),y(ξ,η))=U(ξ,η). How does the partial differential equation (???) transform?
We first need to transform the derivatives of u(x,t). We have
ux=Uξξx+Uηηx,uy=Uξξy+Uηηy,uxx=∂∂x(Uξξx+Uηηx),=Uξξξ2x+2Uξηξxηx+Uηηη2x+Uξξxx+Uηηxx,uyy=∂∂y(Uξξy+Uηηy),=Uξξξ2y+2Uξηξyηy+Uηηη2y+Uξξyy+Uηηyy,uxy=∂∂y(Uξξx+Uηηx),=Uξξξxξy+Uξηξxηy+Uηξξyηx+Uηηηxηy+Uξξxy+Uηηxy.
Inserting these derivatives into Equation (???), we have
g−fU=auxx+2buxy+cuyy+dux+euy=a(Uξξξ2x+2Uξηξxηx+Uηηη2x+Uξξxx+Uηηxx)+2b(Uξξξxξy+Uξηξxηy+Uξηξyηx+Uηηηxη+y+Uξξxy+Uηηxy)+c(Uξξξ2y+2Uξηξyηy+Uηηη2y+Uξξyy+Uηηyy)+d(Uξξx+Uηηx)+e(Uξξy+Uηηy)=(aξ2x+2bξxξy+cξ2y)Uξξ+(2aξxηx+2bξxηy+2bξyηx+2cξyηy)Uξη+(aη2x+2bηxηy+cη2y)Uηη+(aξxx+2bξxy+cξyy+dξx+eξy)Uξ+(aηxx+2bηxy+cηyy+dηx+eηy)Uη=AUξξ+2BUξη+CUηη+DUξ+EUη.
Picking the right transformation, we can eliminate some of the second order derivative terms depending on the type of differential equation. This leads to three types: elliptic, hyperbolic, or parabolic.
For example, if transformations can be found to make A≡0 and C≡0, then the equation reduces
Uξη=lower order terms.
. Such an equation is called hyperbolic. A generic example of a hyperbolic equation is the wave equation.
The conditions that A≡0 and C≡0 give the conditions
aξ2x+2bξxξy+cξ2y=0.aη2x+2bηxηy+cη2y=0.
) We seek ξ and η satisfying these two equations, which are of the same form. Let’s assume that ξ=ξ(x,y) is a constant curve in the xy-plane. Furthermore, if this curve is the graph of a function, y=y(x), then
dξdx=ξx+dydxξy=0.
Then
dydx=−ξxξy.
Inserting this expression in A=0, we have
A=aξ2x+2bξxξy+cξ2y=ξ2y(a(ξxξy)2+2bξxξy+c)=ξ2y(a(dydx)2−2bdydx+c)=0.
This equation is satisfied if y(x) satisfies the differential equation
dydx=b±√b2−aca.
So, for A=0, we choose ξ and η to be constant on these characteristic curves.
Show that uxx−uyy=0 is hyperbolic.
Solution
In this case we have a=1=−c and b=0. Then,
dydx=±1.
This gives y(x)=±x+c. So, we choose ξ and η constant on these characteristic curves. Therefore, we let ξ=x−y, η=x+y.
Let’s see if this transformation transforms the differential equation into a canonical form. Let u(x,y)=U(ξ,η). Then, the needed derivatives become
ux=Uξξx+Uηηx=Uξ+Uη.uy=Uξξy+Uηηy=−Uξ+Uη.uxx=∂∂x(Uξ+Uη)=Uξξξx+Uξηηx+Uηξξx+Uηηηx=Uξξ+2Uξη+Uηη.uyy=∂∂y(−Uξ+Uη)=−Uξξξy−Uξηηy+Uηξξy+Uηηηy=Uξξ−2Uξη+Uηη.
Inserting these derivatives into the differential equation, we have
0=uxx−uyy=4Uξη.
Thus, the transformed equation is Uξη=0. Thus, showing it is a hyperbolic equation.
We have seen that A and C vanish for ξ(x,y) and η(x,y) constant along the characteristics
dydx=b±√b2−aca
for second order hyperbolic equations. This is possible when b2−ac>0 since this leads to two characteristics.
In general, if we consider the second order operator
L[u]=a(x,y)uxx+2b(x,y)uxy+c(x,y)uyy,
then this operator can be transformed to the new form
L′[U]=BUξη
if b2−ac>0. An example of a hyperbolic equation is the wave equation, utt=uxx.
When b2−ac=0, then there is only one characteristic solution, dydx=ba. This is the parabolic case. But, dydx=−ξxξy. So,
ba=−ξxξy,
or
aξx+bξy=0.
Also, b2−ac=0 implies that c=b2/a.
Inserting these expression into coefficient B, we have
B=2aξxηx+2bξxηy+2bξyηx+2cξyηy=2(aξx+bξy)ηx+2(bξx+cξy)ηy=2ba(aξx+bξy)ηy=0.
Therefore, in the parabolic case, A=0 and B=0, and L[u] transforms to
L′[U]=CUηη
when b2−ac=0. This is the canonical form for a parabolic operator. An example of a parabolic equation is the heat equation, ut=uxx.
Finally, when b2−ac<0, we have the elliptic case. In this case we Elliptic case. cannot force A=0 or C=0. However, in this case we can force B=0. As we just showed, we can write
B=2(aξx+bξy)ηx+2(bξx+cξy)ηy.
Letting ηx=0, we can choose ξ to satisfy bξx+cξy=0.
A=aξ2x+2bξxξy+cξ2y=aξ2x−cξ2y=ac−b2cξ2x
C=aη2x+2bηxηy+cη2y=cη2y
Furthermore, setting ac−b2cξ2x=cη2y, we can make A=c and L[u] transforms to
L′[U]=A[Uξξ+Uηη]
when b2−ac<0. This is the canonical form for an elliptic operator. An example of an elliptic equation is Laplace’s equation, uxx+uyy=0.
The second order differential operator
L[u]=a(x,y)uxx+2b(x,y)uxy+c(x,y)uyy,
can be transformed to one of the following forms:
- b2−ac>0. Hyperbolic: L[u]=B(x,y)uxy
- b2−ac=0. Parabolic: L[u]=C(x,y)uyy
- b2−ac<0. Elliptic: L[u]=A(x,y)[uxx+uyy]
As a final note, the terminology used in this classification is borrowed from the general theory of quadratic equations which are the equations for translated and rotated conics. Recall that the general quadratic equation in two variable takes the form
ax2+2bxy+cy2+dx+ey+f=0.
One can complete the squares in x and y to obtain the new form
a(x−h)2+2bxy+c(y−k)2+f′=0.
So, translating points (x,y) using the transformations x′=x−h and y′=y−k, we find the simpler form
ax2+2bxy+cy2+f=0.
Here we dropped all primes.
We can also introduce transformations to simplify the quadratic terms. Consider a rotation of the coordinate axes by θ,
x′=xcosθ+ysinθy′=−xsinθ+ycosθ,
or
x=x′cosθ−y′sinθy=x′sinθ+y′cosθ.
The resulting equation takes the form
Ax′2+2Bx′y′+Cy′2+D=0,
where
A=acos2θ+2bsinθcosθ+csin2θ.B=(c−a)sinθcosθ+b(cos2θ−sinθ).C=asin2θ−2bsinθcosθ+ccos2θ.
Furthermore, one can show that b2−ac=B2−AC. From the form Ax′2+2Bx′y′+Cy′2+D=0, the resulting quadratic equation takes one of the following forms:
- b2−ac>0. Hyperbolic: Ax2−Cy2+D=0.
- b2−ac=0. Parabolic: Ax2+By+D=0.
- b2−ac<0. Elliptic: Ax2+Cy2+D=0.
Thus, one can see the connection between the classification of quadratic equations and second order partial differential equations in two independent variables.