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Mathematics LibreTexts

1.2: PDE’s

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

    \[\dfrac{\partial^2 u(x,t)}{\partial x^2} = \frac{1}{k} \dfrac{\partial u(x,t)}{\partial t}. \label{eq:I:heat}\]

    • It is a PDE since partial derivatives are involved.


    To remind you of what that means: \(\dfrac{\partial}{\partial x}{u(x,t)}\) denotes the differentiation of \(u(x,t)\) w.r.t. \(x\) keeping \(t\) fixed,

    \[\dfrac{\partial}{\partial x} (x^2t+xt^2) = 2xt+t^2.\]

    • Equation \ref{eq:I:heat} called linear since \(u\) and its derivatives appear linearly, i.e., once per term. No functions of \(u\) are allowed. Terms like \(u^2\), \(\sin(u)\), \(u\dfrac{\partial}{\partial x}{u}\), etc., break this rule, and lead to non-linear equations. These are interesting and important in their own right, but outside the scope of this course.
    • Equation \ref{eq:I:heat} is also homogeneous (which just means that every term involves either \(u\) or one of its derivatives, there is no term that does not contain \(u\)). The equation \[\dfrac{\partial^2}{\partial x^2}{u(x,t)} = \frac{1}{k} \dfrac{\partial}{\partial t}{u(x,t)}+\sin(x)\] is called inhomogeneous, due to the \(\sin(x)\) term on the right, that is independent of \(u\).


    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,

    \[\dfrac{\partial^2}{\partial x^2}{u_1(x,t)} - \frac{1}{k} \dfrac{\partial}{\partial t}{u_1(x,t)}= \dfrac{\partial}{\partial t}{u_2(x,t)} - \frac{1}{k} \dfrac{\partial^2}{\partial x^2}{u_2(x,t)}=0, \label{eq:I:heatsol}\]

    any combination is also a solution,

    \[\dfrac{\partial^2}{\partial x^2}{[a u_1(x,t)+bu_2(x,t)]} - \frac{1}{k} \dfrac{\partial}{\partial t}{[au_1(x,t)+b u_2(x,t)]}=0.\]

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

    \[\dfrac{\partial^2}{\partial x^2}{v(x,t)} - \frac{1}{k} \dfrac{\partial}{\partial t}{v(x,t)}=\sin(x).\]

    In that case \(v+au_1\), with \(u_1\) a solution to the homogeneous equation, see Equation \ref{eq:I:heatsol}, is also a solution,

    \[\begin{aligned} \dfrac{\partial^2}{\partial x^2}{[v(x,t)+a u_1(x,t)]} - \frac{1}{k} \dfrac{\partial}{\partial t}{[v(x,t)+a u_1(x,t)]}&=&\nonumber\\ \dfrac{\partial^2}{\partial x^2}{v(x,t)} - \frac{1}{k} \dfrac{\partial}{\partial x}{v(x,t)} +a \left(\dfrac{\partial}{\partial x}{u_1(x,t)} - \frac{1}{k} \dfrac{\partial}{\partial t}{u_1(x,t)}\right)&=& \sin(x).\end{aligned}\]

    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).

    Exercise \(\PageIndex{1}\)

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

    1. \(\dfrac{\partial^2 u}{\partial x^2} + x^2 \dfrac{\partial u}{\partial y} = x^2 + y^2\)
    2. \(y^2\dfrac{\partial^2 u}{\partial x^2}+u\dfrac{\partial u}{\partial x} + x^2\dfrac{\partial^2 u}{\partial y^2} = 0\)
    3. \(\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial x^2}=0\)


    Exercise \(\PageIndex{2}\)

    What is the order of the following equations?

    1. \(\dfrac {\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0\)
    2. \(\dfrac{\partial^2 u}{\partial x^2} - 2\dfrac{\partial^4 u}{\partial^3 x \partial y} + \dfrac{\partial^2 u}{\partial y^2}= 0\)