PDE problem, Solve using Method of Characteristics

  • Thread starter Red_CCF
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  • #1
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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.
 

Answers and Replies

  • #2
532
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So can anyone help me out?

Thanks
 
  • #3
532
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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
hunt_mat
Homework Helper
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So found the characteristics and now you apply the boundary conditions. What is your problem exactly?

Mat
 
  • #5
532
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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
 
  • #6
hunt_mat
Homework Helper
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Just differentiate!
[tex]
\frac{\partial u}{\partial x}(0,y)=2f'(y)+3g'(-y)=3y
[/tex]
Integrating w.r.t. y shows that:
[tex]
2f(y)-3g(-y)=\frac{3y^{2}}{2}
[/tex]
Likewise setting x=0 in the equations shows that:
[tex]
u(0,y)=f(y)+g(-y)=y^{2}
[/tex]
To find f and g just add and subtract.
 
  • #7
532
0
Just differentiate!
[tex]
\frac{\partial u}{\partial x}(0,y)=2f'(y)+3g'(-y)=3y
[/tex]
Integrating w.r.t. y shows that:
[tex]
2f(y)-3g(-y)=\frac{3y^{2}}{2}
[/tex]
Likewise setting x=0 in the equations shows that:
[tex]
u(0,y)=f(y)+g(-y)=y^{2}
[/tex]
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?
 
  • #8
hunt_mat
Homework Helper
1,745
26
Sounds good.
 

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