{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Q8A_130 - Math 415 Quiz 8A Given the heat conduction...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 415 Quiz 8A Given the heat conduction equation in two space dimensions, α 2 ( u xx + u yy ) = u t , and assuming that u ( x, y, t ) = X ( x ) Y ( y ) T ( t ), find ordinary differential equations satisfied by X ( x ), Y ( y ), and T ( t ). Plugging u = XY T into the differential equation gives us α 2 ( X 00 Y T + XY 00 T ) = XY T 0 , and noting that the left hand side has T as a factor, we can separate the equation as X 00 Y + XY 00 XY = T 0 α 2 T . The left side of this equation does not depend on t at all, and the right side depends only on t , so we can let both sides equal a constant, - λ , and we immediately have t ’s differential equation: T 0 + α 2 λT = 0. There are a few equivalent ways to continue with the equation X 00 Y + XY 00 = - λXY , one of which is to write this as X 00 X = - Y 00 - λY Y = μ, where, we have set both sides equal to a constant for the usual reason that we have two functions dependent on different variables which are always equal to each other. This gives us the differential equations T 0 + α 2 λT = 0 , X 00 - μX = 0 , Y 00 + ( λ + μ ) Y = 0 .
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
Image of page 2
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