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Math Help - PDE

  1. #1
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    PDE

    Consider the following boundary value problems for first order PEDs. Sketch both the characteristic curves and the boundary value curve. Describe the set of (x,y) for which the solution is uniquely defined. Solve the boundary value problem.

    (a) xux + yuy = 2u, with u(x,1) = g(x)
    (b) x^2ux + xyuy = u^2, with u(y^2,y) = 1

    could anyone please help me with this problem ?
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  2. #2
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    Quote Originally Posted by jin_nzzang View Post
    Consider the following boundary value problems for first order PEDs. Sketch both the characteristic curves and the boundary value curve. Describe the set of (x,y) for which the solution is uniquely defined. Solve the boundary value problem.

    (a) xux + yuy = 2u, with u(x,1) = g(x)
    (b) x^2ux + xyuy = u^2, with u(y^2,y) = 1

    could anyone please help me with this problem ?
    I'll help with the first. The characteristic curve is where your PDE reduces to an ODE. So

     <br />
\frac{du}{dx} = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \frac{dy}{dx}<br />

    So with x u_x + y u_y = 2u or u_x + u_y \frac{y}{x} = \frac{2u}{x}, then along

    \frac{dy}{dx} = \frac{y}{x} (1)

    your PDE becomes

    \frac{du}{dx} = \frac{2u}{x} (2)

    Solving (1) gives \frac{y}{x} = c_1 - the characteristic curves are straight lines through the origin. To see this more clearly, if we introduce new coordinates

     <br />
r = x,\;\;\; s = \frac{y}{x}<br />

    then

    u_x = u_r r_x + u_s s_x = u_r - \frac{y}{x^2} u_s
    u_y = u_r r_y + u_s s_y = \frac{1}{x} u_s

    and your PDE becomes x u_x + y u_y = 2u \;\; \Rightarrow \;\;r u_r = 2u which is (2)

    The solution of the reduced ODE is

    \frac{u}{r^2} = f(s) and the solution of the original PDE is  <br />
\frac{u}{x^2} = f\left( \frac{y}{x}\right)<br />

    Then use your initial condition: when y = 1 then u = g(x) which gives

     <br />
\frac{g(x)}{x^2} = f\left( \frac{1}{x}\right)<br />
so f(x) = \frac{1}{x^2} g \left( \frac{1}{x} \right)

    which gives as your final solution u = y^2 g\left( \frac{x}{y}\right)
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  3. #3
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    thanks a lot for your help.

    but in part (b),

    i got the characteristic curve as dy/dx = y/x

    hence y = xe^c

    could you please help me to get to the next step?
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  4. #4
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    Quote Originally Posted by jin_nzzang View Post
    thanks a lot for your help.

    but in part (b),

    i got the characteristic curve as dy/dx = y/x

    hence y = xe^c

    could you please help me to get to the next step?
    But c is just a constant so e^c is also a constant - call it C so

    y = C x
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