Pdes Flashcards
When is the equation D/Dr(r^2 DT/Dr) useful?
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
What is the diffusion equation?
dT/dt = alpha del^2 T
Show that B = fn(y/sqrt(alpha * t) solves the diffusion equation.
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.
What is cp? How is it calculated?
cp = omega / k
It is the speed of individual wave crests.
What is cg? How is it calculated?
cg = d omega / dk
It is the speed of energy.
How would you calculate cg for a 3D wave?
Let eta = eta* * exp(j (x.k - omega t)
Solve for omega = fn ( k )
cg = d omega / dk
What is curl (u) in index notation?
epsilon_ijk * d/dxj * u_k
Show that u x u = 0 using index notation
epsilon_ijk * u_j * u_k = - epsilon_ikj * u_j * u_k
= + epsilon_ikj * u_k * u_j
therefore = 0
What is the first step to showing a poisson solution is unique?
Let v = u1 - u2, where u1 & u2 are valid.
In what circumstances is a poisson solution unique?
del^2 (u) = f
subject to u = g OR du/dn = h on S
v = u1 - u2, where u1 and u2 are valid poisson solutions.
How would you start to show that v is zero everywhere?
Take del.(v del v) = v * del^2 (v) + (del(v))^2
= del(v)^2
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?
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.
What is {[A][B]}_ij in index notation?
A_ik B_kj
Show that {[I][B]} = [B] where [I] is the identity matrix.
[I]_ij = delta_ij
IB = I_ik B_kj = delta_ik Bkj = B_ij
What is div(u) in index notation?
divergence applies to a vector, so u is a vector.
div(u) = du_i / dx_i = a scalar.
What is grad(u) in index noation?
grad applies to a scalar, so f is a scalar.
grad(f) = df / dx_i = a vector.
What is grad(abs(x)) in index notation?
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.
What is integration by parts in index notation?
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.
How is the Rayleigh-Ritz method performed?
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.
How is the Galerkin method performed?
Take the weak form of the equation.
Use u = a_1 f_1 + … and w = f_1, then w = f_2
Run the Galerkin method on u’’ - u = -1, using sin(x) as a basis function.
[The weak form equivalent is u’v’ + uv - v]
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
A strong form PDE is u’’ - u = -1. How would you start to perform the Galerkin method?
Find the equivalent weak form. This turns out to be int{ u’v’ + uv - v }
How does the Galerkin method vary from the Rayleigh-Ritz method? Which one is more general?
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.
Under what conditions do the Galerkin and R-R methods agree?
They always agree if the R-R method is possible (ie the variational form exists)
How does the Rayleigh-Ritz method work?
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
The Rayleigh-Ritz method acts on the ____
Variational form
The Galerkin method acts on the ____
Weak form
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)?
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
The strong form of a PDE is u’’ + 1 = 0. How would the Galerkin method be applied using a trial function u = x?
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!)
The strong form of a PDE is u’’ + 1 = 0. How would the Rayleigh-Ritz method be applied using a trial function u = x?
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
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?
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.
What is the Euler-Lagrange equation?
D = \partial
DF/Dy - d/dx ( DF/Dy’ )
In deriving the E-L equation, what is the first step?
I = int { F (y, y’, x) }
DI[ z ] = int{ F(y + z, y’ + z’, x) - F( y, y’, x ) }
D = \partial
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?
Expand out the first term and eliminate thus:
DI [z]= int{ DF/Dy * z + DF/Dy’ * z’ }
D = \partial
In deriving the E-L equation, we have the line:
DI [z]= int{ DF/Dy * z + DF/Dy’ * z’ }
What comes next?
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
In deriving the E-L equation, we have the line:
int{ z * (DF/Dy - d/dx(DF/Dy’))} = 0
What comes next?
This is basically finished, z is arbitrary so the integrand must be zero everywhere, thus:
DF/Dy - d/dx(DF/Dy’) = 0
D = \partial
What is the E-L equation?
DF/Dy - d/dx(DF/Dy’) = 0
D = \partial
Without going in to detail, what are the steps in the E-L derivation?
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
In deriving the Beltrami identity, what is dF/dx expanded to?
dF/dx = DF/Dy * y’ + DF/Dy’ * y’’ + DF/Dx
Don’t forget the last term (and note it is just DF/Dx)!
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?
Equate y’ * DF/Dy terms:
dF/dx - DF/Dy’ * y’’ + DF/Dx = y’ * d/dx(DF/Dy’)
In deriving the Beltrami identity, we have the expression:
dF/dx - DF/Dy’ * y’’ - DF/Dx = y’ * d/dx(DF/Dy’)
what comes next?
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.
Apply the E-L equation to:
I = int{ (u’)^2 + lambda * u^2 - lambda}
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.
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?
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
The diffusion equation is:
\alpha d2T/dx2 = dT / dt
What classification is it?
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
The wave equation is:
\alpha^2 d2u/dx2 = d2u//dt2
What classification is it?
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
The Poisson equation is:
d2u/dx2 + d2u/dy2 = F(x,y)
What classification is it?
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
Give an example of an elliptic equation.
The poisson equation is elliptic.
Give an example of a Parabolic equation.
The diffusion equation is parabolic.
Give an example of a hyperbolic equation.
The wave equation is hyperbolic.
The wave equation is hyperbolic. What is the value of “b^2-ac”?
a = c^2 c = - 1
b^2 - ac > 0 so it is hyperbolic.
The poisson equation is elliptic. What is the value of “b^2 - ac”?
a = c = 1
b^2 - ac < 0 so it is elliptic.
The diffusion equation is parabolic. What is the value of “b^2 - ac”?
a = \alpha ; c = 0
b^2 - ac = 0 so it is parabolic.
Parabolic, hyperbolic and elliptic equations are defined by “b^2 - ac”. In what order do they go?
Hyperbolic: +ve
Parabolic: 0
Elliptic: -ve
Hyperbolic, parabolic and elliptic equations are defined by “b^2 - ac”. In what order do they go?
Hyperbolic: +ve
Parabolic: 0
Elliptic: -ve
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?
a d2u/dx2 + 2b d2u/dxdy + c d2u/dy2 = F(x,y,du/dx,dy/dx)
A second-order DE has the form:
d2u/dx2 + d2u/dxdy + d2u/dy2 + 5 du/dy = 0
What is “b^2 - ac”?
a = 1 ; c = 1
2b = 1 -> b = 0.5
b^2 - ac = 0.5^2 - 1 * 1 < 0
Elliptical Poisson Equation.
A second-order DE has the form:
d2u/dx2 + du/dx = 0
What is “b^2 - ac”?
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
A second-order DE has the form:
d2u/dxdy + du/dx = 0
What is “b^2 - ac”?
2b = 1 -> b = 0.5 a = c = 0
b^2 - a c > 0
so it behaves like a wave equation.
A second-order DE has the form:
d2u/dx2 + d2u/dy2 + 5 du/dy = 0
What is “b^2 - ac”?
a = c = 1 (this is a modified poisson equation)
b^2 - ac < 0 so it is elliptic and similar to poisson.
A second-order DE has the form:
d2u/dx2 + 5 du/dy = d2u/dt2
What is “b^2 - ac”?
a = 1 ; c = -1 (this is a modified wave equation)
b^2 - ac > 0 so it is hyperbolic and similar to wave.
