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Unformatted text preview: Section 3.1 Partial Diﬀerential Equations in Physics and Engineering 29 Solutions to Exercises 3.1
1. uxx + uxy = 2u is a second order, linear, and homogeneous partial diﬀerential
equation. ux(0, y) = 0 is linear and homogeneous.
5. ut ux + uxt = 2u is second order and nonlinear because of the term utux . u(0, t)+
ux (0, t) = 0 is linear and homogeneous.
9. (a) Let u(x, y) = eax eby . Then
ux
uy =
= aeax eby
beax eby uxx
uyy
uxy =
=
= a2eax eby
b2eax eby
abeax eby . So
Auxx + 2Buxy + Cuyy + Dux + Euy + F u = 0
⇔ Aa2eax eby + 2Babeax eby + Cb2eax eby
+Daeax eby + Ebeax eby + F eax eby = 0
⇔ eax eby Aa2 + 2Bab + Cb2 + Da + Eb + F = 0
⇔ Aa2 + 2Bab + Cb2 + Da + Eb + F = 0,
because eax eby = 0 for all x and y.
(b) By (a), in order to solve
uxx + 2uxy + uyy + 2 ux + 2 uy + u = 0,
we can try u(x, y) = eax eby , where a and b are solutions of
a2 + 2ab + b2 + 2a + 2b + 1 = 0.
But
a2 + 2ab + b2 + 2a + 2b + 1 = (a + b + 1)2.
So a + b + 1 = 0. Clearly, this equation admits inﬁnitely many pairs of solutions
(a, b). Here are four possible solutions of the partial diﬀerential equation:
a = 1, b = −2 ⇒ u(x, y) = ex e−2y a = 0, b = −1 ⇒ u(x, y) = e−y
a = −1/2, b = −1/2 ⇒ u(x, y) = e−x/2 e−y/2
a = −3/2, b = 1/2 ⇒ u(x, y) = e−3x/2 ey/2
13. We follow the outlined solution in Exercise 12. We have
A(u) = ln(u), φ(x) = ex , ⇒ A(u(x(t)), t)) = A(φ(x(0))) = ln(ex(0) ) = x(0). So the characteristic lines are
x = tx(0) + x(0)
So u(x, t) = f (L(x, t)) = f
ex and so x
t+1 ⇒ x(0) = L(x t) = x
.
t+1 . The condition u(x, 0) = ex implies that f (x) =
x u(x, t) = e t+1 .
x x x
Check: ut = −e t+1 (t+1)2 , ux = e t+1
x ut + ln(u)ux = −e t+1 1
t+1 ,
x
x
1
x
+
e t+1
= 0.
(t + 1)2 t + 1
t+1 ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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