Linear algebra and differential equations (Math 54): Lecture 25 - PowerPoint PPT Presentation

Linear algebra and differential equations (Math 54): Lecture 25 Vivek Shende April 30, 2019

Hello, and welcome to class!

Hello, and welcome to class! Last time

Hello, and welcome to class! Last time We introduced the notion of Fourier series, and discussed how to expand a function into one.

Hello, and welcome to class! Last time We introduced the notion of Fourier series, and discussed how to expand a function into one.

Hello, and welcome to class! Last time We introduced the notion of Fourier series, and discussed how to expand a function into one. This time

Hello, and welcome to class! Last time We introduced the notion of Fourier series, and discussed how to expand a function into one. This time Having developed this tool, we return to studying the heat equation.

Heat equation review

Heat equation review The heat equation in one variable is: ∂ t u ( x , t ) = β ∂ 2 ∂ ∂ x 2 u ( x , t )

Heat equation review The heat equation in one variable is: ∂ t u ( x , t ) = β ∂ 2 ∂ ∂ x 2 u ( x , t ) We saw that some solutions are given by u ( x , t ) = e λ t ( A λ e x √ λ/β + B λ e − x √ λ/β ) λ > 0 u ( x , t ) = A λ + B λ x λ = 0 u ( x , t ) = e λ t ( A λ cos( x � � − λ/β )+ B λ sin( x − λ/β )) λ < 0

Heat equation review Last week, we considered a wire of length L ,

Heat equation review Last week, we considered a wire of length L , whose endpoints were kept at temperature zero.

Heat equation review Last week, we considered a wire of length L , whose endpoints were kept at temperature zero. In other words, we imposed boundary conditions u (0 , t ) = 0 = u ( L , t )

Heat equation review Of our above solutions, the only ones which take this form are the e λ t sin( x � � − λ/β ) when − λ/β = N π/ L

Heat equation review Of our above solutions, the only ones which take this form are the e λ t sin( x � � − λ/β ) when − λ/β = N π/ L This led to a general solution of the form ∞ � N π � 2 t sin c N e − β ( N π L ) � u ( x , t ) = L x N =1

Initial conditions and Fourier expansion Finally suppose we are given the initial temperature in the form of some function u ( x , 0).

Initial conditions and Fourier expansion Finally suppose we are given the initial temperature in the form of some function u ( x , 0). Our job now is to express ∞ � N π � � u ( x , 0) = c N sin L x N =1

Initial conditions and Fourier expansion Finally suppose we are given the initial temperature in the form of some function u ( x , 0). Our job now is to express ∞ � N π � � u ( x , 0) = c N sin L x N =1 In other words, to find values c N making the above formula true.

Initial conditions and Fourier expansion Finally suppose we are given the initial temperature in the form of some function u ( x , 0). Our job now is to express ∞ � N π � � u ( x , 0) = c N sin L x N =1 In other words, to find values c N making the above formula true. Because then the solution will be given by ∞ � N π � 2 t sin c N e − β ( N π L ) � u ( x , t ) = L x N =1

Initial conditions and Fourier expansion The expression ∞ � N π � � u ( x , 0) = c N sin L x N =1 looks much like a Fourier expansion.

Initial conditions and Fourier expansion The expression ∞ � N π � � u ( x , 0) = c N sin L x N =1 looks much like a Fourier expansion. Two differences from last time:

Initial conditions and Fourier expansion The expression ∞ � N π � � u ( x , 0) = c N sin L x N =1 looks much like a Fourier expansion. Two differences from last time: first, the function is only defined on the interval [0 , L ],

Initial conditions and Fourier expansion The expression ∞ � N π � � u ( x , 0) = c N sin L x N =1 looks much like a Fourier expansion. Two differences from last time: first, the function is only defined on the interval [0 , L ], and second, we want to expand it only in sin rather than in sin and cos.

Fourier series review The Fourier series of a function f ( x ) defined on [ − L , L ] is ∞ a n cos n π x + b n sin n π x a 0 � 2 + L L n =1 � L a n = 1 f ( x ) cos n π x L dx L − L � L b n = 1 f ( x ) sin n π x L dx L − L

Initial conditions and Fourier expansion To absorb these differences, we extend the function f to [ − L , L ] simply by defining f ( − x ) = − f ( x ).

Initial conditions and Fourier expansion To absorb these differences, we extend the function f to [ − L , L ] simply by defining f ( − x ) = − f ( x ). This has the virtue of ensuring that the extension is an odd function,

Initial conditions and Fourier expansion To absorb these differences, we extend the function f to [ − L , L ] simply by defining f ( − x ) = − f ( x ). This has the virtue of ensuring that the extension is an odd function, which therefore has a Fourier expansion consisting only of sin n π x waves, L

Initial conditions and Fourier expansion To absorb these differences, we extend the function f to [ − L , L ] simply by defining f ( − x ) = − f ( x ). This has the virtue of ensuring that the extension is an odd function, which therefore has a Fourier expansion consisting only of sin n π x waves, exactly as we wanted. L

Initial conditions and Fourier expansion To absorb these differences, we extend the function f to [ − L , L ] simply by defining f ( − x ) = − f ( x ). This has the virtue of ensuring that the extension is an odd function, which therefore has a Fourier expansion consisting only of sin n π x waves, exactly as we wanted. L In interpreting the answer, we just ignore the value of the function on [ − L , 0].

Example Consider a wire of length L and diffusivity β in which the initial temperature is described by the function � 0 ≤ x ≤ L / 2 x u ( x ) = L − x L / 2 ≤ x ≤ L and in which the temperature at the endpoints is kept at zero.

Example Consider a wire of length L and diffusivity β in which the initial temperature is described by the function � 0 ≤ x ≤ L / 2 x u ( x ) = L − x L / 2 ≤ x ≤ L and in which the temperature at the endpoints is kept at zero. Let us determine the temperature in the wire as a function of time.

Example Consider a wire of length L and diffusivity β in which the initial temperature is described by the function � 0 ≤ x ≤ L / 2 x u ( x ) = L − x L / 2 ≤ x ≤ L and in which the temperature at the endpoints is kept at zero. Let us determine the temperature in the wire as a function of time. First, we should expand out u ( x ) into its sin Fourier series.

Example That is, we want to compute � L 2 � n π x � b n = f ( x ) sin dx L L 0 �� L / 2 � L � 2 � n π x � n π x � � = x sin dx + ( L − x ) sin dx L L L L / 2 0 � L � 2 �� n π/ 2 � π � 2 = u sin u du + ( n π − u ) sin u du n π L 0 n π/ 2

Example That is, we want to compute � L 2 � n π x � b n = f ( x ) sin dx L L 0 �� L / 2 � L � 2 � n π x � n π x � � = x sin dx + ( L − x ) sin dx L L L L / 2 0 � L � 2 �� n π/ 2 � π � 2 = u sin u du + ( n π − u ) sin u du n π L 0 n π/ 2 � Noting u sin udu = sin u − u cos u ,

Example That is, we want to compute � L 2 � n π x � b n = f ( x ) sin dx L L 0 �� L / 2 � L � 2 � n π x � n π x � � = x sin dx + ( L − x ) sin dx L L L L / 2 0 � L � 2 �� n π/ 2 � π � 2 = u sin u du + ( n π − u ) sin u du n π L 0 n π/ 2 � Noting u sin udu = sin u − u cos u , this is � L � 2 �� � n π/ 2 n π n π � � � 2 � � � � � sin u − u cos u sin u − u cos u n π cos u − − � � � L n π � � � n π/ 2 n π/ 2 0

Example � L � 2 �� n π/ 2 n π n π � � � � 2 � � � � � sin u − u cos u sin u − u cos u n π cos u − − � � � n π L � � � 0 n π/ 2 n π/ 2

Example � L � 2 �� n π/ 2 n π n π � � � � 2 � � � � � sin u − u cos u sin u − u cos u n π cos u − − � � � n π L � � � 0 n π/ 2 n π/ 2 � L � 2 � = 2 2 sin n π ( n π ) 2 sin n π 4 L � = L n π 2 2

Example � L � 2 �� n π/ 2 n π n π � � � � 2 � � � � � sin u − u cos u sin u − u cos u n π cos u − − � � � n π L � � � 0 n π/ 2 n π/ 2 � L � 2 � = 2 2 sin n π ( n π ) 2 sin n π 4 L � = L n π 2 2 Thus the Fourier expansion of the original function u ( x ) is � � u ( x ) = 4 L sin( x ) − 1 9 sin 3 x + 1 25 sin 5 x − 1 49 sin 7 x + · · · π 2

Example And finally, having written the initial condition as u ( x ) = 4 L � sin( x ) − 1 9 sin 3 x + 1 25 sin 5 x − 1 � 49 sin 7 x + · · · π 2

Example And finally, having written the initial condition as u ( x ) = 4 L � sin( x ) − 1 9 sin 3 x + 1 25 sin 5 x − 1 � 49 sin 7 x + · · · π 2 we see that the time evolution is given by u ( x , t ) = 4 L � 2 t sin( x ) − 1 2 t sin 3 x + 1 � 2 t sin 5 x − · · · e − β ( π L ) 9 e − β ( 3 π L ) 25 e − β ( π 5 L ) π 2