15CCM211A and 6CCM211B

Outline solutions to Part I of Assignment 1

[1]

From the lecture notes we know that

u(x; y) = f(2x 3y)

for any dierentiable f : R1 ! R1 is a solution to the PDE. The boundarycondition requires that u(x; 3x) = ex and hence

f(7x) = ex:

Hence f is the function f(w) = ew=7. Hence

u(x; y) = e37y 2

7x:

[2]

(a) If v(x; y) = f(x)g(y) is a solution of the PDE then f 0 g = f g. So if

f 0 = f

then this function v is a solution for any function g : R1 ! R1 That is, iff(x) = Cex for a constant C. Absorbing that constant into the arbitrary

function g we see that

v(x; y) = g(y)ex

is a solution for any function g of 1-variable.

(b) We have to solve@u

@x+

@u

@y= u (1)

With the change of variable

= (x; y) := x+ y; = (x; y) := x y

the Chain Rule gives that

@

@x=

@

@x

@

@

=x+y=xy

+@

@x

@

@

=x+y=xy

=@

@+

@

@

=x+y=xy

and@

@y=

@

@y

@

@

=x+y=xy

+@

@y

@

@

=x+y=xy

=@

@ @

@

=x+y=xy

2so @@x

+ @@y

= 2 @@

=x+y=xy

. Hence giving a solution v(; ) to the PDE in

(; )-coordinates

2@v

@= v (2)

determines a solution

u(x; y)) = v((x; y); (x; y)) = v(x+ y; x y)

to the PDE in equation (1). But by the same reasoning as in part (a), eqn

(2) has solution v(; ) = g() e12 for an arbitrary g : R1 ! R1. Hence

u(x; y) = g(x y) e 12 (x+y)

is a solution to the PDE in (1). The given boundary condition requires

that g(x) = e12xejxj and so the required solution of the PDE satisfying the

boundary condition is

u(x; y) = e12(xy)ejxyje

12(x+y) = eyjxyj:

[3] Dierentiate the given function u(x; y) twice with respect to x. Then

dierentiate the given function u(x; y) twice with respect to y. Summing

them together yields the constant function v(x; y) = 4, as required. When

x2 + y2 = 1, then log(x2 + y2) = 0 and x2 + y2 1 = 0, so the boundarycondition is satised.