3.36pt CVT Sajid Ali PDEs Classi cation Solutions

Sajid Ali

Classification

Solutions

3.36pt

Sajid Ali

Classification

Solutions

Intro to Partial Differential Equations

Sajid Ali

SEECS-NUST

December 14, 2017

Sajid Ali

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.

Sajid Ali

Classification

Solutions

Concepts

Q. Does it tell the complete story?

NO !!!

Sajid Ali

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.

Sajid Ali

Classification

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Partial Differential Equations

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

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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.

Sajid Ali

Classification

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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.

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Classification

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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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Examples

1. PDE of a Sphere

Sajid Ali

Classification

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Examples

1. PDE of a SphereThink of the following function

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

Sajid Ali

Classification

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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.

Sajid Ali

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

=⇒ uux = −2x

which is a partial differential equation.

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Classification

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Examples

2. PDE of a ParaboloidA paraboloid

u = x2 + y2

has the corresponding PDE given by

uxx − uyy = 0.

Sajid Ali

Classification

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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

√1− x2

we can find a partial differential equation ?

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Natural Way of Obtaining PDEs

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Solutions

PDEs Everywhere !!!

Sajid Ali

Classification

Solutions

PDEs Everywhere !!!

Sajid Ali

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

due to E. Schrodinger.

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PDEs Everywhere !!!

Sajid Ali

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.

Sajid Ali

Classification

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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.

Sajid Ali

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Classification

Type Equation Phenomenon

Wave utt = c2uxx Wave behavior

Heat ut = c2uxx Heat distribution

Laplace uxx + uyy = 0 Steady state temperatures

Sajid Ali

Classification

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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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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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Classification

Type Definition Example

Hyperbolic ac − b2 < 0 Wave

Parabolic ac − b2 = 0 Heat

Elliptic ac − b2 > 0 Laplace

Sajid Ali

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).

Sajid Ali

Classification

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.

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Classification

Solutions

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

Sajid Ali

Classification

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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Classification

Solutions

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

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Solutions

ut = −kc sin(kx) sin(kct)

⇒ utt = −k2c2 sin (kx) cos (kct),

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Classification

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ux = k cos(kx) sin(kct)

⇒ uxx = −k2 sin (kx) cos (kct)

Therefore, utt = c2uxx .

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Classification

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.

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Classification

Solutions

From above examples it is quite clear that

All PDEs have infinitely many solutions !!!

Sajid Ali

Classification

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.

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Classification

Solutions

Practice:

Sajid Ali

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Solutions

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

2−y2sin(2xy) satisfy the heat

and Laplace equations (verify !!!).

Sajid Ali

Classification

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?

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Classification

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.

Sajid Ali

Classification

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 ,

Sajid Ali

Classification

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

=⇒ u = c1(y)ex + c2(y)e−x

Sajid Ali

Classification

Solutions

Examples:2. Let us solve this PDE

uxy = −ux .

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Classification

Solutions

uxy = −ux .

Sol. Let ux = p, then

Sajid Ali

Classification

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)

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

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Classification

Solutions

w′′

+ 16w = 0

=⇒ w = c1 cos(4y) + c2 sin(4y)

Sajid Ali

Classification

Solutions

w′′

+ 16w = 0

=⇒ w = c1 cos(4y) + c2 sin(4y)

Therefore,

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

Sajid Ali

Classification

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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