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Problems and solutions for
Book:

Partial Differential Equations

Chapter: 1
...
Strauss
Aleksi Kristian Winst´en
February 17, 2023

1

If not stated otherwise a, b are coefficients, and u, v are functions with multiple variables and subscript notation ux is a shorthand for partial derivative
∂
∂x u ≡ ux
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Verify the linearity and nonlinearity of the eight examples of PDEs
given in the text, by checking whether or not equations (3) are valid
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The equation (3) is about linear operator L
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ux + uy = 0

(transport)

2
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ux + uuy = 0

(shockwave)

4
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utt − uxx + u3 = 0
6
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utt + uxxxx = 0
8
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1
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2
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The thrill should be clear now
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A linear because partial differentiation is a linear operation
5
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The dependent variable
is raised to the third power
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A linear
7
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A linear

Problem 2
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The idea is to input a linear combination of two functions to the
operator and see wether or not the result is a linear combination of the operators
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L (au + bv) = (au + bv)x + x(au + bv)y
= aL u + bL (b)

b) By applying the methods descriped above one can deduce that the operator L u = ux + uuy is a nonlinear operator
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d) A linear operator
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Thus this is
also a linear operator
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For each of the following equations, state the order and whether
it is nonlinear, linear inhomogeneous, or linear homogeneous; provide reasons
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The order of a partial differential equation is the highest mixed derivative that appears
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The inhomogeneus equations are those that is not equally the zero
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Thus the order is two
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Thus this is a linear second order inhomogeneous
partial differential equation
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c) A nonlinear third order homogeneous PDE
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e) Note that the imaginary unit is also a complex number, which makes this
equation a linear second order homogeneous PDE
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g) A linear first order homogeneous PDE
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This is a
nonlinear fourth order homogeneous PDE
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Show that the difference of two solutions of an inhomogeneous
linear equation L u = g with the same g is a solution of the homogeneous
equation L u = 0
Solution
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Input the difference of the solutions into the linear operator L
L (u − v) = L u − L v = g − g = 0
Thus the difference ω = u − v is a solution to the homogeneous linear equation
Lω = 0

Problem 5
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a) The vectors with b = 0
b) The vectors with b = 1
c) The vectors with ab = 0
d) All the linear combinations of the two vectors [1, 1, 0] and [2, 0, 1]
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Assume that the notation [a, b, c] means a point in the three dimensional euclidean space, and the vector addition and multiplication with scalar
obeys those rules that are familiar from the linear algebra
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4

a) The set to investigate is
V := {[a, b, c] ∈ R3 : b = 0}
This defines the set V which includes all the points from the R3 such that
b = 0
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[0, 0, 0] ∈ V
Also the additive inverse is included, because −[a, 0, c] = [−a, 0, −c]
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b) The situtation is similar to the last one, but this time b = 1, which means
that the zero vector cannot be in the set, so this is not a vector space
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This means that there is no zero divisors
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That implies the zero vector is included
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d) The task here is to find out if the given vectors are linearly independent
and thus span a vector space
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Thus the linear
combination of given vectors span a vector space
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A familiar consept here to think
about is to rename the parameters x = a, y = b and z = c and notice that
the set V is a surface over the xy-plane
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Because the parameter c depends linearly from a
and b the additive inverse is also included
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Are the three vectors [1, 2, 3], [−2, 0, 1], and [1, 10, 17] linearly dependent or independent? Do they span all vectors or not?
Solution
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Now lets try if the given set of vectors fullfills this requirement
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
   
1 −2 1
a1
0
2 0 10 a2  = 0
3 1 17 a3
0
The determinant is zero, which implies that the column-vector space is linearly dependent
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Problem 7
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There is two approaches to solve the problem
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Other would notice that the
given functions are analytic and thus check if their Wronskian is nonzero, which
implies linear independence
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a1 (1 + x) + a2 (1 − x) + a3 (1 + x + x2 ) = 0
Open the brackets and prosper
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For this to be true it must be that a1
and a2 equals zero also
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The other one used the Wronskian to solve this
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1 + x 1 − x 1 + x + x2 


−1
1 + 2x  = −3 − x
W (1 + x, 1 − x, 1 + x + x2 )(x) =  1
 0

0
2
This is never zero for any x ∈ R so the given functions are linearly independent
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Find a vector that, together with the vectors [1, 1, 1] and [1, 2, 1],
forms a basis of R3
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The given vectors are all 3-vectors, so it is easy to visualize their
orientation in the space
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But nobody is so sceptical as a mathematician, so lets verify this
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6

Problem 9
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Find a basis of it
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To show they form a vector space we have to check whether or not the
zero vector and the additive inverse belongs to the set
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The second additive inverse
follows from the fact that the given function is actually a linear combination
of three different functions
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Thus functions (c1 +
c2 sin2 x + c3 cos2 x) form a vector space
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Notice that the set {1, sin2 , cos2 }
is not linearly independent, because sin2 + cos2 = 1
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This element can be removed from the set, and the leftover span the same vector
space
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These are linearly independent,
which can be verified using the wronskian

 2
sin x cos2 x 
 = − sin 2x
W (sin2 , cos2 )(x) = 
sin 2x − sin 2x
which is non-zero except at countable isolated points
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Show that the solutions of the differential equation u′′′ − 3u′′ +
4u = 0 form a vector space
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Solution
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To find the solution for a linear ordinary differential equation (ODE) with
constant coefficients define u := erx and solve for the r
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From these facts deduct that the given ODE equals zero if and
only if the following characteristic equation is true
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So the three solutions are r1 = −1, r2 = 2 and r3 = 2
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The tactic to overcome this problem
is well known and can be found from any differential equations textbook
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This method gives
u(x) = C1 e−x + C2 e2x + C3 xe2x
One should verify this to be a solution for the given ODE
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Also note that the
functions {e−x , e2x , xe2x } are linearly independent solutions and thus span a
vector space, and is the basis of that space
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Verify that u(x, y) = f (x)g(y) is a solution of the PDE uuxy =
ux uy for all pairs of (differentiable) functions f and g of one variable
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The quest is to verify the function u(x, y) = f (x)g(y) to be a solution
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First find out the partial derivatives of the function u
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f (x)g(y)f ′ (x)g ′ (y) = f ′ (x)g(y)f (x)g ′ (y)
The claim is obviously true
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Verify by direct substitution that
un (x, y) = sin(nx) sinh(ny)
is a solution of uxx + uyy = 0 for every n > 0
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Here the notation is littlebit misleading, because the subscript notation here un is not the same as partial derivative with respect to a variable n
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Lets do what was asked to
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