Pdes

Question 1

When is the equation D/Dr(r^2 DT/Dr) useful?

Answer 1

In spherical symmetry, we get the equn

(1/r)^2 * D/Dr(r^2 DT/Dr)

This rearranges to:

1/r * D^2/Dr^2 (rT)

Set A = rT, then DT/Dt = 1/r * DA/Dt

So equn is diffusion equation with A = rT

Question 2

What is the diffusion equation?

Answer 2

dT/dt = alpha del^2 T

Question 3

Show that B = fn(y/sqrt(alpha * t) solves the diffusion equation.

Answer 3

Require that dB/dt = d^2B/dt^2

dB/dt = B’ * -1/2 * alpha^-.5 * -1/2 * t^-1.5

d^2B/dy^2 = d/dy( B’ * 1/sqrt(alpha t) )
= B’’ * 1/alpha t

then B’’ - B’ * eta / 2 = 0

This is solved by B = int(0 -> eta) {exp(-x^2 / 4)} dx

B’ = exp(-eta^2 / 4)
B’’ = -eta / 2 * B’
as required.

Question 4

What is cp? How is it calculated?

Answer 4

cp = omega / k

It is the speed of individual wave crests.

Question 5

What is cg? How is it calculated?

Answer 5

cg = d omega / dk

It is the speed of energy.

Question 6

How would you calculate cg for a 3D wave?

Answer 6

Let eta = eta* * exp(j (x.k - omega t)

Solve for omega = fn ( k )

cg = d omega / dk

Question 7

What is curl (u) in index notation?

Answer 7

epsilon_ijk * d/dxj * u_k

Question 8

Show that u x u = 0 using index notation

Answer 8

epsilon_ijk * u_j * u_k = - epsilon_ikj * u_j * u_k
= + epsilon_ikj * u_k * u_j

therefore = 0

Question 9

What is the first step to showing a poisson solution is unique?

Answer 9

Let v = u1 - u2, where u1 & u2 are valid.

Question 10

In what circumstances is a poisson solution unique?

Answer 10

del^2 (u) = f

subject to u = g OR du/dn = h on S

Question 11

v = u1 - u2, where u1 and u2 are valid poisson solutions.

How would you start to show that v is zero everywhere?

Answer 11

Take del.(v del v) = v * del^2 (v) + (del(v))^2

= del(v)^2

Question 12

In showing that u is a unique soln to a poisson equation, we have shown that

del.(v del v) = v * del^2 (v) + (del(v))^2
= del(v)^2

What comes next?

Answer 12

Show that
Then perform integration by parts on del.(v del v)

= integral( (v del(v)) . n ) ds

and v(part of boundary) = del(v(part of boundary)) = 0

So del(v) ^2 = 0 so v = 0 everywhere. u1 = u2.

Question 13

What is {[A][B]}_ij in index notation?

Answer 13

A_ik B_kj

Question 14

Show that {[I][B]} = [B] where [I] is the identity matrix.

Answer 14

[I]_ij = delta_ij

IB = I_ik B_kj = delta_ik Bkj = B_ij

Question 15

What is div(u) in index notation?

Answer 15

divergence applies to a vector, so u is a vector.

div(u) = du_i / dx_i = a scalar.

Question 16

What is grad(u) in index noation?

Answer 16

grad applies to a scalar, so f is a scalar.

grad(f) = df / dx_i = a vector.

Question 17

What is grad(abs(x)) in index notation?

Answer 17

d/dx_i ( (xj xj)^0.5 )

This evaluates to:

2 x_j * dx_j / dx_i * 0.5 * (xk xk)^-0.5

= x_i / abs(x)

ie a vector.

Question 18

What is integration by parts in index notation?

Answer 18

d/dxi ( f * u_i ) = df/dxi . u_i + f * d/dxi ( u_i )

lhs is divergence theorem = f * u_i . n_i along the surface.

Question 19

How is the Rayleigh-Ritz method performed?

Answer 19

Must have a functional I_h as a function of u, the solution. Set u = a_1 f_1 + a_2 f_2 + …

Then dI_h / da_1 = 0 etc.

This may give a matrix expression that needs to be inverted.

Question 20

How is the Galerkin method performed?

Answer 20

Take the weak form of the equation.

Use u = a_1 f_1 + … and w = f_1, then w = f_2

Question 21

Run the Galerkin method on u’’ - u = -1, using sin(x) as a basis function.

[The weak form equivalent is u’v’ + uv - v]

Answer 21

u = a sin(x) ; v = sin(x)
u' = a cos(x) ; v' = cos(x)

integral{ acos(x) * cos(x) + a sin(x) * sin(x) - cos(x) } = 0

a * pi/2 - 1 = 0

a = 2/pi

Question 22

A strong form PDE is u’’ - u = -1. How would you start to perform the Galerkin method?

Answer 22

Find the equivalent weak form. This turns out to be int{ u’v’ + uv - v }

Question 23

How does the Galerkin method vary from the Rayleigh-Ritz method? Which one is more general?

Answer 23

The Galerkin method works on the weak form, and so is general. The R-R method works on the variational form, so only works if the variational form exists.

Question 24

Under what conditions do the Galerkin and R-R methods agree?

Answer 24

They always agree if the R-R method is possible (ie the variational form exists)

Question 25

How does the Rayleigh-Ritz method work?

Answer 25

Take the variational form I(u) = integral{}, where DI = 0 Insert u = a_1 f_1 + a_2 f_2 + ... Take derivatives wrt a_1, a_2 etc. If necessary, perform matrix inversion to find a_1, a_2

Question 26

The Rayleigh-Ritz method acts on the ____

Answer 26

Variational form

Question 27

The Galerkin method acts on the ____

Answer 27

Weak form

Question 28

The variational form of a PDE is I = int{u^2+u} between x = 0 and x = pi/2. How would the Rayleigh-Ritz method be performed using the trial function u = sin(x)?

Answer 28

Set u = a1 sin(x). Write out I = int{a1^2 sin^2(x) + a1 sin(x)} Take derivative of I wrt a1: a1^2 * 1/2 + a1 = 0 a1 = -0.5

Question 29

The strong form of a PDE is u'' + 1 = 0. How would the Galerkin method be applied using a trial function u = x?

Answer 29

Find the weak form: int{u' v' + v} = 0 Set u = a1 x ; v = x int{ a1 * 1 + x} = 0 Evaluate integral to get a1 = 4 (or similar!)

Question 30

The strong form of a PDE is u'' + 1 = 0. How would the Rayleigh-Ritz method be applied using a trial function u = x?

Answer 30

Find the weak form int{u' v' + v} = 0 Find the variational form: I = int{1/2(u')^2 + u} Insert u = a1 x ; u' = a1 I = int{1/2 (a1)^2+a1 x} dI/ d(a1) = 0 1/2 * 2a1 * 5 + 5^2/2 = 0 a1 = -5/2

Question 31

In showing that u1 is a unique solution to the Poisson equation, the vector identity: del( v del(v) ) = v * del^2 (v) + (del(v))^2 is used. What does it show?

Answer 31

The first term on the rhs evaluates to zero. Using the divergence theorem on the lhs, this is zero. Therefore, int{(del(v))^2} = 0, so v = 0 everywhere, so u1 = u2 and there is only one unique solution.

Question 32

What is the Euler-Lagrange equation?

Answer 32

D = \partial DF/Dy - d/dx ( DF/Dy' )

Question 33

In deriving the E-L equation, what is the first step?

Answer 33

I = int { F (y, y', x) } DI[ z ] = int{ F(y + z, y' + z', x) - F( y, y', x ) } D = \partial

Question 34

In deriving the E-L equation, we have the line: DI[ z ] = int{ F(y + z, y' + z', x) - F( y, y', x ) } What comes next?

Answer 34

Expand out the first term and eliminate thus: DI [z]= int{ DF/Dy * z + DF/Dy' * z' } D = \partial

Question 35

In deriving the E-L equation, we have the line: DI [z]= int{ DF/Dy * z + DF/Dy' * z' } What comes next?

Answer 35

Integrate the second term by parts to get: int{DF/Dy' * z'} = [DF/Dy' * z] - int{ z * d/dx{ DF/Dy' } } and so: int{ z * (DF/Dy - d/dx(DF/Dy'))} = 0 D = \partial

Question 36

In deriving the E-L equation, we have the line: int{ z * (DF/Dy - d/dx(DF/Dy'))} = 0 What comes next?

Answer 36

This is basically finished, z is arbitrary so the integrand must be zero everywhere, thus: DF/Dy - d/dx(DF/Dy') = 0 D = \partial

Question 37

What is the E-L equation?

Answer 37

DF/Dy - d/dx(DF/Dy') = 0 | D = \partial

Question 38

Without going in to detail, what are the steps in the E-L derivation?

Answer 38

Take functional derivative of variational form. Expand in terms of DF/Dy and DF/Dy'. Integration by parts on the second term. Note that integrand now multiplied by z everywhere, so remove integral sign and say = 0 everywhere. D = \partial

Question 39

In deriving the Beltrami identity, what is dF/dx expanded to?

Answer 39

dF/dx = DF/Dy * y' + DF/Dy' * y'' + DF/Dx Don't forget the last term (and note it is just DF/Dx)!

Question 40

In deriving the Beltrami identity, we have the expressions: dF/dx = DF/Dy * y' + DF/Dy' * y'' + DF/Dx and y' * DF/Dy - y' * d/dx(DF/Dy') = 0 What comes next?

Answer 40

Equate y' * DF/Dy terms: dF/dx - DF/Dy' * y'' + DF/Dx = y' * d/dx(DF/Dy')

Question 41

In deriving the Beltrami identity, we have the expression: dF/dx - DF/Dy' * y'' - DF/Dx = y' * d/dx(DF/Dy') what comes next?

Answer 41

This is basically it. This becomes: - DF/Dx + d/dx( F - y' * DF/Dy' ) Note how the last term would expand using the product rule.

Question 42

Apply the E-L equation to: I = int{ (u')^2 + lambda * u^2 - lambda}

Answer 42

DF/Du - d/dx(DF/Du') = 0 2 * lamda * u - d/dx( 2 u' ) = 0 lamda * u - u'' = 0 Which has a solution that depends on the sign of lambda.

Question 43

A second order PDE has the equation a d2u/dx2 + 2b d2u/dxdy + c d2u/dy2 = F(x,y,du/dx,dy/dx) How would you know its classification?

Answer 43

The value of b^2-ac dictates its type. b^2-ac > 0 hyperbolic / wave b^2-ac = 0 parabolic / diffusion b^2-ac < 0 elliptic / poisson

Question 44

The diffusion equation is: \alpha d2T/dx2 = dT / dt What classification is it?

Answer 44

Rearrange to the form: a d2u/dx2 + 2b d2u/dxdy + c d2u/dy2 = F(x,y,du/dx,dy/dx) i.e. a = \alpha Then b^2 - ac = 0 so parabolic

Question 45

The wave equation is: \alpha^2 d2u/dx2 = d2u//dt2 What classification is it?

Answer 45

Rearrange to the form: a d2u/dx2 + 2b d2u/dxdy + c d2u/dy2 = F(x,y,du/dx,dy/dx) i.e. a = \alpha^2; c = - 1 Then b^2 - ac > 0 so hyperbolic

Question 46

The Poisson equation is: d2u/dx2 + d2u/dy2 = F(x,y) What classification is it?

Answer 46

Rearrange to the form: a d2u/dx2 + 2b d2u/dxdy + c d2u/dy2 = F(x,y,du/dx,dy/dx) i.e. a = c = 1 Then b^2 - ac < 0 so elliptic

Question 47

Give an example of an elliptic equation.

Answer 47

The poisson equation is elliptic.

Question 48

Give an example of a Parabolic equation.

Answer 48

The diffusion equation is parabolic.

Question 49

Give an example of a hyperbolic equation.

Answer 49

The wave equation is hyperbolic.

Question 50

The wave equation is hyperbolic. What is the value of "b^2-ac"?

Answer 50

``` a = c^2 c = - 1 ``` b^2 - ac > 0 so it is hyperbolic.

Question 51

The poisson equation is elliptic. What is the value of "b^2 - ac"?

Answer 51

a = c = 1 b^2 - ac < 0 so it is elliptic.

Question 52

The diffusion equation is parabolic. What is the value of "b^2 - ac"?

Answer 52

a = \alpha ; c = 0 b^2 - ac = 0 so it is parabolic.

Question 53

Parabolic, hyperbolic and elliptic equations are defined by "b^2 - ac". In what order do they go?

Answer 53

Hyperbolic: +ve Parabolic: 0 Elliptic: -ve

Question 54

Hyperbolic, parabolic and elliptic equations are defined by "b^2 - ac". In what order do they go?

Answer 54

Hyperbolic: +ve Parabolic: 0 Elliptic: -ve

Question 55

Hyperbolic, parabolic and elliptic equations are usually identified by the value of "b^2 - ac". To what do the parameters a, b and c refer?

Answer 55

a d2u/dx2 + 2b d2u/dxdy + c d2u/dy2 = F(x,y,du/dx,dy/dx)

Question 56

A second-order DE has the form: d2u/dx2 + d2u/dxdy + d2u/dy2 + 5 du/dy = 0 What is "b^2 - ac"?

Answer 56

a = 1 ; c = 1 2b = 1 -> b = 0.5 b^2 - ac = 0.5^2 - 1 * 1 < 0 Elliptical Poisson Equation.

Question 57

A second-order DE has the form: d2u/dx2 + du/dx = 0 What is "b^2 - ac"?

Answer 57

a = 1 ; b = c = 0 (Note that the second derivative is a first order derivative) b^2 - ac = 0 ie similar to a diffusion equation

Question 58

A second-order DE has the form: d2u/dxdy + du/dx = 0 What is "b^2 - ac"?

Answer 58

``` 2b = 1 -> b = 0.5 a = c = 0 ``` b^2 - a c > 0 so it behaves like a wave equation.

Question 59

A second-order DE has the form: d2u/dx2 + d2u/dy2 + 5 du/dy = 0 What is "b^2 - ac"?

Answer 59

a = c = 1 (this is a modified poisson equation) b^2 - ac < 0 so it is elliptic and similar to poisson.

Question 60

A second-order DE has the form: d2u/dx2 + 5 du/dy = d2u/dt2 What is "b^2 - ac"?

Answer 60

a = 1 ; c = -1 (this is a modified wave equation) b^2 - ac > 0 so it is hyperbolic and similar to wave.