MA 1506 Mathematics II Tutorial 11 - Partial Differential ...Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 3/14 Question 2 Let E(t) and D(t) be the number of Elves

MA 1506Mathematics II

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MA 1506 Mathematics II

Tutorial 11Partial Differential Equations

Groups: B03 & B08April 11, 2012

Ngo Quoc AnhDepartment of Mathematics

National University of Singapore


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Question 1: Classification of systems of ODEs

Given matrix B representing a sys-tem of ODEs, in order to deter-mine the phase portrait of thesystem, we calculate detB andtrace(B). For example, given

B =

(2 −24 0

),

we get that

detB = 8, trace(B) = 2.

Using the picture on the right, we

conclude that the system represents either SPIRAL SOURCEor NODAL SOURCE. Since detB > 1

4(trace(B))2, i.e., thepoint (trace(B),detB) lies above the parabola, we get theanswer SPIRAL SOURCE.

http://en.wikipedia.org/wiki/Phase_plane


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Let

E(t) and D(t) be the number of Elves and Dwarvesrespectively. Let also

BE and DE be the birth and death rates per capita forElves. Similarly,

BD and DD for Dwarves.

We are told that

BE > BD and DE < DD.

In order to formulate ODEs, we still need

PE and PD: constants measuring the prejudice of Elvesand Dwarves respectively.

Moreover, we are told that PE > PD. In the standardMalthusian model, dE

dt is controlled by (BE −DE)E.However, due to the presence of PE , the amount PED must


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be deduced. Thus, we can formulate the following{dEdt = (BE −DE)E − PED,dDdt = (BD −DD)D − PDE.

The above system of ODEs can be rewritten into(dEdtdDdt

)=

(BE −DE −PE−PD BD −DD

)(ED

).

In a concrete case, the given matrix B =

(5 −4−1 2

)satisfies all above conditions.

Since detB = 6, trace(B) = 7, andtrace(B)2 − 4 detB = 25, we have a nodal source in thephase portrait. Given that at a certain time, the number ofDwarves is slightly larger than that of Elves, we now study


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the population of Elves. Fromthe picture on the right, thephase portrait is bisected by theline D = E.

All points above that line willmove along trajectories thateventually hit the D axis (equiv-alently, E = 0). So, if at sometime, there holds D > E, thenElf population may increase for

E

D

a while, but eventually it will reach a maximum and thecollapse to zero.

In other words, Rivendell is completely taken over byDwarves even though BE > BD and DE < DD. The keypoint is that the prejudice of the Elves cancels out theirother advantages and causes them to lose the competition.


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Let us first observe the following

lbs

sec=

lbs

gallons× gallons

sec.

Therefore, in order to solve theproblem, we use the following rule:

Tan

k A T

ank B

25 lbs UF6

Exit

6 gal/min

2 gal/min4 gal/min

Pure water in4 gal.min

Solution to B

Solution to A

Rate of change of the amount of UF6 equalsconcentration in × flow rate in − concentration out ×flow rate out where concentration is the mass of UF6 perunit volume of water.

Let xA and xB be the mass of UF6 in tanks A and Brespectively. By hypothesis, the concentration in the tanks Aand B are xA

100 and xB100 respectively.

Clearly, the amounts of water in both tanks remain constantsince the 4 gallons/min of pure water flowing into the tankA balances the solution that exits from the tank B.


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From the given data, we get

xA = 2xB100︸︷︷︸

in from B

− 6xA100︸︷︷︸

out to B

,

xB = 6xA100︸︷︷︸

in from A

− 2xB100︸︷︷︸

out to A

− 4xB100︸︷︷︸exit

.

Tan

k A T

ank B

25 lbs UF6

Exit

6 gal/min

2 gal/min4 gal/min

Pure water in4 gal.min

Solution to B

Solution to A

For the initial condition, xA(0) = 25 and xB(0) = 0. Inmatrix form, we get(

xAxB

)=

1

100

(−6 26 −6

)(xAxB

).

Remember, for the system x = Ax with x(0) = x0, thesolution is x(t) = x0e

At. Using this fact and by diagonalize

the matrix A = 1100

(−6 26 −6

)appearing in the above


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system, we get that

xA =25

2(eλ1t + eλ2t), xB =

25√3

2(eλ1t − eλ2t)

where

λ1 =−3 +

√3

50, λ2 =

−3−√3

50.

Since A has detA = 2410000 and trace(A) = − 12

100 , we get anodal sink.

xB

xA


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Given the PDE ux + 3uy = 0, in order to check thatu(x, y) = F (y − 3x) is a solution, we simply check that

F (y − 3x)x + 3F (y − 3x)y = 0.

By the chain rule,

F (y − 3x)x+3F (y − 3x)y

= (−3)F ′(t)∣∣t=y−3x + 3F ′(t)

∣∣t=y−3x = 0.

To find the particular solution to the PDE, we need to findthe specific function F . This can be done if we impose someboundary condition. For example,

(a) Suppose u(0, y) = 4 sin y, then F (y) = 4 sin y. Henceu(x, y) = 4 sin(y − 3x).

(b) Suppose u(x, 0) = ex+1, then F (−3x) = ex+1, or

F (x) = e−x3+1. Thus, u(x, y) = e−

y−3x3

+1.


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Question 5: PDEs with derivatives of only one variable

(a) When we look at the PDE uxy = ux, we can see thefollowing (ux)y = ux. Therefore, if we think v = ux thenvy = v. This can be regarded as an ODE w.r.t the variabley. In other words,

dv

v= dy.

By integrating (x is being considered as a parameter), onegets

ln |v| = y + a(x),

or equivalently,

v = ±ea(x)︸︷︷︸b(x)

ey = b(x)ey.

Going back to u, we get

du

dx= b(x)ey,


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Question 5: PDEs with derivatives of only one variable

which implies

u = ey∫b(x)dx︸︷︷︸c(x)

+h(y) = c(x)ey + h(y).

(b) For the PDE ux = 2xyu, the situation remains the same.Think about an ODE of u w.r.t. the variable x, hence y is aparameter. Therefore, we write

du

u= 2xydx.

By integrating both sides,

ln |u| = yx2 + a(y).

Thus,u = ±ea(y)︸︷︷︸

c(y)

eyx2= c(y)eyx

2.


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Question 6: Solving PDEs by the method of separation ofvariables

Let consider the PDE yux − xuy = 0. Our aim was to findnon-trivial solution, i.e., u 6≡ 0. The idea of the method is tosplit u(x, y) into X(x)Y (y) where X and Y are one-variable(single variable) functions. Using this special form of u, wecan easily calculate all derivatives appearing in the PDE

ux = X ′Y, uy = XY ′,

which transforms the PDE to

yX ′Y = xXY ′.

By the no-crossing principle, u 6= 0 at every point. Hence,we can divide both sides by XY to get

1

x

X ′

X=

1

y

Y ′

Y.


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Since the LHS depends only on x while the RHS dependsonly on y, the common value must be constant, say k.Therefore, we have two corresponding ODEs

X ′

X= kx,

Y ′

Y= ky.

By solving, we get that

X = ±ec1e12kx2 , Y = X = ±ec2e

12ky2 .

Thus,

u = ±ec1+c2︸︷︷︸c1

e

c2︷︸︸︷k

2(x2+y2)

= c1ec2(x2+y2).


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The PDE ux = yuy can be solved similarly. For the PDEuxy = u, the transformed equation is X ′Y ′ = XY which isnothing but

X ′

X

Y ′

Y= 1 6= 0.

Therefore,X ′

X=Y

Y ′.

In particular, X′

X = YY ′ = k for some constant k. We are now

in position to solve for X and Y .For the PDE xuxy + 2yu = 0, the transformed equation isxX ′Y ′ + 2yXY = 0. Using the no-crossing principle, eitheru ≡ 0 or u 6= 0 at every point which helps us to write(

xX ′

X

)(− 1

2y

Y ′

Y

)= 1.

Documents

MA 1506 Mathematics II Tutorial 11 - Partial Differential ...Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 3/14 Question 2 Let E(t) and D(t) be the number of Elves