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Transforming a parabolic pde to normal form

  1. Mar 6, 2008 #1
    [SOLVED] transforming a parabolic pde to normal form

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

    The problem is to transform the PDE to normal form.

    The PDE in question is parabolic: U[tex]_{xx}[/tex] - 2U[tex]_{xy}[/tex] + U[tex]_{yy}[/tex] = 0 but I also need to solve other problems for hyperbolic pde's so general advice would be appreciated.

    2. Relevant equations

    The characteristic equation is: Ay'[tex]^{2}[/tex] - 2By' + C = 0

    The new variables should be v=x, w=psi, and the normal form is U[tex]_{ww}[/tex]=F[tex]_{2}[/tex]

    3. The attempt at a solution

    The solutions manual provides: [​IMG]

    I get lost right after we solve the characteristic equation. I don't understand how the variable substitution works or what is going on after that. My textbook only offers 1 example similar to this problem with no explanation of how it goes from step to step...so I'm completely lost. I looked online for information but the limited amount of stuff I did find is too technical(I read through all of them).
     
  2. jcsd
  3. Mar 7, 2008 #2

    HallsofIvy

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    Here's how I would do it (finding the characteristic equation slightly differently):
    From [itex]U_{xx}- 2U_{xy}+ U_{yy}= 0[/itex] we get x2- 2xy+ y2= (x- y)2= 0 which has only x-y= 0 or x= y as solution. Perpendicular to that line is the "characteristic curve" (line in this case) x+y= 0. We change coordinate systems so that line is one of the axes: let v= x+ y. Since that is the only characteristic line, just let u= x be the other. In these new variables, by the chain rule,
    [tex]U_x= U_u\frac{\partial u}{\partial x}+ U_v\frac{\partial v}{\partial x}[/tex]
    Since [itex]\partial u/\partial x= 1[/itex] and [itex]\partial v/\partial x= 1[/itex] that is
    [tex]U_x= U_u+ U_v[/tex]

    [tex]U_y= U_u\frac{\partial u}{\partial y}+ U_v\frac{\partial v}{\partial y}[/tex]
    Since [itex]\partial u/\partial y= 0[/itex] and [itex]\partial v/\partial y= 1[/itex] that is
    [tex]U_y= U_v[/itex]

    Continuing to use the chain rule,
    [tex]U_{xx}= (U_x)_x= (U_u+ U_v)_x= (U_{uu}\frac{\partial u}{\partial x}+ U_{uv}\frac{\partial v}{\partial x})+ (U_{vu}\frac{\partial u}{\partial x}+ U_{vv}\frac{\partial v}{\partial x})[/tex]
    [tex]= U_{uu}+ 2U_{uv}+ U_{vv}[/tex]

    [tex]U_{xy}= (U_x)_y= (U_u+ U_v)_y= (U_{uu}\frac{\partial u}{\partial y}+ U_{uv}\frac{\partial v}{\partial y})+ (U_{vu}\frac{\partial u}{\partial y}+ U_{vv}\frac{\partial v}{\partial y})[/tex]
    [tex]= U_{uv}+ U_{vv}[/tex]

    [tex]U_{yy}= (U_y)_y= (U_u+ U_v)_y= (U_{uu}\frac{\partial u}{\partial y}+ U_{uv}\frac{\partial v}{\partial y})+ (U_{vu}\frac{\partial u}{\partial y}+ U_{vv}\frac{\partial v}{\partial y})[/tex]
    [tex]= U_{vv}[/tex]

    Putting those into the equation [itex]U_{xx}- 2U_{xy}+ U_{yy}= 0[/itex] we get
    [itex](U_{uu}+ 2U_{uv}+ U_{vv})-2(U_{uv}+ U_{vv})+ U_{vv}= U_{uu}= 0[/itex]
    the "normal" form.

    Saying that [itex]U_{uu}= (U_u)_u= 0[/itex] means that [itex]U_u[/itex] is a "constant" with respect to u. It must depend only on v: [itex]U_u= F(v)[/itex]. Integrating with respect to u again, U(u,v)= F(v)u+ G(v) where G(v) is the "constant" of itegration- since we are only integrating with respect to u, G may depend upon v. That is the general solution to the differential equation in terms of u and v. Since u= x and v= x+y, that is the same as U(x,y)= F(x+y)x+ G(x+y). Since F and G, above, were arbitrary functions of one variable, I can choose any such functions and replace the one variable with x+ y. For example, suppose F(x)= x2 and G(x)= sin(x) (chosen pretty much arbitrarily). Then U(x,y)= (x+y)2x+ sin(x+ y)= x3+ 2x2y+ xy2+ sin(x+y). I'll leave it to you to show that this function does, in fact, satisfy the original differential equation.
     
    Last edited: Mar 7, 2008
  4. Mar 7, 2008 #3

    HallsofIvy

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    Take deep breath, here we go again for hyperbolic equations!

    An an example, I will use [itex]U_{xx}- 5U_{xy}+ 6U_{yy}= 0[/itex].

    Convert that to the algebraic equation x2- 5xy+ 6y2= (x- 3y)(x- 2y)= 0. The fact that there are two distinct factors tells us that there are two distinct characteristic curves (lines) and so this equation is hyperbolic. The line, through the origin, perpendicular to x- 3y= 0 is y+ 3x= 0. The line, through the origin, perpendicular to x- 2y= 0 is y+ 2x= 0.

    Let u= 3x+ y and v= y+ 2x. so that
    [tex]\frac{\partial u}{\partial x}= 3[/itex]
    [tex]\frac{\partial v}{\partial x}= 2[/itex]
    [tex]\frac{\partial u}{\partial y}= 1[/itex]
    [tex]\frac{\partial v}{\partial y}= 1[/itex]
    Then
    [tex]U_x= 3U_u+ 2U_v[/tex]
    [tex]U_y= U_u+ U_v[/tex]

    [tex]U_{xx}= 3(3U_{uu}+2U_{vu})+ 2(3U_{uv}+ 2U_{vv})[/tex]
    [tex]= 9U_{uu}+ 12U_{uv}+ 4U{vv}[/tex]

    [tex]U_{yy}= (U_{uu}+ U_{vu})+ (U_{vu}+ U_{vv})[/tex]
    [tex]= U_{uu}+ 2U_{uv}+ U_{vv}[/tex]

    [tex]U_{xy}= (U_x)_y= (3U_{uu}+ 2U_{uv})+ (3U_{uv}+ 2U{vv})[/tex]
    [tex]= 3U_{uu}+ 5U_{uv}+ 2U_{vv}[/tex]

    [tex]U_{xx}= 9U_{uu}+ 12U_{uv}+ 4U{vv}[/tex]
    [tex]-5U_{xy}= -15U_{uu}- 25U_{uv}- 10U{vv}[/tex]
    [tex]6U_{yy}= 6U_{uu}+ 12U+{uv}+ 6U{vv}[/tex]
    Adding, we get
    [tex]U_{xx}- 5U_{xy}+ 6U{yy}= -U_{uv}= 0[/tex]

    That tells us the derivative of [itex]U_u[/itex], with respect to v, is 0 and so [itex]U_u[/itex] is a function of U only: [itex]U_u= f(u)[/itex]. Integrating, with respect to u now, U(u,v)= F(u)+ G(v) where F is the anti-derivative of f with respect to u and G is the "constant of integration" which, since the derivative is with respect to u, may be a function of v. That is the general solution to the differential equation in terms of u and v. Putting it back into x and y, U(x,y)= F(y+ 3x)+ G(y+ 2x).

    Again, I will leave it to you to show that, taking F(x)= x2 and G(x)= sin(x), U(x,y)= (y+ 3x)2+ sin(y+ 2x) satisfies the original differential equation.
     
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