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

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

It is a PDE since partial derivatives are involved.

Review

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,

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

• 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 \[\dfrac{\partial^2}{\partial x^2}{u(x,t)} = \frac{1}{k} \dfrac{\partial}{\partial t}{u(x,t)}+\sin(x) \nonumber \] is called inhomogeneous, due to the \(\sin(x)\) term on the right, that is independent of \(u\).

Linearity

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,

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

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

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

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,

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

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

Answer

TBA

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

Answer

TBA