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Sturm-Liouville Theory

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Solve the Sturm-Liouville problem ddx(exdydx)+λy=0\frac{d}{dx}\left( e^x \frac{dy}{dx} \right) + \lambda y = 0 for y(0)=y(1)=0y(0) = y'(1) = 0.

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The general solution is y(x)=Asinh(λx)+Bcosh(λx)y(x) = A \sinh(\sqrt{\lambda}x) + B \cosh(\sqrt{\lambda}x). To satisfy y(1)=0y'(1)=0, we must have λAcosh(λ)+Bsinh(λ)=0\sqrt{\lambda} A \cosh(\sqrt{\lambda}) + B \sinh(\sqrt{\lambda}) = 0. These solutions are orthogonal with respect to the weight function w(x)=1w(x) = 1 over the interval [0,1][0,1].

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Resolve the equation ddx(sin(x)dydx)+λsin(x)y=0\frac{d}{dx}\left(\sin(x)\frac{dy}{dx}\right) + \lambda \sin(x) y = 0 with boundary conditions y(0)=y(π)=0y(0) = y(\pi) = 0.

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The solution involves Legendre polynomials, with yn(x)=Pn(cos(x))y_n(x) = P_n(\cos(x)), which are orthogonal over [0,π][0,\pi] with respect to weight function w(x)=sin(x)w(x)=\sin(x). The orthogonal functions satisfy 0πPm(cos(x))Pn(cos(x))sin(x)dx=0\int_{0}^{\pi} P_m(\cos(x)) P_n(\cos(x)) \sin(x) dx = 0 for mnm \neq n.

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Determine the eigenfunctions for the Sturm-Liouville problem ddx(exdydx)=λy\frac{d}{dx}\left( e^{-x}\frac{dy}{dx} \right) = \lambda y with y(0)=0y(0) = 0 and y(1)+y(1)=0y'(1) + y(1) = 0.

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The general solution is of the form y(x)=Aex/2sin(λ1/4x)+Bex/2cos(λ1/4x)y(x) = A e^{x/2} \sin(\sqrt{\lambda - 1/4}x) + B e^{x/2} \cos(\sqrt{\lambda - 1/4}x). The eigenfunctions satisfying the boundary conditions are given by yn(x)=ex/2sin(λn1/4(x1))y_n(x) = e^{x/2}\sin(\sqrt{\lambda_n - 1/4}(x-1)) and they're orthogonal with respect to the weight function w(x)=exw(x) = e^{-x} on [0,1][0,1].

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For the boundary value problem ddx(1xdydx)+λy=0\frac{d}{dx}\left( \frac{1}{x} \frac{dy}{dx} \right) + \lambda y = 0 with y(1)=y(e)=0y(1) = y'(e) = 0, determine the orthogonal functions.

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The eigenfunctions are yn(x)=Ansin(nπlogx)y_n(x) = A_n \sin(n \pi \log x), with nn being integers ensuring the boundary conditions. Two functions ym(x)y_m(x) and yn(x)y_n(x) are orthogonal over [1,e][1,e] with respect to the weight function w(x)=xw(x)=x if mnm \neq n.

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Find the general solution of ddx(p(x)dydx)+λw(x)y=0\frac{d}{dx}\left( p(x)\frac{dy}{dx} \right) + \lambda w(x) y = 0 with y(0)=y(π)=0y(0) = y(\pi) = 0, where p(x)=xp(x) = x and w(x)=1w(x) = 1.

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The general solution is given by a series of orthogonal functions yn(x)=Ansin(nxπx)y_n(x) = A_n \sin(\frac{n}{\sqrt{x}} \pi x). These functions are orthogonal with respect to the weight function w(x)=1w(x) = 1 over the interval [0,π][0,\pi].

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Solve the Sturm-Liouville problem x2d2ydx2+xdydx+(x2λ2)y=0x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + (x^2 - \lambda^2)y = 0 for y(0)=0y(0)=0 and y(a)=0y(a)=0, where a>0a > 0.

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The Bessel's equation of order λ\lambda has solutions y(x)=AJλ(x)+BYλ(x)y(x) = AJ_\lambda(x) + BY_\lambda(x). To satisfy the boundary condition y(0)=0y(0)=0, we set B=0B=0. The remaining solution AJλ(x)AJ_\lambda(x) must vanish at x=ax=a, leading to specific eigenvalues. Eigenfunctions are orthogonal with respect to the weight function w(x)=xw(x)=x on [0,a][0,a].

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Analyze the Sturm-Liouville problem given by d2ydx2λy=0\frac{d^2y}{dx^2} - \lambda y = 0 with y(0)=y(L)=0y'(0) = y'(L) = 0, where LL is a positive constant.

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The general solution for this equation is y(x)=Acos(λx)+Bsin(λx)y(x) = A \cos(\sqrt{\lambda}x) + B \sin(\sqrt{\lambda}x). Applying the Neumann boundary conditions ' y(0)=y(L)=0y'(0) = y'(L) = 0 ' leads to A=0A=0 and λ=nπL\sqrt{\lambda} = \frac{n\pi}{L}, where nn is an integer. The eigenfunctions yn(x)=Bsin(nπLx)y_n(x) = B \sin(\frac{n\pi}{L}x) are orthogonal with respect to the weight function w(x)=1w(x)=1.

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Considering ddx(x2dydx)λy=0\frac{d}{dx}\left( x^2 \frac{dy}{dx} \right) - \lambda y = 0 on the interval [1,2][1,2] with boundary conditions y(1)=0y(1) = 0 and (x2y(x))x=2=0(x^2 y'(x))|_{x=2} = 0. Find orthogonal functions.

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The solution for orthogonal functions is of the form yn(x)=Anx1/2sin(nπlogx)y_n(x) = A_n x^{1/2} \sin(n \pi \log x), which are orthogonal on the interval [1,2][1,2] with respect to the weight function w(x)=1/x2w(x) = 1/x^2.

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