Boundary-Value Problems in Other Coordinate Systems Solve ∆u = 0 in a disk x^2 + y^2...

Boundary-Value Problems in Other Coordinate Systems

Solve ∆u = 0 in a disk x^2 + y^2 ≤ 25, where u(5, θ) = 7 sin 3θ − 6 sin 8θ and u is bounded when r = 0.
Solve ∆u = 0 in an annulus 1 ≤ x^2+y^2 ≤ 4, where u(1, θ) = 75 sin θ, u(2, θ) = 60 cos θ.
Find the steady-state temperature distribution in a disk of radius 1 if the upper half of the circumference is kept at 100◦ and the lower half is kept at 0◦ .

Expert Solution

Colby Messinger answered 1 year ago

Solve the following initial value problems y'' + y = 2/cos x , y(0) = y'(0)...

Solve the following initial value problems y'' + y = 2/cos x , y(0) = y'(0) = 2 x^3 y''' − 6xy' + 12y = 20x^4, x > 0, y(1) = 8/3 , y'(1) = 50/3 , y''(1) = 14 x^2 y'' − 2xy' + 2y = x^2, x > 0, y(1) = 3, y'(1) = 5

Let u(x, y) be the harmonic function in the unit disk with the boundary values u(x,...

Let u(x, y) be the harmonic function in the unit disk with the boundary values u(x, y) = x^2 on {x^2 + y^2 = 1}. Find its Rayleigh–Ritz approximation of the form x^2 +C1*(1−x^2 −y^2).

11. Solve numerically the following Boundary Value Problem. X2Y” – X(X+2)Y’ + (X+2)Y = 0 Y(1)...

11. Solve numerically the following Boundary Value Problem. X2Y” – X(X+2)Y’ + (X+2)Y = 0 Y(1) = e and Y(2) = 2e2 The value of e = 2.71828

u'' + sinu = sinx (-1<x<1) u(-1)=1, u'(1)=0 solve this boundary value problem.

Solve x(y^2+U)Ux -y(x^2+U)Uy =(x^2-y^2)U, U(x,-x)=1

Solve the Boundary Value Problem, PDE: Utt-a2uxx=0, 0≤x≤1, 0≤t<∞ BCs: u(0,t)=0 u(1,t)=cos(t) u(x,0)=0 ut(x,0)=0

Solve the initial value problem: y''+2y'+y = x^2 , y(0)=0 , y'(0) = 0

solve the given boundary value problem. y"+6y=24x, y(0)=0, y(1)+y'(1)=0

Solve the following initial value problems: a) ut+xux= -tu, x is in R, t>0; u(x,0) =f(x),...

Solve the following initial value problems: a) ut+xux= -tu, x is in R, t>0; u(x,0) =f(x), x is in R. c) ut+ux=-tu, x is in R, t>0; u(x,0)=f(x), x is in R d)2ut+ux = -2u, x,t in R, t>0; u(x,t)=f(x,t) on the straight line x = t, where f is a given function.

Solve the initial value problem: y'' + y = cos(x) y(0) = 2 y'(0) = -3...

Solve the initial value problem: y'' + y = cos(x) y(0) = 2 y'(0) = -3 y' being the first derivative of y(x), y'' being the second derivative, etc.

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