DOC PREVIEW
TAMU MATH 412 - Lecture 4-6

This preview shows page 1-2-3-25-26-27 out of 27 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 27 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 27 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 27 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 27 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 27 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 27 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 27 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Math 412-501Theory of Partial Differential EquationsLecture 4-6:Review for the final exam.Math 412-501 Fall 2006Sample problems for the final examAny problem may be altered or replaced by a different one!Some possibly useful information• Parseval’s equality for the complex form of the Fourier serieson (−π, π):f (x) =∞Xn=−∞cneinx=⇒Zπ−π|f (x)|2dx = 2π∞Xn=−∞|cn|2.• Fourier sine and cosine transforms of the second derivative:S[f′′](ω) =2πf (0) ω − ω2S[f ](ω),C [f′′](ω) = −2πf′(0) − ω2C [f ](ω).• Laplace’s operator in polar coordinates r , θ:∇2u =∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2.• Any nonzero solution of a regular Sturm-Liouville equation(pφ′)′+ qφ + λσφ = 0 (a < x < b)satisfies the Rayleigh quotient relationλ =−pφφ′ba+Zbap(φ′)2− qφ2dxZbaφ2σ dx.• Some table integrals:Zx2eiaxdx =x2ia+2xa2−2ia3eiax+ C , a 6= 0;Z∞−∞e−αx2ei βxdx =rπαe−β2/(4α), α > 0, β ∈ R;Z∞−∞e−α|x |ei βxdx =2αα2+ β2, α > 0, β ∈ R.Problem 1 Let f (x) = x2.(i) Find the Fourier series (complex form) of f (x) on the interval(−π, π).(ii) Rewrite the Fourier series of f (x) in the real form.(iii) Sketch the function to which the Fourier series converges.(iv) Use Parseval’s equality to evaluateP∞n=1n−4.Problem 1 Let f (x) = x2.(i) Find the Fourier series (complex form) of f (x) on the interval(−π, π).The required series is∞Xn=−∞cneinx, wherecn=12πZπ−πf (x)e−inxdx.In particular,c0=12πZπ−πf (x) dx =12πZπ−πx2dx =12πx33π−π=12π2π33=π23.If n 6= 0 then we have to integrate by parts twice.To save time, we could instead use the table integralZx2eiaxdx =x2ia+2xa2−2ia3eiax+ C , a 6= 0.According to this integral,cn=12πZπ−πx2e−inxdx =12π−x2in+2xn2+2in3e−inxπ−π=12π2π(e−inπ+ einπ)n2=2(−1)nn2.Thusx2∼π23+X−∞<n<∞n6=02(−1)nn2einx.(ii) Rewrite the Fourier series of f (x) in the real form.π23+X−∞<n<∞n6=02(−1)nn2einx=π23+∞Xn=12(−1)nn2(einx+ e−inx)=π23+∞Xn=14(−1)nn2cos nx.Thusx2∼π23+∞Xn=14(−1)nn2cos nx.(iii) Sketch the function to which the Fourier series converges.The series converges to the 2π-periodic function that coincideswith f (x) for −π ≤ x ≤ π.The sum is continuous and piecewise smooth hence theconvergence is uniform.The derivative of the sum has jump discontinuities at pointsπ + 2kπ, k ∈ Z.The graph is a scalloped curve.(iv) Use Parseval’s equality to evaluateP∞n=1n−4.In our case, Parseval’s equality can be written ashf , f i =∞Xn=−∞|hf , φni|2hφn, φni,where φn(x) = einxandhg, hi =Zπ−πg(x)h(x) dx.Since cn=hf ,φnihφn,φniand hφn, φni = 2π for all n ∈ Z, it can bereduced to an equivalent formZπ−π|f (x)|2dx = 2π∞Xn=−∞|cn|2.NowZπ−π|f (x)|2dx =Zπ−πx4dx =x55π−π=2π55,∞Xn=−∞|cn|2=π49+ 2∞Xn=14n4.Therefore12π2π55=π49+ 2∞Xn=14n4.It follows that∞Xn=11n4=18π45−π49=π490.Problem 2 Solve Laplace’s equation in a disk,∇2u = 0 (0 ≤ r < a), u(a, θ ) = f (θ).Laplace’s operator in polar coordinates r , θ:∇2u =∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2.We search for the solution of the boundary value problem as asuperposition of solutions u(r , θ) = h(r )φ(θ) wit h separatedvariables.Solutions with separated variables satisfy periodic boundaryconditionsu(r, −π) = u(r, π),∂u∂θ(r, −π) =∂u∂θ(r, π)and the singular boundary condition|u(0, θ)| < ∞.Separation of variables provides the following solutions:u0= 1, un(r, θ) = rncos nθ, ˜un(r, θ) = rnsin nθ, n = 1, 2, . . .A superposition of these solutions is a seriesu(r, θ) = α0+X∞n=1rn(αncos nθ + βnsin nθ),where α0, α1, . . . and β1, β2, . . . are constants. Substituting theseries into the boundary condition u(a, θ) = f (θ), we getf (θ) = α0+X∞n=1an(αncos nθ + βnsin nθ).The boundary condition is satisfied if the right-hand side coincideswith the Fourier seriesA0+X∞n=1(Ancos nθ + Bnsin nθ)of the function f (θ) on (−π, π).Henceα0= A0, αn= a−nAn, βn= a−nBn, n = 1, 2, . . .andu(r, θ) = A0+X∞n=1ran(Ancos nθ + Bnsin nθ),whereA0=12πZπ−πf (θ) dθ, An=1πZπ−πf (θ) cos nθ dθ,Bn=1πZπ−πf (θ) sin nθ dθ, n = 1, 2, . . .Bonus Problem 7 Find a Green function implementing thesolution of Problem 2.The solution of Problem 2:u(r, θ) = A0+∞Xn=1ran(Ancos nθ + Bnsin nθ),whereA0=12πZπ−πf (θ0) dθ0, An=1πZπ−πf (θ0) cos nθ0dθ0,Bn=1πZπ−πf (θ0) sin nθ0dθ0, n = 1, 2, . . .It can be rewritten asu(r, θ) =Zπ−πG (r , θ; θ0) f (θ0) dθ0,whereG (r , θ; θ0) =12π+1π∞Xn=1ran(cos nθ cos nθ0+ sin nθ si n nθ0).This is the desired Green function. The expression can besimplified:G (r , θ; θ0) =12π+1π∞Xn=1ran(cos nθ cos nθ0+ sin nθ si n nθ0)=12π+1π∞Xn=1rancos n(θ − θ0)=12π+1π∞Xn=1ran·ein(θ−θ0)+ e−in(θ−θ0)2=12π∞Xn=0ra−1ei (θ−θ0)n+12π∞Xn=1ra−1e−i (θ−θ0)nG (r , θ; θ0) =12π 11 − ra−1ei (θ−θ0)+ra−1e−i (θ−θ0)1 − ra−1e−i (θ−θ0)!=12π aa − rei (θ−θ0)+re−i (θ−θ0)a − re−i (θ−θ0)!=12πa2− r2(a − rei (θ−θ0))(a − re−i (θ−θ0))=12πa2− r2a2− 2ar cos(θ −θ0) + r2.(Poisson’s formula)Problem 3 Find Green’s function for the boundary value problemd2udx2− u = f (x) (0 < x < 1), u′(0) = u′(1) = 0.The Green function G (x, x0) should satisfy∂2G∂x2− G = δ(x −x0),∂G∂x(0, x0) =∂G∂x(1, x0) = 0.It follows thatG (x, x0) =(aex+ be−xfor x < x0,cex+ de−xfor x > x0,where constants a, b, c, d may depend on x0. Then∂G∂x(x, x0) =(aex− be−xfor x < x0,cex− de−xfor x > x0.The boundary conditions imply that a = b and ce = de−1.Gluing conditions at x = x0are continuity of the function andjump discontinuity of the first derivative:G (x, x0)x =x0−= G (x, x0)x =x0+,∂G∂xx =x0+−∂G∂xx =x0−= 1.The two conditions imply thataex0+be−x0= cex0+de−x0, cex0−de−x0−(aex0−be−x0) = 1.Now we have 4 equations to determine 4 quantities a, b, c, d.Solution:c =ex0+ e−x02(1 − e2), a =ex0+ e2−x02(1 − e2),d =ex0+ e−x02(e−2− 1), b =ex0+ e2−x02(1 − e2).Finally,G (x, x0) =(ex0+ e2−x0)(ex+ e−x)2(1 − e2)for x < x0,(ex0+ e−x0)(ex+ e2−x)2(1 − e2)for x > x0.Observe that G (x, x0)


View Full Document

TAMU MATH 412 - Lecture 4-6

Download Lecture 4-6
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture 4-6 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture 4-6 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?