{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

kontsyslos090824

# kontsyslos090824 - LUNDS TEKNISKA HGSKOLA MATEMATIK...

This preview shows page 1. Sign up to view the full content.

LUNDS TEKNISKA HÖGSKOLA LÖSNINGAR MATEMATIK Kontinuerliga system 2009–08–24 1. Egenvektorer: ϕ k = e 5 x sin kπx, k = 1 , 2 , . . . Egenvärden: λ k = 25 + k 2 π 2 , k = 1 , 2 , . . . Skalärprodukt: ( u | v ) = integraltext 1 0 u ( x ) v ( x ) e 10 x dx Vid beräkning av ( ϕ 1 | ϕ 2 ) försvinner exponentialfunktionerna. 2. Låt D vara området 0 < x < L, 0 < y < B och sätt A u = - Δ u, D A = braceleftbig u C 2 ( D ) , u (0 , y ) = u ( L, y ) = 0 , u y ( x, 0) = u y ( x, B ) = 0 bracerightbig Egenvektorer: ϕ jk = sin jπx L cos kπy B , j = 1 , 2 , . . . k = 0 , 1 , 2 , . . . Egenvärden: λ jk = parenleftbigg L parenrightbigg 2 + parenleftbigg B parenrightbigg 2 Egenfrekvenser: f jk = c 2 π radicalbig λ jk , c = 335 Lägsta frekvens f 10 = c 2 L = 670 ger L = 0 . 25 och vi ser att f 20 = 2 f 10 = 1340 och f 30 = 3 f 10 = 2010 . Då måste 1498 = f 11 vilket ger B = 0 . 125 . Kontroll visar att f 21 = 1895 och att f 12 är större. 3. Efter homogenisering och udda spegling blir u ( x, t ) = T 0 (1 - erf ( x 4 at )) u ( L, L 2 a ) = T 0 (1 - erf ( 1 2 )) 0 . 5 T 0 4. Tidsoberoende värmeledningsproblem där q = 70 , λ = 0 . 2 och α = (1 , 2) . braceleftBigg - Δ u = q λ δ α , x 2 + y 2 < 25 u = 20
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern