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MATH3078 Mathematics and Statistics
1. (a) For each of the following PDEs, determine whether it is linear, semilinear,
quasilinear or fully nonlinear. Justify your answer.
(i) utt + (ut)2 uxx = 1.
Solution: Semilinear since the di↵erential operator in linear in the
highest order derivatives. Indeed, utt and uxx are raised to the power
one and their coecients are constants. The PDE is not linear because
of the term (ut)2.
(ii) ut ux1 + ex2 ux2 = 0.
Solution: Linear since each term in the PDE contains a partial
derivative of u that is raised to the power one and the coecients
of these partial derivatives, ut, ux1 and ux2 , are either constants or a
function of the independent variable x2 (for the latter derivative).
(iii) ut + ux + uxxx = eu.
Solution: Semilinear since eu is a nonlinear term and the highest
order derivative uxxx is raised to the power one and has a constant
coecient.
(iv) utuxx + ux = t.
Solution: Quasilinear since the coecient of the highest order
derivative uxx depends on the lower order derivative ut.
(v) uxxuyy + (uxy)3 = 0.
Solution: Fully nonliner since the di↵erential operator is nonlinear
in the highest order derivative: the power of uxy is three and also the
coecient of uyy depends on the second order derivative uxx.
(b) Write down the order of each of the above PDEs and determine whether the
PDE is homogeneous or nonhomogenous.
Solution:
(i) Second order (the highest order derivatives are utt and uxx) and non-
homogeneous because the right-hand side is 1;
(ii) First order (all derivatives are of order one) and homogeneous since all
terms involve some partial derivative of u;
(iii) third order (the highest order derivative is uxxx) and homogeneous since
all terms involve u or some partial derivative of u;
(iv) second order (the highest order derivative is uxx) and non-homogeneous
because of the right-hand side t (which is an independent variable);
(v) second order (the highest order derivatives are uxx, uyy and uxy) and
homogeneous since all terms involve some partial derivative of u.
Copyright c 2021 The University of Sydney 1
2. For each of the following PDEs, find the general solution u = u(x, t), where x, t 2 R:
(a) ux = 0.
Solution: Integrating with respect to x, we obtain u(x, t) = f(t), where
f 2 C1(R).
(b) ut = t.
Solution: Integrating with respect to t, we find that u(x, t) = f(x) + t
2
2 ,
where f 2 C1(R).
(c) ut + 3u = 0.
Solution: Multiply the equation by e3t to obtain that @(e
3tu)
@t = 0. Inte-
grating with respect to t, we find that e3tu(x, t) = f(x), where f 2 C1(R).
Thus, u(x, t) = f(x) e3t.
(d) utt + 4u = 0.
Solution: The characteristic equation associated to the equation in (d) is
2 + 4 = 0, the roots of which are 1,2 = ±2i. Thus, the general solution
of (d) is u(x, t) = f1(x) cos(2t) + f2(x) sin(2t), where f1 and f2 are arbitrary
C2(R)-functions.
3. Using the method of characteristics, solve the initial value problem8<:ut + (1 2t)ux = 0 for all (x, t) 2 R⇥ (0,1),u(x, 0) = 1
1 + x2
for all x 2 R.
[Hint: Set z(t) = u(x(t), t) and write the characteristic equations for x(t) and z(t).]
Solution: Fix k 2 R. The characteristic equations for x(t) and z(t) are(
x˙(t) = 1 2t
z˙(t) = ut + (1 2t)ux = 0.
We solve x˙(t) = 1 2t, subject to x(0) = k. This implies that
x(t) = t t2 + x(0) = t t2 + k.
From z˙(t) = 0, we find that z(t) = constant for every t 2 (0,1). Hence,
z(t) = z(0) = u(x(0), 0) = u(k, 0) =
1
1 + k2
, (1)
where we have used the fact that u(x, 0) = 1/(1 + x2) for all x 2 R.
Let (x, t) 2 R⇥ (0,1) be arbitrary. Then, x(t) = x implies that
k = x t+ t2,
which used in (1) gives that
u(x, t) =
1
1 + (x t+ t2)2 for all (x, t) 2 R⇥ (0,1).
2
4. Let a > 0 be a constant. Consider the one-dimensional heat equation
ut a
2
uxx = 0 for (x, t) 2 R⇥ (0,1). (2)
Use the substitution
u(x, t) = exp
✓
v(x, t)
a
◆
to show that the heat equation in (2) becomes the following equation in v:
vt +
1
2
(vx)
2 a
2
vxx = 0.
Solution: By the given substitution, we find that
ut(x, t) = 1
a
exp
✓
v(x, t)
a
◆
vt = u
a
vt,
ux(x, t) = 1
a
exp
✓
v(x, t)
a
◆
vx = u
a
vx.
(3)
Di↵erentiating the second identity in (3) with respect to x, we arrive at
uxx(x, t) = 1
a
(ux vx + u vxx) . (4)
Using in (4) the formula for ux in (3), we obtain
uxx =
u
a2
⇥
(vx)
2 a vxx
⇤
.
From this last identity and the first identity in (3), we see that the heat equation
in (2) becomes
u
a
vt +
1
2
(vx)
2 a vxx
= 0.
Since u(x, t) > 0 and a > 0, by dividing the above equation by u/a, we attain
the desired equation for v.
