Introduction and preliminaries

Chapter Introduction and preliminaries Partial differential equations What is a partial differential equation? ODEs Ordinary Differential Equations) have one variable x). PDEs Partial Differential Equations) have multiple variables x,y,...). For fx) with one variable x, we know f x) = d f dx. For ux,y), we introduce partial derivatives as u x = du dx = x u=u x. y is fixed Similarly, Example. For ux,y)=xy, we have u x = x u=u xx. u x = y, u xx = 0, u y = xy, u yy = x. Example. The equations below are PDEs. u xx u y = 0 u xx u yy = 0 u xx + u yy = 0 the heat equation) the wave equation) aplace s equation) The order of a PDE is the order of the highest-order derivative in the equation. The above examples are second order. The equation u x + u y = 0 is first order. et us write a PDE as u=g, where g is independent of u. In the case of the wave equation, = x y. The equation with = x + y and some function gx,y) is called Poisson s equation. Winter 04 Math 454 Sec Boundary Value Problems for Partial Differential Equations Manabu Machida University of Michigan) This section corresponds to 0. of the textbook.

Math 454 If we have u+v)= u+ v, cu)=c u, for any functions u,v and constant c, then is a linear differential operator and the equation is said to be linear. If g 0, then the equation is called homogeneous. The heat equation, wave equation, and aplace s equation are all linear homogeneous equations. Poisson s equation is a linear nonhomogeneous PDE. Classification of second-order PDEs In this course we will mainly consider second-order equations. In general, secondorder PDEs are written as ax,y)u xx + bx,y)u xy + cx,y)u yy + dx,y)u x + ex,y)u y + fx,y)u=gx,y), with coefficients a,b,c,d,e, f and source term g. We assume a + b + c 0 at least one of a,b,c is nonzero). These equations are classified as follows by the coefficients a, b, c. 4ac b > 0 4ac b = 0 4ac b < 0 elliptic parabolic hyperbolic et α,β,γ be constants. Note that ellipsesx/α) +y/β) = satisfy 4/α )/β ) 0 > 0, parabolas y = 4γx satisfy 4 0 0 = 0, and hyperbolas x/α) y/β) = satisfy 4/α ) /β ) 0 < 0. ) a b/ For constants a,b,c, let us consider the symmetric matrix A=. Then we have b/ c au xx + bu xy + cu yy = i= j= et λ = λ,λ be the eigenvalues of A. a λ b/ det b/ c λ where D=b 4ac. Therefore, A i j u i j = i= j= ) = 0 λ = ) u x A i j = x y )A. i j y [ ] a+c± a+c) + D, elliptic D<0 λ > 0,λ > 0 or λ < 0,λ < 0, parabolic D=0 λ = 0,λ 0 or λ 0,λ = 0, hyperbolic D>0 λ < 0,λ > 0 or λ > 0,λ < 0.

Introduction and preliminaries 3 The heat equation is parabolic and the wave equation is hyperbolic. aplace s equation and Poisson s equation are elliptic. Complex Numbers et us recall Euler s formula e iθ = cosθ + isinθ,.) where i=, i =. We can write cosθ,sinθ using exponential functions: cosθ = eiθ + e iθ Note that coshx and sinhx are defined similarly: Note also the following relations., sinθ = eiθ e iθ. i coshx= ex + e x, sinhx= ex e x. cos θ + sin θ =, cosh x sinh x=, cosiθ)=coshθ, siniθ)=isinhθ, coshix)=cosx, sinhix)=isinx. Review of ODEs ) 3 et us consider ODEs of yx). Separable Sometimes we can separate x and y in the equation. In such a case, the equation is said to be separable. Example 3. et us solve y = 6xy. We can rewrite this as dy y = 6xdx. 3 See A. of the textbook.

4 Math 454 dy We integrate both sides: y = 6x)dx. Thus we obtain ln y = 3x +C, and hence y= Ce 3x, where C =±e C ) is an arbitrary constant. inear first-order equations Consider y x)+ px)yx)=qx), yx 0 )=y 0. This type can always be solved as follows. By multiplying the integrating factor [ exp x x 0 px )dx ), we have dx d xx0 px yx)e )dx ] xx0 px = qx)e )dx. Therefore, [ y=e x x0 px )dx x qx x x px )e )dx 0 dx + y 0 ]. x 0 Example 4. x + )y + 3xy=6x, y0)=3 is solved as yx)=+x + ) 3/. Homogeneous second-order linear equations Consider y + px)y + qx)y=0. The general solution is given by a superposition linear combination) of two linearly independent solutions y,y : yx)= C y x)+ C y x). C y +C y = 0 for all x only when C = C = 0.) Constant Coefficients Consider ay + by + cy=0 a 0). By y=e rx, we obtain ar + br+ c=0. From this characteristic equation, we obtain two solutions y x) = e rx and y x) = e rx, where r,r = b± b 4ac a, and the general solution is written as yx)= C e rx +C e rx. If ar +br+c=0 has complex roots α±iβ, then the general solution can be written as yx)= C e α+iβ)x +C e α iβ)x = e αx c cosβx+c sinβx), where c = C +C, c = ic C ). If the characteristic equation has equal roots r = r = r, then two linearly independent solutions are found as y = e rx and y = xe rx. Thus the general solution is

Introduction and preliminaries 5 given by yx)= C y x)+c y x)=c +C x)e rx. Example 5. et us solve y + y + y=0, y0)=5, y 0)= 3. The characteristic equation is r + r+ = r+ ) = 0. Thus r = r =. The general solution is written as yx)=c +C x)e x. By the conditions at x = 0, we get C = 5 and C =. We obtain yx)=5+x)e x. Separation of variables 4 Many linear PDEs can be reduced to linear ODEs with the method of separation of variables, described below. We look for a separated solution this is an ansatz 5 ) ux,y)=xx)yy). Consider aplace s equation u xx + u yy = 0. We obtain X Y + XY = 0 This implies X /X and Y /Y are constants. X X + Y Y = 0. et λ be a constant and we write X + λx = 0, Y λy = 0. We call λ the separation constant. At this moment λ is arbitrary. Thus the PDE reduced to two ODEs. If λ = 0, then two ODEs have the following linearly independent solutions. X =, x, Y =, y. If λ 0, then two ODEs have the following linearly independent solutions. X = e λx, e λx, Y = e λy, e λy..) In either case, the solution is given by superpositions: 4 This section corresponds to 0. of the textbook. 5 an assumption or a guess to be verified later.

6 Math 454 A x+a )B y+b ), λ = 0, u= A λx e ) + A e λx B λy e ) + B e.3) λy, λ 0, where A,A,B,B are constants. For λ > 0, by writing λ = k k>0) we have ux,y)= A e ikx + A e ikx) B e ky + B e ky),.4) and for λ < 0, by writing λ = l l > 0) we have ux,y)= A e lx + A e lx) B e ily + B e ily)..5) Instead of.) we can also choose ) X = cos λx, sin In this case we have λx ), Y = cosh ) ) λy, sinh λy. ux,y) = A coskx)+a sinkx))b coshky)+b sinhky)),.6) ux,y) = A coshlx)+a sinhlx))b cosly)+b sinly))..7) Note that.6) becomes.4) and.7) becomes.5) by redefining the coefficients. We call solutions such as.3) through.7) separated solutions because they are given in the form ux,y)=xx)yy). The separation constant λ and coefficients A,A,B,B are partially determined by boundary conditions. Suppose that our aplace s equation is considered in the region 0<x<, 0<y< with boundary conditions u0,y)=0, u,y)=0, ux,0)=0. We find that u=a x+a )B y+b ) in.3) satisfies the boundary conditions only when u=0. We then find that.5) and.7) satisfy the conditions u0,y)=u,y)=0 only when A = A = 0. That is, only the solution u=0 satisfies the boundary conditions. Finally.4) and.6) satisfy u0,y)=u,y)=0 when A = 0 and k = nπ/, where n is an integer. Furthermore we find B = 0 by the condition ux,0)=0. That is, the solution A B sinkx)sinhky) with k=nπ/ n=0,±,±,...) satisfies the boundary conditions. Therefore we obtain the following separated solutions of aplace s equation satisfying the boundary conditions. ux,y)=asin nπx nπy sinh,,,...,

Introduction and preliminaries 7 where A is a constant. Note that we still have infinitely many solutions. In general A depends on n. To have a unique solution, we need one more condition and need to use Fourier series. Fourier series 6 et A 0,A,B,... be constants. The series below is called a trigonometric series. A 0 + A n cos nπx + B n sin nπx ). Suppose that a function fx) R, <x<, is given by a trigonometric series: fx)=a 0 + A n cos nπx + B n sin nπx )..8) Example 6. The function fx) = cosπx/), < x <, is given by the trigonometric series with A = and A 0 = A = A 3 = =B=B= =0. For a given fx), we can determine the coefficients A 0,A n,b n making use of orthogonality relations. Definition. The Kronecker delta δ mn is defined as {, m=n, δ mn = 0, m n. Theorem Orthogonality relations). et n, m be integers. We assume > 0. The following orthogonality relations hold. { cos nπx mπx n=m=0), cos dx= δ nm otherwise), { sin nπx mπx sin dx= 0 n=m=0), otherwise), sin nπx mπx cos dx=0 δ nm all n,m). Proof. If n=m=0, then 6 This section corresponds to. of the textbook.

8 Math 454 cos nπx mπx cos dx= dx=. Suppose that at least one of n,m is nonzero. We have cos nπx mπx cos dx= If n m nor n m, then cos nπx mπx cos dx = If n=±m m 0), then = 0. cos n m)πx + cos n+m)πx [ n m)π sinn m)πx cos nπx mπx cos dx= cos 0 πx dx=. Thus the orthogonality relations for onality relations for sin mπx proved. sin nπx cos nπx dx and cos mπx sin nπx ) dx + n+m)π sinn+m)πx dx is proved. The orthog- cos mπx dx are similarly To determine A 0 in.8), we multiply cos 0 πx = on both sides and integrate with respect to x: fx)dx= A 0 dx+ A n cos nπx + B n sin nπx ) dx= A 0 dx=a 0. To determine A m m =,,...) in.8), we multiply cos mπx on both sides and integrate with respect to x: fx)cos mπx dx = = A 0 cos mπx dx+ A n cos nπx A n cos nπx mπx cos dx= A n δ nm = A m. ] + B n sin nπx ) cos mπx dx Similarly we can determine B m m =,,...) in.8) by multiplying sin mπx on both sides and integrate with respect to x: fx)sin mπx dx = = A 0 sin mπx dx+ A n cos nπx B n sin nπx mπx sin dx= B n δ nm = B m. Therefore the Fourier coefficients in.8) are given by + B n sin nπx ) sin mπx dx

Introduction and preliminaries 9 A 0 = fx)dx, A n = fx)cos nπx dx, B n = fx)sin nπx dx..9) In general, for a given function fx) R, <x<, the trigonometric series with coefficients.9) is called the Fourier series of f. Example 7. et us calculate the Fourier series of fx)=x, <x<. We have A 0 = A n = B n = Therefore we obtain xdx=0, [ xcos nπx dx= [ xsin nπx dx= nπ nπ nπx xsin nπx xcos ] sin nπx nπ dx = 0, ] + cos nπx nπ dx = nπ )n+. x= nπ )n+ sin nπx, <x<. Even and odd functions A function fx) is even if f x)= fx) and is odd if f x)= fx). Example 8. x is odd and x is even. But x+x is neither. Example 9. cosx is even and sinx is odd. For constants a,b, sinax)cosbx) is odd because sin ax) cos bx) = [ sinax)][cosbx)] = sinax) cosbx). But sinax) sinbx) is even because sin ax) sin bx) =[ sinax)][ sinbx)] = sinax) sinbx). Note that fx)dx even function), fx)dx= 0 0 odd function). If fx) is even, then fx)=a 0 + A n cos nπx. If fx) is odd, then fx)= B n sin nπx.

0 Math 454 Example 0. et us compute the Fourier series of fx) = x, < x <. Since f is odd, A 0,A,... are zero. We have B n = xsin nπx dx= xsin nπx 0 dx [ = x ] nπx cos nπ + cos nπx nπ dx = Therefore, = cosnπ = nπ nπ )n+. x= π 0 ) n+ n 0 ] [ nπ cosnπ) sin nπx, <x<..0) Note that each term on the right-hand side is zero when x=±. Because of this, the above sum does not converge uniformly to x and there appear oscillations near and in Fig... This is known as the Gibbs phenomenon 7, which shows up when the function has discontinuities..0 term 0 terms 50 terms 0.5 y = x 0.0 0.5.0.0 0.5 0.0 0.5.0 x Fig.. Example 0. x= π N ) n+ n sin nπx, <x<, where =, and N =,0,50. 7 See.3 of the textbook.

Introduction and preliminaries The following formulas are useful. x k sin nπx [ dx= xk x k cos nπx dx= nπx cos nπ [ x k nπx sin nπ ] + k nπ ] k nπ x k cos nπx dx, x k sin nπx dx. Periodic functions A function fx), <x<, is -periodic if fx+)= fx) for x,). Note that sinnπx/) and cosnπx/) are -periodic: sin nπx+) cos nπx+) nπx = sin = cos Therefore any convergent trigonometric series A 0 + ) + nπ = sin nπx, nπx ) + nπ = cos nπx. A n cos nπx + B n sin nπx ) defines a -periodic function on <x<. We divide,) into intervals n )<x<n+) n=0,±,±,...) and we can focus on one of these intervals. In particular we can restrict x to < x<. Example. et us consider the Fourier series of the -periodic function fx), {, n )<x<n, fx)=, n<x<n+), where n = 0,±,... In particular fx) = for < x < 0 and fx) = for 0<x<. Thus f is an odd function. The Fourier series is given by fx)= π ) n n sin nπx, <x<.

Math 454 Fourier sine and cosine series Consider the Fourier series of fx), 0<x<. Since the orthogonality relations are given for the interval <x<, we need to extend f. There are two ways to extend the function. That is, the Fourier sine series and the Fourier cosine series. Fourier sine series The first way is to define the odd extension f O x) as fx), 0<x<, f O x)= 0, x=0, f x), <x<0. We note that f O x) is odd. The Fourier series is given by where B n = f O x)= On the interval 0<x< we have where fx)= B n sin nπx, <x<, f O x)sin nπx dx= fx)sin nπx 0 dx. B n = B n sin nπx, 0<x<,.) This series is called the Fourier sine series. 0 fx)sin nπx dx..) Example. The Fourier sine series of fx)=x, 0<x<, is obtained through the extension f O x). In this case f O x)=x and we obtain the series.0) on 0<x<. Fourier cosine series The second way is to define the even extension f E x) as fx), 0<x<, f E x)= 0, x=0, f x), <x<0.

Introduction and preliminaries 3 We note that f E x) is even. Indeed the value f E 0) is arbitrary and not necessarily zero. The Fourier series is given by where A 0 = A n = f E x)=a 0 + On the interval 0<x< we have where A 0 = fx)=a 0 + 0 f E x)dx= A n cos nπx, <x<, 0 fx)dx, f E x)cos nπx dx= 0 fx)cos nπx dx. A n cos nπx, 0<x<,.3) fx)dx, A n = 0 fx)cos nπx dx..4) This series is called the Fourier cosine series. Thus a function fx), 0 < x <, is expressed either in the Fourier sine series.),.), or the Fourier cosine series.3),.4). Figures. and. show that the convergence rates of these two series are generally different. Example 3. et us obtain the Fourier cosine series of fx) = x, 0 < x <. We extend f as x, 0<x<, f E x)= 0, x=0, x, <x<0. Indeed f E x)= x. A 0 = A n = [ = nπ f E x)dx= xdx= 0, f E x)cos nπx dx= xcos nπx 0 dx ] nπx xsin sin nπx nπ dx 0 0 = nπ) )n ). Therefore, x= + π ) n n cos nπx, 0<x<..5)

4 Math 454.0 term 0 terms 50 terms y = x 0.5 0.0 0.0 0.5.0 x Fig.. Example 3. x= + π N ) n n cos nπx, 0<x<, where =, and N =,0,50. Convergence of Fourier series 8 Definition. For a given fx), let us write where ε > 0. fx+0)= lim ε 0 fx+ε), fx 0)= lim ε 0 fx ε), Definition 3 Piecewise continuous). A function fx), a<x<b, is said to be piecewise continuous if there is a finite set of points a=x 0 < x < <x p < x p+ = b such that fx) is continuous at x x i i =,..., p), fx i + 0) i = 0,..., p) exists, and fx i 0) i=,..., p+) exists. Definition 4 Piecewise smooth). A function fx), a<x<b, is said to be piecewise smooth if fx) and all of its derivatives are piecewise continuous. Example 4. The function fx)= x, <x<, is piecewise smooth. The function fx) = x sin/x), < x <, is piecewise continuous but is not piecewise smooth because lim ε 0 f 0±ε) does not exist. The function fx)=/x ), 8 This section corresponds to. of the textbook.

Introduction and preliminaries 5 < x <, is not piecewise continuous because f +0) and f 0) are not finite. Theorem Convergence theorem). et fx), < x <, be piecewise smooth. Then the Fourier series of f converges for all x to the value [ fx+0)+ fx 0) ], where f is the -periodic extension of f. If fx) is continuous on [,] and f ) = f) in addition to the conditions assumed in the above theorem, then the Fourier series uniformly converges. For example, the Fourier series of fx)= x, <x<, see.5)) uniformly converges. Parseval s Theorem and Mean Square Error 9 Theorem 3 Parseval s theorem). et fx), < x <, be a piecewise smooth function with Fourier series A 0 + [A n cosnπx/)+b n sinnπx/)]. Then, mean square of fx)= fx) dx=a 0+ A n + B ) n. Proof. Direct calculation of the integral using the Fourier series of f. We define the mean square error σ N as where f N x)=a 0 + σn = [ fx) f N x)] dx, N By Parseval s theorem, we obtain σ N = [A n cosnπx/)+b n sinnπx/)]. n=n+ 9 This section corresponds to.4 of the textbook. A n + B ) n.

6 Math 454 Example 5. et us find σn for fx)=x, <x<. From Example 0, we have A 0 = A n = 0 and f N x)= N B n sin nπx, B n = nπ )n+. This is also the Fourier sine series of x, 0<x<, in Example. We obtain Note that N+ σ N = n=n+ x dx= N x+) dx ) nπ )n+ = π n=n+ n N+ n=n+ n. x ) dx= x dx. N We have x+) dx= N+ = N N N + N N 3 + ), N x dx= N. et us introduce the symbol O this is called big O ) to express the order. For some f N, f N = O N ) as N means that there exist a constant C > 0 and a number N 0 such that f N CN for all N > N 0. Therefore we obtain [ )] σn = π +O = O N ), N..6) N N We note that σn goes to zero as N although we know that the sum in.0) does not converge uniformly. This happened because we considered the mean square and took the integral. Note that each term on the right-hand side is zero when x=±. Because of this, the above sum to x and there appear oscillations near and in Fig... Example 6. et us find σn for fx)= x, <x<. From Example 3 we know that B n = 0, A m = 0 m=,,...), and f N x)=a 0 + N m= A m cos m )πx, A 0 =, A 4 m = π m ). This is also the Fourier cosine series of x, 0 < x <, in Example 3. Hence we obtain Note that σn = n=n+a n = m=n+ A m = 8 π 4 m=n+ m ) 4.

Introduction and preliminaries 7 N and HS = Therefore we obtain x+) 4 dx 6N+) 3 = m=n+ m ) 4 x ) 4 dx, 48N 3 + O N 4) and RHS = N 6N ) 3 = 48N 3 + O N 4). σ N = 6π 4 N 3 + O N 4) = O N 3), N..7) Thus the Fourier series of x converges as O/N) and the Fourier series of x converges as O/N 3 ). Equations.6) and.7) explain the difference between Figs.. and.. Complex form of Fourier series 0 Throughout this course, we use a bar to indicate complex conjugate. That is, if c=a+ib, then c=a ib. Suppose fx) R is given. Using Euler s formula.), we can rewrite the Fourier series of f as follows. fx)=a 0 + We define We obtain fx)=α 0 + A n cos nπx + B n sin nπx ) α 0 = A 0, α n = A n ib n = A 0 + α n e inπx/ + α n e inπx/) = An ib n, α n = A n+ ib n. n= e inπx/ + A n+ ib n e ). inπx/ α n e inπx/..8) This is the Fourier series in complex form. Using.9), α n n=0,±,±,...) are given by α n = fx)e inπx/ dx..9) We have the following orthogonality relations. e inπx/ e imπx/ dx= e in m)πx/ dx=δ mn. 0 This section corresponds to.5 of the textbook.

8 Math 454 With the help of the orthogonality relations, we can directly obtain.9) by integrating both sides of.8): ) fx)e inπx/ dx = α n e in πx/ e inπx/ dx n = = α n e in n)πx/ dx= α n δ nn = α n. n = n = We can write Parseval s theorem as follows in complex form. This is seen by the calculation below. fx) dx = A 0+ = A 0+ = α 0 + = α 0 + = fx) dx= α n. n= n= A n + B n) A n ib n α n. A n + ib n α n α n = α 0 + α n + n= α n α n + α n