[PDF] Solving the Heat Equation (Sect. 10.5). Review: The Stationary Heat

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Solving the Heat Equation (Sect. 10.5).

?Review: The Stationary Heat Equation. ?The Heat Equation. ?The Initial-Boundary Value Problem. ?The separation of variables method. ?An example of separation of variables.Review: The Stationary Heat Equation. Review:The Stationary Heat Equation describes the temperature distribution in a solid material in thermal equilibrium. The temperature is time-independent.Problem:The time-independent temperature,T, of a bar of lengthLwith insulated horizontal sides and vertical extremes kept at fixed temperaturesT0,TL, is the solution of the BVP:T ??(x) = 0,x?(0,L),T(0) =T0,T(L) =TL,T 0 xLy z xL insulation insulation 0 T Remark:The heat transfer occurs only along thex-axis.

Solving the Heat Equation (Sect. 10.5).

?Review: The Stationary Heat Equation. ?The Heat Equation. ?The Initial-Boundary Value Problem. ?The separation of variables method. ?An example of separation of variables.The Heat Equation.

Remarks:

?The unknown of the problem isu(t,x), the temperature of the bar at the timetand positionx. ?The temperaturedoes notdepend onyorz.?The one-dimensional Heat Equation is: tu(t,x) =k∂2xu(t,x),wherek>0 is the heat conductivity,units:[k] =(distance)2(time). ?The Heat Equation is a Partial Differential Equation, PDE.0 u(t,x)t t t u = 0u < 0u > 0 t is held constant. xLu

Solving the Heat Equation (Sect. 10.5).

?Review: The Stationary Heat Equation. ?The Heat Equation. ?The Initial-Boundary Value Problem. ?The separation of variables method. ?An example of separation of variables.The Initial-Boundary Value Problem.

Definition

TheIBVP for the one-dimensional Heat Equationis the following: Given a constantk>0 and a functionf: [0,L]→Rwith f(0) =f(L) = 0, findu: [0,∞)×[0,L]→Rsolution of tu(t,x) =k∂2xu(t,x),I.C.:u(0,x) =f(x),B.C.:u(t,0) = 0,u(t,L) = 0.u ( t, L ) = 0 xt 0Lt x2d u = k d u u ( 0, x ) = f ( x )u ( t, 0 ) = 0

Solving the Heat Equation (Sect. 10.5).

?Review: The Stationary Heat Equation. ?The Heat Equation. ?The Initial-Boundary Value Problem. ?The separation of variables method. ?An example of separation of variables.The separation of variables method.

Summary:

?The idea is to transform the PDE into infinitely many ODEs. ?We describe this method in 6 steps.

Step 1:

One looks for solutionsugiven by an infinite series of simpler functions,u n,that is, u(t,x) =∞? n=1c nun(t,x),whereu nis simpler thanuis the sense, u n(t,x) =vn(t)wn(x).Herecnare constants,n= 1,2,···.

The separation of variables method.

Step 2:

Introduce the series expansion foruinto the Heat Equation, tu-k∂2xu= 0? n=1c n?∂tun-k∂2xun?= 0.A sufficient condition for the equation above is:To findu n, for n= 1,2,···,solutions of tun-k∂2xun= 0.Step 3: Findu n(t,x) =vn(t)wn(x)solution of the IBVP tun-k∂2xun= 0.I.C.:u n(0,x) =wn(x),B.C.:u n(t,0) = 0,un(t,L) = 0.The separation of variables method.

Step 4:(Key step.)

Transform the IBVP foru

ninto:(a)IVP forv n;(b)BVP forw n.

Notice:

tun(t,x)=∂t?vn(t)wn(x)?=wn(x)dvndt (t).∂

2(x).Therefore, the equation∂tun=k∂2xunis given by

w n(x)dvndt (t) =k vn(t)d2wndx 2(x)1 k v n(t)dv ndt (t) =1w n(x)d 2wndx

2(x).Depends only ont=Depends only onx.

The separation of variables method.

Recall:1

k v n(t)dv ndt (t) =1w n(x)d 2wndx

2(x).Depends only ont=Depends only onx.?The Heat Equation has the following property:

The left-hand side depends only ont, while the right-hand side depends only onx.?When this happens in a PDE, one can use the separation of variables method on that PDE.?We conclude that for appropriate constantsλ mholds 1 k v n(t)dv ndt (t) =-λn,1 w n(x)d 2wndx

2(x) =-λn.?We have transformed the original PDE into infinitely many

ODEs parametrized byn, positive integer.The separation of variables method. Summary Step 4:The originalIBVPfor the Heat Equation, PDE, is transformed into:(a)The IVP forv n, 1 k v n(t)dv ndt (t) =-λn,I.C.:v n(0) = 1.(b)The BVP forw n, 1 w n(x)d 2wndx

2(x) =-λn,B.C.:w

n(0) = 0,wn(L) = 0.Step 5: (a)Solve the IVP forv n. (b)Solve the BVP forw n.

The separation of variables method.

Step 5(a):Solving the IVP forv

n. v ?n(t) +kλnvn(t) = 0,I.C.:v n(0) = 1.The integrating factor method implies thatμ(t) =ekλnt. e kλntv?n(t) +kλnekλntvn(t) = 0? ?ekλntvn(t)? ?= 0.e kλntvn(t) =cn?vn(t) =cne-kλnt.1 =vn(0) =c?v n(t) =e-kλnt.The separation of variables method.

Step 5(a):Recall:v

n(t) =e-kλnt.

Step 5(b):Eigenvalue-eigenvector problem forw

n: Find the eigenvaluesλnand the non-zero eigenfunctionswn solutions of the BVPw ??n(x) +λnwn(x) = 0B.C.:w n(0) = 0,wn(L) = 0.We know that this problem has solution only forλ n>0.

Denote:λ

n=μ2n.Proposingwn(x) =ernx,we get that p(rn) =r2n+μ2n= 0?r n±=±μniThe real-valued general solution is w n(x) =c1cos(μnx) +c2sin(μnx).

The separation of variables method.

Recall:v

n(t) =e-kλnt,w n(x) =c1cos(μnx) +c2sin(μnx).

The boundary conditions imply,

0 =wn(0)=c1?wn(x) =c2sin(μnx).0 =wn(L)=c2sin(μnL),c

2?= 0,?sin(μnL) = 0.μ

nL=nπ?μn=nπL?λ n=?nπL

2.Choosingc2= 1, we getw

n(x) = sin?nπxL

We conclude that:u

n(t,x) =e-k(nπL )2tsin?nπxL ?,n= 1,2,···.The separation of variables method.

Step 6:Recall:u

n(t,x) =e-k(nπL )2tsin?nπxL Compute the solution to the IBVP for the Heat Equation, u(t,x) =∞? n=1c nun(t,x).u(t,x) =∞? n=1c ne-k(nπL )2tsin?nπxL .By construction, this solution satisfies the boundary conditions, u(t,0) = 0,u(t,L) = 0.Given a functionfwithf(0) =f(L) = 0, the solutionuabove satisfies the initial conditionf(x) =u(0,x) iff holdsf(x) =∞? n=1c nsin?nπxL?

The separation of variables method.

Recall:

u(t,x) =∞? n=1c ne-k(nπL )2tsin?nπxL ?,f(x) =∞? n=1c nsin?nπxL This is a Sine Series forf. The coefficientscnare computed in the usual way.Recall the orthogonality relation L 0 sin?nπxL sin?mπxL dx=? ?0,m?=n, L2 ,m=n.Multiply the equation foruby sin?mπxL nd integrate, n=1c n? L 0 sin?nπxL sin?mπxL dx=? L 0 f(x) sin?mπxL dx.c n=2L L 0 f(x) sin?nπxL dx,u(t,x) =∞? n=1c ne-k(nπL )2tsin?nπxL .The separation of variables method.

Summary:IBVP for the Heat Equation.

Propose:

u(t,x) =∞? n=1c nvn(t)wn(x).where ?v n: Solution of an IVP. ?w n: Solution of a BVP, an eigenvalue-eigenfunction problem. ?c n: Fourier Series coefficients.

Remark:

The separation of variables method does not work for every PDE.

Solving the Heat Equation (Sect. 10.5).

?Review: The Stationary Heat Equation. ?The Heat Equation. ?The Initial-Boundary Value Problem. ?The separation of variables method. ?An example of separation of variables.An example of separation of variables.

Example

Find the solution to the IBVP 4∂tu=∂2xu,t>0,x?[0,2], u(0,x) = 3sin(πx/2),u(t,0) = 0,u(t,2) = 0.Solution:Letun(t,x) =vn(t)wn(x).Then

4wn(x)dvdt

(t) =vn(t)d2wdx 2(x)?

4v?n(t)v

n(t)=w??n(x)w n(x)=-λn.The equations forvnandwnarev ?n(t) +λn4 vn(t) = 0,w ??n(x) +λnwn(x) = 0.We solve forvnwith the initial conditionvn(0) = 1. e n4 tv?n(t) +λn4 eλ n4 tvn(t) = 0? ?eλ n4 tvn(t)??= 0.

An example of separation of variables.

Example

Find the solution to the IBVP 4∂tu=∂2xu,t>0,x?[0,2], u(0,x) = 3sin(πx/2),u(t,0) = 0,u(t,2) = 0.Solution:Recall:? eλ n4 tvn(t)??= 0.Therefore, v n(t) =c e-λn4 t,1 =vn(0)=c?v n(t) =e-λn4 t.Next the BVP:w ??n(x) +λnwn(x) = 0, withw n(0) =wn(L) = 0. Sinceλn>0, introduceλn=μ2n.The characteristic polynomial is p(r) =r2+μ2n= 0?r n±=±μni.The general solution,w n(x) =c1cos(μnx) +c2sin(μnx).

The boundary conditions imply

0 =wn(0) =c1,?wn(x) =c2sin(μnx).An example of separation of variables.

Example

Find the solution to the IBVP 4∂tu=∂2xu,t>0,x?[0,2], u(0,x) = 3sin(πx/2),u(t,0) = 0,u(t,2) = 0.Solution:Recall:v n(t) =e-λn4 t, andw n(x) =c2sin(μnx).

0 =wn(2) =c2sin(μn2),c

2?= 0,?sin(μn2) = 0.Then,μn2 =nπ,that is,μn=nπ2

.Choosingc2= 1, we conclude,λ m=?nπ2

2,wn(x) = sin?nπx2

.u(t,x) =∞? n=1c ne-(nπ4 )2tsin?nπx2

An example of separation of variables.

Example

Find the solution to the IBVP 4∂tu=∂2xu,t>0,x?[0,2],quotesdbs_dbs20.pdfusesText_26

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