Part 5 Laplace Equation jmb/lectures/pdelecture5.pdf · PDF file Steady state stress...

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Transcript of Part 5 Laplace Equation jmb/lectures/pdelecture5.pdf · PDF file Steady state stress...

  • Laplace’s Equation

    • Separation of variables – two examples • Laplace’s Equation in Polar Coordinates

    – Derivation of the explicit form – An example from electrostatics

    • A surprising application of Laplace’s eqn – Image analysis – This bit is NOT examined

  • 0 yx 2 2

    2

    2

    = ∂ ∂

    + ∂ ∂ φφ

    02 =∇ φ

    Laplace’s Equation

    In the vector calculus course, this appears as where

    ⎥ ⎥ ⎥ ⎥

    ⎢ ⎢ ⎢ ⎢

    ∂ ∂ ∂ ∂

    =∇

    y

    x

    Note that the equation has no dependence on time, just on the spatial variables x,y. This means that Laplace’s Equation describes steady state situations such as:

    • steady state temperature distributions

    • steady state stress distributions

    • steady state potential distributions (it is also called the potential equation

    • steady state flows, for example in a cylinder, around a corner, …

  • Steady state stress analysis problem, which satisfies Laplace’s equation; that is, a stretched elastic membrane on a rectangular former that has prescribed out-of-plane displacements along the boundaries

    y

    x

    w = 0

    w = 0 w = 0

    a xsinww 0 π

    =

    a

    b

    w(x,y) is the displacement in z-direction

    x

    y z

    0 y w

    x w

    2

    2

    2

    2

    = ∂ ∂

    + ∂ ∂

    Stress analysis example: Dirichlet conditions

    Boundary conditions To solve:

    axx a

    wbxw

    byyaw axxw byyw

    ≤≤=

    ≤≤= ≤≤= ≤≤=

    0sin),(

    00),( 00)0,( 00),0(

    0 for ,

    for , for , for ,

    π

  • Solution by separation of variables

    k Y Y

    X X

    Y Y

    X X

    YXYX yYxXyxw

    = ′′

    −= ′′

    = ′′

    + ′′

    =′′+′′ =

    0

    0 )()(),(

    from which

    and so

    as usual …

    where k is a constant that is either equal to, >, or < 0.

  • Case k=0 )()(),()( DCyyYBAxxX +=+=

    )(),(:0 0),(0)(0

    000),0(

    DCyAxyxwB yxwyYDC

    DCByw

    +== ≡≡==

    ===⇒=

    withContinue so , then , if

    or

    ACxyyxw DwA

    ADxxw

    = =≡=

    =⇒=

    ),( 0)00

    00)0,(

    withContinue or (so Either

    0),(0000),( ≡⇒==⇒=⇒= yxwCAACayyaw or

    That is, the case k=0 is not possible

  • Case k>0

    )sincos)(sinhcosh(),( ,2

    yDyCxBxAyxw k

    αααα α

    ++= =

    that so that Suppose

    )sincos(sinh),(0 0),(0 0)sincos(0),0(

    00sinh,10cosh

    yDyCxByxwA yxwDC yDyCAyw

    ααα

    αα

    +=⇒= ≡⇒== =+⇒=

    ==

    withContinue

    that Recall

    yxBDyxwC yxwB

    xBCxw

    αα

    α

    sinsinh),(0 0),(0

    0sinh0)0,(

    =⇒= ≡⇒=

    =⇒=

    withContinue

    0),(00 0sinsinh0),(

    ≡⇒== =⇒=

    yxwDB yaBDyaw

    or either so αα

    Again, we find that the case k>0 is not possible

  • Final case k

  • Solution

    a yn

    a xnKyxw

    n n

    ππ sinhsin),( 1 ∑ ∞

    =

    =Applying the first three boundary conditions, we have

    a bsinh

    wK 01 π=We can see from this that n must take only one value, namely 1, so that

    which gives: a

    bn a

    xnK a xw

    n n

    πππ sinhsinsin 1

    0 ∑ ∞

    =

    =

    and the final solution to the stress distribution is

    a y

    a x

    a b

    wyxw πππ sinhsinsinh ),( 0=

    a xsinw)b,x(w 0 π

    =The final boundary condition is:

    Check out: http://mathworld.wolfram.com/HyperbolicSine.html

  • a bn

    a xnKxfw

    n n

    ππ sinhsin)( 1

    0 ∑ ∞

    =

    = Then

    and as usual we use orthogonality formulae/HLT to find the Kn

    y

    x

    w = 0

    w = 0 w = 0

    )x(fww 0=

    a

    b

    More general boundary condition

  • Types of boundary condition

    1. The value is specified at each point on the boundary: “Dirichlet conditions”

    2. The derivative normal to the boundary is specified at each point of the boundary: “Neumann conditions”

    3. A mixture of type 1 and 2 conditions is specified

    ),( yxφ

    ),( yx n∂ ∂φ

    Johann Dirichlet (1805-1859) http://www-gap.dcs.st- and.ac.uk/~history/Mathematicians/Dirichlet.html

    Carl Gottfried Neumann (1832 -1925) http://www-history.mcs.st-

    andrews.ac.uk/history/Mathematicians/Neumann_Carl.html

  • A mixed condition problem y

    x

    w = 0

    w = 0

    x a

    ww 2

    sin0 π

    =

    a

    b

    0= ∂ ∂

    x w

    0 y w

    x w

    2

    2

    2

    2

    = ∂ ∂

    + ∂ ∂

    To solve:

    Boundary conditions

    axx a

    wbxw

    by x w

    axxw byyw

    ax

    ≤≤=

    ≤≤= ∂ ∂

    ≤≤= ≤≤=

    =

    0 2

    sin),(

    00

    00)0,( 00),0(

    0 for ,

    for ,

    for , for ,

    π

    A steady state heat transfer problem

    There is no flow of heat across this boundary; but it does not necessarily have a constant temperature along the edge

  • Solution by separation of variables

    k Y Y

    X X

    Y Y

    X X

    YXYX yYxXyxw

    = ′′

    −= ′′

    = ′′

    + ′′

    =′′+′′ =

    0

    0 )()(),(

    from which

    and so

    as usual …

    where k is a constant that is either equal to, >, or < 0.

  • Case k=0 )()(),()( DCyyYBAxxX +=+=

    )(),(:0 0),(0)(0

    000),0(

    DCyAxyxwB yxwyYDC

    DCByw

    +== ≡≡==

    ===⇒=

    withContinue so , then , if

    or

    ACxyyxw DwA

    ADxxw

    = =≡=

    =⇒=

    ),( 0)00

    00)0,(

    withContinue or (so Either

    0),(0000 ≡⇒==⇒=⇒= ∂ ∂

    =

    yxwCAACy x w

    ax

    or

    That is, the case k=0 is not possible

  • Case k>0

    )sincos)(sinhcosh(),( ,2

    yDyCxBxAyxw k

    αααα α

    ++= =

    that so that Suppose

    )sincos(sinh),(0 0),(0 0)sincos(0),0(

    00sinh,10cosh

    yDyCxByxwA yxwDC yDyCAyw

    ααα

    αα

    +=⇒= ≡⇒== =+⇒=

    ==

    withContinue

    that Recall

    yxBDyxwC yxwB

    xBCxw

    αα

    α

    sinsinh),(0 0),(0

    0sinh0)0,(

    =⇒= ≡⇒=

    =⇒=

    withContinue

    0),(00

    0sincosh0

    ≡⇒==

    =⇒= ∂ ∂

    =

    yxwDB

    yaBD x w

    ax

    or either so

    ααα

    Again, we find that the case k>0 is not possible

  • Final case k

  • Solution y

    a nx

    a nKyxw

    n n 2

    )12(sinh 2

    )12(sin),( 1

    ππ −− =∑

    =

    Applying the first three boundary conditions, we have

    b a

    wK

    2 sinh

    0 1 π=We can see from this that n must take only one value, namely 1, so that

    which gives: b a

    nx a

    nK a xw

    n n 2

    )12(sinh 2

    )12(sin 2

    sin 1

    0 πππ −−

    =∑ ∞

    =

    and the final solution to the stress distribution is

    a y

    a x

    a b

    wyxw πππ sinhsinsinh ),( 0=

    a xwbxw

    2 sin),( 0

    π =The final boundary condition is:

    Check out: http://mathworld.wolfram.com/HyperbolicSine.html

  • PDEs in other coordinates…

    • In the vector algebra course, we find that it is often easier to express problems in coordinates other than (x,y), for example in polar coordinates (r,Θ)

    • Recall that in practice, for example for finite element techniques, it is usual to use curvilinear coordinates … but we won’t go that far

    We illustrate the solution of Laplace’s Equation using polar coordinates*

    *Kreysig, Section 11.11, page 636

  • A problem in electrostatics

    Thin strip of insulating material

    0V

    r θ

    Radius a halfupper on the ),,( UzrV =θ

    011 2 2

    2

    2

    22

    2 2 =

    ∂ ∂

    + ∂ ∂

    + ∂ ∂

    + ∂ ∂

    =∇ z VV

    rr V

    rr VV

    θ

    I could simply TELL you that Laplace’s Equation in cylindrical polars is:

    This is a cross section of a charged cylindrical rod.

    … brief time out while I DERIVE this

  • 2D Laplace’s Equation in