SOLUTIONS MANUAL
Essential Partial Differential Equations:
Analytical and Computational Aspects
Solutions to all exercises
David F. Griffiths, John W. Dold and David J. Silvester
Springer International Publishing Switzerland, 2015
Solutions to all exercises are available to approved instructors
by contacting the publishers.
Springer Undergraduate Mathematics Series
ISBN: 978-3-319-22568-5, e-ISBN: 978-3-319-22569-2
, Essential Partial Differential Equations:
Analytical and Computational Aspects
Solutions to all exercises
David F. Griffiths, John W. Dold and David J. Silvester
Exercises
1 Introduction 2
2 Boundary and initial data 6
3 Origins of PDEs 9
4 Classification of PDEs 10
5 Boundary value problems in R1 19
6 Finite difference methods in R1 30
7 Maximum principles and energy methods 46
8 Separation of variables 50
9 The method of characteristics 66
10 Finite difference methods for elliptic PDEs 86
11 Finite difference methods for parabolic PDEs 102
12 Finite difference methods for hyperbolic PDEs 115
13 Projects 131
1
,Exercises 1 Introduction
1.1
Function Comment Conclusion
u(x, y) = A(y) uy = A′ (y) False
u(x, y) = A(y) uxy = 0 True
uxy = 0 concerns different
u(x, t) = A(x)B(t) False
independent variables!
u(x, t) = A(x)B(t) uuxt = ABA′ B ′ = u xut True
u(x, t, y) = A(x, y) ut = ∂t A(x, y) = 0 True
u(x, t) = A(x+ct) + B(x—ct) utt + c2uxx = 2c2(A′′ + B′′) False
1.2
These are not the only possible cases; you might find other PDEs:
(a) u(x, t) = et cos x: ∂n u = et cos x = u (any n), u x = —et sin x, u xx = —et cos x so ∂n u = u
t t
or ∂tnu + uxx = 0, etc.
(b) u(x, y) = x2 + y2: uxx = 2, uyy = 2 so uxx — uyy = 0. Also uxy = 0, etc.
(c) u(x, t) = x2t: ux = 2xt, ut = x2, u tx = 2x so 2tut — xux = 0 or ux = tutx, etc.
(d) u(x, t) = x2t2: utx = 4xt, uttxx = 4 so (utx)2 = 16u or utttxx = 0, etc.
2 2 2
(e) u(x, y) = e— x : ux = —2xe— x , uxx = (4x2 — 2)e— x , uy = 0 so uy = 0 or u xx = (4x2 — 2)u,
etc.
2
(f) u(x, y) = ln(x2 + y2 ): ux = 2x
x2+y 2 , uy = 2y
x2+y 2 , u xx = 2
x2+y 2 — (x24x
+y 2) 2 , uyy =
2
x2+y 2 —
4y2
(x2+y 2) 2 so yux — xuy = 0, uxx + uyy = 0, etc.
1.3
These are not the only possible cases; you might find other PDEs:
(a) u(x, t) = A(x+ct) + B(x —ct): ut = cA′(x+ct) — cB′(x— ct), ux = A′ (x+ct) + B′(x —ct),
utt = c2 A′′ (x+ct) + c2B′′(x —ct), uxx = A′′(x+ct) + B′′(x —
ct) so u tt —c2uxx = 0 (wave
equation).
(b) u(x, t) = A(x) + B(t): ut = B′(t) so u tx = 0.
(c) u(x, t) = A(x)/B(t): ln u = ln A(x) — ln B(x) so (ln u)tx = 0 or uu tx — ut ux = 0.
(d) u(x, t) = A(xt): ut = xA′(xt), ux = tA′ (xt), so tu t — xux = 0.
(e) u(x, t) = A(x2t): ut = x2A′ (x2t), ux = 2xtA′(x2t) so 2tut — xux = 0.
2
(f) u(x, t) = A(x2/t): ut = — xt2 A′(x2/t), u x = — 2xt A′ (x2/t) so 2tu t + xux = 0.
2
, 1.4
With u(x, y) = f (2x + y2) + g(2x — y2 ) we find, using the chain rule
ux = 2f ′(2x + y2 ) + 2g ′(2x — y2 ), uy = 2yf ′ (2x + y2) — 2yg′ (2x — y2)
uxx = 4f ′′(2x + y2) + 4g′′ (2x — y2 )
uyy = 2f ′ (2x + y2 ) — 2g′ (2x — y2) + 4y2 f ′′(2x + y2 ) + 4y2 g′′ (2x — y2 )
and the result follows.
1.5 √
Suppose that u(x, t) = 1 c sech2 (z), where z = 1 c(x—ct—x0 ). Then zt = — 1 c3 /2 and zx = 1 c1/2
2 2 2 2
so, by the chain rule, we find
sinh z sinh z
∂ u(x, t) = 1 c5/2 , ∂ u(x, t) = — 1 c3/2
t 2 x
cosh3 z cosh 3 z 2
cosh (z) — 3 2
∂tu + 6u∂xu = 21 c5/2(sinh z) = —∂3xu(x, t)
cosh5 z
and so ut + 6uux + u xxx = 0.
1
−5 −4 −3 −2 −1 0 1 2 3 4 5 x
Figure 1: Soliton solutions of the KdV equation with c = 2 (solid) and c = 4 (dashed) for
Exercise 1.5 travel with speed c.
1.6
With u defined by (1.2) we may write
∫ ∞
1
u(x, t) = √ f (x — s, t) ds
4π —∞
2
and f (x, t) = t—1/2e—x /4t is the function used in Example 1.3. The partial derivatives of u are
given by
∫ ∞ ∫ ∞ ∫ ∞
ut = f t(x — s, t) ds, ux = f x (x — s, t) ds, ux = f xx (x — s, t) ds,
—∞ —∞ —∞
and ut — uxx = 0 follows since ft = fxx for each s (see Example 1.3).
1.7
Since u = —2∂xφ = —2φx /φ, the partial derivatives are
φxt φx φt φxx (φx)2
ut = —2 — 2 φ2 , ux = —2 φ + 2 2 ,
φ φ
φxxx φxxφx φx φxx 2
(φx)
uxx = —2 +2 + 4 — 2
φ φ2 φ φ φ
3
Essential Partial Differential Equations:
Analytical and Computational Aspects
Solutions to all exercises
David F. Griffiths, John W. Dold and David J. Silvester
Springer International Publishing Switzerland, 2015
Solutions to all exercises are available to approved instructors
by contacting the publishers.
Springer Undergraduate Mathematics Series
ISBN: 978-3-319-22568-5, e-ISBN: 978-3-319-22569-2
, Essential Partial Differential Equations:
Analytical and Computational Aspects
Solutions to all exercises
David F. Griffiths, John W. Dold and David J. Silvester
Exercises
1 Introduction 2
2 Boundary and initial data 6
3 Origins of PDEs 9
4 Classification of PDEs 10
5 Boundary value problems in R1 19
6 Finite difference methods in R1 30
7 Maximum principles and energy methods 46
8 Separation of variables 50
9 The method of characteristics 66
10 Finite difference methods for elliptic PDEs 86
11 Finite difference methods for parabolic PDEs 102
12 Finite difference methods for hyperbolic PDEs 115
13 Projects 131
1
,Exercises 1 Introduction
1.1
Function Comment Conclusion
u(x, y) = A(y) uy = A′ (y) False
u(x, y) = A(y) uxy = 0 True
uxy = 0 concerns different
u(x, t) = A(x)B(t) False
independent variables!
u(x, t) = A(x)B(t) uuxt = ABA′ B ′ = u xut True
u(x, t, y) = A(x, y) ut = ∂t A(x, y) = 0 True
u(x, t) = A(x+ct) + B(x—ct) utt + c2uxx = 2c2(A′′ + B′′) False
1.2
These are not the only possible cases; you might find other PDEs:
(a) u(x, t) = et cos x: ∂n u = et cos x = u (any n), u x = —et sin x, u xx = —et cos x so ∂n u = u
t t
or ∂tnu + uxx = 0, etc.
(b) u(x, y) = x2 + y2: uxx = 2, uyy = 2 so uxx — uyy = 0. Also uxy = 0, etc.
(c) u(x, t) = x2t: ux = 2xt, ut = x2, u tx = 2x so 2tut — xux = 0 or ux = tutx, etc.
(d) u(x, t) = x2t2: utx = 4xt, uttxx = 4 so (utx)2 = 16u or utttxx = 0, etc.
2 2 2
(e) u(x, y) = e— x : ux = —2xe— x , uxx = (4x2 — 2)e— x , uy = 0 so uy = 0 or u xx = (4x2 — 2)u,
etc.
2
(f) u(x, y) = ln(x2 + y2 ): ux = 2x
x2+y 2 , uy = 2y
x2+y 2 , u xx = 2
x2+y 2 — (x24x
+y 2) 2 , uyy =
2
x2+y 2 —
4y2
(x2+y 2) 2 so yux — xuy = 0, uxx + uyy = 0, etc.
1.3
These are not the only possible cases; you might find other PDEs:
(a) u(x, t) = A(x+ct) + B(x —ct): ut = cA′(x+ct) — cB′(x— ct), ux = A′ (x+ct) + B′(x —ct),
utt = c2 A′′ (x+ct) + c2B′′(x —ct), uxx = A′′(x+ct) + B′′(x —
ct) so u tt —c2uxx = 0 (wave
equation).
(b) u(x, t) = A(x) + B(t): ut = B′(t) so u tx = 0.
(c) u(x, t) = A(x)/B(t): ln u = ln A(x) — ln B(x) so (ln u)tx = 0 or uu tx — ut ux = 0.
(d) u(x, t) = A(xt): ut = xA′(xt), ux = tA′ (xt), so tu t — xux = 0.
(e) u(x, t) = A(x2t): ut = x2A′ (x2t), ux = 2xtA′(x2t) so 2tut — xux = 0.
2
(f) u(x, t) = A(x2/t): ut = — xt2 A′(x2/t), u x = — 2xt A′ (x2/t) so 2tu t + xux = 0.
2
, 1.4
With u(x, y) = f (2x + y2) + g(2x — y2 ) we find, using the chain rule
ux = 2f ′(2x + y2 ) + 2g ′(2x — y2 ), uy = 2yf ′ (2x + y2) — 2yg′ (2x — y2)
uxx = 4f ′′(2x + y2) + 4g′′ (2x — y2 )
uyy = 2f ′ (2x + y2 ) — 2g′ (2x — y2) + 4y2 f ′′(2x + y2 ) + 4y2 g′′ (2x — y2 )
and the result follows.
1.5 √
Suppose that u(x, t) = 1 c sech2 (z), where z = 1 c(x—ct—x0 ). Then zt = — 1 c3 /2 and zx = 1 c1/2
2 2 2 2
so, by the chain rule, we find
sinh z sinh z
∂ u(x, t) = 1 c5/2 , ∂ u(x, t) = — 1 c3/2
t 2 x
cosh3 z cosh 3 z 2
cosh (z) — 3 2
∂tu + 6u∂xu = 21 c5/2(sinh z) = —∂3xu(x, t)
cosh5 z
and so ut + 6uux + u xxx = 0.
1
−5 −4 −3 −2 −1 0 1 2 3 4 5 x
Figure 1: Soliton solutions of the KdV equation with c = 2 (solid) and c = 4 (dashed) for
Exercise 1.5 travel with speed c.
1.6
With u defined by (1.2) we may write
∫ ∞
1
u(x, t) = √ f (x — s, t) ds
4π —∞
2
and f (x, t) = t—1/2e—x /4t is the function used in Example 1.3. The partial derivatives of u are
given by
∫ ∞ ∫ ∞ ∫ ∞
ut = f t(x — s, t) ds, ux = f x (x — s, t) ds, ux = f xx (x — s, t) ds,
—∞ —∞ —∞
and ut — uxx = 0 follows since ft = fxx for each s (see Example 1.3).
1.7
Since u = —2∂xφ = —2φx /φ, the partial derivatives are
φxt φx φt φxx (φx)2
ut = —2 — 2 φ2 , ux = —2 φ + 2 2 ,
φ φ
φxxx φxxφx φx φxx 2
(φx)
uxx = —2 +2 + 4 — 2
φ φ2 φ φ φ
3
