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Linearity of PDE

  1. Feb 22, 2015 #1
    This isn't a homework problem so hopefully this section is fine.

    I came across something that's bothering me while reviewing PDEs.
    Take something like: [tex]u_{x}(x,t) = 1.[/tex] which has the general solution: [tex]u(x,t) = c_{1}(t) + x.[/tex] Wolfram says this is linear but if I take a different solution: [tex]v(x,t) = c_{2}(t) + x[/tex] and add it to u it's not also a solution: [tex]u + v = c_{1}(t) + c_{2}(t) + 2x\\ (u + v)_{x} = 2 \neq 1[/tex]

    I must be missing something really simple here.
     
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  3. Feb 22, 2015 #2

    Orodruin

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    It is a linear differential equation, but it is not homogeneous.
     
  4. Feb 22, 2015 #3
    Ahh, so the sum of solutions being a solution generally only applies to homogeneous equations?
     
  5. Feb 22, 2015 #4

    BruceW

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    yeah... you can think about it another way. Your operator is not linear. And so a sum of solutions is not necessarily a solution. Your solution is linear in the 'x' argument, but the operator is not linear on its 'argument' i.e. the function it is acting on.

    edit: uh... wait, I did not say this properly... If you define your operator as ##\partial_x## then the operator is linear, but as you said, the operator is mapping to something nonzero (i.e. 1). And if you can somehow rewrite the equation so that it is mapping to zero, then the operator would no longer be a linear operator. But now that I think about it, maybe it is not possible to re-write this as a map going to 'zero'. So maybe ignore my post hehe sorry about that.
     
    Last edited: Feb 22, 2015
  6. Feb 22, 2015 #5

    BruceW

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    In terms of homomorphisms, let's call the linear operator ##\partial_x## a homomorphism from the vector space of functions to itself. (i.e. we have functions as the vectors and real numbers as the scalars of this vector space). Then the kernel of this homomorphism is simply the set of all functions such that the linear operator gives zero when it acts on them. And, generally the kernel of a homomorphism is a vector space itself. Therefore, the set of solutions are a vector space. Which is why adding two solutions gives another solution.

    Instead, if we consider the set of functions such that the linear operator gives the constant function 1, what can we say about this set of functions? They all map to the same function. So this means that you can get one function by adding an element of the kernel to the other function. This is what we know more intuitively as particular solutions and homogeneous solutions. If you find one particular solution, then you can just add any solution of the homogeneous solution, to get the general solution.

    So in your case, the kernel of the homomorphism is given by the set of all solutions to the equation ##\partial_x(f(x,t))=0## (which are of the form ##u(t)##). Also, one possible function which gets mapped to 1, is simply 'x'. Therefore, the set of all functions which get mapped to 1 are of the form ##u(t) + x##. I hope this explanation helped a bit.
     
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