2 Partial differential equations
In all the above examples we had a function $y$ of a single variable $x,y$ being the solution of an ordinary differential equation.
In engineering and science ODEs arise as models for systems where there is one independent variable (often $x$ ) and one dependent variable (often $y$ ). Obvious examples are lumped electrical circuits where the current $i$ is a function only of time $t$ (and not of position in the circuit) and lumped mechanical systems (such as the simple harmonic oscillator referred to above) where the displacement of a moving particle depends only on $t$ .
However, in problems where one variable, say $u$ , depends on more than one independent variable, say both $x$ and $t$ , then any derivatives of $u$ will be partial derivatives such as $\frac{\partial u}{\partial x}$ or $\frac{{\partial}^{2}u}{\partial {t}^{2}}$ and any differential equation arising will be known as a partial differential equation. In particular, onedimensional (1D) timedependent problems where $u$ depends on a position coordinate $x$ and the time $t$ and twodimensional (2D) timeindependent problems where $u$ is a function of the two position coordinates $x$ and $y$ both give rise to PDEs involving two independent variables. This is the case we shall concentrate on. A twodimensional timedependent problem would involve 3 independent variables $x,y,t$ as would a threedimensional timeindependent problem where $x,y,z$ would be the independent variables.
Example 1
Show that $u=sinxcoshy$ satisfies the PDE $\frac{{\partial}^{2}u}{\partial {x}^{2}}+\frac{{\partial}^{2}u}{\partial {y}^{2}}=0$ .
This PDE is known as Laplace’s equation in two dimensions and it arises in many applications e.g. electrostatics, fluid flow, heat conduction.
Solution
$u=sinxcoshy\phantom{\rule{2em}{0ex}}\Rightarrow \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial u}{\partial x}=cosxcoshy\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\partial u}{\partial y}=sinxsinhy$
Differentiating again gives $\phantom{\rule{1em}{0ex}}\frac{{\partial}^{2}u}{\partial {x}^{2}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=sinxcoshy\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{{\partial}^{2}u}{\partial {y}^{2}}=sinxcoshy$
Hence
$\phantom{\rule{2em}{0ex}}\frac{{\partial}^{2}u}{\partial {x}^{2}}+\frac{{\partial}^{2}u}{\partial {y}^{2}}=sinxcoshy+sinxcoshy=0$
so the given function $u\left(x,y\right)$ is indeed a solution of the PDE.
Task!
Show that $u={e}^{2{\pi}^{2}t}sin\pi x$ is a solution of the PDE $\frac{{\partial}^{2}u}{\partial {x}^{2}}=\frac{1}{2}\frac{\partial u}{\partial t}$
First find $\frac{\partial u}{\partial t}$ and $\frac{\partial u}{\partial x}$ :
$\frac{\partial u}{\partial t}=2{\pi}^{2}{e}^{2{\pi}^{2}t}sin\pi x\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{\partial u}{\partial x}=\pi {e}^{2{\pi}^{2}t}cos\pi x$
Now find $\frac{{\partial}^{2}u}{\partial {x}^{2}}$ and complete the Task:
$\frac{{\partial}^{2}u}{\partial {x}^{2}}={\pi}^{2}{e}^{2{\pi}^{2}t}sin\pi x,$ and we see that $\frac{{\partial}^{2}u}{\partial {x}^{2}}=\frac{1}{2}\frac{\partial u}{\partial t}$ as required.
The PDE in the above Task has the general form
$\phantom{\rule{2em}{0ex}}\frac{{\partial}^{2}u}{\partial {x}^{2}}=\frac{1}{k}\frac{\partial u}{\partial t}$
where $k$ is a positive constant. This equation is referred to as the onedimensional heat conduction equation (or sometimes as the diffusion equation ). In a heat conduction context the dependent variable $u$ represents the temperature $u\left(x,t\right)$ .
The third important PDE involving two independent variables is known as the onedimensional wave equation . This has the general form
$\frac{{\partial}^{2}u}{\partial {x}^{2}}=\frac{1}{{c}^{2}}\frac{{\partial}^{2}u}{\partial {t}^{2}}$
(Note that both partial derivatives in the wave equation are secondorder in contrast to the heat conduction equation where the time derivative is first order.)
Example 2
 Verify that $u\left(x,t\right)={u}_{0}sin\left(\frac{\pi x}{\ell}\right)\phantom{\rule{1em}{0ex}}cos\left(\frac{\pi ct}{\ell}\right)$ (where ${u}_{0},\phantom{\rule{1em}{0ex}}\ell $ and $c$ are constants) satisfies the onedimensional wave equation.
 Verify the boundary conditions i.e. $u\left(0,t\right)=u\left(\ell ,t\right)=0$ .
 Verify the initial conditions i.e. $\frac{\partial u}{\partial t}\left(x,0\right)=0$ and $u\left(x,0\right)={u}_{0}sin\left(\frac{\pi x}{\ell}\right)$ .
 Give a physical interpretation of this problem.

By straightforward partial differentiation of the given function
$u\left(x,t\right)$
:
$$\begin{array}{rcll}\frac{\partial u}{\partial x}& =& {u}_{0}\frac{\pi}{\ell}cos\left(\frac{\pi x}{\ell}\right)cos\left(\frac{\pi ct}{\ell}\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{{\partial}^{2}u}{\partial {x}^{2}}={u}_{0}{\left(\frac{\pi}{\ell}\right)}^{2}sin\left(\frac{\pi x}{\ell}\right)cos\left(\frac{\pi ct}{\ell}\right)& \text{}\\ & & & \text{}\\ \frac{\partial u}{\partial t}& =& {u}_{0}\left(\frac{\pi c}{\ell}\right)sin\left(\frac{\pi x}{\ell}\right)sin\left(\frac{\pi ct}{\ell}\right)\phantom{\rule{2em}{0ex}}\frac{{\partial}^{2}u}{\partial {t}^{2}}={u}_{0}{\left(\frac{\pi c}{\ell}\right)}^{2}sin\left(\frac{\pi x}{\ell}\right)cos\left(\frac{\pi ct}{\ell}\right)& \text{}\end{array}$$
We see that $\frac{{\partial}^{2}u}{\partial {x}^{2}}=\frac{1}{{c}^{2}}\frac{{\partial}^{2}u}{\partial {t}^{2}}$ which completes the verification.

Putting
$x=0$
, and leaving
$t$
arbitrary, in the given solution for
$u\left(x,t\right)$
gives
$\phantom{\rule{2em}{0ex}}u\left(x,0\right)={u}_{0}sin0\phantom{\rule{1em}{0ex}}cos\left(\frac{\pi ct}{\ell}\right)=0\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{forall}\phantom{\rule{1em}{0ex}}t$
Similarly putting $x=\ell ,\phantom{\rule{1em}{0ex}}t$ arbitrary: $\phantom{\rule{1em}{0ex}}u\left(\ell ,0\right)={u}_{0}sin\pi \phantom{\rule{1em}{0ex}}cos\left(\frac{\pi ct}{\ell}\right)=0\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{forall}\phantom{\rule{1em}{0ex}}t$

Evaluating
$\frac{\partial u}{\partial t}$
firstly for general
$x$
and
$t$
$\phantom{\rule{2em}{0ex}}\frac{\partial u}{\partial t}={u}_{0}\left(\frac{\pi c}{\ell}\right)sin\left(\frac{\pi x}{\ell}\right)sin\left(\frac{\pi ct}{\ell}\right)$
Now putting $t=0$ leaving $x$ arbitrary
$\phantom{\rule{2em}{0ex}}\frac{\partial u}{\partial t}\left(x,0\right)={u}_{0}\left(\frac{\pi c}{\ell}\right)sin\left(\frac{\pi x}{\ell}\right)sin0=0.$
Also, putting $t=0$ in the expression for $u\left(x,t\right)$ gives
$\phantom{\rule{2em}{0ex}}u\left(x,0\right)={u}_{0}sin\left(\frac{\pi x}{\ell}\right)cos0={u}_{0}sin\left(\frac{\pi x}{\ell}\right).$

Mathematically we have now proved that the given function
$u\left(x,t\right)$
satisfies the
$1$
D wave equation specified in 1., the two
boundary conditions
specified in 2. and the two
initial conditions
specified in 3.
One possible physical interpretation of this problem is that $u\left(x,t\right)$ represents the displacement of a string stretched between two points at $x=0$ and $x=\ell $ . Clearly the position of any point $P$ on the vibrating string will depend upon its distance $x$ from one end and on the time $t$ .
The boundary conditions 2. represent the fact that the string is fixed at these endpoints.
The initial condition $u\left(x,0\right)={u}_{0}sin\left(\frac{\pi x}{\ell}\right)$ represents the displacement of the string at $t=0$ .
The initial condition $\frac{\partial u}{\partial t}\left(x,0\right)=0$ tells us that the string is at rest at $t=0$ .
Figure 1
Note that it can be proved formally that if $T$ is the tension in the string and if $\rho $ is the mass per unit length of the string then $u$ does, under certain conditions, satisfy the $1$ D wave equation with ${c}^{2}=\frac{T}{\rho}$ .
Key Point 4
The three PDEs of greatest general interest involving two independent variables are:

The twodimensional Laplace equation
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{{\partial}^{2}u}{\partial {x}^{2}}+\frac{{\partial}^{2}u}{\partial {y}^{2}}=0$

The onedimensional heat conduction equation:
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{{\partial}^{2}u}{\partial {x}^{2}}=\frac{1}{k}\frac{\partial u}{\partial t}$

The onedimensional wave equation:
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{{\partial}^{2}u}{\partial {x}^{2}}=\frac{1}{{c}^{2}}\frac{{\partial}^{2}u}{\partial {t}^{2}}$