Download H*Here we solve the partial differential equation for steady state

H*Here we solve the partial differential equation for steady state temperature
distribution in the semi-infinite slab with the Laplace equation “2 T@x,yD=
0 with the Boundary Conditions HBCL THx,0L=T0 =100, T@x,¶D=0, T@0,yD=0,
T@L,yD=0 where L=10. Assuming a solution of the form T@x,yD=
X@xD*Y@yD and using the 2D Laplacian “2 =
∂2
∂x2
+
∂2
∂y2
which gives X"@xDY@yD+X@xDY"@yD=
0. Dividing by X@xDY@yD≠0 we get X"@xDêX@xD+Y"@yDêY@yD=0,
which can only be satisfied if both terms are the same constant or X"@xDêX@xD=-Y"@yDêY@yD=
-k2 , where k is the separation constant. We can solve these two diffy-Q's using DSolve.*L
DSolve@X ''@xD + k ^ 2 * X@xD ã 0, X@xD, xD
88X@xD Ø C@1D Cos@k xD + C@2D Sin@k xD<<
DSolve@Y ''@yD - k ^ 2 * Y@yD ã 0, Y@yD, yD
99Y@yD Ø ‰k y C@1D + ‰-k y C@2D==
H*That is X@xD=+A*Sin@k*xD+B*Cos@k*xD and Y@yD=C*Exp@k*yD+D*Exp@-k*yD,
as in the notes. Demanding the BC's X@0D=0,
X@LD=0 gives B=0 and k=kn =npêL Hthe "quantization condition"L respectively,
where L=10. Demanding Y@¶D=
0 gives C=0. That gives us a final solution of T@x,yD=X@xD*Y@yD=A*D*Exp@-kn *yD*Sin@kn *xD,
where bn =A*D is a constant Hto be determinedL that depends on n = 0,
1,2,.... However the last BC,
that T@x,0D=X@xD*Y@0D=T0 =100=bn *Sin@kn *xD= can not be satisfied for any single choice of bn
since Sin@kn *xD is not a non-zero constant for any choice of kn or bn . The key
point is that since any solution of the form T@x,yD=X@xD*Y@yD=bn *Exp@-kn *yD*
Sin@kn *xD solves the partial diffy-q but no single one satisfies the BC,
we use superposition and linearity to construct a more general solution:
T@x,yD=X@xD*Y@yD=ڦ
n=0 bn *Exp@-kn *yD*Sin@kn *xD,
with the hope that the sum of the individual solutions solves the DiffyQ Hby defaultL AND the final BC.
The final BC for this new solution reads:
¶
T@x,0D=X@xD*Y@0D=ڦ
n=0 bn *Exp@-kn *0D*Sin@kn *xD=⁄n=0 bn *Sin@kn *xD=T0 =100.
This is the Fourier Sine Series for the function:
f@xD=T0 =ڦ
n=0 bn *Sin@kn *xD
Note the functions Sin@kn *xD=Sin@n*p*xêLD,
for n=0,2,3,.... are orthogonal on the interval xe@0,LD.*L
Integrate@Sin@n * Pi * x ê LD * Sin@m * Pi * x ê LD, 8x, 0, L<,
Assumptions Ø 8Element@n, IntegersD, Element@m, IntegersD, L > 0<D
L n Cos@n pD Sin@m pD - L m Cos@m pD Sin@n pD
m 2 p - n2 p
H*Which is zero if m≠n. For m=n we have:*L
2
13.2.nb
Integrate@Sin@n * Pi * x ê LD * Sin@n * Pi * x ê LD, 8x, 0, L<,
Assumptions Ø 8Element@n, IntegersD, Element@m, IntegersD, L > 0<D
1
L 2-
4
Sin@2 n pD
np
H*Which is Lê2 if n is an integer. Hence the functions §n\ = Sqrt@2êLD*Sin@n*Pi*xêLD
form an orthonormal set, such that Xn§m\
dnm . We can use these to find bn . Rewriting the BC
in terms of the orthonormal Sine functions we get:
¶
f@xD=T0 =ڦ
n=0 HSqrt@Lê2DL*bn M*HSqrt@2êLDSin@kn *xDL=⁄n=0 cn *•n],
where cn =Sqrt@Lê2DM*bn in the Dirac notation. Hence we can then use:
¶
¶
Xn•f]=⁄¶
m=0 cm *Xn•m]=⁄m=0 cm *Xn•m]=⁄m=0 cm *dnm =cn .
Hence we can now compute cn using f=T0 to get,
L
cn =Xn•T0 >ªŸ0 HSqrt@2êLDSin@kn *xDL*HT0 L*„x,
*L
cn = FullSimplify@Integrate@Sqrt@2 ê LD Sin@n * Pi * x ê LD * T0 , 8x, 0, L<D, 8n > 0, L > 0<D
-
2
L H- 1 + Cos@n pDL T0
np
bn = FullSimplify@Sqrt@2 ê LD * cn , 8n > 0, L > 0<D
-
2 H- 1 + Cos@n pDL T0
np
H*Note the conditional assumption
8n>0,L>0< forces n and L to be treated at positive real numbers,
simplifying the result. This is the result in the notes. Notice that this is zero if neven, and bn =4*T0 êHn pL if n-odd so we replace nØ2m+1 to get: b2m+1 =2*T0 *LêHH2*m+1L*PiL
which gives us T@x,yD=H4*T0 êpL*⁄¶
m=0 Exp@-H2*m+1L*pê10*yD*Sin@H2*m+1Lpê10*xDEë
H2m+1L. We can sum the first few terms by recalling T0 =
100 and defining a Fourier series partial sum.*L
s@M_, x_, y_D :=
400 ê Pi * Sum@Exp@- H2 * m + 1L * Pi ê 10 * yD * Sin@H2 * m + 1L p ê 10 * xD ê H2 * m + 1L, 8m, 0, M<D
H*We can plot the first 100 terms as a 3D plot:*L
13.2.nb
Plot3D@s@100, x, yD, 8x, 0, 10<, 8y, 0, 5<, PlotRange Ø 80, 100<, AxesLabel Ø 8x, y, T@x, yD<D
H*Click on the plot and you can rotate it with your
mouse. The plot in the x direction shows the oscillations die out
quickly with more terms but also that the function dives to zero at x=
0 and x=10 and clearly it is a step function at y=0 where T=
100 and then drops off exponentially with increasing y. There are no real oscillations
in the Laplace equation “2 T@x,yD=0 since their is no term proportional to T. For
oscillations we would solve the wave equation intead “2 T@x,yD+K^2*T@x,yD=0.*L
3