Mathematical Methods of Physics Collection of problems and solutions Maxim Zabzine, Joel Ekstrand Department of Physics and Astronomy, Uppsala University 2010
1
Vector fields, fields, cu curved rved coordinates, coordinates, operators, operators, integrals integrals Sho Show that that u v is divergence-free if u and v are rotation-
∗
×
1.1 (Yngve, vektoranalys, 1.8.2) free.
Show that that is divergence-free if is a constant v = r is ω ω is 1.2 (Yngve, vektoranalys, 1.8.4) Sho vector field ( is the position vector). r is
∗
×
1.3 (Yngve, vektoranalys, 2.4.7 )
Use cylindrical cylindrical coordinates coordinates and determine determine
ω ω is 1. is a constant angle velocity, and is the position v = r , where v is the velocity, r is vector.
×
v , i.e. ∇ 2. the rotati rotation on of
× v.
1.4 (Yngve, vektoranalys,1.8.16 )
The potential potential from an electric electric dipol p is given by
ψ ( r ) = Determine E = =
∗
1.5
p r
·
4π ε 0 r 3
.
at the point r . −∇ψ at 0
( Exam FFM NV 060322, 1)
Show Show that that the vector vector field field
= 20 x9 yz xˆ + (2 x10 z + 10 y9 ) yˆ + (2 x10 y + 25 z24 ) zˆ F = is conserved conserved and there exists a scalar scalar potential. Calculate Calculate the line integral integral
d for the curve F r for
C
starting starting at the point ( 0, 0, 0) and finishing at (1, 1, 1). 1.6 (Yngve, vektoranalys, 2.4.13) F x =
− x
y
The The vect vector or field field has the cartesian components F has
, 2
2 + y
F y =
x x2 + y2
,
F y = 0.
1. Determine Determine the components components of in cylindrical coordinates. F in 2. What is the rotation rotation of F , i.e. ∇
× F ?
3. Calcul Calculate ate W = F d r , where the integration is along the unit circle.
·
1
4. Is the the result resultss of 2 and 2 and 3 3 in in agreement with Stokes theorem? 1.7 (Yngve, vektoranalys, 1.10.1)
Calcul Calculate ate the contour contour integr integral al
where the co− (( ,, )) F · d r where 44 11
ordinates are cartesian ( x, y)-coordinates. has the cartesian components F has F x =
−kx,
F y =
−ky,
F y = 0,
where k is is a constant. Calculate the integral along different contours: 1. Along Along ( 1, 1) 2. Along Along x = y .
∗
→ (4, 1) → (4, 4) in the xy-plane.
1.8
( Homework 1 FFM NV 2006, 2)
1.9
Use Gauss Gauss law to show show that
1.10
∗
Calc Calcul ulat atee ∇
S ∇u
m r r 3
, where m is constant vector.
= 0 if Laplace equation is fulfilled. · d S =
( Homework 1 FFM QX 2008, 1)
∇
× ×
Calc Calcul ulat atee
·
m r
×
r 7
+ 3 r ,
= x xˆ + y yˆ + z zˆ. where m is constant vector and r = 1.11
∗
The The vect vector or field field is defined by F is
( Homework 1 FFM QX 2008, 2)
= F = ( yz + 2 xz2 ) xˆ + xz yˆ + ( xy + 2 x2 z) zˆ. Calculate the line integral
d where C is is any curve you wish starting at point ( 0, 0, 0) and F r where
·
C
finishes at point ( 1, 2, 1).
. B d S , where B = ∇ A 1.12 Calculate 1.13
∗
·
×
( Homework 2 FFM NV 2006, 1)
The vector vector field field is is defin defined ed by
= F = 2 xz xˆ + 2 yz2 yˆ + ( x2 + 2 y2 z Show that for any curve C the the integral
− 1) zˆ.
= 0 F d r =
·
C
1.14
∗
( Homework 2 FFM NV 2006, 2)
Evalu Evaluate ate the surf surface ace integr integral al over the sphere r d S over
· S
x2 + y2 + z2 = R2 . 01 1012, 1) 1.15 ( Exam 011012, scribed by r (ϕ ) = 2a(1
×
Calcul Calculate ate the contour contour inte integra grall C r d r , where C is the contour decos ϕ )( ˆ sin ϕ ), a is a constant and ϕ [0, 2π ] )( xˆ cos ϕ + + y
− ∈ r × d r = = 2 A zˆ, where A is the area bounded by the curve C , and C is is in the Show that
1.16 xy-plane.
C
2
1.17
∗
( Exam FMM QX 061023, 061 023, 1)
Conside Considerr the vector vector field field r F = r , = 3 + 5 r
where r = = x xˆ + y yˆ + z zˆ,
r = = r .
||
Show that is conserved and there exists a scalar potential. Calculate the line integral F is
F d r
C
for the curve starting at the point with the cartesian coordinate (1, 0, 0) and finishing at the point with the cartesian coordinate ( 1, 1, 1).
2
Heat and wave wave equation. equation. One dimensional dimensional problems. problems.
∗
Determine the steady-state steady-state concentrati concentration on distributi distribution on 2.1 ( Homework 3 FFM NV 2006, 1) Determine outside the sphere of radius R , if the concentration on the sphere is c 0 . Hint: use the diffusion equation and think about the correct boundary condition at infinity. Assumee that that the the tempe tempera ratu ture re is sphe spheri rica call lly y symme symmetr tric ic,, u(r , t ). 2.2 ( Exam FFM NV 060322, 2) Assum Consider the heat flow (without sources) between two concentric spheres of radii R1 and R2 , R2 > R1 . Determine the steady-state temperature distribution, if the temperature of outer sphere is u 2 and the inner sphere is u 1 . Assuming Assuming the constant specific specific heat c and the constant mass density ρ , calculate the total heat energy for this state-state solution.
∗
∗
Consider the diffusion diffusion of a chemical chemical pollutan pollutantt described described 2.3 ( Homework 2 FFM QX 2008 ) Consider by the diffusion equation without sources: 1. Determine Determine the steady-state steady-state concentration concentration distribution distribution between two concentric concentric spheres with radii R 1 and R 2 , R 1 < R 2 . The concentratio concentration n on the outer sphere sphere is c 2 and on the inner sphere is c 1 . 2. Calculate Calculate the total mass of the pollutant between between two spheres for this steady-state steady-state concentration distribution. 3. Calculate Calculate the total flow of pollutant pollutant through the outer sphere. sphere.
∗
Considerr a steadysteady-sta state te tempera temperatur turee distribu distributio tion n in a 2.4 ( Homework 3 FFM NV 2006, 2) Conside volume V with the boundary given by surface S . Please Please calculate calculate the total heat flow through through surface S and and give possible physical explanation for the answer. Assume me that that the the tempe tempera ratur turee is sphe spheri rica call lly y symme symmetr tric ic,, u(r , t ). 06 1023, 2) Assu 2.5 ( Exam FFM QX 061023, Consider the heat flow (without sources) between two concentric spheres of radii R1 and R2 , R2 > R1 . Determine the steady-state temperature distribution, if the temperature of outer sphere is u 2 and the inner sphere is perfectly insulated. Assuming the constant specific heat c and the constant mass density ρ , calculate the total heat energy for this steady-state solution.
∗
3
3
Two and and three three dimensional dimensional problems. problems. Separation Separation of variables variables..
∗
3.1
( Homework 3 FFM QX 2007 )
Solve Solve the follo followin wing g PDE
∂ u ∂ 2 u ∂ u = 2 + 2 + u , ∂ t ∂ x ∂ x x x with the boundary conditions u (0, t ) = 0 , u( L, t ) = 0 . The initial condition is u ( x, 0) = f ( x). Considerr the vibrat vibration ion u (r , θ , t ) of a circular membrane of 3.2 ( Exam FFM NV 060322, 3) Conside radius a which is described by the two-dimensional wave equation
∗
∂ 2 u = c 2 ∆u 2 ∂ t Find a solution solution which satisfies u (a, θ , t ) = 0, u(r , θ , 0) = 0, ∂ ∂ ut (r , θ , 0) = f (r ) sin3θ .
∗
Laplace equation equation insider insider the sphere of the radius 3.3 ( Exam FFM NV 060322, 4) Solve the Laplace a with the following boundary condition u(a, φ , θ ) = f (φ ) cos θ
Hint: use the following conventions for the Laplace operator in the spherical coordinates: 1 ∂ ∂ u ∂ 2 u sin φ r + 2 + 2 2 . ∆u = 2 ∂ r ∂ φ r ∂ r r sin φ ∂ φ r sin φ ∂ θ 2 1 ∂
∗
3.4
2 ∂ u
( Exam FFM NV 060322, 5)
1
On the the inte interrval [0, π ] solv solvee the the heat heat equa equati tion on with with the the sour source ce
∂ u ∂ 2 u = 2 + 10sin(6 x)e−t x ∂ t ∂ x with the boundary conditions u(0, t ) = 1,
and the initial condition
u(π , t ) = 3
2 u( x, 0) = 1 + x + 6sin6 x + 3sin x
π
∗
07 0109, 4) 3.5 ( Exam FFM QX 070109, radius b
Consider Consider the two-dimensio two-dimensional nal heat heat equation equation on the disk of
∂ u = 2∆u ∂ t
Find a solution solution u (ρ , φ , t ) which satisfies u(b, φ , t ) = 0,
u(ρ , φ , 0) = f (ρ ) sin3φ .
4
∗
Laplace equation equation outside outside of sphere sphere of the radius 07 0109, 5) Solve the Laplace 3.6 ( Exam FFM QX 070109, a with the following boundary condition u(a, θ , φ ) = P31 (cos θ ) sin φ
∗
07 0109, 3) 3.7 ( Exam FFM QX 070109, for the PDE
Using the the separati separation on of variabl variables es write write the the general general solution solution
∂ u ∂ 2 u = 2 + 5u ∂ t ∂ x x
where u ( x, t ) satisfies the following boundary conditions u(0, t ) = 0,
u(1, t ) = 0 2
u Solvee the the PDE PDE ∂ ∂ ut = k ∂ 3.8 ( Homework 4 FFM NV 2006 ) Solv α u, where k and α are con∂ x x2 stants stants.. The boundar boundary y condit conditions ions are u(0, t ) = 0, u( L, t ) = 0 and and the the initi initial al condi conditi tion on is u( x, 0) = f ( x). What happens with the solution when t and α .] ∞? [ Hint : the limit depends from k and
∗
−
→
∗
3.9
( Homework 5 FFM NV 2006 )
Solve Solve the heat heat equati equation on
∂ u = k ∆u ∂ t inside a cylinder (of radius a and height H ) subject to the initial condition, u(r , θ , z, 0) = f (r , z)
conditions are independent of θ θ , if the boundary conditions u(r , θ , 0, t ) = 0 u(r , θ , H , t ) = 0
∂ u (a, θ , z, t ) = 0 ∂ r [ Hint : try to use the independence of f from θ as soon as possible, this will simplify some of the calculations]
∗
3.10 ( Homework 4 FFM QX 2007 ) of radius b
Consider Consider the two-dimensio two-dimensional nal heat heat equation equation on the disk
∂ u = 4∆u . ∂ t
Find a solution solution u (ρ , φ , t ) which satisfies u(b, φ , t ) = 0 , u(ρ , φ , 0) = f (ρ ) sin(7φ ) + g(ρ ) sin(3φ ) .
5
3.11
∗
( Homework 5 FFM QX 2007 )
Solve Solve the Laplac Laplacee equat equation ion ∆u = 0
between two concentric spheres of the radius a and b , a < b. The boundary conditions conditions are u(a, θ , φ ) = f (θ ) cos φ , u(b, θ , φ ) = g(θ ) cos φ .
3.12
∗
( Homework 6 FFM NV 2006 )
Solve Solve the wave wave equation equation
∂ 2 u = c 2 ∆u 2 ∂ t inside a sphere of radius a subject to boundary conditions u(a, θ , φ , t ) = 0 and the initial conditions u (r , θ , φ , 0) = F (r ) sin2θ , ∂ ∂ ut (r , θ , φ , 0) = 0. 3.13
∗
( Homework 3 FFM QX 2008 )
Solve Solve the follow following ing PDE
∂ u ∂ 2 u = k 2 x ∂ t ∂ x
− α u,
where k and and α are constants. constants. The boundary conditions conditions are
∂ u (0, t ) = 0, x ∂ x
u( L, t ) = 0
and the initial condition is u( x, 0) = f ( x).
What happens with the solution when t
∗
3.14 ( Homework 4 FFM QX 2008 ) of radius 1
and α .] → ∞? [ Hint : the limit depends from k and Consider Consider the two-dimensio two-dimensional nal heat heat equation equation on the disk
∂ u = ∆u . ∂ t
Find a solution solution u (ρ , φ , t ) which satisfies the following boundary and initial conditions u(1, φ , t ) = 0 ,
3.15
∗
+ g(ρ ) sin φ . u(ρ , φ , 0) = f (ρ ) sin2φ +
( Homework 5 FFM QX 2008 )
Solve Solve the wave wave equati equation on
∂ 2 u = c 2 ∆u 2 ∂ t outside of a sphere of a radius a subject to the following boundary conditions u(a, θ , φ , t ) = 0 ,
and the initial conditions u(r , θ , φ , 0) = 0 ,
∂ u (r , θ , φ , 0) = f (r ) sin2θ . ∂ t 6
4
Inhomo Inhomogen geneou eouss pr probl oblems ems..
∗
4.1
( Homework 7 FFM NV 2008,1)
Solve the heat equation equation with with the source
∂ u ∂ 2 u = 2 + sin7 xe−3t ∂ t ∂ x x on the interval [ 0, π ] subject to u(0, t ) = 1,
u(π , t ) = 0,
7
u( x, 0) = 0
Solutions
S 1.1
Use ∇ ( a b) = b (∇
· × a) − a · (∇ × b)
S 1.2
¯ = = ω z zˆ We are free free to choose a coordinate system so ω
· ×
r ) = ω z ∇ [ zˆ ∇ (ω z zˆ y x ∂ y ∂ x ω z ( xˆ + yˆ) = 0. ∂ x ∂ y x y
·
· × ( x xˆ + y yˆ + z zˆ)] = ω ∇ · (− y xˆ + x yˆ) =
×
z
−
S 1.5
The vector vector field field has a potential potential f if the following is satisfied
= F = ∇ f . In components it can be written as the following set of PDEs
∂ f f = 20 x9 yz x ∂ x f ∂ f = 2 x10 z + 10 y9 ∂ y y
∂ f f = 2 x10 y + 25 z24 ∂ z z These equations have the solution f ( x, y, z) = 2 x10 yz + y10 + z25 + constant
The line integral is calculates as
d F r = =
C
S 1.8
The answer is ∇
d f = f (1, 1, 1)
C
− f (0, 0, 0) = 4
× ( × ) = 3 · r − .
S 1.10 The answer is ∇
S 1.11
m r r 3
m r r 5
m r 3
· × m r r 7
r = 9 . + 3
Let us show show that is conserved vector field, namely F is
= F = ∇ f =
f f f ∂ f ∂ f ∂ f xˆ + yˆ + zˆ. ∂ x ∂ y ∂ z x y z
8
This implies the following set of PDEs f ∂ f = yz + 2 xz2 x ∂ x
∂ f f = xzy ∂ y y ∂ f f = xy + 2 x2 z z ∂ z which have the following solution f ( x, y, z) = xyz + x2 z2 + constant
Since the vector field is conserved the line integral is independent from the concrete curve C and it is given by the following expression
·
= r = ∇ f d
d = F r =
C
S 1.13
·
C
d f = f (1, 2, 1)
C
− f (0, 0, 0) = 3 .
Let us show show that is conserved vector field, namely F is
∂ f ∂ f ∂ f f f f xˆ + yˆ + zˆ. x y z ∂ x ∂ y ∂ z
= = F ∇ f = This implies the following set of PDEs
f ∂ f = 2 xz ∂ x x
∂ f f = 2 yz2 y ∂ y f ∂ f = x 2 + 2 y2 z ∂ z z
−1
which have the following solution f ( x, y, z) = x2 z + y2 z2
− z + constant ∇ f · d r = = ∇ × ( ∇ f ) · d s , where we have used that ∇ × Using the Stockes theorem we have C
( ∇ f ) = 0. S 1.14
S
Using the the Gauss Gauss theorem theorem we get
·
r )dV = 3 ( ∇
r d s =
S
V
·
9
dV
V
Next we have to calculate the volume of the ball of radius R ( x2 + y2 + z2 in the spherical coordinates R π 2π
θ d φ φ r 2 sin θ = drd θ
= dV =
V
0 0 0
2
≤ R ). It is easy to do
4 3 π R , 3
where we use the following formula
θ d φ φ . dV = = h1 h2 h3 du 1 du 2 du 3 = r 2 sin θ drd θ The final answer is
·
r d s = 4π R3
S
S 1.17
The vector vector field field has a potential f if the following is satisfied
= F = ∇ f . In components it can be written as the following set of PDEs f x ∂ f = 2 + 5 x ∂ x x ( x + y2 + z2 )3/2 f y ∂ f = 2 + 5 y y ∂ y ( x + y2 + z2 )3/2
∂ f f z = 2 + 5 z ∂ z z ( x + y2 + z2 )3/2 These equations have the solution f ( x, y, z) =
−
1
x2 + y2 + z2
5 2
+ ( x2 + y2 + z2 ) + constant
Alternati Alternatively vely one can find the function f in spherical coordinates. coordinates. In spherical coordninates coordninates F is written as = = r −2 + 5r r F rˆ
and the gradient of f in spherical coordinates has a form
∂ f 1 ∂ f 1 ∂ f f f ˆ f ˆ + θ + φ . ˆ+ r ∇ f = r ∂ θ r sin θ ∂ φ ∂ r Thus we arrive to the following equation f ∂ f = r −2 + 5r ∂ r
and thus f is f (r ) =
− 1r + 52 r , 2
10
which is the same as above when we rewrite rewrite it in the cartesian cartesian coordinates. coordinates. The line integral is calculates as
d = F r =
C
S 2.1
d f = f (1, 1, 1)
C
− f (1, 0, 0) = 6 − √ 13
The diffusion equation is
∂ c = k ∆c, ∂ t where c ( diffusion coefficient coefficient.. We have to look for a steadysteadyr , t ) is concentration and k is the diffusion ∂ c state solution, ∂ t = 0 and thus c satisfies the Laplace equation, ∆c = 0.
Moreover the problem has spherical symmetry and therefore c depends only on the radial coordinate ρ . Using the form of ∆ in spherical coordinates ∆c =
1 ∂
ρ2 ∂ ρ
ρ
2 ∂ c
∂ρ
1 ∂ ∂ c ∂ 2 c sin φ + 2 + 2 2 ρ sin φ ∂ φ ∂ φ ρ sin φ ∂ θ 2
1
we arrive at the following equation 1 d
ρ 2 d ρ
ρ2
dc d c
d ρ
= 0.
This equation has the following solution c(ρ ) = B
− Aρ ,
where B and A are constants which can be fixed by imposing the correct boundary conditions. We look for the solution outside the sphere, R ρ ∞. On the sphere we have
≤ ≤
c( R) = c 0 .
At infinite distance from the sphere we would expect that there will be no any chemical and thus 0. ρ c(ρ ) ∞,
→
→
We can conclude that the solution of the problem is c(ρ ) =
11
c0 R
ρ
.
To find a spherically symmetric steady-state we have have to solve
S 2.2
1 d 2 du (r ) = 0, r 2 dr dr which is the Laplace equation for u (r ) written in the spherical coordinates. The equation has a solution A + B, u(r ) = r
−
where A and B are constants. The constants constants can be fixed by imposing imposing the boundary boundary conditions u( R1 ) = u1 ,
u( R2 ) = u2 .
Thus we get
− u ) R R , R − R ( u − u ) R R u(r ) = ( R − R )r
A =
( u2
1
2
and
1 2
2
2
cρ
= c ρ udV =
V
S 2.3
π
R2
R1
2π
R2
+
u 2 R2 R2
− u R − R
1 1 1
− u R − R
1 1 1
− Ar + B) = 2π cρ A( R − R ) + 43π cρ B( R − R )
φ sin sin θ r 2 ( d φ
θ d θ
0
1 2
1
dr
u2 R2
1
1
The total heat energy is
B =
0
2 1
2 2
3 2
3 1
1. The diffusion diffusion equation equation is
∂ c = k ∆c, ∂ t where c ( diffusion coefficient coefficient.. We have to look for a steadysteadyr , t ) is concentration and k is the diffusion ∂ c state solution, ∂ t = 0 and thus c satisfies the Laplace equation, ∆c = 0.
Moreover the problem has spherical symmetry and therefore c depends only on the radial coordinate r . Using the form of ∆ in spherical coordinates 1 ∂ 1 1 ∂ c ∂ ∂ c ∂ 2 c sin θ r 2 + 2 + 2 2 ∆c = 2 ∂ r ∂ θ r ∂ r r sin θ ∂ θ r sin θ ∂ φ 2
we arrive at the following equation 1 d r 2 dr
r 2
dc
dr
= 0.
This equation has the following solution c(r ) = B
12
− Ar ,
where B and A are constants which can be fixed by imposing the correct boundary conditions. We look for the solution between two concentric spheres, R 1 r R2 . On the sphere we have
≤ ≤
c( R1 ) = c 1 ,
c( R2 ) = c 2 .
We can conclude that the solution of the problem is c(r ) = A =
( c2 c1 ) R1 R2 c2 R2 c1 R1 + . R2 R1 ( R1 R2 )r
−
( c1
− −
−
− c ) R R R − R 2
1 2
1
,
B =
c2 R2 R2
2
− c R − R
1 1
.
1
2. The mass of pollutant is given by the volume integral over the space between two concentric spheres
c dV d V =
V
π
R2
2π
d θ θ
dr
R1
0
−
2
A
d φ φ r sin θ B
0
r
B
− − 4π A2 ( R − R ) .
= 4π ( R32 R31 ) 3
2 2
2 1
3. The flux vector vector is proportional proportional to the gradient gradient of the concentration concentration c
= J =
−K ∇c .
= = J
−K Ar ˆr r ..
0
In the spherical coordinates we have 0 2
The total heat flow through the outer sphere S is is given by surface integral π
·
d s = J
S
S 2.4
−K
0
2π
θ d θ
0
φ d φ
0
A R22
= R22 sin θ =
−4π AK . 0
The steady-state temperature distribution satisfies the Laplace Laplace equation ∆u = 0.
The heat flow vector is proportional to the gradient of the temperature
φ = =
−K ∇u. 0
The total heat flow through the surface S is is given by surface integral
·
s = φ d
S
−K
0
s = ∇u d
S
·
−K
0
∇2 u dV = 0,
V
where we have applied the Gauss theorem and used the fact that u satisfies the Laplace equation (remember that ∆ = ∇2 = ∇ ∇). Thus for the steady state temperature distribution the total flow through any closed surface is zero. zero. This This is natural natural since the total heat energy energy in V does not change in time (there are no sources).
·
13
S 2.5
To find a spherically symmetric steady-state we have have to solve 1 d r 2
(r 2
dr
du dr
) = 0,
which is the Laplace equation for u (r ) written in the spherical coordinates. The equation has a solution A + B, u(r ) = r where A and B are constants. The constants constants can be fixed by imposing imposing the boundary boundary conditions
−
du dr
( R1 ) = 0,
u( R2 ) = u2 ,
where the first condition corresponds to the perfectly insulated boundary. Thus we get B = u2
and u(r ) = u2
The total heat energy is cρ
udV = cρ
V
π
R2
2π
R1
sin θ r 2 u2 = d φ φ sin
d θ θ
dr
0
0
4π cρ u2 ( R32 3
3 1
− R )
equation should be solved by the method method of separation separation of variables. variables. Let us look for S 3.1 The equation the solutions of the form u( x, t ) = φ ( x)G(t ) . Plugging Plugging this into the equation we get dG d G
φ
dt
= G
and 1 dG G dt
=
1
φ
Thus we get two ODEs
dx 2
d 2 φ dx 2
dG dt d 2 φ
d 2 φ
+2
+2
+2
φ d φ dx
+ φ
φ d φ dx
+ φ =
+ λ G = 0 , d φ φ
+ φ = =
−λ φ ,
dx 2 dx which should be supplemented supplemented by the boundary conditions
φ (0) = 0 ,
φ ( L) = 0 .
The equation for G is solved very easily G(t ) = Ae−λ t t .
14
,
−λ .
For φ we can look for the solutions of the form φ = = eα x , where
(α + + 1)2 = λ
−
and we get
√ ±i λ − 1 .
= α = If λ λ > 0 then the solution can be written as
φ ( x) = e− x
√
√
A cos( λ x x) + B sin( λ x x)
Imposing Imposing the boundary conditions we get
φ ( x) = Ae− x sin
π nx nx
, λ =
L
There will be no non-trivial solutions for λ The general solution solution of the problem will be ∞
∑
u( x, t ) =
π n
2
L
,
.
n = 1, 2, 3,...
≤ ≤ 0 (see the lectures notes for the derivation).
π 2 An e−( ) t e− x sin n L
n=1
π nx nx
L
.
Using the orthogonality of sin sin ’s we get
An =
S 3.2
2 L
L
x
dx f ( x)e sin
0
π nx nx
L
.
We have to solve the following equation
∂ 2 u = c 2 ∂ t 2
1 ∂ 2 u ∂ u r + 2 r ∂ r r ∂ θ 2 ∂ r
1 ∂
Assuming that u(t , r , θ ) = G(t )F (r )g(θ )
we get the following following set of ODEs d 2 g
θ 2 d θ r 2
d 2 F dr 2
dF
+ r
dr
=
−µ g,
+ (λ r 2
g(θ + + 2π ) = g(θ )
= 0, − µ )F = d 2 G dt 2
F (a) = 0,
|F (0)| < ∞
= λ c2 G
−
The first equation has a solution solution g(θ ) = A cos(mθ ) + B sin(mθ ),
15
µ = = m2 , m = 0, 1, 2,...
and the second
F (r ) = CJ m (
The solution for G is
λ n r ),
λ n =
G(t ) = D cos(c
zmn
2
a
, J m ( zmn ) = 0, n = 1, 2,...
λ nt ) + E sin(c
λ nt ),
where we can drop cos in order to satisfy the first initial condition. condition. Thus we get ∞
u(t , r , θ ) =
∞
Bmn sin(mθ ) + Amn cos(mθ )] sin(c ∑ ∑ [ B m=0 n=1
λ nt ) J m (
λ n r )
Imposing the remaining initial condition we arrive at the solution of the problem ∞
u(t , r , θ ) =
∑ Bn sin(c n=1
with
λ n r )
√
a
λ nt ) sin(3θ ) J 3 (
dr f (r )rJ 3 ( λ n )
0
Bn =
√
√
a
c λ n drrJ 32 ( λ n r )
0
S 3.3
We have have to solve
∂ ∂ u ∂ 2 u 1 ∂ 1 1 2 ∂ u + 2 + = 0 sin φ r r 2 ∂ r r sin φ ∂ φ ∂ r ∂ φ r 2 sin2 φ ∂ θ 2
The problem is solved by the separation separation of the variables variables u(r , φ , θ ) = f (r )h(φ )g(θ )
We get the following set of ODEs d 2 g
θ 2 d θ d
φ d φ
=
−µ g,
d h
sin φ
φ d φ
g(θ + + 2π ) = g(θ )
2 2 d f r dr 2
sin φ + λ sin d f
+ 2r
dr
m2
− sin φ
h = 0
− λ f f = 0
The first equation has a solution solution g(θ ) = A sin(mθ ) + B cos(mθ ),
= m2 , m = 0, 1, 2,... µ =
For the second equation in order to have a finite solution everywhere we have to require λ = l (l + 1) with l m h(φ ) = CPlm (cos φ )
≥
16
For f we have f (r ) = Dr l + Er −l −1
and we keep only the solution finite at the origin. Thus the solution of Laplace equation inside a sphere is ∞
u(r , θ , φ ) =
∞
l
m
Aml cos(mθ ) + Bml sin(mθ )]Pl ∑ ∑ r [ A
(cos φ )
m=0 l =m
Imposing Imposing the boundary boundary conditions we arrive at the final solution of the problem problem ∞
l
∑ a Al cos θ Pl1 ( cos φ )
u(r , θ , φ ) =
l =1
with
π
Al =
f (φ )Pl1 ( cos φ ) sin φ d φ φ
0
π
al (Pl1 ( cos φ ))2 sin φ d φ φ 0
S 3.4
First we would would like to switch switch to the homogeneou homogeneouss BC 2 u( x, t ) = 1 + x + v( x, t )
π
where v (0, t ) = 0, v (π , t ) = 0 and v satisfies the same equation as u . We can look for for v in the form ∞
∑ an (t ) sin(nx)
v( x, t ) =
n=1
Substituting Substituting this into PDE we get da n dt da 6 dt
= n2 an ,
n = 6
−
= 36a6 + 10e−t
−
with the initial conditions a6 (0) = 6,
a1 (0) = 3,
an (0) = 0, n = 1, 6
So the final solution is
2 u( x, t ) = 1 + x + 3e−t sin x + (6
π
S 3.5
10
− 35 )e−
36t
+
10 35
We have to solve the following equation
∂ u 1 ∂ = 2 ∂ t ρ ∂ρ
∂ u ρ ∂ρ
17
+
1 ∂ 2 u
ρ 2 ∂ φ 2
−
e
t
sin6 x
Assuming that u(ρ , φ , t ) = G(t )F (ρ )g(φ )
we get the following following set of ODEs d 2 g d φ φ 2
=
−µ g,
2 dF 2 d F ρ + ρ + (λ ρ 2 2 d ρ d ρ
+ 2π ) = g(φ ) g(φ +
= 0, − µ )F = dG dt
F (b) = 0,
|F (0)| < ∞
= 2λ G
−
The first equation has a solution solution
= m2 , m = 0, 1, 2,... µ =
g(θ ) = A cos(mφ ) + B sin(mφ ),
and the second
√
F (ρ ) = CJ m ( λ ρ ),
= λ =
The solution for G is
zmn
2
b
, J m ( zmn ) = 0, n = 1, 2,...
G(t ) = De−2λ t t .
Combing all we get the following following solution of heat equation together with boundary conditions ∞
u(ρ , φ , t ) =
∑
∞
Bmn sin(mφ ) + Amn cos(mφ )]e−( ∑ [ B
zmn b
) 2t J ( zmn ρ ) m 2
m=0 n=1
b
Imposing the initial condition we arrive at the solution of the problem ∞
u(ρ , φ , t ) =
− ∑ Bn e (
z3n b
n=1
with
b
Bn =
2
2t
z3n sin(3φ ) J 3 ( ρ) b
d ρ f (ρ )ρ J 3 ( zb3n ρ )
0
b
0
S 3.6
)
d ρ ρ ( J 3 ( zb3n ρ ))2
We have have to solve 1 ∂ 1 1 ∂ ∂ u ∂ 2 u 2 ∂ u sin r + + = 0 θ ∂ r ∂ θ r 2 ∂ r r 2 sin θ ∂ φ r 2 sin2 θ ∂ φ 2
The problem is solved by the separation separation of the variables variables u(r , θ , φ ) = f (r )h(θ )g(φ )
18
We get the following set of ODEs d 2 g
φ 2 d φ d
θ d θ
=
−µ g,
d h
sin θ
θ d θ
g(φ + + 2π ) = g(φ )
sin θ + λ sin
2 2 d f r dr 2
m2
− sin θ
h = 0
d f
+ 2r
− λ f f = 0
dr
The first equation has a solution solution
= m 2 , m = 0, 1, 2,... µ =
g(φ ) = A sin(mφ ) + B cos(mφ ),
For the second equation in order to have a finite solution everywhere we have to require λ = l (l + 1) with l m h(θ ) = CPlm (cos θ )
≥
For f we have f (r ) = Dr l + Er −l −1
and we keep only the solution finite at the infinity. Thus the solution of Laplace equation inside a sphere is ∞
u(r , φ , θ ) =
∞
−l − Aml cos(mφ ) + Bml sin(mφ )]Pm(cos θ ) ∑ ∑ r 1 [ A l m=0 l =m
Imposing the boundary conditions and using the orthogonality of associated Legendre functions (together (together with the orthogonality orthogonality of sin and cos) we arrive arrive at the final solution of the problem u(r , θ , φ ) =
S 3.7
We have have to solve
4
a
r
P31 (cos θ ) sin φ
∂ u ∂ 2 u = 2 + 5u. ∂ t ∂ x x
Assuming that u ( x, t ) = G(t )φ ( x) we get dG d G
φ
dt
= G
d 2 φ x2 ∂ x
+ 5φ
which implies the following ODEs dG
∂ t d 2 φ dx 2
= λ G
−
+ (λ + + 5)φ = 0
with the last equation supplemented supplemented by the boundary conditions
φ (0) = 0,
φ (1) = 0. 19
The equation for φ is solved by
= (π n)2 λ =
nx), φ ( x) = A sin(π nx
− 5,
n = 1, 2,...
It is important to notice that λ > 0 for all n = 1 , 2, .. since for π 2 > 5. The equati equation on for G is solved by G(t ) = Ae−λ t t Combing all together we get the following solution ∞
2
nx) ∑ An e(5−(π n) )t sin(π nx
u( x, t ) =
n=1
S 3.8 We have a linear homogeneous PDE with the linear homogeneous boundary conditions and thus we can solve the problem by the separation of variables. Let us solve PDE ∂ u ∂ 2 u = k 2 α u x ∂ t ∂ x assuming assuming the following following form of the solution solution
−
u( x, t ) = h(t )φ ( x).
PDE reduces to the following equation 1 d 2 φ = k h dt φ dx 2
1 dh
− α
Using the standard argument for the separation of variables we get d 2 φ dx 2
=
−λ φ
with λ being a constant and dh dt
= (λ k + α )h.
−
The last equation can be easily solved h(t ) = Ae−(λ k +α )t .
For φ we have to specify the boundary conditions
φ (0) = 0,
φ ( L) = 0.
The system for φ has been considered many times during the lectures and it has the following solution nx π nx π n 2 n = 1, 2, 3,... , λ = , φ ( x) = A sin L L The solution of the original PDE can be written as
∞
π 2 u( x, t ) = ∑ An e−(( ) k +α )t sin n L
n=1
20
π nx nx
L
Finally we have to impose the initial conditions ∞
u( x, 0) =
∑ An sin n=1
nx π nx
L
= f ( x)
Using the orthogonality of sines we get the following expressions for A’s
An =
Now about the limit t
2 L
L
f ( x) sin
0
nx π nx
L
dx
> 0 then → ∞. If λ λ k + α > u( x, t )
→0
Otherwise u ( x, t ) will go to infinity.
PDE with the linear homogeneous boundary conditions. S 3.9 We have a linear homogeneous PDE Therefore we can apply the method of separation of variables. Assume our solution of the form u(r , θ , z, t ) = G(t )φ (r , θ , z)
then we arrive to the followin problems dG dt
+ λ φ = = 0, ∆φ +
= λ kG
−
φ (r , θ , 0) = 0, φ (r , θ , H ) = 0,
∂ φ (a, θ , z) = 0. ∂ r
First equation equation can be easily solved by G(t ) = Ae−λ kt
and second problem should be solved solved by the separation separation of variables variables.. Assuming Assuming that
φ (r , θ , z) = f (r )g(θ )h( z) we arrive to the following systems of equations d 2 h dz 2 d 2 g d θ θ 2 r 2
d 2 f dr 2
=
−σ g, d f
+ r
dr
=
−µ h,
h(0) = 0,
h( H ) = 0 dg
+ 2π ) = g(θ ), , g(θ +
+ (λ
2
dg
(θ + + 2π ) = (θ ) d θ d θ θ θ
− µ )r − σ f = 0, | f (0)| < ∞,
d f dr
First two equations can be solved easily as follows h( z) = A sin
nz π nz
H
,
= µ = 21
π n
2
H
,
n = 1, 2, 3,...
(a) = 0
= m2 , σ =
g(θ ) = A sin(mθ ) + B cos(mθ ),
m = 0, 1, 2,....
However we can observe at this stage that the initial conditions do not have any dependence on and therefore the solution is independent from θ . Thus we can conclude conclude that σ = = 0. Thus the θ and equation for f becomes r 2
where λ = λ +
d 2 f dr 2
π n 2 . H
d f
+ r
dr
+ λ r 2 f = 0,
d f
| f (0)| < ∞,
dr
(a) = 0
= 0) is Assuming that λ > 0 the solution (which is finite at r =
√
f (r ) = CJ 0 ( λ r ).
= a . We use the following Next we have to impose the boundary conditions at r = following property of Bessel functions (see the lecture notes) J 1 ( x) =
Thus we get that
√
λ =
J 1 ( λ a) = 0,
l
. − dJ dx 0
z1l
2
a
, l = 1, 2, 3,...
where z1l are zeros of J solution exists due to the properties properties of modified Bessel J 1 . For λ 0 no solution functions (no zeros). Thus the solution of the PDE with the boundary conditions is
≤
∞
∞
π nz nz
u(r , θ , z, t ) = ∑ ∑ Anl e−λ kt sin nl nl
n=1 l =1
where
λ nl nl = λ l + Finally we have to impose the initial conditions ∞
∞
∑
u(r , θ , z, 0) =
∑ Anl sin
n=1 l =1
π n
H
J 0 (
λ l r ),
2
H
π nz nz
H
J 0 (
λ l r ) = f (r , z)
Using the orthogonality orthogonality of sines and J 0 we get H
a
Anl =
0
0
π nz nz
2 dz dr r f (r , z) sin
H
J 0 (
λ l r )
H
H dr rJ 02 ( 0
λ l r )
details of the solution which which have been presented presented on the lectures lectures is not written S 3.10 (Some details out). We have to solve the heat equation in cylindrical cylindrical coordinates coordinates ignoring the dependence from z . We solve the following equation
∂ u 1 ∂ = 4 ∂ t ρ ∂ρ
∂ u ρ ∂ρ
22
+
1 ∂ 2 u
ρ 2 ∂ φ 2
Assuming that u(ρ , φ , t ) = G(t )F (ρ )g(φ )
we get the following following set of ODEs d 2 g d φ φ 2
=
−µ g,
2 dF 2 d F ρ + ρ + (λ ρ 2 2 d ρ d ρ
+ 2π ) = g(φ ) g(φ +
= 0, − µ )F = dG dt
F (b) = 0,
|F (0)| < ∞
= 4λ G
−
The first equation has a solution solution
= m2 , m = 0, 1, 2,... µ =
g(φ ) = A cos(mφ ) + B sin(mφ ),
and the second
F (ρ ) = CJ m (
λ mn mn ρ ),
λ mn mn =
zmn
2
a
, J m ( zmn ) = 0, n = 1, 2, ... ... ,
where we dropped the Bessel function of second kind since it goes to infinity at ρ = 0. 0. The The solution for G is G(t ) = De−4λ t t . Thus we get ∞
u(ρ , φ , t ) =
∑
∞
mn t Bmn sin(mφ ) + Amn cos(mφ )]e−4λ mn J m ( λ mn mn ρ ) ∑ [ B
m=0 n=1
Imposing the remaining initial condition we arrive at the solution of the problem ∞
u(t , r , θ ) =
∑
B7n e−4λ 7 t sin(7φ ) J 7 ( n
n=1
with
λ 7n ρ ) + B3n e−4λ 3 t sin(3φ ) J 3 ( n
√
a
d ρ f (ρ )ρ J 7 ( λ 7n ρ )
B7n =
0
a
√
d ρ ρ ( J 7 ( λ 7n ρ ))2
0
and
√
a
d ρ g(ρ )ρ J 3 ( λ 3n ρ )
B3n =
0
a
√
d ρ ρ ( J 3 ( λ 3n ρ ))2
0
23
λ 3n ρ )
S 3.11
We have have to solve
∂ ∂ u ∂ 2 u 1 1 sin θ r + 2 + 2 2 = 0 ∂ r ∂ θ r 2 ∂ r r sin θ ∂ θ r sin θ ∂ φ 2 1 ∂
2 ∂ u
The problem is solved by the separation separation of the variables variables u(r , φ , θ ) = f (r )h(θ )g(φ )
We get the following set of ODEs d 2 g d φ φ 2 d d θ θ
=
−µ g,
d h
sin θ
d θ θ
r 2
+ 2π ) = g(φ ) g(φ +
+ λ sin sin θ
d 2 f dr 2
m2
− sin θ
h = 0
d f
+ 2r
− λ f f = 0
dr
The first equation has a solution solution
= m 2 , m = 0, 1, 2,... µ =
g(φ ) = A sin(mφ ) + B cos(mφ ),
For the second equation in order to have a finite solution everywhere we have to require λ = l (l + 1) with l being a positive integer such that l m
≥
h(φ ) = CPlm (cos θ )
For f we have f (r ) = Dr l + Er −l −1
and we keep keep both both the solutions solutions,, since since we solve the problem problem betwee between n two spheres spheres.. Thus Thus the solution of Laplace equation between two spheres is ∞
∞
Aml cos(mφ ) + Bml sin(mφ )]Plm (cos θ )+ ∑ ∑ r −l−1 [ A
u(r , θ , φ ) =
m=0 l =m ∞
+
∞
l
m
Dml cos(mφ ) + C ml ml sin(mφ )]Pl ∑ ∑ r [ D
(cos θ )
m=0 l =m
Imposing Imposing the boundary boundary conditions we arrive at the final solution of the problem problem ∞
u(r , θ , φ ) =
−− ∑ r
l 1
l
Al + r Bl Pl1 ( cos θ ) cos φ
l =1
with
π
a−l −1 Al + al Bl =
f (θ )Pl1 ( cos θ ) sin θ d θ θ
0
π
0
(Pl1 ( cos θ ))2 sin θ d θ θ
24
π
b−l −1 Al + bl Bl =
θ g(θ )Pl1 ( cos θ ) sin θ d θ
0
0
Finally the coefficients are π
bl f (θ )
0
Al =
(bl a−l −1 π
− − b
Bl =
l 1
(Pl1 ( cos θ ))2 sin θ d θ θ
l
− a g(θ )
θ Pl1 ( cos θ ) sin θ d θ
π
1 l
− a b− − ) (P ( cos θ )) sin θ d θ θ l
l 1
0
f (θ )
0
(b−l −1 al
.
π
− a− − g(θ ) l 1
π
1 l
2
Pl1 ( cos θ ) sin θ d θ θ
− a− − b ) (P ( cos θ )) sin θ d θ θ l 1 l
0
2
homogeneous PDE with linear homogeneous homogeneous BC and thus we can S 3.12 We have a linear homogeneous apply the method of separation separation of variables. variables. Our equation is 1 ∂ 2 u
1 ∂ ∂ u ∂ 2 u sin φ r = 2 + 2 + 2 2 ∂ r ∂ φ c2 ∂ t 2 r ∂ r r sin φ ∂ φ r sin φ ∂ θ 2 1 ∂
2 ∂ u
1
We could observe that initial conditions require that u is independent of φ . Thus we can simplify the equation 1 ∂ 2 u 1 ∂ 1 ∂ 2 u 2 ∂ u r = 2 + 2 2 . c2 ∂ t 2 r ∂ r ∂ r r sin φ ∂ θ 2
However although we assumed the independence of the solution from φ the right hand side depends depends on φ . Thus we can proceed only if
∂ 2 u = 0 ∂ θ 2 and this will contradict contradict the IC unless F (r ) = 0. The problem has a trivial trivial solution only if F otherwise there is no solution. F (r ) = 0 otherwise Alternatively you could separate the variables u(r , φ , θ , t ) = G(t ) f (r )h(φ )g(θ )
(see the lecture notes) and find among other things h(φ ) = Plm (cos φ ),
g(θ ) = A cos(mθ ) + B sin(mθ )
with m l . The initial initial conditions conditions require that m = 2 and at the same time there is no φ dependence. pendence. So we will arrive arrive at the same conclusion, that there exists exists only trivial trivial solution when F (r ) = 0.
≤
25
S 3.13 We have a linear homogeneous PDE with the linear homogeneous boundary conditions and thus we can solve the problem by the separation of variables. Let us solve PDE ∂ u ∂ 2 u = k 2 α u , ∂ t ∂ x x
−
assuming assuming the following following form of the solution solution u( x, t ) = h(t )φ ( x) .
PDE reduces to the following equation 1 dh 1 d 2 φ = k h dt φ dx 2
− α . .
Using the standard argument for the separation of variables we get d 2 φ dx 2
=
−λ φ
with λ being a constant and dh dt
= (λ k + α )h .
−
The last equation can be easily solved h(t ) = Ae−(λ k +α )t .
For φ we have to specify the boundary conditions
φ d φ dx
φ ( L) = 0 .
(0) = 0 ,
Let’s consider the different cases. For the case λ > 0 the general solution is
√
√
x) + B cos( λ x x) . φ ( x) = A sin( λ x
Imposing the boundary conditions we arrive at the solution
φ ( x) = A cos
π (n + 12 ) x L
, λ =
π (n + 12 ) L
2
,
n = 0, 1, 2, 3,...
For the case λ 0 there are no solutions solutions satisfying the boundary conditions. conditions. Thus the solution of the original PDE can be written as
≤ ≤
∞
u( x, t ) =
n (( π ( +1/2) )2 k +α )t
∑ An e−
L
cos
n=0
π (n + 12 ) x L
Finally we have to impose the initial conditions ∞
u( x, 0) =
∑ An cos n=0
π (n + 12 ) x
26
L
= f ( x)
.
Using the orthogonality of cos’s we get the following expressions for A’s
An =
Now about the limit t
2 L
→ ∞. If
L
k
f ( x) cos
0
π (n + 12 ) L
π (n + 12 ) x L
dx
2
+ α > > 0
(for all n ’s) then u( x, t )
→0
Otherwise u ( x, t ) will go to infinity.
S 3.14 We have to solve the heat equation in cylindrical coordinates ignoring the dependence from z . We solve the following equation
∂ u = ∂ t
1 ∂
ρ ∂ρ
∂ u ρ ∂ρ
+
1 ∂ 2 u
ρ 2 ∂ φ 2
Assuming that u(ρ , φ , t ) = G(t )F (ρ )g(φ )
we get the following following set of ODEs d 2 g d φ φ 2
=
−µ g,
2 dF 2 d F + + (λ ρ 2 ρ ρ 2 d ρ d ρ
g(φ + + 2π ) = g(φ )
= 0, − µ )F = dG dt
=
F (1) = 0,
|F (0)| < ∞
−λ G
The first equation has a solution solution g(φ ) = A cos(mφ ) + B sin(mφ ),
= m2 , m = 0, 1, 2,... µ =
and the second F (ρ ) = CJ m ( zmn ρ ),
2 ... , λ mn mn = ( zmn ) , J m ( zmn ) = 0 , n = 1 , 2, ...
where we dropped the Bessel function of second kind since it goes to infinity at ρ = 0. 0. The The solution for G is G(t ) = De−λ t t . Thus we get ∞
u(ρ , φ , t ) =
∞
Bmn sin(mφ ) + Amn cos(mφ )]e−( z ∑ ∑ [ B
mn )
m=0 n=1
27
2
t
J m ( zmn ρ )
Imposing the remaining initial condition we arrive at the solution of the problem ∞
u(t , r , θ ) =
∑
2 2 B1n e−( z1 ) t sin(φ ) J 1 ( z1n ρ ) + B2n e−( z2 ) t sin(2φ ) J 2 ( z2n ρ ) n
n
n=1
with
a
d ρ g(ρ )ρ J 1 ( z1n ρ )
B1n =
0
a
d ρ ρ ( J 1 ( z1n ρ ))2
0
and
a
d ρ f (ρ )ρ J 2 ( z2n ρ )
B2n =
0
a
d ρ ρ ( J 2 ( z2n ρ ))2
0
the problem problem inside of a sphere sphere first. first. S 3.15 Solve the We have a linear homogeneous PDE with linear homogeneous BC and thus we can apply the method of separation of variables. Our equation is 1 ∂ 2 u 1 ∂ 1 1 ∂ ∂ u ∂ 2 u 2 ∂ u sin r = + + θ ∂ r ∂ θ c2 ∂ t 2 r 2 ∂ r r 2 sin θ ∂ θ r 2 sin2 θ ∂ φ 2
We could observe that initial conditions require that u is independent of φ . Thus we can simplify the equation 1 ∂ 2 u 1 ∂ 1 ∂ ∂ u 2 ∂ u sin r = + . θ ∂ r ∂ θ c2 ∂ t 2 r 2 ∂ r r 2 sin θ ∂ θ
We separate the variables u(r , φ , θ , t ) = G (t ) f (r )h(θ )
and arrive at the following equations and BC d
θ d θ
d h
sin θ
d dr
r 2
θ d θ
d f
+ µ sin θ h = 0 ,
+ (λ r 2
dr
|h(0)| < ∞ , |h(π )| < ∞ ,
− µ ) f (r ) = 0 ,
d 2 G dt 2
f (a) = 0 ,
f < ∞
+ λ c2 G = 0 ,
The solutions of the first equation are h(φ ) = Pl (cos φ ) ,
l = 0, 1, 2,...
and the second equation is solved by f (r ) = r −1/2 J
l + 12
( λ lk lk r ) , λ lk lk = 28
z(l +1/2)k
a
2
,
k = = 1, 2, 3,...
where z(l +1/2)k are the zeros of J follows J l + 1 . The last equation is solved as follows 2
√
√
G(t ) = A cos( λ ct ) + B sin( λ ct ) .
Imposing the one of IC
∂ u (r , θ , φ , t ) = 0 ∂ t we get
∞
∞
2
l =0 k =1
Imposing the second IC we get a
Alk =
−/ ∑ Alk r 1 2 J l+ 1 (
∑
u(r , θ , t ) =
π
0 a
0
2
π
0
λ lk lk r )Pl (cos θ )
))2 (Pl (cos θ ))2 λ lk lk r ))
dr d θ θ r sin θ ( J l + 1 ( 2
λ lk lk ct ) .
θ F (r ) sin2θ r 3/2 sin θ J dr d θ J l + 1 (
0
λ lk lk r )Pl (cos θ ) cos(
Now let’s let’s comment on the solution solution outside of a sphere. sphere. In this case the solution for f will be
−1/2 J f lk lk (r ) = r
l + 12 (
where λ lk lk is now defined as
J l + 1 ( 2
−1/2Y λ lk lk r ) + r
λ lk lk r ) ,
l + 12 (
λ lk lk a) + Y l + 1 ( 2
λ lk lk a) = 0 .
Due to the general properties of the SL problem there is the following orthogonality condition ∞
d r r flk l k (r ) f ln ln (r ) = 0 ,
k = n .
a
S 4.1
First we have have to switch to the homogeneous BC. We We can do it by the following following change v( x, t ) = u( x, t )
− 1 + π x .
v satisfies the following PDE
∂ v ∂ 2 v = 2 + sin(7 x)e−3t t x ∂ ∂ x on the interval [ 0, π ] subject to v(0, t ) = 0,
v(π , t ) = 0,
v( x, 0) =
We look for the solution in the form ∞
v( x, t ) =
∑ an (t ) sin(nx) n=1
29
x
− 1.
π
which satisfies satisfies BC. Substituting Substituting this solution into the PDE we get the following following set of ODEs da n
= n2 an ,
n = 7,
−
dt
da 7 dt
= 49a7 + e−3t .
−
These ODEs have the solutions in the form 2
an (t ) = e−n t an (0),
a7 (t ) = e−49t a7 (0)
Imposing Imposing the IC
n = 7
1
− 46
∞
v( x, 0) =
+
1 46
e−3t .
x
∑ an (0) sin(nx) = π − 1
n=1
we get an (0) =
2
π
π
sin(nx)
0
x
− π
1 dx .
Thus the final solution is u( x, t ) = 1
x
1
− π + 46
∞
− − − e
3t
e
49t
30
2
sin(7 x) + ∑ an (0)e−n t sin(nx) n=1