Introduction to Partial Differential Equations
Exercise Sheet 3

Ross Ogilvie      16th September, 2024
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7.
Royale with Cheese

Recall Burgers’ equation from Example 1.5 of the lecture script:

u˙ + uxu = 0,

for u : × . In this question we will apply the method of characteristics to solve this equation for the initial condition g(x) = 2x.

(a)
According to Theorem 1.4, there is a unique C1 solution to this initial value problem, at least when t is small. For how long does the theorem guarantee that the solution exists uniquely? (1 point)
(b)
Suppose that u is a solution to this equation and suppose that (x(s),t(s)) is a path in the domain of u. What is the s derivative of u along this path? What constraints should we place on the derivatives of x and t? (2 points)
(c)
On an (x,t)-plane draw the characteristics. (1 point)
(d)
Finally, derive the following solution to the initial value problem: (2 points)
u(x,t) = 2x 1 + 2t.
8.
Linear Partial Differential Equations Consider a PDE of the form F(u(x),u(x),x) = 0. Suppose that F is linear in the derivatives and has continuously differentiable coefficients. That is, it can be written in the form
F(p,z,x) = b(z,x) p + c(z,x)

with b and c continuously differentiable. Show that the characteristic curves (x(s),z(s)) for z(s) := u(x(s)) can be described by ODEs that are independent of p(s) := u(x(s)). (4 points)

9.
Solving PDEs Solve the initial value problems of the following PDEs using the method of characteristics. You may assume that g is continuously differentiable on the corresponding domain.
(a)
x21u x12u = u on the domain x1,x2 > 0, with initial condition u(x1,0) = g(x1).
(3 points)
(b)
x11u + 3x22u + 3u = 2u on x1,x2 , x3 > 0, with initial condition u(x1,x2,0) = g(x1,x2).
(3 points)
(c)
u1u + 2u = 1 on the domain x1,x2 > 0, with initial condition u(x1,x1) = 1 2x1.
(4 points)