General second order linear PDE

We now define the general partial differential equation referred to as Euler's equation given by:

where $u=u(x, y)$ and $a, b, c$ are constants. Note that $x$ and $y$ denote the independent variables now. Motivated by the change of variables carried out in Sec. 2.6 [refer to Eqs. (2.172)], we solve Eq. (2.196) by introducing the following:

where $\lambda$ and $\mu$ are unknown constants to be determined. Introducing the change of variables into the original equation, yields:

At this point let us note the following:

• We need $X$ and $Y$ to be different from each other $\Rightarrow \lambda$ and $\mu$ need to be different.
• In addition, we need to choose the constants $\lambda$ and $\mu$ such that we obtain a simplified version of Eq. (2.198). The aforementioned can be achieved by determining the roots of the quadratic equation given by Eq. (2.199):

whose solution gives two roots $\left(r_{1}\right.$ and $\left.r_{2}\right)$ where we can set $r_{1}=\lambda$ and $r_{2}=\mu$. Note that Eq. (2.199) with roots at $\lambda$ and $\mu$ ensures that the coloured terms in Eq. (2.198), vanish. It follows that (2.198) simplifies significantly to:

Equation (2.200) may be solved using the method indicated in Sec. 2.6 in particular, refer back to Eqs. (2.178) -(2.181). However, given $a, b, c$ coefficients in the general equation (2.196), Eq. (2.199) can result in the roots being of the following nature:

1. real + distinct (as discussed above)
2. complex conjugate pair
3. repeated (multiplicity 2)

The three cases listed above motivates the classification of the linear PDE (2.196), as discussed next.

Classification

Depending on the nature of the roots of Eq. (2.199), we obtain the following 3 types of linear PDEs:

1. real + distinct roots $\Rightarrow$ Hyperbolic
2. complex conjugates $\Rightarrow$ Elliptic
3. real + repeated $\Rightarrow$ Parabolic

Note that we have already encountered a PDE of each type: the wave equation (hyperbolic), Laplace's equation (elliptic) and the heat equation (parabolic). The general solution of Eq. (2.196) therefore varies depending on the type of PDE we are interested in and it is important to know that each type results in qualitatively different solutions. In what follows, we consider each case separately.

Hyperbolic PDEs

For real and distinct roots, the discriminant $b^{2}-a c>0$; we obtain the unknowns $\lambda$ and $\mu$ as:

Following Sec. 2.6 [Eqs. (2.178)-(2.181)], we can write down the general solution to (2.196) as:

where, $\lambda$ and $\mu$ are given by Eqs. (2.201) and $F$ and $G$ are arbitrary functions which can be determined from boundary conditions.

Example 2.4 Find the general solution to:

Solution First, by solving the quadratic equation:

which has roots 2 and $-1 / 3$. The PDE is hyperbolic since the roots are real and distinct. By introducing:

where, note that $\lambda$ and $\mu$ in Eqs. (2.197) have been replaced by the roots 2 and $-1 / 3$, reduces PDE (2.203) to:

Integrating with respect to $X$ gives $u_{Y}=f(Y)$ and, integrating with respect to $Y$ yields $u=F(Y)+G(X)$ where $F(Y)=\int f(Y) d Y$ and $G(X)$ are arbitrary functions. The general solution to Eq. (2.203), in terms of the original variables $x$ and $y$ is:

Elliptic PDEs

For complex roots, the discriminant $b^{2}-a c<0$ which results in $\lambda$ and $\mu$ being complex:

Of course we note that if we were to simply use $\lambda$ and $\mu$ as defined in Eqs. (2.208) in the general solution (2.202), our solution would be described by solutions which are not real. Since we know that Laplace's equation (given by an elliptic PDE) represents a physical situation, we require that all solutions be real.

The solution given by Eq. (2.202) is still the solution for the case of elliptic PDEs; it is real if $F$ and $G$ are related. By related, we mean that $F$ and $G$ are complex conjugates; then their sum gives real solutions.

Parabolic PDEs

For real and repeated roots, we need the discriminant $b^{2}-a c=0$ and the only root is $r_{1}=r_{2}=-b / c$. We can choose $\lambda$ to be arbitrary and, without loss of generality, take it to be zero. Then, $\mu=-b / c$ and Eq. (2.198) is rendered independent of $u_{X Y}$ and $u_{Y Y}$ terms. In fact, with our choice of $\lambda$ and $\mu$, we are left with:

Equation (2.209) gives $u=X F(Y)+G(Y)$ which, in terms of the original variables, gives the general solution as:

where, again, $F$ and $G$ are arbitrary functions to be determined with the use of boundary conditions. Note that we can have equivalent forms of the general solution (2.210) corresponding to different choices of $\lambda$.