Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 6 pps

To determine a suitable set of eigenfunctions, wesubstitute the separation of variables ux, y = XxY y into the homogeneous partial differential equation.. We expand f x, y in a Fourier s

Solution 37.19

uxx+ uyy = f (x, y), 0 < x < a, 0 < y < b,u(0, y) = u(a, y) = 0, uy(x, 0) = uy(x, b) = 0

We will solve this problem with an eigenfunction expansion in x To determine a suitable set of eigenfunctions, wesubstitute the separation of variables u(x, y) = X(x)Y (y) into the homogeneous partial differential equation

uxx+ uyy = 0(XY )xx+ (XY )yy = 0

λn = nπ

a , Xn= sin

nπxa

, n ∈ Z+

We expand u(x, y) in a series of the eigenfunctions

2

un(y) sinnπx

a

+ u00n(y) sinnπx

We expand f (x, y) in a Fourier sine series.

f (x, y) sinnπx

a

dx

We obtain the ordinary differential equations for the coefficients in the expansion

u00n(y) −nπ

a

2

un(y) = fn(y), u0n(0) = u0n(b) = 0, n ∈ Z+

We will solve these ordinary differential equations with Green functions

Consider the Green function problem,

a

satisfy the left and right boundary conditions, respectively We compute the Wronskian of these two solutions

W (y) =

cosh(nπy/a) cosh(nπ(y − b)/a)

nπ

a sinh(nπy/a) nπa sinh(nπ(y − b)/a)

= nπa

cosh

nπya

sinh nπ(y − b)



The Green function is

gn(y; η) = −a cosh(nπy</a) cosh(nπ(y>− b)/a)

The solutions for the coefficients in the expansion are

un(y) =

Z b 0

gn(y; η)fn(η) dη

Solution 37.20

utt+ a2uxxxx= 0, 0 < x < L, t > 0,u(x, 0) = f (x), ut(x, 0) = g(x),u(0, t) = uxx(0, t) = 0, u(L, t) = uxx(L, t) = 0,

We will solve this problem by expanding the solution in a series of eigen-solutions that satisfy the partial differentialequation and the homogeneous boundary conditions We will use the initial conditions to determine the coefficients inthe expansion We substitute the separation of variables, u(x, t) = X(x)T (t) into the partial differential equation

Here we make the assumption that 0 ≤ arg(λ) < π/2, i.e., λ lies in the first quadrant of the complex plane Note that

λ4 covers the entire complex plane We have the ordinary differential equation,

T00= −a2λ4T,and with the boundary conditions at x = 0, L, the eigenvalue problem,

X0000 = λ4X, X(0) = X00(0) = X(L) = X00(L) = 0

For λ = 0, the general solution of the differential equation is

X = c1+ c2x + c3x2+ c4x3

Only the trivial solution satisfies the boundary conditions λ = 0 is not an eigenvalue For λ 6= 0, a set of linearlyindependent solutions is

{eλx, eıλx, e−λx, e−ıλx}

Another linearly independent set, (which will be more useful for this problem), is

{cos(λx), sin(λx), cosh(λx), sinh(λx)}

Both sin(λx) and sinh(λx) satisfy the left boundary conditions Consider the linear combination c1cos(λx)+c2cosh(λx).The left boundary conditions impose the two constraints c1 + c2 = 0, c1 − c2 = 0 Only the trivial linear combination

of cos(λx) and cosh(λx) can satisfy the left boundary condition Thus the solution has the form,

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Only the trivial solution satisfies the boundary conditions λ = is not an eigenvalue For λ 6= 0, a set of linearlyindependent... class="text_page_counter">Trang 25< /span>

This will be maximized if

Since this is an inhomogeneous partial differential equation, we will expand the solution... eigenfunctions in

x for which the coefficients are functions of t The solution for u has the form,



Substituting this expression into the inhomogeneous partial differential equation