This article covers solutions and explanations for questions related to Heaviside step function, Fourier transform, inner product, and Sturm-Liouville problem in advanced mathematics.
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INSTITUTIONAL AFFILIATION FACULTY OR DEPARTMENT COURSE ID & NAME TITLE: ADVANCED MATHEMATICS STUDENT NAME STUDENT ID NUMBER PROFESSOR (TUTOR) DATE OF SUBMISSION
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QUESTION 4 The Heaviside step function is piecewise-defined as H(τ)={0,τ<0 1,τ≥0 (i)GraphH(τ) (ii)GraphH(t−τ)for both fixed t<0 and fixed t>0 (iii)Show your work to confirm that ∫ −∞ ∞ H(t−τ)H(τ)dτ=H(τ)∫ 0 t dτ (iv)Use a convolution to find the inverse Fourier transform of the function ¿1 (1+iω)(2+iω) Solution, For the case ofH(τ), considering a unit step=1, For the case ofH(t−τ) At t<0, 1
At t>0, The area between 0 and t is of interest to the function. Therefore, Convolving the two functions, ∫ −∞ ∞ H(t−τ)H(τ)dτ The region is under 1 from[0,t]. As a result, ∫ 0 t 1∗dτ=∫ 0 t dτ QUESTION 5 Apply the Fourier transform to the differential equation y+6y'+5y=δ(t-3 To obtain the solutiony=f(t) Solving the differential equation using Laplace transform L{y+6 y'+5 y\} =(s^2 +6s+5)Y(s)-y'(0 With no initial boundary conditions, assumption on all to be unity. (s2+6s+5)Y(s)−y'(0)=0 Lety'(0)=1, 2
(s2+6s+5)Y(s)−1=e−cs Y(s)=1 s2+6s+5 Finding the roots, Y(s)=1 (s+1)(s+5) On the other hand of the equation, δ(t−3) Using Laplace transformation, δ(t−3)→e−3s Using the partial fractions, Y(s)=a s+1+b s+5 e−3s=a(s+5)+b(s+1) ats=−1,1=a(4)+b(0)=1 a=1 4 ats=−5,1=a(0)+b(−4)=1 b=−1 4 Y(s)=1 4(1 s+1)−1 4(1 s+5)performing an Inverse Laplace Transform, Y(s)=e−3s (1 4F(s)−1 4G(s)) Solving in time domain, y(t)=1 4u3(t)f(t−3)−g(t) QUESTION 6 3
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Iffn(x)=cos(nπx L), what is the value of the inner product(fn,fn)with weightω(x)=1on the interval(0,L)forn=0,1,2,… Solution For a norm squared, the inner product is obtained from multiplying the function with itself, such that, (fn,fn)=||f|| 2 =∫ a b r(x)f2(x)dx fn(x)=cos(nπx L)∧r(x)=w(x)=1 The interval is given as [0, L] Therefore, ¿∫ 0 L 1∗cos2 (nπx L)dx Integrating a cosine function results in a sinusoidal function which is an orthogonal function, Letu=nπx L→dx=l πndu ¿l πn∫cos2(u)du Applying the reduction formula, ∫cos2(u)du=n−1 n∫cosn−2(u)du+cosn−1(u)sin(u) n Values can be obtained for different values of n. At n=2, ∫cos2(u)du=1 2∫cos(2u)du+1 2∫1du Letv=2u→du=1 2dv ∫cos2(u)du=1 2∫cos(v)dv Replacing in the equation above, 4
∫cos2(u)du=1 2sin(v) ∫cos2(u)du=sin(2u) 2 Solving for∫1du=u Replacing in the main equation, l 4πn∫cos2(u)du=l 4πnsin(2u)+lu 2πn Replacingu=nxπ l, l 4πnsin(2πnx l)+x 2+C,∀:l>0 QUESTION 7 Sturm-Liouville problem on(0,L), Part a The function is given as,X”(x)=−λX(x),0<x<L A regular Sturm-Liouville problem is an eigen value problem that has an unknown value. It is defined as a two-point boundary value problem such that, p(x)y'−q(x)y+λr(x)y=00<x<l α1y(0)+α2y'(0)=0 β1y(l)+β2y'(l)=0 p(x)>0∧r(x)>0 All are S-L variables. 5
Part b Case: Neumann boundary conditions, p(x)=1,q(x)=0,r(x)=1 Part b & Part c The eigen values are λn=n2π2 L2 The eigen functions are yn(x)=sin(nπ Lx) Supposing thatp(x),p'(x),q(x),∧r(x)are continuous on [a,b] and supposep(x)>0andr(x)>0 for allx∈[a,b]. Then the Sturm-Liouville regular problem has an increasing sequence of eigen values such that, λn=λ1>λ2>λ3 lim n→∞ λn=+∞ All the eigenvalues are non-negative as predicted by the theorem.λnis a constant multiple of a single eigenfunctionyn(x). Further, the general solution is given as, 6
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X(x)=Acos(√λx)+Bsin(√λx)ifλ>0 X(x)=Ax+Bifλ=0 To satisfy,hB−A=0,A=0,henceB=0 note:(h>0) λ=0,notaneigenvalue→noeigenfunction(nononzerosolution) Part d Testing forλ>0 Using the boundary conditions, 0=hA−√λB 0=−A√λsin(√λ)+B√λcos(√λ) AtA=0,B=0hence both are non-zeros. B=hA √λ∧0=−A√λsin(√λ)+hA √A√λcos(√λ) A≠0(solutionwouldbetoolarge) 0=−√λsin(√λ)+h√λcos(√λ) h √λ=tan√λ Part e The appropriate eigen function, assuming A=1,B=h √λ Such that, Xn(x)=cos(√λnx)+h √λn sin(√λnx) Part f 7
Thesinnxterms of the function are orthogonal for distinct n on the range[0,π]. Ther(x)term is the weight function on[a,b].The orthogonality of a function is based on the weight function, r(x). The inner product is given as, ⟨f|g⟩¿ ¿∫ a b f(x)g(x)r(x)dx 8