Partial Differential Equations

[JunkieJim]

Homework Statement

U_tt-U_xx+2U_xy-U_yy=0
with the conditions:
U(1,x,y)=cos(x)+e^y
U_t(1,x,y)=sin(x)-y²

Homework Equations

Not using separation of variables to solve.

The Attempt at a Solution

I've gotten the general equation to be of the form:

U(t,x,y)=ψ(x+t,y-t)+ζ(x-t,y+t)
but solving for the initial conditions is giving me a problem:

ψ(x+1,y-1)= sin(x)+e^y-ζ(x-1,y+1)

ψ₁(x+1,y-1)=-cos(x)-ζ₁(x-1,y+1)

ψ₂(x+1,y-1)=-e^y-ζ₂(x-1,y+1)

subbing that into the U_t and simplifying gives me this:

2 ζ ₁ +2 sin(X) = 2ζ₂-e^y +y²

which I'm not sure what to do with...

help greatly apreciated

 

[mighty2000]

Thank you for posting your question. It seems like you are on the right track with your solution approach. However, there are a few things that need to be clarified in order to solve the problem.

Firstly, it is important to note that the given partial differential equation (PDE) is a linear homogeneous equation. This means that the general solution can be expressed as a linear combination of two independent solutions. In your attempt at a solution, you have already identified the two independent solutions as ψ(x+t,y-t) and ζ(x-t,y+t). However, in order to fully solve the PDE, we need to find the specific forms of these solutions.

To do this, we can use the method of characteristics. This involves finding a set of curves in the x-y plane, called characteristic curves, along which the solution remains constant. These curves can be determined by solving the following system of ordinary differential equations:

dx/dt = 1, dy/dt = -1, dU/dt = 0

Solving this system gives us the characteristic curves as x-y = C, where C is a constant. This means that the solution can be written as U(x,y) = ψ(C) + ζ(C), where C is a function of x and y.

Next, we can use the initial conditions to find the specific forms of ψ and ζ. Substituting the given initial conditions into the general solution, we get:

U(1,x,y) = ψ(x+1,y-1) + ζ(x-1,y+1) = cos(x) + e^y

Ut(1,x,y) = ψ1(x+1,y-1) + ζ1(x-1,y+1) = sin(x) - y^2

From the first equation, we can see that ψ(x+1,y-1) = cos(x) and ζ(x-1,y+1) = e^y. This means that ψ(x,y) = cos(x-1) and ζ(x,y) = e^(y+1). Similarly, from the second equation, we can find ψ1(x+1,y-1) = sin(x) and ζ1(x-1,y+1) = -2y. This gives us the specific forms of ψ and ζ as:

ψ(x,y) = cos(x