PDE problem, Solve using Method of Characteristics

[Red_CCF]

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.

 

[Red_CCF]

So can anyone help me out?

Thanks

 

[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.

 

[hunt_mat]

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

Mat

 

[Red_CCF]

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]

Just differentiate!
<br /> \frac{\partial u}{\partial x}(0,y)=2f&#039;(y)+3g&#039;(-y)=3y<br />
Integrating w.r.t. y shows that:
<br /> 2f(y)-3g(-y)=\frac{3y^{2}}{2}<br />
Likewise setting x=0 in the equations shows that:
<br /> u(0,y)=f(y)+g(-y)=y^{2}<br />
To find f and g just add and subtract.

 

[Red_CCF]

Just differentiate!
<br /> \frac{\partial u}{\partial x}(0,y)=2f&#039;(y)+3g&#039;(-y)=3y<br />
Integrating w.r.t. y shows that:
<br /> 2f(y)-3g(-y)=\frac{3y^{2}}{2}<br />
Likewise setting x=0 in the equations shows that:
<br /> u(0,y)=f(y)+g(-y)=y^{2}<br />
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]

Sounds good.