# PDE problem, Solve using Method of Characteristics

Hi

This problem occurred on my final and I could not figure it out.

## Homework Statement

The problem was a partial differential equation (I forgot the exact equation) but the solution was a hyperbolic function in the form of u(x,y)= f(x+y) + g (x+y), it was part b that gave me the problem.

Part b asked to solve the initial value problem give u(0,y)=y and u_x_(0,y)=y^2 (I'm not 100% sure what they were, but they were functions of y). I just want to know how I would solve an initial value problem for these equations as we never covered it in class.

Thanks.

So can anyone help me out?

Thanks

So I ordered my exam, it turns out the actual question was:

Solve u_xx+u_xy-6u_yy = 0. The solution which I solved correctly was a hyperbolic function u (x,y)= f(2x+y)+g(3x-y).

The IVP is u(0,y) = y^2 and u_x_(0,y)=3y.

Thanks.

hunt_mat
Homework Helper
So found the characteristics and now you apply the boundary conditions. What is your problem exactly?

Mat

So found the characteristics and now you apply the boundary conditions. What is your problem exactly?

Mat

Hi

I just don't know how to solve the IVP u(0,y) = y^2 and u_x_(0,y)=3y when u (x,y)= f(2x+y)+g(3x-y).

I found a solved example online (http://www.mathsman.co.uk/DEnotes[2].pdf, the IVP is solved beginning page 48) but I'm very confused on their methodology, so if someone can explain this or do it some different way I would really appreciate it.

Thanks

hunt_mat
Homework Helper
Just differentiate!
$$\frac{\partial u}{\partial x}(0,y)=2f'(y)+3g'(-y)=3y$$
Integrating w.r.t. y shows that:
$$2f(y)-3g(-y)=\frac{3y^{2}}{2}$$
Likewise setting x=0 in the equations shows that:
$$u(0,y)=f(y)+g(-y)=y^{2}$$
To find f and g just add and subtract.

Just differentiate!
$$\frac{\partial u}{\partial x}(0,y)=2f'(y)+3g'(-y)=3y$$
Integrating w.r.t. y shows that:
$$2f(y)-3g(-y)=\frac{3y^{2}}{2}$$
Likewise setting x=0 in the equations shows that:
$$u(0,y)=f(y)+g(-y)=y^{2}$$
To find f and g just add and subtract.

Hi

Just one question, when we take derivative of f and g, what is it with respect to? I substituted w = 2x+y and v = 3x-y, so u_x = df/dw*dw/dx + dg/dv*dv/dx and when I integrated w.r.t. y I let dw=dy and dv = -dy which gave me the same equation you had, is this correct?

hunt_mat
Homework Helper
Sounds good.