Partial Differential Equations - An introduction to PDES

Partial DifferentialEquations

Dr. David RobertGrimes

Revision

Introduction

Direct Integration

Separation ofVariables

1D Heat Equation

1D Wave Equation

Principle ofsuperposition

Recap of method

Summary

Partial Differential EquationsAn introduction to PDES

Dr. David Robert Grimes1

1Old Road Campus, OX3 7DQUniversity of Oxford

Web: http://users.ox.ac.uk/~donc0074/

E-mail: [email protected]

University of Oxford Doctoral Training Centre, 2015

http://users.ox.ac.uk/~donc0074/



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Ordinary Differential EquationsRecall that earlier in the lecture series we explored ODEs andsystems of ODEs, and investigated some methods to solvethese. Let’s recall two common types.

I First order ODEs: When the derivative is first order,then we can solve it explicitly provided we’re given avalue of the function at some point. In physicalsystems, the argument is often time, and the seed valueis referred to as the initial condition. Often, we’ll seefirst order ODE problems refered to as Initial Valueproblems for this reason.

For example, if y ′ = 0.5y and y(0) = 5, then generalsolution: y = Ce0.5t . Evaluating at I.V. gives us completesolution y = 5e0.5t

I We have looked at ways of solving these (directly,integrating factor, etc.) previously.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Ordinary Differential Equations

I Second order ODEs: When the derivative is secondorder, we require at least two conditions to solve itcompletely. Physically, this often corresponds to spatialproblems where we know the function value at givenpoints. Problems like this are referred to as BoundaryValue problems, with boundary conditions (B.C.s).

For example, if y ′′ + 9y = 0 and y(0) = 5, y(π6 ) = 2 thengeneral solution: y = A cos(3x) + B sin(3x). Evaluating B.C.gives us complete solution:

y = 5 cos(3x) + 2 sin(3x)

I We have looked at ways of solving second-order ODEsalready in earlier lectures.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Partial Differential Equations

So far, we’ve confined ourselves to systems with onevariable. But in reality, there are many systems where wecannot take such a neat approach. When we want to modelsuch systems we need to use partial differential equations.

I PDEs describe multi-dimensonal systems withmulti-variate functions and their derivatives. Inconstrast, ODEs focus on a single variable.

I Why are PDEs important? Many things vary in bothspace and time, including physical systems such as heat,diffusion, electrodynamics and fluid flow.

I To model and understand physical systems, we need tohave a handle on PDEs.



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Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Examples of PDEs

I Diffusion equation (Fick’s second law): describes howconcentration changes with time.

∂C∂t = D∇2C

I Wave Equation: describes the behaviour of waves andwave motion.

∂2u∂t2 = D∇2u

I Schrodinger equation: describes how quantum states ofphysical systems evolve with time.

i~ ∂∂tϕ(r , t) = Hϕ(r , t)

Many more: All of Maxwell’s equations in electrodynamics,Poisson and Laplace’s equations in potential theory etc.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Examples of PDEs



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

PDEs - a word of caution

Because PDEs deal with functions of several variables, theyare intrinsically trickier than the ODEs we have thus farencountered. It’s worth noting the following:

I In general, PDEs often don’t have explicit analyticalsolutions, nor is there any neat way to predict this apriori. The reality is that many PDEs are not directlysolvable and trying to do so can be a Sisyphean task!

I For complicated PDEs, we can take a numerical oralgorithmic approach to finding a solution. Manysoftware packages are available for this purpose withdedicated solvers, including MATLAB andMathematica, which you’ll encounter later.

I For now, we’ll confine ourselves to techniques forsolving simple PDEs - direct integration andSeparation of Variables



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Direct Integration

Some PDEs are amenable to a direct assault. Consider:

∂2u

∂x2= 6xe−t

We can integrate this with respect to x to yield

∂u

∂x= 3x2e−t + f (t)

Notice that the unknown constant of integration that weencountered in ODEs has been replaced by an unknownfunction of integration f (t) in PDEs.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Direct Integration

We can integrate this again to get the general solution:

u(x , t) = x3e−t + xf (t) + g(t).

If we know initial conditions, we can find exact solutions. Ifwe were told that u(0, t) = t and ∂u

∂x (0, t) = e−t , then wecan show exact solution is

u(x , t) = x3e−t + xe−t + t.

However, only very simple PDEs are solvable by a directapproach. We usually need more sophisticated tools!



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Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Separation of Variables

In general, PDEs are hard to solve whilst ODEs arecomparatively easy. But it is sometimes possible to reduce aPDE to a series of easier ODEs. Separation of variables canbe employed when the variables are assumed to beindependent of one another, and can be written in the form

u(x , t) = X (x)T (t)

where X (x) and T (t) are factorized solutions, independentof one another. There are a number of conditions on this,but first let’s see an example of how to use SOV.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Heat equation

Consider 1D heat equation of the form ∂u∂t = 2∂

2u∂x2 . This

function is defined on the spatial domain 0 ≤ x ≤ 3 andt > 0. BCs: u(0, t) = u(3, t) = 0, IC: u(x , 0) = 5 sin 4πx .Assuming SOV possible, we get

XT ′ = 2X ′′T

Notice that these can be re-arranged to pull all of X and Tterms to one side of the equation.

T ′

2T=

X ′′

X

This equation must always hold - this means both sidesmust equal a constant value. We’ll call this value K .



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Heat equation

The PDE is now separated, so we write

∂T

∂t− 2KT = 0

∂2X

∂x2− KX = 0

We don’t yet know K - all we know is that it’s constant.There are three distinct cases affecting general solution.

I K < 0 (Negative coefficient)

I K = 0 (Null coefficient)

I K > 0 (Positive coefficient)

Let’s see how these choices play out...



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Heat equation

First, let’s assume K < 0. For clarity, we’ll re-writeK = −λ2 to emphasize that the coefficient is negative.

T ′ + 2λ2T = 0

X ′′ + λ2X = 0

which yields the following parametrized general solutions:

T = Ae−2λ2t

X = B cos(λx) + C sin(λx)

thus u = XT = e−2λ2t(A∗ cos(λx) + B∗ sin(λx))

Observation: This general form decays with time. Note thatA∗ and B∗ are just reduced constants; we’ll see shortly thatit’s a good idea to reduce the number of arbitrary constants.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Heat equation

What if K = 0 ? Then it follows that

T ′ = 0

X ′′ = 0


T = AX = Bx + C

thus u = XT = A∗x + B∗

Observation: Linear function with no explicit timedependence or second order derivatives.



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Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Heat equation

If K > 0 then K = λ2, and thus

T ′ − 2λ2T = 0

X ′′ − λ2X = 0


T = Ae2λ2t

X = Beλx + Ce−λx

thus u = XT = e2λ2t(A∗eλx + B∗e−λx)

Observation: This form exponentially increases with time.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Heat equation

We have three potential forms with respective generalsolutions. How can we choose between them? We test themagainst IC/BC - consider the case K = 0

Recall in this case, u = Ax + B, with condition thatu(0, t) = u(3, t) = 0. So applying the boundary conditions,we find that A∗ = B∗ = 0. This yields a solution of

u(x , t) = 0

This might be true, but it’s a trivial solution and notespecially enlightening, so we reject possibility that K = 0.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Heat equation

Now consider the case K > 0

Thus, u(x , t) = e2λ2t(A∗eλx + B∗e−λx) with condition thatu(0, t) = u(3, t) = 0. So applying the boundary conditions,we find that yet again the only way this can be true is ifλ = 0. If this is true, then we’re right back to the trivalsolution of

u(x , t) = 0

Again, we must reject possibility that K > 0.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Heat equation

Finally, we look at K < 0;

u = e−2λ2t(A∗ cos(λx) + B∗ sin(λx)). We can deduce fromfirst boundary conditions that A∗ = 0 and that

B∗ sin(3λ) = 0

For this to be true, either (i) B∗ = 0 or (ii) 3λ = nπ. Wereject (i) because this yields the trivial solution again. Thus,after applying these conditions we get a solution of form

u(x , t) = B∗n sin(nπx

3

)e−

2n2π2t9

where have have written B∗n to allow for possibility thatconstant varies with n. Finally, we introduce initial conditionto solve..



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Heat equation

Finally, recall initial condition; u(x , 0) = 5 sin 4πx . Wesimply force our solution to agree with this;

B∗n sin(nπx

3

)= 5 sin 4πx

which gives us B∗ = 5 and n = 12, so solution is

u(x , t) = 5 sin(4πx)e−32π2t

This is the complete solution to the question. Note thatwhen examining K , we might have used our physicalintuition to select the physically realistic case of heat decaybut we worked through it on a case-by-case basis toillustrate the method.



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Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Heat equationAs we might get from exponent in terms of t, heat rapidlydecays to zero in a matter of milliseconds!



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Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Wave EquationConsider a wave on a guitar string. In the simpliest case∂2u∂t2 = ∂2u

∂x2 .The string is clamped at both ends, and as aresult we know BCs: u(0, t) = u(L, t) = 0. Furthermore, wealso know ICs: u(x , 0) = sinπx and ux(x , 0) = 0.

XT ′′ = X ′′T

T ′′

T=

X ′′

X= K

Looking at the boundary conditions, we conclude K < 0 andthus we get form for X (x) of

X = A cos(λx) + B sin(λx)

Applying boundary conditions gives us

λ =nπ

L

X = Bn sin(nπx

L)



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Wave Equation

T is similar, and thus general form is

u = sin(nπx

L

)(A∗ cos

(nπtL

)+ B∗ sin

(nπtL

)).

Now we apply initial conditions. From first initial condition,we deduce A∗ = 1. From second, we can state B∗ = 0.Thus, we can yield an expression for u(x , t) of

u = sin(nπx

L

)cos(nπt

L

).

Note that n can be any integer; for n = 1 we get thefundamental frequency, but there are an infinity of solutions,each corresponding to a harmonic.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

1D Wave Equation



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Principle of superposition

I The previous result brings us to an important pointabout separation of variables. For the 1D waveequation, we saw there was an infinity of solutions.

I This is in fact the case for most SOV type problems,where the solutions are linear.

I In general, if Xi (x)Ti (t) (i = 1, 2, 3...) solve a linearequation, then any combination of these

∑∞n=1 AiXiTi

is also a solution where Ai is an arbitrary constant.

I This is usually related to the function describing theinitial condition. For example, let’s revisit our 1D heatequation to see this further.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary


u(x , t) = An sin(nπx

3

)e−

2n2π2t9

I This is the general form solved with boundaryconditions; we might view this as our building blocksolution - in general, the linear combinations for allsolutions will be of the form

u(x , t) =∞∑n=1

An sin(nπx

3

)e−

2n2π2t9

I and initial condition becomes

u(x , 0) =∞∑n=1

An sin(nπx

3

)



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary


u(x , 0) = f (x) =∞∑n=1

An sin(nπx

3

)

The initial condition can be further simplified in many casesusing a Fourier series. This can yield an explicit expressionfor A. However, this is beyond the scope of this lecture, butmany of you will encounter it further on. For now, it issimply important to realise that there can be an infinity ofsolutions, and that any linear combination of solutions is initself a solution.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Recap of method

When using separation of variables method, we take thefollowing steps

I Write the equation in the form u = X (x)T (t).

I Separate the variables on each side.

I Determine sign of K with boundary conditions .

I Find building block solution.

I If possible, solve with initial condition.

I Otherwise, write as an infinite sum over n.



Revision

Introduction

Direct Integration


1D Heat Equation

1D Wave Equation


Recap of method

Summary

Summary

I Partial Differential Equations involve the rates ofchange of multi-variate functions and are vital indescribing many physical phenomena.

I Some simple PDEs are amenable to direct integration.

I Others can be solved with Separation of Variables.

I Linear solutions obey the principle of superposition.

I It is important to remember that many PDEs cannot besolved analytically, and numerical methods are requiredto find solutions for these.



Appendix

Further reading andresources

For Further Reading I

Riley et al.Mathematical Methods for Physics and Engineering: AComprehensive Guide - 3rd Edition .Cambridge, 2006.

D.J. Griffiths.Introduction to Electrodynamics.Prentice Hall, 1999.

Various Authors.DTC Online resources.University of Oxford DTC

Documents

Partial Differential Equations - An introduction to PDES