3.36pt CVT Sajid Ali PDEs Classi cation Solutions

CVT

Sajid Ali

PDEs

Classification

Solutions

3.36pt

CVT

Sajid Ali

PDEs

Classification

Solutions

Intro to Partial Differential Equations

Sajid Ali

SEECS-NUST

December 14, 2017

CVT

Sajid Ali

PDEs

Classification

Solutions

Concepts

Basic Idea. We have studied ordinary differential equa-tions (ODEs) that basically carry the information of how somechange occurs in a physical phenomenon.

CVT

Sajid Ali

PDEs

Classification

Solutions

Concepts


Q. Does it tell the complete story?

NO !!!

CVT

Sajid Ali

PDEs

Classification

Solutions

Concepts


Q. Does it tell the complete story?

NO !!!

The reason is very simple. The changes that we observearound us normally occur in a three dimensional space inwhich we live in. Therefore a physical phenomenon is bet-ter represented by equations that contain functions which areof more than one variables, like u = u(x , y) or u = u(x , y , z).However we have studied in our course (MVC) that for suchfunctions, change is represented by partial derivatives insteadof ordinary derivatives.

CVT

Sajid Ali

PDEs

Classification

Solutions

Partial Differential Equations

A partial differential equation (PDE) is an equation that con-tains partial derivatives (e.g., ux , uy , uxx , uyy , ...) in it.

CVT

Sajid Ali

PDEs

Classification

Solutions



For a function of two variables u = u(x , y), a general first-order PDE can be expressed as follows:

E (x , y , u, ux , uy ) = 0.

CVT

Sajid Ali

PDEs

Classification

Solutions



For a function of two variables u = u(x , y), a general first-order PDE can be expressed as follows:

E (x , y , u, ux , uy ) = 0.

Similarly a general second-order PDE can be written as

E (x , y , u, ux , uy , uxx , uxy , uyy ) = 0.

In this course we will focus on those PDEs which are of atmost second-order.

CVT

Sajid Ali

PDEs

Classification

Solutions

Natural Way of Obtaining PDEs

While reviewing ordinary differential equations we studied howto find a simple differential equation of a known function y =f (x). For example, we obtained the ODEs of the followingwell-studied functions.

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Sajid Ali

PDEs

Classification

Solutions

Examples

1. PDE of a Sphere

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Sajid Ali

PDEs

Classification

Solutions

Examples

1. PDE of a SphereThink of the following function

u = f (x , y) =√a2 − x2 − y2, a = fixed

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Sajid Ali

PDEs

Classification

Solutions

Examples


u = f (x , y) =√a2 − x2 − y2, a = fixed

which looks complicated but if we play with it, we get

u2 = a2 − x2 − y2

=⇒x2 + y2 + u2 = a2,

which can be identified as a sphere in x , y , u coordinates.

CVT

Sajid Ali

PDEs

Classification

Solutions

Examples


u = f (x , y) =√a2 − x2 − y2, a = fixed

which looks complicated but if we play with it, we get

u2 = a2 − x2 − y2

=⇒x2 + y2 + u2 = a2,

which can be identified as a sphere in x , y , u coordinates.Let us differentiate u, w.r.t. x

ux =−2x√

a2 − x2 − y2= −2x

u

=⇒ uux = −2x

which is a partial differential equation.

CVT

Sajid Ali

PDEs

Classification

Solutions

Examples

2. PDE of a ParaboloidA paraboloid

u = x2 + y2

has the corresponding PDE given by

uxx − uyy = 0.

CVT

Sajid Ali

PDEs

Classification

Solutions

Examples

2. PDE of a ParaboloidA paraboloid

u = x2 + y2

has the corresponding PDE given by

uxx − uyy = 0.

3. PDE of an Ellipsoid (Practice)Similarly for an ellipsoid

u =

√1− x2

a2+

y2

b2

we can find a partial differential equation ?

CVT

Sajid Ali

PDEs

Classification

Solutions

Natural Way of Obtaining PDEs

CVT

Sajid Ali

PDEs

Classification

Solutions

PDEs Everywhere !!!

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Sajid Ali

PDEs

Classification

Solutions

PDEs Everywhere !!!

The equation that describes the evolution of the stateψ(x , t), of a quantum particle is a PDE

i∂ψ

∂t=∂2ψ

∂x2

due to E. Schrodinger.

CVT

Sajid Ali

PDEs

Classification

Solutions

PDEs Everywhere !!!

CVT

Sajid Ali

PDEs

Classification

Solutions

PDEs Everywhere !!!

The equation that describes the dynamics of our universe isa system of PDEs and in a matrix form are compiled as

Rµν −1

2Rgµν = κTµν

due to A. Einstein.

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Sajid Ali

PDEs

Classification

Solutions

Classification

We will confine our study on second-order PDEs of the normalform

auxx + 2buxy + cuyy = F (x , y , u, ux , uy )

There are three famous second-order linear PDEs known fortheir physical characteristics.

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Sajid Ali

PDEs

Classification

Solutions

Classification




Type Equation Phenomenon

Wave utt = c2uxx Wave behavior

Heat ut = c2uxx Heat distribution

Laplace uxx + uyy = 0 Steady state temperatures

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Sajid Ali

PDEs

Classification

Solutions

Classification




Type Equation Phenomenon

Wave utt = c2uxx Wave behavior

Heat ut = c2uxx Heat distribution

Laplace uxx + uyy = 0 Steady state temperatures

Note that in the first two PDEs the dependent variable isu = u(x , t), because it corresponds to a quantity that changeswith respect to time as well as distance.

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Sajid Ali

PDEs

Classification

Solutions

Classification

We will confine our study on second-order PDEs of thenormal form

auxx + 2buxy + cuyy = F (x , y , u, ux , uy ) −→ (1)

which are classified into three main branches

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Sajid Ali

PDEs

Classification

Solutions

Classification




Type Definition Example

Hyperbolic ac − b2 < 0 Wave

Parabolic ac − b2 = 0 Heat

Elliptic ac − b2 > 0 Laplace

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Sajid Ali

PDEs

Classification

Solutions

Classification




Type Definition Example

Hyperbolic ac − b2 < 0 Wave

Parabolic ac − b2 = 0 Heat

Elliptic ac − b2 > 0 Laplace

We normally obtain the solution of a second-order PDE bylooking at its type and comparing it with one of the equation(wave, heat or Laplace).

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions

The solution u = u(x , y), of a partial differential equation isa function defined in a domain D, of xy−plane that satisfiesthe equation identically.

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions


Examples:1. The ellipsoid u(t, x) = t2 + 4x2, satisfy the wave equation

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Sajid Ali

PDEs

Classification

Solutions

Solutions


Examples:1. The ellipsoid u(t, x) = t2 + 4x2, satisfy the wave equation

u(x , t) = t2 + 4x2

=⇒ ut = 2t, utt = 2

=⇒ ux = 8x , uxx = 8

=⇒ utt −1

4uxx = 0, c = 1/2

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Sajid Ali

PDEs

Classification

Solutions

Solutions

Examples:2. The function u = sin (kx) cos (kct), satisfy the waveequation.

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Sajid Ali

PDEs

Classification

Solutions

Solutions


ut = −kc sin(kx) sin(kct)

⇒ utt = −k2c2 sin (kx) cos (kct),

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Sajid Ali

PDEs

Classification

Solutions

Solutions




ux = k cos(kx) sin(kct)

⇒ uxx = −k2 sin (kx) cos (kct)

Therefore, utt = c2uxx .

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions




ux = k cos(kx) sin(kct)

⇒ uxx = −k2 sin (kx) cos (kct)

Therefore, utt = c2uxx .

3. The function u = (x − ct)2, also satisfy wave equation.

4. The function u = ex−ct , also satisfy wave equation.

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions


From above examples it is quite clear that

All PDEs have infinitely many solutions !!!

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions


From above examples it is quite clear that

All PDEs have infinitely many solutions !!!

How are we going to deal with this plethora of solutions fora particular partial differential equation? Although you mayfind an answer to this question very difficult but it turns outthat we can explicitly determine generic forms of thesolutions for some PDEs.

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions

Practice:

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Sajid Ali

PDEs

Classification

Solutions

Solutions

Practice:1. The functionsu = e−2kt cos(8x) & u = ex

2−y2sin(2xy) satisfy the heat

and Laplace equations (verify !!!).

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions

Practice:1. The functionsu = e−2kt cos(8x) & u = ex

2−y2sin(2xy) satisfy the heat

and Laplace equations (verify !!!).

2. For what value of k , does the functionu(x , y) = x2 + ky2, satisfy Laplace equation?

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions

ODE Vs PDE:For some PDEs which do not involve partial derivatives withrespect to both x & y , then we can solve them by solvingtheir ODE analogues.

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions

ODE Vs PDE:For some PDEs which do not involve partial derivatives withrespect to both x & y , then we can solve them by solvingtheir ODE analogues.Examples:1. Consider u = u(x , y) and its PDE does not involve partialderivative w.r.t y , i.e.,

uxx − u = 0.

Sol. Since no y−derivative exists we can compare it with anODE, so

w′′ − w = 0

which has the solution (verify !!!) w = c1ex + c2e

−x ,

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions

ODE Vs PDE:For some PDEs which do not involve partial derivatives withrespect to both x & y , then we can solve them by solvingtheir ODE analogues.Examples:1. Consider u = u(x , y) and its PDE does not involve partialderivative w.r.t y , i.e.,

uxx − u = 0.

Sol. Since no y−derivative exists we can compare it with anODE, so

w′′ − w = 0

which has the solution (verify !!!) w = c1ex + c2e

−x ,therefore the solution of PDE is

=⇒ u = c1ex + c2e

−x

=⇒ u = c1(y)ex + c2(y)e−x

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions

Examples:2. Let us solve this PDE

uxy = −ux .

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Sajid Ali

PDEs

Classification

Solutions

Solutions


uxy = −ux .

Sol. Let ux = p, then

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Sajid Ali

PDEs

Classification

Solutions

Solutions


uxy = −ux .

Sol. Let ux = p, then

py = −p

−→ pyp

= −1

−→ ln p = −y + c1(x)

−→ p = ec1(x)−y −→ p = c(x)e−y

−→ ux = c(x)e−y

−→ u = e−y

∫c(x)dx + g(y)

−→ u = f (x)e−y + g(y)

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions

Examples:3. Let us solve PDE uyy + 16u = 0, by using

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Sajid Ali

PDEs

Classification

Solutions

Solutions


w′′

+ 16w = 0

=⇒ w = c1 cos(4y) + c2 sin(4y)

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Sajid Ali

PDEs

Classification

Solutions

Solutions


w′′

+ 16w = 0

=⇒ w = c1 cos(4y) + c2 sin(4y)

Therefore,

=⇒ u(x , y) = c1(x) cos(4y) + c2(x) sin(4y)

CVT

Sajid Ali

PDEs

Classification

Solutions

Solutions

Practice:1. Solve

uyy = 0.

2. Also integrate the PDE

uyy = 4xuy .

3.36pt CVT Sajid Ali PDEs Classi cation Solutions

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