Deck 14: Integral Transform Method

Question 1

The solution of the integral equation $\int _ { 0 } ^ { - \infty } f ( x ) \cos ( a x ) d x = F ( \alpha )$ is

A) $f ( x ) = \int _ { 0 } ^ { \infty } F ( \alpha ) \cos ( \alpha x ) d \alpha / \pi$
B) $f ( x ) = 2 \int _ { 0 } ^ { \infty } F ( \alpha ) \cos ( \alpha x ) d \alpha / \pi$
C) $f ( x ) = 2 \int _ { 0 } ^ { \infty } F ( \alpha ) \cos ( \alpha x ) d \alpha$
D) $f ( x ) = \int _ { 0 } ^ { \infty } F ( \alpha ) \cos ( \alpha x ) d \alpha$
E) none of the above

Answer 1

$f ( x ) = 2 \int _ { 0 } ^ { \infty } F ( \alpha ) \cos ( \alpha x ) d \alpha / \pi$

Question 2

In the three previous problems, the solution for $u ( x , t )$ is

A) $u = \sin ( \pi ) u ( t - x / a )$
B) $u = \sin ( \pi ( t - x / a ) ) u ( t - x / a )$
C) $u = \sin ( \pi x ) \cosh ( \pi t )$
D) $u = \cos ( \pi x ) \sinh ( \pi t )$
E) $u = \cos ( \pi ( t - x / a ) ) u ( t - x / a )$

Answer 2

$u = \sin ( \pi ( t - x / a ) ) u ( t - x / a )$

Question 3

Let $f ( x ) = \left\{ \begin{array} { c c c } 0 & \text { if } & x < 0 \\2 & \text { if } & 0 \leq x \leq 3 \\0 & \text { if } & x > 3\end{array} \right\}$ . The Fourier integral representation of f is

A) $f ( x ) = \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha / \pi$
B) $f ( x ) = 2 \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha$
C) $f ( x ) = 2 \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha / \pi$
D) $$f ( x ) = \int _ { 0 } ^ { \infty } \{ [ \sin ( a x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha$$
E) none of the above

Answer 3

$f ( x ) = 2 \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha / \pi$

Question 4

In the previous problem, the solution for $U ( \alpha , t )$ is

A) $e ^ { k a t } / \left( 1 + \alpha ^ { 2 } \right)$
B) $e ^ { k \alpha ^ { 2 } t } / \left( 1 + \alpha ^ { 2 } \right)$
C) $2 e ^ { k c s } / \left( 1 + \alpha ^ { 2 } \right)$
D) $2 e ^ { - k c e } / \left( 1 + \alpha ^ { 2 } \right)$
E) $2 e ^ { - k \alpha ^ { 2 } t } / \left( 1 + \alpha ^ { 2 } \right)$

Question 5

In the previous problem, the integral converges at $x = 0$ to the value

A) 0
B) $1 / 2$
C) 1
D) 2
E) $e$

Question 6

The value of $\mathcal { L } \left\{ \mathrm { e } ^ { t } \operatorname { erf } ( \sqrt { t } ) \right\}$ is

A) $1 / ( \sqrt { s } ( \sqrt { s } + 1 ) )$
B) $1 / ( \sqrt { s } ( \sqrt { s } - 1 ) )$
C) $1 / ( s ( \sqrt { s } + 1 ) )$
D) $1 / ( s ( \sqrt { s } - 1 ) )$
E) none of the above

Question 7

The value of $\mathcal { L } \{ \operatorname { erfc } ( \sqrt { t } ) \}$ is

A) $( 1 + 1 / \sqrt { s + 1 } ) / s$
B) $( 1 + 1 / \sqrt { s - 1 } ) / s$
C) $( 1 - 1 / \sqrt { s + 1 } ) / s$
D) $( 1 - 1 / \sqrt { s - 1 } ) / s$
E) none of the above

Question 8

In the previous two problems, the integral converges for $x < 0$ to the function

A) $e ^ { - \mathrm { x } }$
B) $e ^ { x }$
C) $x$
D) 1
E) none of the above

Question 9

In the previous three problems, the solution for $u ( x , t )$ is

A) $\int _ { - \infty } ^ { \infty } \left[ e ^ { - i \alpha x } e ^ { k at } / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / ( 2 \pi )$
B) $\int _ { - \infty } ^ { \infty } \left[ e ^ { - i \alpha x } e ^ { - k \alpha ^ { 2 } t } / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / \pi$
C) $\int _ { - \infty } ^ { \infty } \left[ e ^ { - i \alpha \alpha } e ^ { k \alpha t } / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / \pi$
D) $\int _ { - \infty } ^ { \infty } \left[ e ^ { - i \alpha x } e ^ { - k \alpha t } / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / \pi$
E) $\int _ { - \infty } ^ { \infty } \left[ e ^ { - i \alpha \alpha } e ^ { k \alpha ^ { 2 } t } / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / ( 2 \pi )$

Question 10

The complementary error function is defined as

A) $\operatorname { erfc } ( x ) = \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u$
B) $\operatorname { erfc } ( x ) = \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u$
C) $\operatorname { erfc } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \pi$
D) $\operatorname { erfc } ( x ) = 2 \int _ { x } ^ { \infty } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$
E) $\operatorname { erfc } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$

Question 11

In the previous problem, assume that $f ( x ) = \sin ( \pi t )$ . The solution for $U ( x , s )$ is

A) $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } e ^ { s x / \alpha }$
B) $$U = \frac { s } { s ^ { 2 } - \pi ^ { 2 } } e ^ { - s x / \alpha }$$
C) $U = \frac { \pi } { s ^ { 2 } - \pi ^ { 2 } } e ^ { s x / \alpha }$
D) $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } e ^ { - s x / \alpha }$
E) $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sinh ( - s x / \alpha )$

Question 12

Apply a Fourier transform in $x$ in the previous problem. The resulting equation for $U ( \alpha , t ) = \mathcal { F } \{ u ( x , t ) \}$ is

A) $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$
B) $k \alpha U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$
C) $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$
D) $- k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$
E) $- k \alpha U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$

Question 13

The Fourier cosine integral of a function f defined on $[ 0 , \infty ]$ is

A) $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha / \pi$
B) $\int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha / \pi$
C) $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha$
D) $\int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d \alpha \right] \cos ( \alpha x ) d x$
E) $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d \alpha \right] \cos ( \alpha x ) d x / \pi$

Question 14

In the previous problem, the integral converges at $x = 0$ to the value

A) 0
B) 1
C) 2
D) 3
E) 4

Question 15

In the previous problem, if $F ( \alpha ) = e ^ { - \alpha } \text {, then } f ( x )$ is

A) $x$
B) $1$
C) $1 / \left( \pi \left( 1 + x ^ { 2 } \right) \right)$
D) $2 / \left( \pi \left( 1 + x ^ { 2 } \right) \right)$
E) 0

Question 16

Consider the temperature, $u ( x , t )$ , in an infinite rod $( - \infty < x < \infty )$ , with an initial temperature of $f ( x ) = e ^ { - | x | }$ . The mathematical model for this is

A) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( x , 0 ) = e ^ { - | x | }$
B) $k \frac { \partial u } { \partial x } = \frac { \partial u } { \partial t } , u ( x , 0 ) = e ^ { - | x | }$
C) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( x , 0 ) = e ^ { - | x | }$
D) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( x , 0 ) = e ^ { - | x | }$
E) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( x , 0 ) = e ^ { - | x | }$

Question 17

In the previous problem, apply the Laplace transform. The resulting equation for $U ( x , s ) = \mathcal { L } \{ u ( x , t ) \}$ is

A) $a ^ { 2 } \frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } + s ^ { 2 } U = - s f ( x )$
B) $a ^ { 2 } \frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } + s ^ { 2 } U = 0$
C) $a ^ { 2 } \frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = - f ( x )$
D) $a ^ { 2 } \frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = s f ( x )$
E) $a ^ { 2 } \frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = 0$

Question 18

Consider a semi-infinite, elastic, vibrating string, with zero initial position and velocity, driven by a vertical force at $x = 0$ , so that $u ( 0 , t ) = f ( t )$ . Assume that $\lim _ { x \rightarrow \infty } u ( x , t ) = 0$ . The mathematical model for the deflection, $u ( x , t )$ , is

A) $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = 0$
B) $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = f ( t ) , u _ { t } ( x , 0 ) = 0$
C) $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = 0$
D) $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = 0$
E) $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = f ( t ) , u _ { t } ( x , 0 ) = 0$

Question 19

The value of $\lim _ { x \rightarrow \infty } \operatorname { erfc } ( x )$ is

A) 0
B) 1
C) $\pi / 2$
D) $\sqrt { \pi / 2 }$
E) $\sqrt { \pi } / 2$

Question 20

The Fourier cosine integral of $f ( x ) = e ^ { - x } , x \geq 0$ is

A) $f ( x ) = 2 \int _ { 0 } ^ { \infty } [ \alpha \cos ( \alpha x ) / ( 1 + \alpha ) ] d \alpha$
B) $f ( x ) = \int _ { 0 } ^ { \infty } \left[ \alpha \cos ( \alpha x ) / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / \pi$
C) $f ( x ) = \int _ { 0 } ^ { \infty } \left[ \alpha \cos ( \alpha x ) / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha$
D) $f ( x ) = 2 \int _ { 0 } ^ { \infty } \left[ \cos ( \alpha x ) / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / \pi$
E) $f ( x ) = 2 \int _ { 0 } ^ { \infty } [ \cos ( \alpha x ) / ( 1 + \alpha ) ] d \alpha / \pi$

Question 21

In the previous problem, assume that $f ( x ) = \sin ( \pi x )$ . The solution for $U ( x , s )$ is

A) $U = \sinh ( s x ) + \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$
B) $U = \cosh ( s x ) + \frac { s } { s ^ { 2 } - \pi ^ { 2 } } \sin ( \pi x )$
C) $U = \frac { 1 } { s ^ { 2 } - \pi ^ { 2 } } \sin ( \pi x )$
D) $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$
E) $U = \frac { 1 } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$

Question 22

The value of $\mathcal { L } \{ \operatorname { erf } ( \sqrt { t } ) \}$ is

A) $1 / ( s \sqrt { s + 1 } )$
B) $1 / ( s \sqrt { s - 1 } )$
C) $1 / ( ( s + 1 ) \sqrt { s } )$
D) $1 / ( ( s - 1 ) \sqrt { s } )$
E) none of the above

Question 23

The error function is defined as

A) $\operatorname { erf } ( x ) = \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u$
B) $\operatorname { erf } ( x ) = \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u$
C) $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \pi$
D) $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$
E) $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$

Question 24

Consider the problem of finding the temperature in a semi-infinite rod with zero initial temperature and a fixed constant temperature, $u _ { 0 }$ , at $x = 0$ . The mathematical model for this problem is

A) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( x , 0 ) = 0 , u ( 0 , t ) = u _ { 0 }$
B) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( x , 0 ) = 0 , u ( 0 , t ) = u _ { 0 }$
C) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( x , 0 ) = 0 , u ( 0 , t ) = u _ { 0 }$
D) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( x , 0 ) = u _ { 0 } , u ( 0 , t ) = 0$
E) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( x , 0 ) = u _ { 0 } , u ( 0 , t ) = 0$

Question 25

The value of $\mathcal { L } \left\{ \mathrm { e } ^ { t } \operatorname { erf } ( \sqrt { t } ) \right\}$ is

A) $1 / ( \sqrt { s } ( s + 1 ) )$
B) $1 / ( \sqrt { s } ( s - 1 ) )$
C) $1 / ( s ( s + 1 ) )$
D) $1 / ( s ( \sqrt { s } - 1 ) )$
E) $1 / ( s ( \sqrt { s } + 1 ) )$

Question 26

The value of $\lim _ { x \rightarrow \infty } \operatorname { erf } ( x )$ is

A) 0
B) 1
C) $\pi / 2$
D) $\sqrt { \pi / 2 }$
E) $\sqrt { \pi } / 2$

Question 27

The Laplace transform of a function f is

A) $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { t } f ( t ) d t$
B) $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { - t } f ( t ) d t$
C) $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { s t } f ( t ) d t$
D) $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { - s t } f ( t ) d t$
E) $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { s } e ^ { s t } f ( t ) d t$

Question 28

The Fourier integral representation of a function f is given by

A) $\int _ { - \infty } ^ { \infty } f ( x ) \cos ( \alpha x ) d x$
B) $\int _ { - \infty } ^ { \infty } f ( x ) \sin ( \alpha x ) d x$
C) $\int _ { 0 } ^ { \infty } \left[ \int _ { - \infty } ^ { \infty } f ( t ) \cos ( \alpha t ) d t \cos ( \alpha x ) + \int _ { - \infty } ^ { \infty } f ( t ) \sin ( \alpha t ) d t \sin ( \alpha x ) \right] d \alpha$
D) $\int _ { 0 } ^ { \infty } \left[ \int _ { - \infty } ^ { \infty } f ( t ) \cos ( \alpha t ) d t \cos ( \alpha x ) + \int _ { - \infty } ^ { \infty } f ( t ) \sin ( \alpha t ) d t \sin ( \alpha x ) \right] d \alpha / \pi$
E) $\int _ { 0 } ^ { \infty } \left[ \int _ { - \infty } ^ { \infty } f ( t ) \cos ( \alpha t ) d t \cos ( \alpha x ) + \int _ { - \infty } ^ { \infty } f ( t ) \sin ( \alpha t ) d t \sin ( \alpha x ) \right] d \alpha / ( 2 \pi )$

Question 29

In the three previous problems, the solution for the temperature $u ( x , t )$ is

A) $u = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha$
B) $u = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$
C) $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha$
D) $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$
E) $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$

Question 30

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} , \text { then } \left\{ \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } \right\}$ is

A) $U _ { s } ( x , s )$
B) $s U ( x , s )$
C) $s ^ { 2 } U ( x , s ) - u ( x , 0 )$
D) $s ^ { 2 } U ( x , s ) - u ( x , 0 ) - s u _ { t } ( x , 0 )$
E) $s ^ { 2 } U ( x , s ) - s u ( x , 0 ) - u _ { t } ( x , 0 )$

Question 31

In the previous problem, apply the Laplace transform. The resulting equation for $U ( x , s ) = \mathcal { L } \{ u ( x , t ) \}$ is

A) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } + s ^ { 2 } U = - s f ( x )$
B) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } + s ^ { 2 } U = f ( x )$
C) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = - f ( x )$
D) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = s f ( x )$
E) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = - s f ( x )$

Question 32

Suppose $\mathcal { F } \{ f ( x ) \} = F ( \alpha ) , \mathcal { F } \{ g ( x ) \} = G ( \alpha )$ . In the convolution theorem, the formula for the Fourier transform is Select all that apply.

A) $\int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}$
B) $\int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}$
C) $\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )$
D) $\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )$
E) none of the above

Question 33

The complex form of the Fourier integral of a function f is

A) $\int _ { - \infty } ^ { \infty } f ( t ) e ^ { i \alpha t} d t$
B) $\int _ { - \infty } ^ { \infty } \left( \int _ { - \infty } ^ { \infty } f ( t ) e ^ { i \alpha t } d t \right) e ^ { - i \alpha \alpha } d \alpha / ( 2 \pi )$
C) $\int _ { - \infty } ^ { \infty } f ( t ) e ^ { - i \alpha t } d t$
D) $\int _ { - \infty } ^ { \infty } \left( \int _ { - \infty } ^ { \infty } f ( t ) e ^ { - i \alpha t } d t \right) e ^ { i \alpha x } d \alpha / \pi$
E) $\int _ { - \infty } ^ { \infty } \left( \int _ { - \infty } ^ { \infty } f ( t ) e ^ { \alpha t } d t \right) e ^ { - i \alpha x } d \alpha / \pi$

Question 34

In the previous problem, the integral representation converges at $x = 1$ to the value

A) 0
B) 1
C) $1 / 2$
D) -1
E) none of the above

Question 35

In the three previous problems, the solution for $u ( x , t )$ is

A) $u = \sin ( \pi x ) \sin ( \pi t )$
B) $u = \sin ( \pi x ) \cos ( \pi t )$
C) $u = \sin ( \pi x ) \cosh ( \pi t )$
D) $u = \cos ( \pi x ) \sinh ( \pi t )$
E) $u = \cos ( \pi x ) \cos ( \pi i )$

Question 36

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} \text {, then } \mathcal { L } \left\{ \frac { \partial u } { \partial x } \right\}$ is

A) $s U ( x , s ) - u ( x , 0 )$
B) $s U ( x , s ) - u _ { x } ( x , 0 )$
C) $s U ( x , s ) - s u ( x , 0 )$
D) $U _ { x } ( x , s )$
E) $$U _ { s } ( x , s )$$

Question 37

Let $f ( x ) = \left\{ \begin{array} { c c c } 0 & \text { if } & x < - 1 \\1 & \text { if } & - 1 < x < 1 \\0 & \text { if } & x > 1\end{array} \right\}$ . The Fourier integral representation of f is $f ( x ) = \int _ { 0 } ^ { \infty } [ A ( \alpha ) \cos ( \alpha x ) + B ( \alpha ) \sin ( \alpha x ) ] d \alpha / \pi$ , where

A) $A ( \alpha ) = \sin \alpha / \alpha \text { and } B ( \alpha ) = \cos \alpha / \alpha$
B) $$A ( \alpha ) = 2 \cos \alpha / \alpha \text { and } B ( \alpha ) = 2 \sin \alpha / \alpha$$
C) $A ( \alpha ) = 2 \sin \alpha / \alpha \text { and } B ( \alpha ) = 2 \cos \alpha / \alpha$
D) $A ( \alpha ) = 0 \text { and } B ( \alpha ) = 2 \sin \alpha / \alpha$
E) $A ( \alpha ) = 2 \sin \alpha / \alpha \text { and } B ( \alpha ) = 0$

Question 38

Consider the problem of a vibrating string, tightly-stretched between $x = 0$ and $x = 1$ , with a fixed initial position, $f ( x )$ , and zero initial velocity. The mathematical problem for the deflection, $u ( x , t )$ , is $( \text { with } u ( 0 , t ) = 0 , u ( 1 , t ) = 0 )$

A) $\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( x , 0 ) = f ( x ) , u _ { t } ( x , 0 ) = 0$
B) $\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( x , 0 ) = f ( x ) , u _ { t } ( x , 0 ) = 0$
C) $\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( x , 0 ) = f ( x ) , u _ { t } ( x , 0 ) = 0$
D) $\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( x , 0 ) = f ( x ) , u _ { t } ( x , 0 ) = 0$
E) $\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = f ( x )$

Question 39

In the previous problem, apply a Fourier sine transform in x. The new problem for the transform $U ( a , t )$ is

A) $k \frac { \partial U } { \partial t } + \alpha ^ { 2 } U = k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0$
B) $k \frac { \partial U } { \partial t } - \alpha ^ { 2 } U = k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0$
C) $\frac { \partial U } { \partial t } + k \alpha ^ { 2 } U = k \alpha \psi _ { 0 } , U ( \alpha , 0 ) = 0$
D) $\frac { \partial U } { \partial t } - k \alpha ^ { 2 } U = k \alpha \psi _ { 0 } , U ( \alpha , 0 ) = 0$
E) $\frac { \partial U } { \partial t } + \alpha ^ { 2 } U = k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0$

Question 40

In the previous problem, the solution is

A) $U = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right)$
B) $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right)$
C) $U = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right) / \alpha$
D) $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha$
E) $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha ^ { 2 }$