Description

Haoran Sun (haoransun@link.cuhk.edu.cn)
Problem 1 (P5 Q3).
(b) Second-order linear homogeneous, since the equation Lu = g has
L = ∂t − ∂xx + x
which is a linear operator and g = 0.
(c) Third-order nonlinear, since there is a uux term.
(d) Second-order linear nonhomogeneous, since the operator
L = ∂tt − ∂xx
while g = −x2 = 0̸ .
(h) Forth-order nonlinear, since there is a √u +1 term.
Problem 2 (P10 Q3). Note that the set of characteristic curve h(x,y) = c has the properties that
dy 1
= dx 1+ x2
Then we have
y = arctanx + C
Then
u(x,y) = g(C) = g(y −arctanx)
Sketch of the characteristic curves:
x
Problem 3 (P10 Q7).
HW#01 Haoran Sun

(a) Characteristic curve satisfies
dy x 2 2
= ⇒ x − y = C dx y
Then
u(x,y) = g(x2 − y2)
Plug in u(0,y) = g(−y2) = e−y2, we have u(x,y) = ex2−y2.
(b) Whole xy plane.
Problem 4 (P10 Q10). Change the variable by
x′ = x + y y′ = −x + y
Then ux = ux′ − uy′, uy = ux′ + uy′, and therefore
ux + uy + u = 2ux′ + u = e(3x′+y′)/2
Solve the homogeneous case ux′ + u = 0, we get kernel ϕ(x′,y′)
ϕ(x′,y′) = C(y′)e−x′/2
Suppose a specific solution of the equation is in the form
u(x′,y′) = ae(3x′+y′)/2 + bx′e(3x′+y′)/2
Easy to obtain that a = 1/4 and b = 0. Then the general solution of the equation would be

Apply the boundary condition u(x,0) = 0, we get

Then our solution would be

Problem 5 (P28 Q5). (a) Let y = 0, then we have
ux(x,0) = 0
which means along the line (x,0), u is a constant. This contradicts with the boundary condition u(x,0) = ϕ(x) = x since u(x,0) is not a constant along the line (x,0). Therefore, there is no solution.
HW#01 Haoran Sun

(b) Applying the technique of characteristic curve, we know that u(x,y) is in the form of
u(x,y) = f(ye−x)
Applying the boundary condition, we have
u(x,0) = f(0) = 0
There are different f = f(x) satisfies f(0) = 0. Then there are multiple solutions.
Problem 6 (P31 Q1).
(a) The equation is uxx −4uxy + uyy +4u = 0
1 2
A = [− − ]
2 1
Since detA < 0, the equation is hyperbolic.
(b) Since

and detA = 0, the equation is parabolic.