Title: Default for 25.1 Partial Differential Equations
Course: HELM Workbooks 2023-2024

Introduction

You have already studied ordinary differential equations (ODEs) and have learnt how to obtain the solution of certain types. Since a knowledge of the solution of certain ODEs (i.e. those with constant coefficients) will be required in solving partial differential equations (PDEs), we will begin this unit reminding you of some important results.

Key Point

The first order ODE \[\dfrac{dy}{dx}=ky\] has general solution \[y=Ae^{kx}\] Here \(k\) is a constant which can be positive or negative and \(A\) is an arbitrary constant.

In the above Key Point the quantity \(A\) in the general solution is a constant. To obtain the value of \(A\) we have to know the value of \(y\) at some value of \(x\), perhaps \(x=0\). In other words, we need to know an initial condition.

Exercise

Find \(y\) as a function of \(x\) if \[ \dfrac{\mathrm{d}y}{\mathrm{d}x}=[value]y \] and the initial condition is \(y(0)=[value].\)

\(y=\)
Feedback: From the above Key Point with \(k=[value]\) we have the general solution \[y=Ae^{[value]x}.\] Putting \(x=0\) and \(y=[value]\) into this we obtain \([value]=Ae^0\) i.e. \(A=[value]\) so the solution to the given initial value problem is \[y=[value].\]

We shall also need to be familiar with solutions to second order, homogeneous, constant coefficient ODEs, summarised in the Key Point below.

Key Point

A second order ODE of the form

\[a\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+b\dfrac{\mathrm{d}y}{\mathrm{d}x}+cy=0\] | \((1)\)

where \(a,b,c\) are constants, has an auxiliary equation

\[am^2+bm+c=0\] | \((2)\)

obtained by inserting the trial solution \(y=e^{mx}\) in \((1)\).

The general solution of \((1)\) then depends on the solutions (or roots) of the quadratic equation \((2)\).

1. If \((2)\) has real, distinct roots \(m=m_1\) and \(m=m_2\) then \[y=Ae^{m_1x}+Be^{m_2x}\]
2. If \((2)\) has a repeated root \(m=m_1\) then \[y=(A+Bx)e^{m_1x}\]
3. If \((2)\) has complex roots (which will be a conjugate pair) \(m=\alpha\pm j \beta\) then \[ y=e^{\alpha x}(A\cos \beta x+B\sin \beta x)\]

Note that in each of these cases (a) to (c) the general solution is a linear combination of two particular solutions:
- For (a) they are \(e^{m_1x}\) and \(e^{m_2x}\).
- For (b) they are \(e^{m_1x}\) and \(xe^{m_1x}\).
- For (c) they are \(e^{\alpha x}\cos\beta x\) and \(e^{\alpha x}\sin \beta x\).

Exercise

Use the Key Point above to find the general solution of \[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - [value]y = 0.\]

1. First write down the auxiliary equation:
2. Now find the roots of the auxiliary equation:
\(m =\) and \(m =\)
3. Finally give the general solution to the ODE:
\(y =\)
Feedback: The auxiliary equation for this second order ODE is \(m^2 - [value]=0\), which has roots \( m = \pm [value]\). By the previous Key Point, the general solution is \(y=A e^{[value]x}+Be^{-[value]x}\) (since the roots of the auxiliary equation are real and distinct).

Exercise

Use the Key Point above to find the general solution of \[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - [value]y = 0.\]

1. First write down the auxiliary equation:
2. Now find the roots of the auxiliary equation:
\(m =\) and \(m =\)
3. Finally give the general solution to the ODE:
\(y =\)
Feedback: The auxiliary equation for this second order ODE is \(m^2 + [value]=0\), which has roots \( m = \pm [value]i\). By the previous Key Point, the general solution is \(y=A\cos([value]x)+B\sin([value]x) \) (since the roots of the auxiliary equation are complex conjugates with real part \(\alpha=0\) and imaginary part \(\beta=[value]\)).

The two Exercises above can be generalised as in the following Key Point.

Key Point

1. The general solution to: \(\dfrac{d^2y}{dx^2}-n^2y=0\) is \[ y=Ae^{nx}+B e^{-nx}\] or, equivalently using hyperbolic functions, \[ y=C\cosh nx+D\sinh nx.\]
2. The general solution to: \(\dfrac{d^2y}{dx^2}+n^2y=0\) is \[ y=A\cos nx+B\sin nx.\]

Those of you who are familiar with elementary dynamics will recognise the second differential equation in the above Key Point as modelling simple harmonic motion.

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 \(\dfrac{\partial u}{\partial x}\) or \(\dfrac{\partial^2u}{\partial t^2}\) 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

Show that \(u=\sin x\cosh y\) satisfies the PDE \[ \dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}=0.\]

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

\[u=\sin x\cosh y\qquad\Rightarrow\qquad \dfrac{\partial u}{\partial x}=\cos x\cosh y\quad \text{and} \quad \dfrac{\partial u}{\partial y}=\sin x\sinh y.\] Differentiating again gives \( \displaystyle \dfrac{\partial^2u}{\partial x^2}\ \ =-\sin x\cosh y\) and \(\dfrac{\partial^2u}{\partial y^2}=\sin x \cosh y\).

Hence \[\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}=-\sin x\cosh y+\sin x\cosh y=0\] so the given function \(u(x,y)\) is indeed a solution of the PDE.

Exercise

Show that \(u= [value]\) is a solution of the PDE \(\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}[value]\dfrac{\partial u}{\partial t}\).

First find \(\dfrac{\partial u}{\partial t}\) and \(\dfrac{\partial u}{\partial x}\):

\(\dfrac{\partial u}{\partial t}=\)

\(\dfrac{\partial u}{\partial x}=\)

Now find \(\dfrac{\partial^2u}{\partial x^2}\)

\(\dfrac{\partial^2u}{\partial x^2}=\)
Feedback: We have that \[ \dfrac{\partial u}{\partial t}=[value]\qquad \text{and} \qquad \dfrac{\partial u}{\partial x}=[value].\] We also have that \[\dfrac{\partial^2u}{\partial x^2}=[value],\] and we see that \(\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{2}\dfrac{\partial u}{\partial t}\) as required.

The PDE in the previous Exercise has the general form \[\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{k}\dfrac{\partial u}{\partial t}\] 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 \(u(x,t)\).

The third important PDE involving two independent variables is known as the one-dimensional wave equation. This has the general form \[\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{c^2}\dfrac{\partial^2u}{\partial t^2}\] (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

1. Verify that \( u(x,t)=u_0\sin\left(\dfrac{\pi x}{\ell}\right)\ \cos\left(\dfrac{\pi c t}{\ell}\right)\) (where \(u_0, \ell\) and \(c\) are constants) satisfies the one-dimensional wave equation.
2. Verify the boundary conditions i.e. \(u(0,t)=u(\ell, t)=0\).
3. Verify the initial conditions i.e. \(\dfrac{\partial u}{\partial t}(x,0)=0\) and \(u(x,0)=u_0\sin\left(\dfrac{\pi x}{\ell}\right)\).
4. Give a physical interpretation of this problem.

Solution

1. By straightforward partial differentiation of the given function \(u(x,t)\): \[\begin{aligned} \dfrac{\partial u}{\partial x} = u_0\dfrac{\pi}{\ell}\cos\left(\dfrac{\pi x}{\ell}\right)\cos\left(\dfrac{\pi c t}{\ell}\right), \\ \dfrac{\partial^2u}{\partial x^2}=-u_0\left(\dfrac{\pi}{\ell}\right)^2\sin\left(\dfrac{\pi x}{\ell}\right)\cos\left(\dfrac{\pi c t}{\ell}\right), \\ \dfrac{\partial u}{\partial t} = -u_0\left(\dfrac{\pi c}{\ell}\right)\sin\left(\dfrac{\pi x}{\ell}\right)\sin\left(\dfrac{\pi c t}{\ell}\right), \\ \dfrac{\partial^2u}{\partial t^2}=-u_0\left(\dfrac{\pi c}{\ell}\right)^2\sin\left(\dfrac{\pi x}{\ell}\right)\cos\left(\dfrac{\pi c t}{\ell}\right). \end{aligned}\] We see that \(\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{c^2}\dfrac{\partial^2u}{\partial t^2}\) which completes the verification.
2. Putting \(x=0\), and leaving \(t\) arbitrary, in the given solution for \(u(x,t)\) gives \[ u(x,0)=u_0\sin 0\ \cos\left(\dfrac{\pi c t}{\ell}\right)=0\ \ \text{for all}\ t\] Similarly putting \(x=\ell,\ t\) arbitrary: \[ u(\ell,0)=u_0\sin \pi\ \cos\left(\dfrac{\pi ct}{\ell}\right)=0\ \text{for all}\ t\]
3. Evaluating \(\dfrac{\partial u}{\partial t}\) firstly for general \(x\) and \(t\) \[\dfrac{\partial u}{\partial t}=-u_0\left(\dfrac{\pi c}{\ell}\right)\sin\left(\dfrac{\pi x}{\ell}\right) \sin\left(\dfrac{\pi ct}{\ell}\right)\] Now putting \(t=0\) leaving \(x\) arbitrary \[\dfrac{\partial u}{\partial t}(x,0)=-u_0\left(\dfrac{\pi c}{\ell}\right)\sin\left(\dfrac{\pi x}{\ell}\right)\sin 0=0.\] Also, putting \(t=0\) in the expression for \(u(x,t)\) gives \[ u(x,0)=u_0\sin\left(\dfrac{\pi x}{\ell}\right) \cos 0=u_0\sin\left(\dfrac{\pi x}{\ell}\right).\]
4. Mathematically we have now proved that the given function \(u(x,t)\) satisfies the \(1\)-D wave equation specified in (a), the two boundary conditions specified in (b) and the two initial conditions specified in (c).
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 \(x=\ell\). 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 (b) represent the fact that the string is fixed at these end-points.
The initial condition \(u(x,0)=u_0\sin\left(\dfrac{\pi x}{\ell}\right)\) represents the displacement of the string at \(t=0\).
The initial condition \(\dfrac{\partial u}{\partial t}(x,0)=0\) tells us that the string is at rest at \(t=0\).
[Image: A graph showing how the function u represents a vibrating stretched string.]

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 \(c^2=\dfrac{T}{\rho}\).

Key Point

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

1. The two-dimensional Laplace equation \[\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}=0\]
2. The one-dimensional heat conduction equation: \[\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{k}\dfrac{\partial u}{\partial t}\]
3. The one-dimensional wave equation: \[\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{c^2}\dfrac{\partial^2u}{\partial t^2}\]