# PDE problem, Solve using Method of Characteristics

1. Jun 20, 2011

### Red_CCF

Hi

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

1. The problem statement, all variables and given/known data

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.

2. Jun 30, 2011

### Red_CCF

So can anyone help me out?

Thanks

3. Jul 24, 2011

### Red_CCF

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.

4. Jul 25, 2011

### hunt_mat

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

Mat

5. Jul 26, 2011

### Red_CCF

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

6. Jul 27, 2011

### hunt_mat

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.

7. Jul 30, 2011

### Red_CCF

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?

8. Jul 30, 2011

Sounds good.