Partial Differential equations: The Method of ...
Partial Differential equations: The Method of Characteristics II
April 2, 2014
0.1 0.1.1
Question 1 i)
Figure 1: The Three characteristic projections of point passing through (−2, 0),(0, 0), (5, 0) Consider the PDE 2u ux + ut = 0
(1)
The characteristics can be attained by solving the ODE dx ds dt ds du ds
=
2u
(2)
=
1
(3)
=
0
(4)
while the initial condition u(x, 0) = 3x
(5)
The initial conditions can be then written as x(τ, 0)
=
τ
(6)
t(τ, 0)
=
0
(7)
u(τ, 0)
=
3τ
(8)
Hence the characteristics can be attained from the ODE. Let us start with u(τ, s) and t(τ, s) first t(τ, s)
= s + c1 (τ )
u(τ, s)
= c3 (τ ) 1
(9) (10)
using the initial conditions we have: 0
=
t(τ, 0) = c1 (τ )
(11)
3τ
=
u(τ, 0) = c3 (τ )
(12)
hence t(τ, s) u(τ, s) now we are only missing x(τ, s), since
= s
(13)
=
(14)
du ds
3τ
= 0, u does not depend on s so that
x(τ, s) = 2u s + c2 (τ )
(15)
x(τ, s) = 6τ s + c2 (τ )
(16)
hence
now using the initial condition τ = x(τ, 0) = c2 (τ )
(17)
x(τ, s) = 6τ s + τ
(18)
x = 6τ t + τ
(19)
we end up with
or
This is just the projection characteristic of the PDE 2u ux + ut = 0
(20)
To plot the projections required in (x, t) let us write the projection characteristic x−τ 6τ
t=
(21)
The solution passing through (0, 0) is just τ = 0 and thus x=0
(22)
This is rendered in fig1 The solution passing through (−2, 0) is found by setting τ = −2 t=−
x 1 − 12 6
(23)
This is rendered in fig1 The solution passing through (5, 0) is found by setting τ = 5 t=
x 1 − 30 6
This is rendered in fig1
2
(24)
0.1.2
ii)
Since the 3 solutions we have found all cross at t = −1/6, as rendered in fig1, we expect u(x, t) not to be a differentiable solution for t < −1/6. Hence c = 1/6.
0.1.3
iii)
To yield the unique solution consider: x = 6τ t + τ
(25)
x 6t + 1
(26)
hence τ= now since u(τ, s)
=
3τ
(27)
3x 6t + 1
(28)
we have the explicit and unique solution u(x, t)
=
for the PDE 2u ux + ut = 0 This exists at least for small |t|. Notice and t 6= −1/6
3
(29)
0.2
Question 2
0.2.1
i)
Consider the PDE u ux + ut = 0
(30)
As in q1 the characteristics can be attained by solving the ODE dx ds dt ds du ds
= u
(31)
=
1
(32)
=
0
(33)
while the initial condition u(x, 0) = |x|
(34)
The initial conditions can be then written as x(τ, 0)
= τ
(35)
t(τ, 0)
=
(36)
u(τ, 0)
= |τ |
0
Hence the characteristics can be attained from the ODE. Again since does not depend on s, so that
(37) du ds
= 0, u
x(τ, s)
= u s + c2 (τ )
(38)
t(τ, s)
= s + c1 (τ )
(39)
u(τ, s)
= c3 (τ )
(40)
but using the initial conditions we have: 0
=
t(τ, 0) = c1 (τ )
(41)
|τ | =
u(τ, 0) = c3 (τ )
(42)
hence t(τ, s)
=
s
(43)
u(τ, s)
=
|τ |
(44)
hence we can write x(τ, s) as x(τ, s) = u s + c2 (τ ) = |τ |s + c2 (τ )
4
(45)
Figure 2: The characteristic projections point passing through (−3, 0),(−2, 0), (−1, 0) (0, 0),(1, 0), (2, 0),(3, 0) but again using the initial condition τ = x(τ, 0) = c2 (τ )
(46)
x(τ, s) = |τ | s + τ
(47)
we end up with
since s = t the projections of the characteristics can be written as: x(τ, s) = |τ | t + τ
(48)
Now for the point (−3, 0) gives τ = −3 and so t=
x +1 3
(49)
for the point (−2, 0) gives τ = −2 and so t=
x +1 2
(50)
for the point (−1, 0) gives τ = −1 and so t=x+1
(51)
for the point (0, 0) gives τ = 0 and so x=0
(52)
for the point (1, 0) gives τ = 1 and so t=x−1 5
(53)
for the point (2, 0) gives τ = 2 and so t=
x −1 2
(54)
for the point (3, 0) gives τ = 3 and so t=
x −1 3
(55)
these straight lines are all plotted in fig.2. Notice the solution x = 0 that is the t-axis should be also considered. These straight lines start to cross on the t-axis is 2 different points that is t = 1 and t = −1 and we will keep finding intersections among them for t > 1 or t < −1.
0.2.2
ii)
Figure 3: Notice how the characteristic projections point passing through (−3, 0),(−2, 0), (−1, 0) (0, 0),(1, 0), (2, 0),(3, 0) never cross for −1 < t < 1. While they do in fact cross for t ≥ 1 or t ≤ −1. In fig.4, we have slightly zoomed out of fig.2 and colored the straight lines representing the projections of the characteristics to show that the red lines t = 1 and t = −1 define the Domain of definition |t| < 1. Outside this domain |t| < 1 and thus for |t| ≥ 1, where the straight lines are painted in blue, the projections of the characteristics start to cross over and we expect to only be able to find a weak solution. 6
In the zone |t| < 1 of the (x, t) space where the straight lines are painted in black we may expect to find in a neighborhood of t = 0 a single solution but we need to check if it is also a differentiable solution. To check whether we can find a differentiable solution in |t| < 1 we can actually write the solutions. From the projections of characteristics x(τ, s) = |τ | t + τ
(56)
so for τ > 0 τ
>
x =
0
(57)
τt+τ
(58)
or τ
>
0
(59)
τ
=
x t+1
(60)
τ
0
(72)
=
x >0 t+1
(73)
gives that this solution is valid if (x > 0 t > −1) or (x < 0 t < −1) likewise τ τ
1) or (x < 0 t < 1) so if we pick up the pieces that give |t| < 1 that is −1 < t < 1, we actually have a piecewise solution x
>
0
x u(x, y) = t+1 x < 0 x u(x, y) = t−1
(76) (77) (78) (79) (80)
Clearly this function is not differentiable with respect to the variable x at x = 0, because x ux (x, y)
>
0
(81)
=
1 t+1
(82) (83)
and x
April 2, 2014
0.1 0.1.1
Question 1 i)
Figure 1: The Three characteristic projections of point passing through (−2, 0),(0, 0), (5, 0) Consider the PDE 2u ux + ut = 0
(1)
The characteristics can be attained by solving the ODE dx ds dt ds du ds
=
2u
(2)
=
1
(3)
=
0
(4)
while the initial condition u(x, 0) = 3x
(5)
The initial conditions can be then written as x(τ, 0)
=
τ
(6)
t(τ, 0)
=
0
(7)
u(τ, 0)
=
3τ
(8)
Hence the characteristics can be attained from the ODE. Let us start with u(τ, s) and t(τ, s) first t(τ, s)
= s + c1 (τ )
u(τ, s)
= c3 (τ ) 1
(9) (10)
using the initial conditions we have: 0
=
t(τ, 0) = c1 (τ )
(11)
3τ
=
u(τ, 0) = c3 (τ )
(12)
hence t(τ, s) u(τ, s) now we are only missing x(τ, s), since
= s
(13)
=
(14)
du ds
3τ
= 0, u does not depend on s so that
x(τ, s) = 2u s + c2 (τ )
(15)
x(τ, s) = 6τ s + c2 (τ )
(16)
hence
now using the initial condition τ = x(τ, 0) = c2 (τ )
(17)
x(τ, s) = 6τ s + τ
(18)
x = 6τ t + τ
(19)
we end up with
or
This is just the projection characteristic of the PDE 2u ux + ut = 0
(20)
To plot the projections required in (x, t) let us write the projection characteristic x−τ 6τ
t=
(21)
The solution passing through (0, 0) is just τ = 0 and thus x=0
(22)
This is rendered in fig1 The solution passing through (−2, 0) is found by setting τ = −2 t=−
x 1 − 12 6
(23)
This is rendered in fig1 The solution passing through (5, 0) is found by setting τ = 5 t=
x 1 − 30 6
This is rendered in fig1
2
(24)
0.1.2
ii)
Since the 3 solutions we have found all cross at t = −1/6, as rendered in fig1, we expect u(x, t) not to be a differentiable solution for t < −1/6. Hence c = 1/6.
0.1.3
iii)
To yield the unique solution consider: x = 6τ t + τ
(25)
x 6t + 1
(26)
hence τ= now since u(τ, s)
=
3τ
(27)
3x 6t + 1
(28)
we have the explicit and unique solution u(x, t)
=
for the PDE 2u ux + ut = 0 This exists at least for small |t|. Notice and t 6= −1/6
3
(29)
0.2
Question 2
0.2.1
i)
Consider the PDE u ux + ut = 0
(30)
As in q1 the characteristics can be attained by solving the ODE dx ds dt ds du ds
= u
(31)
=
1
(32)
=
0
(33)
while the initial condition u(x, 0) = |x|
(34)
The initial conditions can be then written as x(τ, 0)
= τ
(35)
t(τ, 0)
=
(36)
u(τ, 0)
= |τ |
0
Hence the characteristics can be attained from the ODE. Again since does not depend on s, so that
(37) du ds
= 0, u
x(τ, s)
= u s + c2 (τ )
(38)
t(τ, s)
= s + c1 (τ )
(39)
u(τ, s)
= c3 (τ )
(40)
but using the initial conditions we have: 0
=
t(τ, 0) = c1 (τ )
(41)
|τ | =
u(τ, 0) = c3 (τ )
(42)
hence t(τ, s)
=
s
(43)
u(τ, s)
=
|τ |
(44)
hence we can write x(τ, s) as x(τ, s) = u s + c2 (τ ) = |τ |s + c2 (τ )
4
(45)
Figure 2: The characteristic projections point passing through (−3, 0),(−2, 0), (−1, 0) (0, 0),(1, 0), (2, 0),(3, 0) but again using the initial condition τ = x(τ, 0) = c2 (τ )
(46)
x(τ, s) = |τ | s + τ
(47)
we end up with
since s = t the projections of the characteristics can be written as: x(τ, s) = |τ | t + τ
(48)
Now for the point (−3, 0) gives τ = −3 and so t=
x +1 3
(49)
for the point (−2, 0) gives τ = −2 and so t=
x +1 2
(50)
for the point (−1, 0) gives τ = −1 and so t=x+1
(51)
for the point (0, 0) gives τ = 0 and so x=0
(52)
for the point (1, 0) gives τ = 1 and so t=x−1 5
(53)
for the point (2, 0) gives τ = 2 and so t=
x −1 2
(54)
for the point (3, 0) gives τ = 3 and so t=
x −1 3
(55)
these straight lines are all plotted in fig.2. Notice the solution x = 0 that is the t-axis should be also considered. These straight lines start to cross on the t-axis is 2 different points that is t = 1 and t = −1 and we will keep finding intersections among them for t > 1 or t < −1.
0.2.2
ii)
Figure 3: Notice how the characteristic projections point passing through (−3, 0),(−2, 0), (−1, 0) (0, 0),(1, 0), (2, 0),(3, 0) never cross for −1 < t < 1. While they do in fact cross for t ≥ 1 or t ≤ −1. In fig.4, we have slightly zoomed out of fig.2 and colored the straight lines representing the projections of the characteristics to show that the red lines t = 1 and t = −1 define the Domain of definition |t| < 1. Outside this domain |t| < 1 and thus for |t| ≥ 1, where the straight lines are painted in blue, the projections of the characteristics start to cross over and we expect to only be able to find a weak solution. 6
In the zone |t| < 1 of the (x, t) space where the straight lines are painted in black we may expect to find in a neighborhood of t = 0 a single solution but we need to check if it is also a differentiable solution. To check whether we can find a differentiable solution in |t| < 1 we can actually write the solutions. From the projections of characteristics x(τ, s) = |τ | t + τ
(56)
so for τ > 0 τ
>
x =
0
(57)
τt+τ
(58)
or τ
>
0
(59)
τ
=
x t+1
(60)
τ
0
(72)
=
x >0 t+1
(73)
gives that this solution is valid if (x > 0 t > −1) or (x < 0 t < −1) likewise τ τ
1) or (x < 0 t < 1) so if we pick up the pieces that give |t| < 1 that is −1 < t < 1, we actually have a piecewise solution x
>
0
x u(x, y) = t+1 x < 0 x u(x, y) = t−1
(76) (77) (78) (79) (80)
Clearly this function is not differentiable with respect to the variable x at x = 0, because x ux (x, y)
>
0
(81)
=
1 t+1
(82) (83)
and x
