7162019 3705Test4 Sol

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MATH3705 A mdash Test 4 Wed 235pmndash325pm Mar 20

Name and Student Number

Total points 20 No partial marks for Questions 1-2

Closed book Non-programmer calculators are allowed

1 (3 points) Consider the Laplace equation uxx + uyy = 0 0 lt x lt 1 0 lt y lt 1 withboundary conditions u(x 0) = x u(x 1) = 0 u(0 y) = 0 u(1 y) = 1 minus y Let u(x y) bethe polynomial solution Find u(05 06)

(a) 12 (b) 25 (c) 3 (d) 0 (e) 02

Solution (e) Let u(x y) = ax + by + cxy + d u(x y) = x minus xy rArr u(05 06) = 02

2 (3 points) The solution of Laplacersquos equation uxx + uyy = 0 0 lt x lt L 0 lt y lt M withthe boundary conditions u(x 0) = u(x M ) = u(0 y) = 0 u(L y) = f (y) is given by

u(x y) =infinsumn=1

an sinh983080nπx

M

983081sin

983080nπy

M

983081

Find the solution u(x y) of Laplacersquos equation uxx + uyy = 0 0 lt x lt 2 0 lt y lt 2π with

boundary conditions u(x 0) = u(x 2π) = u(0 y) = 0 u(2 y) = 3 sin(y)

(a) 3sinh(2)

sinh (2x)sin(y) (b) 6sinh(2)

sinh (x)sin(y) (c) 3sinh(2)

sinh (x)sin1048616xy

2

1048617(d) 3

sinh(2) sinh (x)sin

10486162yx

1048617 (e) 3

sinh(2) sinh (x)sin(y)

Solution (e) L = 2 M = 2π

u(x y) =infinsumn=1

an sinh983080nπx

2π

983081sin

983080nπy

2π

983081 =

infinsumn=1

an sinh983080nx

2

983081sin

983080ny

2

983081

u(2 y) = 3 sin(y) =infinsumn=1

an sinh (n)sin983080ny

2

983081

Thus when n = 2 3sin(y) = a2 sinh (2) sin10486162y2

1048617rArr a2 = 3

sinh(2) Other an = 0 Hence

u(x y) = 3

sinh(2) sinh (x)sin(y)

7162019 3705Test4 Sol

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3 (7 points) The solution of the wave equation utt = c2uxx 0 lt x lt L t gt 0 subjectto the boundary conditions u(0 t) = 0 u(L t) = 0 and the initial conditions u(x 0) =f (x) ut(x 0) = g(x) is given by

u(x t) =

infin

sumn=1

983131an cos(

nπct

L ) + bn sin(

nπct

L )983133

sin(

nπx

L )

Find the solution of the wave equation utt = 4uxx 0 lt x lt 2 t gt 0 satisfying the condi-tions u(0 t) = u(2 t) = 0 and u(x 0) = 3 sin(2πx) ut(x 0) = 16 sin(4πx)

Solution Note that L = 2 c = 2 We have

u(x t) =infinsumn=1

[an cos(nπt) + bn sin(nπt)] sin(nπx

2 )

we imply that

u(x 0) =infinsumn=1

an sin(nπx

2 ) = 3 sin(2πx) rArr

a4 = 3 an = 0 (n = 4)

Note that

ut(x t) =infinsumn=1

[minusnπan sin(nπt) + nπbn cos(nπt)] sin(nπx

2 ) rArr

ut(x 0) =infinsumn=1

nπbn sin( nπx2

) = 3 sin(4πx) rArr

8πb8 = 16 bn = 0 (n = 8) rArr b8 = 2

π

Thus

u(x t) = 3cos(4πt) sin(2πx) + 2

π sin(8πt) sin(4πx)

4 (7 points) The bounded solution of the Laplace equation urr + 1rur +

1r2 uθθ = 0 outside

the circle r = a subject to the boundary condition u(a θ) = f (θ) is given by

u(r θ) = a0

2 +

infinsumn=1

rminusn[an cos(nθ) + bn sin(nθ)]

where f is continuous 2π-periodic f prime is piecewise continuous

Find the bounded solution of urr + 1r

ur + 1r2

uθθ = 0 outside the circle r = 2 subject tothe boundary condition u(2 θ) = 2 cos2(θ) minus 3sin(3θ)

7162019 3705Test4 Sol

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Solution a = 2 The solution of the PDE above is

u(r θ) = a0

2 +

infinsumn=1

rminusn[an cos(nθ) + bn sin(nθ)]

Note that 2 cos2

(θ) = 1 + cos(2θ) Thus

u(2 θ) = 1 + cos(2θ) minus 3sin(3θ) = a0

2 +

infinsumn=1

2minusn[an cos(nθ) + bn sin(nθ)]

which implies that

a0 = 2 1 = 2minus2a2 an = 0 for n = 0 2 minus3 = 2minus3b3 bn = 0 for n = 3 rArr

a0 = 2 a2 = 4 an = 0 for n = 0 2 b3 = minus24 bn = 0 for n = 3 Hence

u(r θ) = 1 + 4r

minus2

cos(2θ) minus 24r

minus3

sin(3θ)

7162019 3705Test4 Sol

httpslidepdfcomreaderfull3705test4-sol 23

3 (7 points) The solution of the wave equation utt = c2uxx 0 lt x lt L t gt 0 subjectto the boundary conditions u(0 t) = 0 u(L t) = 0 and the initial conditions u(x 0) =f (x) ut(x 0) = g(x) is given by

u(x t) =

infin

sumn=1

983131an cos(

nπct

L ) + bn sin(

nπct

L )983133

sin(

nπx

L )

Find the solution of the wave equation utt = 4uxx 0 lt x lt 2 t gt 0 satisfying the condi-tions u(0 t) = u(2 t) = 0 and u(x 0) = 3 sin(2πx) ut(x 0) = 16 sin(4πx)

Solution Note that L = 2 c = 2 We have

u(x t) =infinsumn=1

[an cos(nπt) + bn sin(nπt)] sin(nπx

2 )

we imply that

u(x 0) =infinsumn=1

an sin(nπx

2 ) = 3 sin(2πx) rArr

a4 = 3 an = 0 (n = 4)

Note that

ut(x t) =infinsumn=1

[minusnπan sin(nπt) + nπbn cos(nπt)] sin(nπx

2 ) rArr

ut(x 0) =infinsumn=1

nπbn sin( nπx2

) = 3 sin(4πx) rArr

8πb8 = 16 bn = 0 (n = 8) rArr b8 = 2

π

Thus

u(x t) = 3cos(4πt) sin(2πx) + 2

π sin(8πt) sin(4πx)

4 (7 points) The bounded solution of the Laplace equation urr + 1rur +

1r2 uθθ = 0 outside

the circle r = a subject to the boundary condition u(a θ) = f (θ) is given by

u(r θ) = a0

2 +

infinsumn=1

rminusn[an cos(nθ) + bn sin(nθ)]

where f is continuous 2π-periodic f prime is piecewise continuous

Find the bounded solution of urr + 1r

ur + 1r2

uθθ = 0 outside the circle r = 2 subject tothe boundary condition u(2 θ) = 2 cos2(θ) minus 3sin(3θ)

7162019 3705Test4 Sol

httpslidepdfcomreaderfull3705test4-sol 33

Solution a = 2 The solution of the PDE above is

u(r θ) = a0

2 +

infinsumn=1

rminusn[an cos(nθ) + bn sin(nθ)]

Note that 2 cos2

(θ) = 1 + cos(2θ) Thus

u(2 θ) = 1 + cos(2θ) minus 3sin(3θ) = a0

2 +

infinsumn=1

2minusn[an cos(nθ) + bn sin(nθ)]

which implies that

a0 = 2 1 = 2minus2a2 an = 0 for n = 0 2 minus3 = 2minus3b3 bn = 0 for n = 3 rArr

a0 = 2 a2 = 4 an = 0 for n = 0 2 b3 = minus24 bn = 0 for n = 3 Hence

u(r θ) = 1 + 4r

minus2

cos(2θ) minus 24r

minus3

sin(3θ)

7162019 3705Test4 Sol

httpslidepdfcomreaderfull3705test4-sol 33

Solution a = 2 The solution of the PDE above is

u(r θ) = a0

2 +

infinsumn=1

rminusn[an cos(nθ) + bn sin(nθ)]

Note that 2 cos2

(θ) = 1 + cos(2θ) Thus

u(2 θ) = 1 + cos(2θ) minus 3sin(3θ) = a0

2 +

infinsumn=1

2minusn[an cos(nθ) + bn sin(nθ)]

which implies that

a0 = 2 1 = 2minus2a2 an = 0 for n = 0 2 minus3 = 2minus3b3 bn = 0 for n = 3 rArr

a0 = 2 a2 = 4 an = 0 for n = 0 2 b3 = minus24 bn = 0 for n = 3 Hence

u(r θ) = 1 + 4r

minus2

cos(2θ) minus 24r

minus3

sin(3θ)