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 ∂ u ∂ x or ∂ 2 u ∂ 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 = sin x cosh y satisfies the PDE   ∂ 2 u ∂ x 2 + ∂ 2 u ∂ 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 = sin x cosh y ⇒ ∂ u ∂ x = cos x cosh y and ∂ u ∂ y = sin x sinh y

Differentiating again gives ∂ 2 u ∂ x 2 = − sin x cosh y and ∂ 2 u ∂ y 2 = sin x cosh y

Hence

∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = − sin x cosh y + sin x cosh y = 0

so the given function u ( x , y ) is indeed a solution of the PDE.

Task!

Show that   u = e − 2 π 2 t sin π x  is a solution of the PDE   ∂ 2 u ∂ x 2 = 1 2 ∂ u ∂ t

First find ∂ u ∂ t and ∂ u ∂ x :

∂ u ∂ t = − 2 π 2 e − 2 π 2 t sin π x ∂ u ∂ x = π e − 2 π 2 t cos π x

Now find ∂ 2 u ∂ x 2 and complete the Task:

∂ 2 u ∂ x 2 = − π 2 e − 2 π 2 t sin π x ,  and we see that  ∂ 2 u ∂ x 2 = 1 2 ∂ u ∂ t  as required.

The PDE in the above Task has the general form

∂ 2 u ∂ x 2 = 1 k ∂ u ∂ 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 ( x , t ) .

The third important PDE involving two independent variables is known as the one-dimensional wave equation . This has the general form

∂ 2 u ∂ x 2 = 1 c 2 ∂ 2 u ∂ 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 ( x , t ) = u 0 sin π x ℓ cos π c t ℓ (where u 0 , ℓ and c are constants)  satisfies the one-dimensional wave equation.
  2. Verify the boundary conditions i.e. u ( 0 , t ) = u ( ℓ , t ) = 0 .
  3. Verify the initial conditions i.e. ∂ u ∂ t ( x , 0 ) = 0 and u ( x , 0 ) = u 0 sin π x ℓ .
  4. Give a physical interpretation of this problem.
  1. By straightforward partial differentiation of the given function u ( x , t ) : ∂ u ∂ x = u 0 π ℓ cos π x ℓ cos π c t ℓ ∂ 2 u ∂ x 2 = − u 0 π ℓ 2 sin π x ℓ cos π c t ℓ ∂ u ∂ t = − u 0 π c ℓ sin π x ℓ sin π c t ℓ ∂ 2 u ∂ t 2 = − u 0 π c ℓ 2 sin π x ℓ cos π c t ℓ

    We see that ∂ 2 u ∂ x 2 = 1 c 2 ∂ 2 u ∂ t 2 which completes the verification.

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

    u ( x , 0 ) = u 0 sin 0 cos π c t ℓ = 0 for all t

    Similarly putting x = ℓ , t arbitrary: u ( ℓ , 0 ) = u 0 sin π cos π c t ℓ = 0 for all t

  3. Evaluating ∂ u ∂ t firstly for general x and t

    ∂ u ∂ t = − u 0 π c ℓ sin π x ℓ sin π c t ℓ

    Now putting t = 0 leaving x arbitrary

    ∂ u ∂ t ( x , 0 ) = − u 0 π c ℓ sin π x ℓ sin 0 = 0.

    Also, putting t = 0 in the expression for u ( x , t ) gives

    u ( x , 0 ) = u 0 sin π x ℓ cos 0 = u 0 sin π x ℓ .

  4. Mathematically we have now proved that the given function u ( x , t ) 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 ( x , t ) represents the displacement of a string stretched between two points at x = 0 and x = ℓ . 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 ( x , 0 ) = u 0 sin π x ℓ  represents the displacement of the string at t = 0 .

    The initial condition   ∂ u ∂ t ( x , 0 ) = 0  tells us that the string is at rest at t = 0 .

    Figure 1

    A graph showing a series of curved lines representing the solution u(x, t) to a partial differential equation, with time increasing from an initial state at t = 0.

    Long description

    Note that it can be proved formally that if T is the tension in the string and if ρ is the mass per unit length of the string then u does, under certain conditions, satisfy the 1 -D wave equation with c 2 = T ρ .

Key Point 4

The three PDEs of greatest general interest involving two independent variables are:

  1. The two-dimensional Laplace equation

    ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0

  2. The one-dimensional heat conduction equation:

    ∂ 2 u ∂ x 2 = 1 k ∂ u ∂ t

  3. The one-dimensional wave equation:

    ∂ 2 u ∂ x 2 = 1 c 2 ∂ 2 u ∂ t 2