Assignment Solutions of Partial Difierential Equations

1 Assignment Solutions of Partial Differential EquationsWeijiu LiuDepartment of MathematicsUniversity of Central Arkansas201 Donaghey Avenue, Conway, AR 72035, USA1 Assignment Derive the heat equation for a rod assuming constant thermal properties with variablecross-sectional areaA(x) assuming no byAthe the cross-sectional quantities: Thermal energy densitye(x, t) = the amount of thermal energy per unit volume. Heat flux (x, t) = the amount of thermal energy flowing across boundaries per unitsurface area per unit time. Heat sourcesQ(x, t) = 0. Temperatureu(x, t). Specific heatc= the heat energy that must be supplied to a unit mass of a substanceto raise its temperature one unit. Mass density (x) = mass per unit volume. Fourier s Law: the heat flux is proportional to the temperature gradient = K0 u.(1)Conservation of heat energy:Rate of change of heat energy in time = Heat energy flowing across boundaries per unittime + Heat energy generated insider per unit time heat energy =e(x, t)A(x) x.

2 Heat energy flowing across boundaries per unit time = (x, t)A(x) (x+ x, t)A(x+ x).Then t[e(x, t)A(x) x] = (x, t)A(x) (x+ x, t)A(x+ x).Dividing it by xand letting xgo to zero giveA(x) e t= A(x) x (x) A x.(2)Heat energy per unit mass =c(x)u(x, t) A x. Soe(x, t)A(x) x=c(x)u(x, t) A(x) x,and thene(x, t) =c(x)u(x, t) .2It then follows from Fourier s law thatc A(x) u t=A(x) x(K0 u x)+K0 A x u x.(3)and then the heat equationA(x) u t=k(A(x) 2u x2+ A x u x),(4)wherek=K0c is the thermal Consider a thin one-dimensional rod without source of thermal energy whose lateralsurface is not insulated. Letw(x, t) dente the heat energy flowing out of the lateral sidesper unit surface area per unit time. Assume thatw(x, t) is proportional to the temperaturedifference between the rodu(x, t) and a known outside temperature (x, t). Derive theequation for the byAthe the cross-sectional area, andPthe lateral quantities: Thermal energy densitye(x, t) = the amount of thermal energy per unit volume.

3 Heat flux (x, t) = the amount of thermal energy flowing across boundaries per unitsurface area per unit time. Temperatureu(x, t). Specific heatc= the heat energy that must be supplied to a unit mass of a substanceto raise its temperature one unit. Mass density (x) = mass per unit volume. Fourier s Law: the heat flux is proportional to the temperature gradient = K0 u.(5)Conservation of heat energy:Rate of change of heat energy in time = Heat energy flowing across boundaries per unittime + Heat energy generated insider per unit time heat energy =e(x, t)A x. Heat energy flowing across boundaries per unit time = (x, t)A (x+ x, t)A. Heat energy flowing out of the lateral sides per unit time =w(x, t)P x= [u(x, t) (x, t)]h(x)P x, whereh(x) is a t[e(x, t)A(x) x] = (x, t)A(x) (x+ x, t)A(x+ x) [u(x, t) (x, t)]h(x)P it byA xand letting xgo to zero give e t= x PA[u(x, t) (x, t)]h(x).(6)Heat energy per unit mass =c(x)u(x, t) A x. Soe(x, t)A x=c(x)u(x, t) A x,and thene(x, t) =c(x)u(x, t).

4 It then follows from Fourier s law thatc u t= x(K0 u x) PA[u(x, t) (x, t)]h(x).(7)42 Assignment Determine the equilibrium temperature distribution for a one-dimensional rod withconstant thermal properties with the following sources and boundary conditions:(a)Q= 0, u(0) = 0, u(L) =T.(f)Q=K0x2, u(0) =T, u (L) = (a) Equilibrium satisfiesu (x) = 0,whose general solution isu=c1+ boundary conditionu(0) = 0 impliesc1= 0 andu(L) =Timpliesc2=T/Lso thatu=Tx/L.(f) In equilibrium,usatisfiesu (x) = Q/K0= x2,whose general solution (by integrating twice) isu= x4/12 +c1+ boundary conditionu(0) =Tyieldsc1=T,whileu (L) = 0 yieldsc2=L3/3. Thusu= x4/12 +L3x/3 + Suppose u t= 2u x2+x, u(x,0) =f(x), u x(0, t) = , u x(L, t) = 7.(a) Calculate the total thermal energy in the one-dimensional rod (as a function of time).(b) From part (a), determine a value of for which an equilibrium exists. For this valueof , determine limt u(x, t).Solution. (a) Integrating the equation, we obtain:ddt L0u(x, t)dx= L0( 2u x2+x)dx= u x L0+12L2= 7 + intfrom 0 tot, we obtain the total thermal energy L0u(x, t)dx= L0f(x)dx+(7 +12L2)t.

5 (8)(b) In order for an equilibrium to exist,(7 +12L2)tmust be 0. So = 7 + equilibrium satisfies (x) +x= general solution (after integrating twice) is = 16x3+c1+ boundary condition yieldsc2= 7 + = 16x3+c1+(7 +12L2) u(x, t) = (x),using (8), we obtain L0f(x)dx= L0u(x, t)dx= limt L0u(x, t)dx= L0 (x)dx= L0( 16x3+c1+(7 +12L2)x)dx= 124L4+c1L+12(7 +12L2) it givesc1= L0f(x)dx 72L2 524L4L,and then = 16x3+(7 +12L2)x+ L0f(x)dx 72L2 For conduction of thermal energy, the heat flux vector is = K0 inaddition the molecules move at an average velocityV, a process called convection, then = K0 u+c the corresponding equation for heat flow, including both conductionand convection of thermal energy (assuming constant thermal properties with no sources). quantities: Thermal energy densitye(x, t) = the amount of thermal energy per unit volume. Heat flux (x, t) = the amount of thermal energy flowing across boundaries per unitsurface area per unit Heat sourcesQ(x, t) = heat energy per unit volume generated per unit time.

6 Temperatureu(x, t). Specific heatc= the heat energy that must be supplied to a unit mass of a substanceto raise its temperature one unit. Mass density (x) = mass per unit of heat energy:Rate of change of heat energy in time = Heat energy flowing across boundaries per unittime + Heat energy generated insider per unit time heat energy = Re(x, t)dV. Heat energy flowing across boundaries per unit time = ndS. Heat energy generated insider per unit time = RQ(x, t)dV= t Re(x, t)dV= divergence theorem give t Re(x, t)dV= R then e t= .(9)Heat energy per unit volume =c(x)u(x, t) . Soe(x, t) =c(x)u(x, t) .It then follows thatc u t= (c uV) + (K0 u).(10)and then the convection-diffusion equation u t+ (uV) =k 2u,(11)wherek=K0c is called the thermal Assignment (d) Find the eigenvalues and the corresponding eigenfunctions of the eigenvalue prob-lem d2 dx2= , (0) = 0,d dx(L) = 0.(i) If >0, =c1cos( x) +c2sin( x). (0) = 0 impliesc1= 0, whiled dx(L) = 0impliesc2 cos( L) = 0.

7 Thus L= 2+n (n= 1,2, ). Then the eigenvaluesare n=( 2+n )2/L2and the corresponding eigenfunctions are n= sin(( 2+n )xL)(n= 1,2, ).(ii) If = 0, =c1+c2x. (0) = 0 impliesc1= 0, whiled dx(L) = 0 impliesc2= 0. Thus = 0 is not an eigenvalue.(ii) If <0, =c1exp( x) +c2exp( x). (0) = 0 impliesc1+c2= 0, whiled dx(L) = 0 impliesc1 exp( L) c2 exp( L) = 0. Solving this system forc1, c2givesc1=c2= 0. Thus <0 is not an (c) Solve the initial boundary value problems: u t=k 2u x2,(12)u(0, t) = 0, u(L, t) = 0,(13)u(x,0) = 2 cos(3 xL).(14)Solution. The problem has infinite series solutionu(x, t) = n=1cnsin(n xL)e kn2 initial condition yields2 cos(3 xL)=u(x,0) = n=1cnsin(n xL).Socn=2L L02 cos(3 xL)sin(n xL) (a) The equation for the equilibrium iskd2udx2 u= 0,(15)u(0) = 0, u(L) = 0.(16)8 The general solution isu=c1exp( kx) +c2exp( kx).u(0) = 0 impliesc1+c2= 0,whileu(L) = 0 impliesc1exp( kL) +c2exp( kL) = 0. Solving this system forc1, c2givesc1=c2= 0.

8 Thusu= 0.(b) Separation of variable,u= (x)G(t), yields two ODEs:dGdt= ( k+ )Gand d2 dx2= , (0) = 0, (L) = has solutionG(t) =Ce te eigenvalue problem has the eigenvalues n=n2 2L2, n= 1,2, (17)and the corresponding eigenfunctions n= sin(n xL), n= 1,2, (18)Thus by superposition,u(x, t) =e t n=1cnsin(n xL)e kn2 initial condition givesf(x) =u(x,0) = n=1cnsin(n xL),which givescn=2L L0f(x) sin(n xL) e kn2 2tL2= 0,limt e t= 0,we havelimt u(x, t) = (x, t) converges to the only equilibrium Assignment (a). The solution isu(x, t) =a0+ n=1ancos(n xL)e kn2 LL/2dx=12,an=2L LL/2cos(n xL)dx=2L Ln sin(n xL) LL/2= 2n sin(n 2). (b). The solution isu(x, t) =a0+ n=1ancos(n xL)e kn2 L0(6 + 4 cos(3 xL))dx= 6,a3=2L L0(6 + 4 cos(3 xL))cos(n xL)dx= 4,and others are u t=k 2u x2,(19) udx(0, t) = 0, u(L, t) = 0,(20)u(x,0) =f(x).(21)Look for a solution of the form of separation of variables:u(x, t) = (x)G(t),(22)Substitute the above expression into the equation (19), we obtain (x)G (t) =k (x)G(t),and thenG (t)kG(t)= (x) (x)= ,(23)10where is constant to be determined.

9 The boundary condition (20) yields that udx(0) = (L) = then have an eigenvalue problem d2 dx2= , dx(0) = 0, (L) = Equations :m2= . Case 1: <0. Distinct real rootsm1= andm2= : (x) =c1e x+c2e boundary conditions imply thatc1=c2= 0. So no non-zero Solutions exist andthen <0 is not an eigenvalue. Case 2: = 0. Repeated real rootsm1=m2= 0: =c1+ boundary conditions imply thatc1=c2= 0. So no non-zero Solutions exist andthen <0 is not an eigenvalue. Case 3: >0. Conjugate complex rootsm1=i andm2= i : =c1cos( x) +c2sin( x). (0) = 0 implies thatc2= 0. (L) = 0 givescos( L) = L= 2+n (n= 0,1,2, ) and then we obtain the eigenvalues n=( 2+n )2L2, n= 0,1,2, (24)and the corresponding eigenfunctions n= cos(( 2+n )xL), n= 0,1,2, (25)11On the other hand, it follows from (23) thatdGdt= kG,(26)which has solutionsG(t) =ce kt=ce ( 2+n ) then derive the infinite series solution:u(x, t) = n=1ancos(( 2+n )xL)e ( 2+n ) initial condition givesf(x) =u(x,0) = n=1ancos(( 2+n )xL).

10 (27)To determinean, we multiply (27) by cos(( 2+n )xL)and integrate from 0 toL. We thenfindan=2L L0f(x) cos(( 2+n )xL)dx, n 1.(28) (c) Solve Laplace s equation 2u x2+ 2u y2= 0,(29) u x(0, y) = 0, u(L, y) =g(Y), u(x,0) = 0, u(x, H) = 0,(30)Look for a solution of the form of separation of variables:u(x, y) =h(x) (y),(31)Substitute the above expression into the equation (29), we obtain (y)h (x) + (y)h(x) = 0,and thenh (x)h(x)= (y) (y)= ,(32)where is constant to be determined. The boundary condition yields that (0) = (H) = 0, h (0) = then have an eigenvalue problem12 d2 dx2= , (0) = 0, (H) = 0,which has the eigenvalues n=n2 2H2, n= 1,2, (33)and the corresponding eigenfunctions n= sin(n yH), n= 1,2, (34)On the other hand, it follows from (32) thath (x) =n2 2H2h(x), h (0) = 0.(35)The general Solutions areh(x) =c1en xH+c2e n boundary conditionh (0) = 0 givesc1n H c2n H= (y) =c1(en xH+e n xH).We then derive the infinite series solution:u(x, t) = n=1ansin(n yH)(en xH+e n xH).