### 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, one-dimensional (1-D) time-dependent problems where $u$ depends on a position coordinate $x$ and the time $t$ and two-dimensional (2-D) time-independent 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 two-dimensional time-dependent problem would involve 3 independent variables $x,y,t$ as would a three-dimensional time-independent 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

$u=sinxcoshy\phantom{\rule{2em}{0ex}}⇒\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\partial u}{\partial x}=cosxcoshy\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\partial u}{\partial y}=sinxsinhy$

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.

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 one-dimensional 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 one-dimensional 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 second-order in contrast to the heat conduction equation where the time derivative is first order.)

##### Example 2
1. 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 one-dimensional wave equation.
2. Verify the boundary conditions i.e. $u\left(0,t\right)=u\left(\ell ,t\right)=0$ .
3. Verify the initial conditions i.e. $\frac{\partial u}{\partial t}\left(x,0\right)=0$ and $u\left(x,0\right)={u}_{0}sin\left(\frac{\pi x}{\ell }\right)$ .
4. Give a physical interpretation of this problem.
1. 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.

2. Putting $x=0$ , and leaving $t$ arbitrary, in the given solution for $u\left(x,t\right)$ gives

Similarly putting $x=\ell ,\phantom{\rule{1em}{0ex}}t$ arbitrary:

3. 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

$\phantom{\rule{2em}{0ex}}u\left(x,0\right)={u}_{0}sin\left(\frac{\pi x}{\ell }\right)cos0={u}_{0}sin\left(\frac{\pi x}{\ell }\right).$

4. 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 end-points.

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:

1. The two-dimensional Laplace equation

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

2. The one-dimensional heat conduction equation:

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{{\partial }^{2}u}{\partial {x}^{2}}=\frac{1}{k}\frac{\partial u}{\partial t}$

3. The one-dimensional wave equation:

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{{\partial }^{2}u}{\partial {x}^{2}}=\frac{1}{{c}^{2}}\frac{{\partial }^{2}u}{\partial {t}^{2}}$