{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw3S - Partial Dierential Equations Math 442 C13/C14 Fall...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Partial Differential Equations – Math 442 C13/C14 Fall 2009 Homework 3 Solutions 1. Here we will prove that solutions to the heat equation satisfy (some of) the invariance principles mentioned in class, or in the book in § 2.4. That is, if u ( x, t ) is a solution to u t = ku xx for x R , t > 0, then so are (a) u ( x - y, t ) for any fixed y , (b) u x , u t , (c) v ( x, t ) = integraltext −∞ u ( x - y, t ) g ( y ) dy where g has finite support, (d) v ( x, t ) = u ( ax, at ) for any a > 0. Solution: (a) Let v ( x, t ) = u ( x - y, t ). Then ∂v ∂t ( x, t ) = ∂u ∂t ( x - y, t ) · 1 = ∂u ∂t ( x - y, t ) , ∂v ∂x ( x, t ) = ∂u ∂x ( x - y, t ) · 1 = ∂u ∂x ( x - y, t ) , 2 v ∂x 2 ( x, t ) = ∂x ∂v ∂x ( x, t ) = ∂x ∂u ∂x ( x - y, t ) = 2 u ∂x 2 ( x - y, t ) . Then ∂v ∂t ( x, t ) - k 2 v ∂x 2 ( x, t ) = ∂u ∂t ( x - y, t ) - k 2 u ∂x 2 ( x - y, t ) = 0 , since u solves the heat equation. (b) We compute for u x , the other is similar. Denoting v = u x gives v t = ( u x ) t = u xt , v xx = ( u x ) xx = u xxx . Then v t - kv xx = u xt - u xxx = u tx - u xxx = ( u t - u xx ) x = 0 x = 0 . (c) Since g has compact support, we can exchange derivatives and integration (see e.g. Theorem A.3.2 from Strauss), and thus we have ∂v ∂t = ∂t integraldisplay −∞ u ( x - y, t ) g ( y ) dy = integraldisplay −∞ ∂u ∂t ( x - y, t ) g ( y ) dy and 2 v ∂x 2 = 2 ∂x 2 integraldisplay −∞ u ( x - y, t ) g ( y ) dy = integraldisplay −∞ 2 u ∂x 2 ( x - y, t ) g ( y ) dy. But then v t - kv xx = integraldisplay −∞ ∂u ∂t ( x - y, t ) g ( y ) dy - integraldisplay −∞ 2 u ∂x 2 ( x - y, t ) g ( y ) dy = integraldisplay −∞ parenleftbigg ∂u ∂t ( x - y, t ) - 2 u ∂x 2 ( x - y, t ) parenrightbigg g ( y ) dy = integraldisplay −∞ 0 dy = 0 . 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(d) We have ∂v ∂t ( x, t ) = a ∂u ∂t ( ax, at ) , ∂v ∂x ( x, t ) = a ∂u ∂x ( ax, at ) , 2 v ∂x 2 ( x, t ) = a 2 u ∂x 2 ( ax, at ) . So then v t - kv xx = au t ( ax, at ) - au xx ( ax, at ) = a ( u t ( ax, at ) - u xx ( ax, at )) = a · 0 = 0 . 2. (Strauss 2.4.1.) Solve the heat equation with initial condition φ ( x ) = braceleftBigg 1 , | x | < L, 0 , | x | ≥ L.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern