312Differential Equations 2301312: ISE Program, Chulalongkorn Univer- sity: First semester 2007
Partial Differential Equations (PDE)
Linear second order with constant coefficients
a∂2u
∂x2 +b ∂2u
∂x∂y +c∂2u
∂y2 +d∂u
∂x +e∂u
∂y =f(x, y) (1) Classification:
(1.) Ifb2−4ac >0, Eq. (1) is called hyperbolic equation.
Example: Wave equationsutt =uxx; vibration of elastic string.
(2.) Ifb2−4ac= 0, Eq. (1) is called parabolic equation.
Example: Heat equation ut=uxx; heat conduction.
(3.) Ifb2−4ac <0, Eq. (1) is called elliptic equation.
Example: Laplace equation uxx+uyy = 0; potential equation.
Method of separating variables for PDE We seek for special solution u(x, y) in the form
u(x, y) =X(x)Y(y). (2)
Substituting (2) back to the given PDE will give us two ordinary differential equations (ODEs). Solve these ODEs to get the answer.
Example: Laplace equation uxx+uyy = 0.
Using the form (2) we get uxx = X00Y, and uyy = XY00, where primes refer to ordinary differentiation w.r.t. the independent variables, whether x or y. Substituting back to Laplace equation:
X00Y +XY00= 0 =⇒ X00
X =−Y00 Y .
Since x and y are independent, the above equation is valid only when XX00 =
−YY00 = const. = σ, then we have two ODEs:
X00−σX = 0 and Y00+σY = 0
Heat Conduction Problem Model Equation
α2uxx =ut, 0< x < L, t >0, (3) whereu=u(x, t) istemperature of a straight bar at positionxat timet,αis thermal diffusivity. α2 = ρsκ has value depending on materials, κ = thermal conductivity, s= specific heat of material, and ρ= density.
Initial condition: (initial temperature distribution)
u(x,0) =f(x) 0≤x≤L; (4)
Boundary conditions: (temperature at the ends are fixed at 0)
u(0, t) = 0, u(L, t) = 0, t >0. (5) Heat conductivity problem consists of Eqs. (3), (4), and (5).
Solving Heat Conduction Problem
We seek nontrivial solutions using method of separating variables, 1
We get two ODES for X(x) and T(t),
X00+σX = 0, (6)
T0+α2σT = 0, (7)
Boundary conditions give, for all t >0,
u(0, t) =X(0)T(t) = 0 =⇒ X(0) = 0, u(L, t) =X(L)T(t) = 0 =⇒ X(L) = 0.
Boundary value problem forX(x):
( X00+σX = 0, 0< x < L
X(0) =X(L) = 0. (8)
We consider 3 possibilities for the constant σ:
1. σ = 0: We have
X00= 0 =⇒ X(x) =k1x+k2, 0≤x≤L.
Boundary conditions X(0) = X(L) = 0 give trivial solution X ≡0.
2. σ < 0: Set σ =−λ2 where λ >0, we get
X00−λ2X = 0 =⇒ X(x) =k1coshλx+k2sinhλx.
Again, boundary conditions give k1 = k2 = 0, and we have only trivial solution X(x)≡0 for this case.
3. σ > 0: Set σ =λ2 where λ >0, and get
X00+λ2X = 0 =⇒ X(x) = k1cosλx+k2sinλx.
Boundary conditions give
0 = X(0) =k1, 0 =X(L) = k2sin(λL).
To have nontrivial solution we need k2 6= 0 and sin(λL) = 0. This requires that λ = nπL, n= 1,2,3, . . .. The values of
σ =λ2 = n2π2
L2 (n= 1,2,3, . . .) 2
are called eigenvalues of the problem (8). The nontrivial solution Xn(x) := sin(nπx
L ) (n = 1,2,3, . . .) are called eigenfunctions of (8).
Substitutingσ back to Eq. (7), we get T0+ n2π2α2
L2 T = 0 =⇒ T(t) =C
:=Tn(t)
z }| { e−
n2π2α2t L2
Combine Xn(x) and Tn(t) we get un(x, t) := Xn(x)Tn(t) = e−
n2π2α2t
L2 sin(nπx
L ) (n = 1,2,3, . . .).
The functions un(x, t) satisfy the heat equation (3) and boundary conditions (5) for all n= 1,2,3, . . .. By superposition principle
u(x, t) :=
X∞ n=1
cnun(x, t) = X∞ n=1
cne−
n2π2α2t
L2 sin(nπx
L ) (9)
also satisfies Eqs (3) and (5). To be a solution for the heat conduction problem, (3)-(5), we use the initial condition (4) to get
f(x) = u(x,0) = X∞ n=1
cnsin(nπx
L ), 0< x < L.
Note: We solve for cn by looking at the Fourier sine series forf.
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