1.2 PDE’s

Rather than giving a strict mathematical definition, let us look at an example of a PDE, the heat equation in 1 space dimension

{{∂}^{2}u(x,t)\over ∂{x}^{2}} = {1\over k} {∂u(x,t)\over ∂t} .
(1.5)

Why is all that so important? A linear homogeneous equation allows superposition of solutions. If {u}_{1} and {u}_{2} are both solutions to the heat equation,

{{∂}^{2}{u}_{1}(x,t)\over ∂{x}^{2}} −{1\over k} {∂{u}_{1}(x,t)\over ∂t} = {{∂}^{2}{u}_{2}(x,t)\over ∂{x}^{2}} −{1\over k} {∂{u}_{2}(x,t)\over ∂t} = 0,
(1.8)

any combination is also a solution,

{{∂}^{2}[a{u}_{1}(x,t) + b{u}_{2}(x,t)]\over ∂{x}^{2}} −{1\over k} {∂[a{u}_{1}(x,t) + b{u}_{2}(x,t)]\over ∂t} = 0.
(1.9)

For a linear inhomogeneous equation this gets somewhat modified. Let v be any solution to the heat equation with a \mathop{sin}\nolimits (x) inhomogeneity,

{{∂}^{2}v(x,t)\over ∂{x}^{2}} −{1\over k} {∂v(x,t)\over ∂t} =\mathop{ sin}\nolimits (x).
(1.10)

In that case v + a{u}_{1}, with {u}_{1} a solution to the homogeneous equation, see Eq. (1.8), is also a solution,

\begin{eqnarray} {{∂}^{2}[v(x,t) + a{u}_{1}(x,t)]\over ∂{x}^{2}} −{1\over k} {∂[v(x,t) + a{u}_{1}(x,t)]\over ∂t} & =& %& \\ {{∂}^{2}v(x,t)\over ∂{x}^{2}} −{1\over k} {∂v(x,t)\over ∂t} + a\left ({{∂}^{2}{u}_{1}(x,t)\over ∂{x}^{2}} −{1\over k} {∂{u}_{1}(x,t)\over ∂t} \right )& =& \mathop{sin}\nolimits (x).%&(1.11) \\ \end{eqnarray}

Finally we would like to define the order of a PDE as the power in the highest derivative, even it is a mixed derivative (w.r.t. more than one variable).

Quiz Which of these equations is linear? and which is homogeneous?

  1. {{∂}^{2}u\over ∂{x}^{2}} + {x}^{2}{∂u\over ∂y} = {x}^{2} + {y}^{2}.
    (1.12)
  2. {y}^{2}{{∂}^{2}u\over ∂{x}^{2}} + u{∂u\over ∂x} + {x}^{2}{{∂}^{2}u\over ∂{y}^{2}} = 0.
    (1.13)
  3. { \left ({∂u\over ∂x}\right )}^{2} + {{∂}^{2}u\over ∂{y}^{2}} = 0.
    (1.14)

What is the order of the following equations?

  1. {{∂}^{3}u\over ∂{x}^{3}} + {{∂}^{2}u\over ∂{y}^{2}} = 0.
    (1.15)
  2. {{∂}^{2}u\over ∂{x}^{2}} − 2 {{∂}^{4}u\over ∂{x}^{3}∂y} + {{∂}^{2}u\over ∂{y}^{2}} = 0.
    (1.16)