Homework #1 Solutions

Homework #1 Solutions
Math 182, Spring 2014
Instructor: Dr. Doreen De Leon
1
Exercises 1.1: 1, 4(b), 5
1. Find the order of each PDE.
(a) The convection or advection equation, ut + cux = 0
The highest order derivative is a first derivative, so the PDE is first order.
(b) The wave equation, utt = c2 uxx
The highest order derivative is a second derivative, so the PDE is second order.
(c) The eikonal equation, u2x + u2y = 1
Both derivatives are first partials, so the PDE is first order.
(d) The Euler-Bernoulli beam equation, utt + α4 uxxxx = 0
In this case, the highest order partial derivative is a fourth derivative of u with respect to x,
so the PDE is fourth order.
(e) uxxyyz − u8 + u6xx = 0
In this equation, the highest order derivative is in the term uxxyyz , which is a fifth derivative,
so the PDE is fifth order.
4. Consider the one-dimensional wave equation utt = c2 uxx , where c is a constant.
(b) Show that u = g(x + ct) also satisfies the wave equation for “any” function g.
Solution: We need to compute the second derivatives of this function with respect to x and
t.
∂
ux =
g(x + ct)
∂x
∂
= g 0 (x + ct) ·
(x + ct)
∂x
= g 0 (x + ct).
∂
ut = g(x + ct)
∂t
∂
= g 0 (x + ct) · (x + ct)
∂t
= cg 0 (x + ct).
1
Then,
uxx =
∂
ux
∂x
∂
(x + ct)
∂x
= g 00 (x + ct).
∂
utt = ut
∂t
∂
= cg 00 (x + ct) (x + ct)
∂t
2 00
= c g (x + ct).
= g 00 (x + ct)
We then plug this into the PDE to observe:
c2 g 00 (x + ct) = utt = c2 uxx = c2 g 00 (x + ct),
and, therefore, u = g(x + ct) solves the PDE.
5. Consider the eikonal equation u2x + u2y = 1.
(a) Show that u = x and u = y are solutions.
Solution: We simply need to verify that these functions satisfy the equation.
u = x: ux = 1 and uy = 0. Therefore, u2x + u2y = 12 + 02 = 1. X
u = y: ux = 0 and uy = 1. Therefore, u2x + u2y = 02 + 12 = 1. X
(b) Are u = 3x and u = −4y solutions?
Solution: We can easily check as we did above that neither function solves the given eikonal
equation.
(c) Is u = x + y a solution?
Solution: No, it is not a solution. To verify this, note that ux = 1 and uy = 1, so u2x + u2y =
2 6= 1.
(d) Find all solutions of the form u = ax + by, where a and b are constants.
Solution: In this case, ux = a and uy = b, so a and b must satisfy
a2 + b2 = 1.
2
Exercises 1.2: 4, 5, 12
Find the general solution of each PDE. The solution u is a function of the variables which appear, unless
otherwise stated.
2
4. uyy = x2 y
Solution:
Z
uy =
x2 y dy
1
= x2 y 2 + f (x),
2
where f is an arbitrary function of x.
Z
u=
uy dy
Z 1 2 2
=
x y + f (x) dy
2
1
= x2 y 3 + f (x)y + g(x),
6
where f and g are arbitrary functions of x.
5. uxy = x − y
Solution:
Z
ux =
(x − y) dy
1
= xy − y 2 + f1 (x),
2
where f1 is an arbitrary function of x.
Z 1 2
u=
xy − y + f1 (x) dx
2
1 2
1 2
= x y − xy + f (x) + g(y),
2
2
where f is an arbitrary function of x and g is an arbitrary function of y.
R
Note: We can replace f1 (x) dx by f (x) because if f1 is an arbitrary function of x, so is its integral.
12. ux + 3u = ex , u = u(x, y)
Solution: If we look at this as a linear differential equation in x, we need to determine the
integrating
R factor, µ(x), and multply both sides of the equation by µ(x). The integrating factor is
µ(x) = e 3dx = e3x , so we have
e3x ux + 3e3x u = ex e3x
∂
e3x u = e4x
∂x
Z
e3x u =
e4x dx
1
e3x u = e4x + f (y)
4
1
u = ex + f (y)e−3x ,
4
where f is an arbitrary function of y.
3
3
Exercises 1.3: 5, 6
5. For which values of the constant L is the following boundary-value problem well-posed?
y 00 + 4y = 8x, y(0) = A, y(L) = B, 0 ≤ x ≤ L
Solution: We will first obtain the general solution and then determine under what conditions we
may solve for the arbitrary constant. This is a nonhomogeneous second order ODE.
• Homogeneous: yh00 + 4yh = 0
Characteristic equation: r2 + 4 = 0 =⇒ r = ±2i. Therefore, the complementary solution is
yh = c1 cos(2x) + c2 sin(2x).
• Particular: yp00 + 4yp = 8x
Try yp = a1 x + a0 . Then, yp0 = a1 and yp00 = 0, so plugging into the equation we obtain
4(a1 x + a0 ) = 8x =⇒ a1 x + a0 = 2x.
Equating coefficients gives a1 = 2 and a0 = 0, so yp (x) = 2x.
The general solution is then
y = yh + yp = c1 cos(2x) + c2 sin(2x) + 2x.
Now, we need to determine under what conditions we can solve for c1 and c2 .
y(0) = A =⇒ c1 = A.
y(L) = B =⇒ c1 cos(2L) + c2 sin(2L) + 2L = B =⇒ c1 cos(2L) + c2 sin(2L) = B − 2L.
Writing this system of equations in matrix form gives
c1
1
0
A
,
=
cos(2L) sin(2L)
c2
B − 2L
which has a unique solution provided the coefficient matrix is invertible; i.e., provided
1
0 cos(2L) sin(2L) = sin(2L) 6= 0,
which is true as long as
L 6=
nπ
, n ∈ Z.
2
6. Explain why the following problem is not well-posed.
uxx = 0, u(0, y) = y 2 , u(1, y) = 3y, u(x, 0) = x + 2, x ≥ 0, 0 ≤ y ≤ 1
Solution: We first need to find a general solution of the PDE, and then determine if we can solve
for the arbitrary functions.
Z
ux = 0 dx
= f (y).
Z
u = f (y) dx
= xf (y) + g(y).
4
Now, we want to use the given initial and boundary conditions to solve for f (y) and g(y).
u(0, y) = y 2 =⇒ y 2 = g(y).
u(1, y) = 3y =⇒ 3y = f (y) + y 2
f (y) = −y 2 + 3y.
u(x, 0) = xf (0) + g(0) = x =⇒ 0 = x + 2, or x = −2.
However, the domain of our problem is x ≥ 0. Therefore, the problem has no solution, and thus, it
is not wel-posed..
4
Exercises 1.4: 2, 4, 6
In Exercises 1-7, determine whether the PDE is linear or nonlinear, and prove your result. If it is linear,
decide if it is homogeneous or nonhomogeneous. If it is nonlinear, point out the term or terms which
make it nonlinear.
2. uxxy − (sin x)uyy + x − y = 0
Solution: Since the equation is linear in u and its derivatives, the PDE is linear. Now, we will
prove this. First, rewrite the PDE as
uxxy − (sin x)uyy = −x + y.
L[u] = uxxy − (sin x)uyy .
So,
L[c1 u1 + c2 u2 ] = (c1 u1 + c2 u2 )xxy − (sin x)(c1 u1 + c2 u2 )yy
= c1 u1xxy + c2 u2xxy − c1 (sin x)u1yy − c2 (sin x)u2yy
= c1 u1xxy − (sin x)u1yy + c2 u2xxy − (sin x)u2yy
= c1 L[u1 ] + c2 L[u2 ].
Therefore, the PDE is linear. From our rewritten PDE, we see that the PDE is nonhomogeneous.
4. uxx = sin u
Solution: We can see that due to the sin u on the right-hand side, this PDE is nonlinear. Now,
we shall prove it. First, rewrite the equation as uxx − sin u = 0. Then
L[u] = uxx − sin u.
L[c1 u1 + c2 u2 ] = (c1 u1 + c2 u2 )xx − sin(c1 u1 + c2 u2 )
= c1 u1xx + c2 u2xx − sin(c1 u1 ) cos(c2 u2 ) − cos(c1 u1 ) sin(c2 u2 )
6= c1 L[u1 ] + c2 L[u2 ].
The nonlinearity is caused by the term sin u.
5
6. Poisson’s equation in two dimensions (in polar coordinates),
1
1
urr + ur + 2 uθθ = f (r, θ).
r
r
Solution: In this case, we see that the equation is linear in u and its derivatives, so the PDE is linear. In addition, since the right-hand side is not 0 (as far as we know), the PDE is nonhomogeneous.
Now, let’s prove that the PDE is linear.
1
1
L[u] = urr + ur + 2 uθθ .
r
r
1
1
L[c1 u1 + c2 u2 ] = (c1 u1 + c2 u2 )rr + (c1 u1 + c2 u2 )r + 2 (c1 u1 + c2 u2 )θθ
r
r
1
1
1
1
= c1 u1rr + u1r + 2 u1θθ + c2 u2rr + u2r + 2 u2θθ
r
r
r
r
= c1 L[u1 ] + c2 L[u2 ].
5
Exercises 1.5: 5, 7
5. Use (1.12) to show that the function u(x, t) =
∞
X
2
cn e−n t sin nx is a solution of the heat equation
n=1
ut = uxx (whenever the series converges, of course).
Solution: We may rewrite the heat equation as ut − uxx = 0, so L[u] = ut − uxx , where L is a
linear operator. Then, by Equation (1.12), we have
"∞
#
∞
h
i
X
X
2
−n2 t
L cn e−n t sin nx
L
cn e
sin nx =
n=1
=
=
n=1
∞ X
n=1
∞
X
2
cn e−n t sin nx
2
t
2
− cn e−n t sin nx
xx
2
−cn n2 e−n t sin nx + cn n2 e−n t sin nx
n=1
(since (sin nx)0 = n cos nx and (sin nx)00 = (n cos nx)0 = −n2 sin nx)
∞
X
=
0
n=1
= 0.
Therefore, u(x, t) =
∞
X
2
cn e−n t sin nx solves the heat equation.
n=1
7. We may also prove theorems for solutions of nonhomogeneous PDEs that are analogous to those
for ODEs. Prove that the general solution of the nonhomogeneous PDE L[u] = f is u = uh + up ,
where up is any one particular solution of L[u] = f , and uh is the general solution of the associated
homogeneous PDE L[u] = 0, as follows:
6
(a) First, prove that uh + up is always a solution of L[u] = f .
Solution: Since L is linear, we have
L[uh + up ] = L[uh ] + L[up ]
=0+f
= f.
Therefore, u = uh + up solves the PDE.
(b) Next, prove that if u is any particular solution of L[u] = f , then we can always write
u = uh0 + up ,
where uh0 is a particular case of the solution uh .
Solution: We look at the function u − up . Then, since L is linear,
L[u − up ] = L[u] − L[up ]
=f −f
= 0.
Therefore, uh0 = u − up solves the associated homogeneous PDE.
7

Download Report