Author Topic: Problem 2  (Read 1070 times)

Emily Deibert

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Problem 2
« on: September 29, 2015, 08:39:43 PM »
2.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.2
a) 2nd order linear homogeneous
    b) 2nd order nonlinear (quasilinear) homogeneous
    c) 3rd order linear homogeneous
    d) 3rd order nonlinear (quasilinear) homogeneous
    e) 4th order linear homogeneous
    f) 4th order linear homogeneous
    g) 4th order linear inhomogeneous
    h) 4th order linear homogeneous 4th order semilinear
« Last Edit: October 01, 2015, 03:44:38 PM by Emily Deibert »

Yaodong Gao

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Re: Problem 2
« Reply #1 on: September 30, 2015, 12:19:27 AM »
I think the answers for question 2g) and 2h) should be nonlinear, since the linearity condition L(au + bv) = aL(u) +bL(v) failed in both questions.

Andrew Lee Chung

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Re: Problem 2
« Reply #2 on: September 30, 2015, 03:50:16 PM »
My take on qu2:
@Emily
e) 4th order linear inhomogeneous (sin(x) is not a dependent variable, u, nor any of its partial derivatives)
g) 4th order linear homogeneous (u occurs linearly in the equation)
h) 4th order semi linear (sin(x)sin(u) is a function f(x,u))

Andrew Lee Chung

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Re: Problem 2
« Reply #3 on: September 30, 2015, 04:08:31 PM »
@Yaodong
Lu is a partial differential expression linear with respect to unknown function u
It could be the 0 order wrt u

From the book, the example given had this form
\begin{equation} Lu:= a_{11} u_{xx} +  2a_{12} u_{xy}+ a_{22}  u_{yy} + a_{1}  u_{x} + a_{2} u_{y} + au = f(x) \end{equation}
So it does include u

Emily Deibert

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Re: Problem 2
« Reply #4 on: October 01, 2015, 03:43:11 PM »
Andrew, for me question e appears to be \begin{equation} u_{tt} + u_{xxxx} = 0 \end{equation} so I believe this would be linear homogeneous? I do not see any sine term (unless I am somehow reading the wrong question!)

For me (g) is \begin{equation} u_{tt} + u_{xxxx} + \sin(x) = 0 \end{equation} so I don't think this is homogeneous, because of the sine term.

For h I made a mistake! You are right about this.

I think somehow our equations e and g have gotten mixed up...maybe the HA somehow appears differently to us....

Victor Ivrii

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Re: Problem 2
« Reply #5 on: October 03, 2015, 05:12:50 AM »
Emily is correct.

Yaodong: there is a notion of linear operator; $L(\alpha u+\beta v)= \alpha L(u)+\beta L(v)$ but talking about equations or systems and not necessarily differential, may be algebraic we have
$Lu=0$ is linear homogeneous
$Lu=f$ with $f$ (which does not depend on $u$!) is linear inhomogeneous  (linear algebra!!)

So linearity test applies only to $L$.

Remark: $Lu=0$ describes a linear subspace, $Lu=f$ (if has solution) describes affine subspace (former linear space but shifted, so we have no longer a linear space!)