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Homework Help: General solution to PDE

  1. Apr 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the general solution f = f(x,y) of class C2 to the partial differential equation
    [tex]\frac{\partial^2 f}{\partial x^2}+4\frac{\partial^2 f}{\partial x \partial y}+\frac{\partial f}{\partial x}=0[/tex]
    by introducing the new variables u = 4x - y, v = y.

    2. Relevant equations
    With "by introducing the new variables" I assume they mean something like this
    [tex]f=f(x,y)=f(u(x,y), y)=f(4x-y, y)[/tex]
    [tex]\frac{\partial u}{\partial x}=u_1(x,y)=u_1=4[/tex]
    [tex]\frac{\partial u}{\partial y}=u_2(x,y)=u_2=-1[/tex]
    Which means the second derivatives of u = 0.

    3. The attempt at a solution
    I'm not sure if this is the right way to approach the problem, but I'll show what I got so far.

    I calculated the partial derivatives according to the first equation and ended up with
    u1 = 4, u2 = -1 gives
    [tex]4f_{21}+f_{1}=4\frac{\partial^2 f}{\partial x \partial y}+\frac{\partial f}{\partial x}=0[/tex]
    From here, I'm uncertain of how to proceed (if I even used the right method to being with).

    NOTE: We haven't gone through PDE's in our calculus course yet (just going through the basics of multivariable calculus at the moment), so I'm only going to assume this can be solved without much/any experience with PDE's.
  2. jcsd
  3. Apr 22, 2014 #2
    First look at the equation: [tex]\frac{\partial^2 f}{\partial x^2}+4\frac{\partial^2 f}{\partial x \partial y}+\frac{\partial f}{\partial x}=0[/tex]

    We can factor a [itex]\frac{∂}{∂x}[/itex] out and write it as such: [itex]\frac{∂}{∂x}[/itex]([itex]\frac{∂f}{∂x}[/itex]+4[itex]\frac{∂f}{∂y}[/itex]+f(x,y)) = 0.

    Now what this implies is [itex]\frac{∂f}{∂x}[/itex]+4[itex]\frac{∂f}{∂y}[/itex]+f(x,y) = g(y), where g(y) is some function depending only on y and maybe some constants. From here utilize the fact that [itex]\frac{∂f}{∂x}[/itex] = [itex]\frac{∂f}{∂u}[/itex][itex]\frac{∂u}{∂x}[/itex] and [itex]\frac{∂f}{∂y}[/itex] = [itex]\frac{∂f}{∂u}[/itex][itex]\frac{∂u}{∂x}[/itex]+[itex]\frac{∂f}{∂v}[/itex][itex]\frac{∂v}{∂y}[/itex].
  4. Apr 22, 2014 #3
    That makes sense. The only thing I don't quite understand is why it equals a function depending only on y.
  5. Apr 22, 2014 #4
    Since you factored out a [itex]\frac{∂}{∂x}[/itex], this implies the right hand side is constant with respect to a change in x. If this were an ODE, it's implied the value is constant, but because you have a function of two variables, it is implied to be a function that is constant with respect to a change in x:

    [itex]\frac{∂g(y)}{∂x}[/itex] = 0
  6. Apr 22, 2014 #5
    I found a mistake here, [itex]\frac{∂f}{∂y}[/itex] = [itex]\frac{∂f}{∂u}[/itex][itex]\frac{∂u}{∂y}[/itex]+[itex]\frac{∂f}{∂v}[/itex][itex]\frac{∂v}{∂y}[/itex], not what it says in the quote.
  7. Apr 23, 2014 #6
    But how would I go about determining the function g(y) from this?
    The equation above doesn't seem like a sufficient answer.
  8. Apr 23, 2014 #7


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    The general solution of a pde of nth order has n undetermined functions. This is analogous to an ode of nth order which has n arbitrary constants.

    One should also be a bit more careful with the variable substitution. The idea is to write
    where [itex]u(x,y)[/itex] and [itex]v(x,y)[/itex] are the given functions. Then you have
    [tex]\partial_x f =\partial_u F(u,v) \partial_x u + \partial_v F \partial_x v,[/tex]
    [tex]\partial_y f =\partial_u F(u,v) \partial_y u + \partial_v F \partial_y v.[/tex]
    This you plug into the remaining ODE and express everything in terms of the independent variables [itex]u[/itex] and [itex]v[/itex], hoping that everything becomes more simple than the original equation, which is indeed the case in this example.
  9. Apr 23, 2014 #8
    Plugging in the values for the partial derivatives gives

    [itex]\frac{\partial}{\partial x} \left[4\frac{\partial F}{\partial v}+F(u,v) \right] = 0 = \frac{\partial}{\partial x}g(y).[/itex]

    Does that mean the general solution is

    [itex]F(u,v) = g(y)-4\frac{\partial F}{\partial v}[/itex]

    where [itex]g(y)[/itex] and [itex]\frac{\partial F}{\partial v}[/itex] are the undetermined functions?
  10. Apr 23, 2014 #9


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    Yes, with u= 4x- y, v= y, [itex]u_x= 4[/itex], [itex]u_y= -1[/itex], [itex]v_x= 0[/itex], and [itex]v_y= 1[/itex].

    Then, by the chain rule, [itex]f_x= f_u u_x+ f_v v_x= 4f_u[/itex], so [itex]f_{xx}= (4f_u)_x= 4(4f_u)_u= 16f_{uu}[/itex] and [itex]f_{xy}= (4f_u)_y= (4f_u)_u u_y+ (4f_u)_v v_y= -4f_{uu}+ 4f_{uv}[/itex]

    So [itex]f_{xx}+ 4f_{xy}+ f_x= 16f_{uu}+ 4(-4f_{uu}+ 4f_{uv})+ 4f_u= 16f_{uv}+ 4f_u= 0[/itex]

    We can think of this as [itex](4f_v+ f)_u= 0[/itex] which says that [itex]4f_v+ f[/itex] is not a function of u. It must be equal to some function of v only: [itex]4f_v+ f= g(v)[/itex]. Since that in v only, we can solve it like an ordinary differential equation: the solution to the "homogeneous part" is [itex]Ce^{-v/4}[/itex] and a particular solution to the entire equation is some unknown (because g(v) is unknown) function of v: [itex]Ce^{-v/4}+ G(u)[/itex]. Of course, because we are treating u as a constant, the "C" may be a function of u: [itex]f(u,v)= F(u)e^{-v/4}+ G(v)[/itex] where F(u) can be any twice differentiable function of u and G(v) can be any twice differentiable function of v.

    Going back to the original x, y variables, since x= 4u- v and y= v, x= 4u- y, 4u= x+ y, and u= (x+ y)/4.
    Replacing u with (x+ y)/4 and v with y, [itex]f(u,v)= F(u)e^{-v/4}+ G(v)[/itex] becomes [itex]f(x, y)= F((x+y)/4)e^{-y/4}+ G(y)[/itex].
  11. Apr 23, 2014 #10
    I understand your point, but the particular solution is still confusing to me. Would you mind further explaining how G(u) becomes G(v)?
  12. Apr 23, 2014 #11
    I think he just made a typo.
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