x2uxx --2xyuxy -- 3y2uyy = 0 u(x,1) = x, uy(x,1) = 1

Expert Solution

Colby Messinger answered 2 years ago

Consider the equation uux + uy = 0 with the initial condition u(x, 0) = h(x)...

Consider the equation uux + uy = 0 with the initial condition u(x, 0) = h(x) = ⇢ 0 for x > 0 uo for x < 0, with uo< 0. Show that there is a second weak solution with a shock along the line x = uo y / 2 The solution in both mathematical and graphical presentation before and after the shock.

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

1. Calculate the Transversality Conditions 2. and then solve the ff: a.) Ux +Uy =1-U U (X,...

1. Calculate the Transversality Conditions 2. and then solve the ff: a.) Ux +Uy =1-U U (X, X+X^2) = SIN X b) XUx + YUy= 4U with initial condition u=1 on the unit circle x^2 +y^2 =1 ( you will need to parameterize the circle in terms of parameter r).

Please solve the following: ut=uxx, 0<x<1, t>0 u(0,t)=0, u(1,t)=A, t>0 u(x,0)=cosx, 0<x<1

Use separation of variables to solve uxx+2uy+uyy=0, u(x, 0) = f(x), u(x, 1) = 0, u(0,...

Use separation of variables to solve uxx+2uy+uyy=0, u(x, 0) = f(x), u(x, 1) = 0, u(0, y) = 0, u(1, y) = 0.

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

We consider the Boundary Value Problem : u'(x)+u(x)=f(x), 0<x<1 u(0)-eu(1)=a ,a is real number kai f...

We consider the Boundary Value Problem : u'(x)+u(x)=f(x), 0<x<1 u(0)-eu(1)=a ,a is real number kai f is continue in [0,1]. 1. Find a a necessary and sufficient condition ,that Boundary value problem is solvabled. 2. Solve the Boundary value problem with a=0.

Consider a semi-infinite square well: U(x)=0 for 0 ≤ x ≤ L, U(x)=U0 for x >...

Consider a semi-infinite square well: U(x)=0 for 0 ≤ x ≤ L, U(x)=U0 for x > L, and U(x) is infinity otherwise. Determine the wavefunction for E < Uo , as far as possible, and obtain the transcendental equation for the allowable energies E. Find the necessary condition(s) on E for the solution to exist.

Solve the nonhomogeneous heat equation: ut-kuxx=sinx, 0<x<pi, t>0 u(0,t)=u(pi,t)=0, t>0 u(x,0)=0, 0<x<pi

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

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