Separation of Variables sdfgsdfg PDF

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MATH2021 Introduction to Applied Mathematics | Workshop Solutions Week 9

2

∂ 2u = k 2 e−kt cos nt cos mx + 2nke−kt sin nt cos mx − n2 e−kt cos nt cos mx ∂t2 ∂ 2u ∂u = −k 2 e−kt cos nt cos mx − n2 e−kt cos nt cos mx = −(k 2 + n2 )u ⇒ + 2k 2 ∂t ∂t and hence the equation requires that ⇒

u = (h2 − k 2 − n2 )✚ u −c2 m2✚

k 2 + n 2 = c 2 m 2 + h2

⇒

e) uuxy = ux uy , u(x, y) = f (x)g (y) for any differentiable functions f and g of one variable. Here ux = ∂x u = ∂x (f (x)g(y)) = f ′ (x)g(y) uy = ∂y u = ∂y (f (x)g(y)) = f (x)g ′ (y) uxy = uyx = ∂x (∂y u) = ∂x (f (x)g ′ (y)) = f ′ (x)g ′ (y) Hence, uuxy − ux uy = f (x)g(y)f ′ (x)g ′ (y) − f ′ (x)g(y)f (x)g ′ (y) = 0 2. Consider the partial differential equation ∂u ∂u + =0 ∂x ∂y Using the method of separation of variables, u(x, y) = X(x)Y (y), obtain ordinary differential equations for X(x) and Y (y) which are coupled via a free parameter λ. Solve these equations and hence find the solution to the partial differential equation which satisfies the boundary condition u(x, 0) = 2e3x We have

∂u ∂u = X ′ (x)Y (y) and = X(x)Y ′ (y), so ∂x ∂y

X ′ (x)Y (y) + X (x)Y ′ (y) = 0

Y ′ (x) X ′ (x) =− X(x) Y (x)

⇒

The left hand side is a function of x only and the right hand side is a function of y only. This can only be true if both equal a constant value, say λ. That is Y ′ (x) X ′ (x) =λ=− Y (x) X(x)

dX = λX, dx

⇒

dY = −λY dy

The solutions of these equations are X(x) = C1 eλx,

Y (y) = C2 e−λy

⇒

u(x, y) = Ceλ(x−y) ,

C = C1 C2

The auxiliary condition gives Ceλx = 2e3x

⇒

C = 2,

λ = 3,

⇒

u(x, y) = 2e3(x−y)

MATH2021 Introduction to Applied Mathematics | Workshop Solutions Week 9

3

3. Use the method of separation of variables to solve Laplace’s equation ∂ 2u ∂ 2u + =0 ∂x2 ∂y 2 subject to boundary conditions u(x, 0) = 0, u(x, b) = 0, Assume u(x, y) = X(x)Y (y). The PDE gives

u(0, y) = 0,

X ′′(x)Y (y) + X (x)Y ′′ (y) = 0

u(a, y ) = f (y). Y ′′(y) X ′′(x) =λ=− Y (y) X(x)

⇒

where λ is the separation constant. We address the differential equation for which we have homogeneous auxiliary conditions, in this case, that for Y (y). Proceeding as in the lectures and assuming λ > 0 (oscillatory solutions), we have Yn (y) = sin(

p

p

λn y),

λn =

n2 π 2 nπ ⇐⇒ λn = 2 , b b

n = 1, 2, · · ·

and for each there exists a corresponding Xn (x) solution: Xn′′ (x) − λn Xn (x) = 0

√ λn x

⇒

Xn (x) = cn e

√ λn x

+ dn e−

The x = 0 boundary condition gives u(0, y) = 0 ⇒ X (0)Y (y) = 0 ⇒ X(0) = 0 → dn = −cn so √ √ − λn x λn x Xn (x) = cn (e −e ). Hence, the solution of the PDE is u(x, y) = = =

∞ X

n=1 ∞ X

n=1 ∞

X

n=1

where sinh z =

√ λn x

cn (e

√ λn x

− e−

) sin(

p

cn (enπx/b − e−nπx/b ) sin 2cn sinh

 nπx b

sin

λn y)

 nπy

b  nπy  b

 1 z e − e−z . Now we fit the other boundary condition u(a, y) = f (y): 2 ∞  nπy  X   cn enπa/b − e−nπa/b sin = f (y) b n=1

Thus, Z b  nπy 2 f (y) sin dy cn = b b (enπa/b − e−nπa/b ) 0 Z b  nπy  1  =  nπa f (y) sin dy b b sinh b 0 4. Use the method of separation of variables to solve the initial-boundary value problem ∂u ∂ 2u ∂u + 5u, +2 = 2 ∂x ∂x ∂t

∂u (π, t) = 0, ∂x u(x, 0) = sin x, 0 < x < π u(0, t) = 0,

t>0...