Laboratorijske vaje Matematika 4

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5. Vaja parcialne diferencialne enaˇcbe

B. Jurˇciˇc Zlobec¹

1Univerza v Ljubljani, Fakulteta za Elektrotehniko 1000 Ljubljana, Trˇzaˇska 25, Slovenija

Matematika 4 FE, Ljubljana, 10. maj 2013

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Laboratorijske vaje Matematika 4 Parcialne diferencialne enaˇcbe

Loˇ cene spremenljivke

Poiˇsˇci omejene reˇsitve, ki imajo loˇcene spremenljivke

∂²U(x,t)

∂x² = 1 c²

∂²U(x,t)

∂t² U[x_,t_] = X[x] T[t];

{D[U[x, t], {x, 2}], 1/c^2*D[U[x, t], {t, 2}]}

%/U[x, t]// Flatten Map[# == -omega^2 &, %]

MapThread[DSolve[#1,#2,#3]&,{%,{X[x],T[t]},{x,t}}]

sol=%//Flatten

{X[x] -> C[1] Cos[omega x] + C[2] Sin[omega x], T[t] -> C[1] Cos[c omega t] + C[2] Sin[c omega t]}

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Lastna nihanja

Poiˇsˇci lastna nihanja strune pri danih robnih pogojih.

∂²U(x,t)

∂x² = ∂²U(x,t)

∂t² , u(0,t) = 0, u(π,t) = 0, ut(x,0) = 0;

Vzamemo reˇsitevsoliz prejˇsnje strani in upoˇstevamo robne pogoje.

Clear[U, x, t, n]

U[x_, t_, n_] := Sin[n x] Cos[n t];

Plot[U[x, 0, 3], {x, 0, Pi}, Axes -> False, PlotRange -> {-1, 1}]

Manipulate[

Plot[U[x, t, n], {x, 0, Pi}, Axes -> False,

PlotRange -> {-1, 1}], {t, 0, 4 Pi}, {n, 1, 4, 1}]

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Slika

0.5 1.0 1.5 2.0 2.5 3.0

-1.0 -0.5 0.5 1.0

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Laboratorijske vaje Matematika 4 Robni problemi

Sploˇsna nihanja vpete strune

Reˇsi diferencialno enaˇcbo.

∂²U(x,t)

∂x² = 1 c²

∂²U(x,t)

∂t² ,

u(0,t) = 0, u(L,t) = 0, u(x,0) =f(x), ut(x,0) = 0;

Clear[f,g,x,u,t]; M=8; L=1; c=1;

f[x_]=6 x^3 (1 - x); g[x_]=0;

lam=Table[n Pi/L , {n, 1, M}];

A = 2/L Integrate[f[x] Sin[lam x], {x, 0, L}];

B = 1/(lam c) 2/L Integrate[g[x] Sin[lam x], {x, 0, L}];

u[x_,t_]=(A Cos[lam c t]+B Sin[lam c t]).Sin[lam x];

Plot[u[x, 0], {x, 0, L}]

Animate[Plot[u[x, t],{x, 0, L},PlotStyle->Thick, PlotRange->{{0,L},{-1,1}}],{t,0,(2L)/c,(2L)/(100c)}]

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Reˇsi Laplaceovo diferencialo enaˇ cbo na kolobarju.

∆u =urr +1 rur + 1

r²u_φφ,

∆u(r, φ) = 0, u(a, φ) =f(φ), u(b, φ) =g(φ), a<b, u[r, φ] =R(r)Φ(φ),→ R⁰⁰

R + 1 r

R⁰ R + 1

r² Φ⁰

Φ = 0 Φ⁰

Φ =−ω², r²R⁰⁰+rR⁰−ω²R= 0

Φn(φ) =A⁰_ncos(nφ)+B_n⁰ sin(nφ), Rn(r) =C_n⁰rⁿ+D_n⁰r⁻ⁿ, n∈N Φ0(φ) =A⁰₀, R0(r) =C0+D0log(r);

un(r, φ) = (Anrⁿ+Bnr⁻ⁿ) cos(nφ) + (Cnrⁿ+Dnr⁻ⁿ) sin(nφ), u₀(r, φ) =A₀+B₀log(r)

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u(r, φ) =

∞

X

n=0

un(r, φ) u(a, φ) =f(φ), u(b, φ) =g(x) f(φ) =A⁰₀+

∞

X

n=1

A⁰_ncos(nφ) +B_n⁰sin(nφ)

g(φ) =A⁰⁰₀ +

∞

X

n=1

A⁰⁰_ncos(nφ) +B_n⁰⁰sin(nφ)

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Razvoj v fourierevo vrsto

A⁰₀=A₀+B₀log(a) = 1 2π

Z _π

−π

f(φ)dφ

A⁰⁰₀ =A0+B0log(b) = 1 2π

Z π

−π

g(φ)dφ

A⁰_n=Anaⁿ+Bna⁻ⁿ= 1 π

Z π

−π

cos(nφ)f(φ)dφ

B_n⁰ =Cnaⁿ+Dna⁻ⁿ= 1 π

Z π

−π

sin(nφ)f(φ)dφ

A⁰⁰_n =Anbⁿ+Bnb⁻ⁿ= 1 π

Z π

−π

cos(nφ)g(φ)dφ

B_n⁰⁰ =Cnbⁿ+Dnb⁻ⁿ= 1 π

Z π

−π

sin(nφ)g(φ)dφ

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Sistema enaˇ cb

Reˇsimo sistem enaˇcb

A⁰₀=A0+B0log(a), A⁰⁰₀ =A0+B0log(b) A⁰_n=Anaⁿ+Bna⁻ⁿ, A⁰⁰_n=Anbⁿ+Bnb⁻ⁿ B_n⁰ =Cnaⁿ+Dna⁻ⁿ, B_n⁰⁰=Cnbⁿ+Dnb⁻ⁿ

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Primer

a= 1, b= 2,f(φ) = sin[φ] ing(φ) = sin(2φ) Funkciji sta lihi zato A⁰_n= 0 in A⁰⁰_n= 0 za vse n.

∞

X

n=1

B_n⁰ sin(nφ) = sin(φ),

∞

X

n=1

B_n⁰⁰sin(nφ) = sin(2φ) Obe vrsti sta konˇcni v prvi jeB₁⁰ = 1 B_n⁰ = 0 zan6= 1, v drugi je B₂⁰⁰= 1 in B_n⁰⁰= 0 zan 6= 2

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Reˇsitev

a=1;b=2;

s1=Solve[{C1 a+D1/a==1,C1 b+D1/b==0},{C1,D1}];

s2=Solve[{C2 a^2+D2/a^2==0,C2 b^2+D2/b^2==1},{C2,D2}];

U[r_,phi_] =((C1 r + D1 r^(-1)) Sin[phi]/.s1) + ((C2 r^2 + D2 r^(-2)) Sin[2phi]/.s2);

ParametricPlot3D[{r Cos[phi],r Sin[phi],U[r,phi]}, {r,a,b},{phi,-Pi,Pi}]

DensityPlot[U[Sqrt[x^2+y^2],ArcTan[x,y]],{x,-2,2}, {y,-2,2},

ColorFunction->Function[{u},RGBColor[u/2,u,1-u]], RegionFunction->Function[{x, y},1<Sqrt[x^2+y^2]<2]]