Pemrograman Komputer Persamaan Konduksi Panas Dimensi Satu

s t c o = ⇔ 2 . Jadi x k s o e s t s x u − = , ~ . Jadi L = , t x u -1 } , ~ { s x u = ⇔ , t x u L -1 ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ − x k s o e s t o t t x u = ⇔ , L -1 ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ − x k s e s 1 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ⇔ kt x erfc t t x u o 2 , . Jadi ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = kt x erfc t t x u o 2 , .

D. Pemrograman Komputer Persamaan Konduksi Panas Dimensi Satu

Berikut ini adalah pemrograman Maple untuk persamaan konduksi panas dimensi satu. Interval setengah tak terbatas pada kasus parabolik restart;withPDEtools: p1:=D2yx+yx=fx; := p1 = + D 2 y x y x f x p2:=yx=Axexpm[1]x+Bxexpm[2]x; := p2 = y x + A x e m 1 x B x e m 2 x s1:=DAxexpm[1]x+DBxexpm[2]x=0; := s1 = + D A x e m 1 x D B x e m 2 x s2:=m[1]DAxexpm[1]x+m[2]DBxexpm[2] x=fx; := s2 = + m 1 D A x e m 1 x m 2 D B x e m 2 x f x sol1:=solve{s1,s2},{DAx,DBx}; := sol1 ⎧ ⎩ ⎪⎪ ⎪⎪⎪⎨ ⎫ ⎭ ⎪⎪ ⎪⎪⎪⎬ , = D A x f x e m 1 x − + m 2 m 1 = D B x − f x e m 2 x − + m 2 m 1 sol2:=sol1[1]; := sol2 = D A x f x e m 1 x − + m 2 m 1 sol3:=sol1[2]; := sol3 = D B x − f x e m 2 x − + m 2 m 1 solu1:=dsolve{sol2},Ax; := solu1 = A x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ e −m 1 x f x − + m 2 m 1 x _C1 solu2:=dsolve{sol3},Bx; := solu2 = B x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − f x e −m 2 x − + m 2 m 1 x _C1 solu3:=subs_C1=_C2,solu2; := solu3 = B x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − f x e −m 2 x − + m 2 m 1 x _C2 pers1:=subssolu1,solu3,p2; := pers1 = y x + ⎛ ⎝ ⎜⎜ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟ ⎟⎟⎟ + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ e −m 1 x f x − + m 2 m 1 x _C1 e m 1 x ⎛ ⎝ ⎜⎜ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟ ⎟⎟⎟ + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − f x e −m 2 x − + m 2 m 1 x _C2 e m 2 x simplifypers1; y x e m 1 x d ⌠ ⌡ ⎮ ⎮ ⎮ e −m 1 x f x x e m 1 x _C1 m 2 e m 1 x _C1 m 1 e m 2 x d ⌠ ⌡ ⎮ ⎮ ⎮ f x e −m 2 x x − + − ⎛ ⎝ ⎜⎜ ⎜ = e m 2 x _C2 m 2 e m 2 x _C2 m 1 − + − + m 2 m 1 ⎞ ⎠ ⎟⎟ ⎟ restart;withinttrans: p:=diffux,s,x2-skux,s=0; := p = − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ ∂ ∂ 2 x 2 u , x s s u , x s k p1:=dsolvep,ux,s; := p1 = u , x s + _F1 s e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k _F2 s e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k p2:=subs_F1s=Ax,_F2s=Bx,p1; := p2 = u , x s + A x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k B x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k s1:=DAxexpsqrtskx+DBxexp- sqrtskx=0; := s1 = + D A x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x D B x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x s2:=sqrtskDAxexpsqrtskx- sqrtskDBxexp-sqrtskx=-fxk; := s2 = − s k D A x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x s k D B x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x − f x k s3:=solve{s1,s2},{DAx,DBx}; := s3 ⎧ ⎩ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪⎪ ⎨ ⎫ ⎭ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪⎪ ⎬ , = D A x − 1 2 f x s k e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k = D B x 1 2 f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x s k k sol1:=s3[1]; := sol1 = D A x − 1 2 f x s k e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k sol2:=s3[2]; := sol2 = D B x 1 2 f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x s k k so1:=dsolve{sol1},Ax; := so1 = A x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − 1 2 f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x s k k x _C1 so2:=dsolve{sol2},Bx; := so2 = B x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ 1 2 f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x s k k x _C1 so3:=subs_C1=_C2,so2; := so3 = B x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ 1 2 f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x s k k x _C2 pers1:=subsso1,so3,p2; pers1 u , x s = := + ⎛ ⎝ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − 1 2 f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x s k k x _C1 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k ⎛ ⎝ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ 1 2 f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x s k k x _C2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k pers2:=simplifypers1; pers2 u , x s 1 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x x 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k _C1 s k k − + ⎛ ⎝ ⎜⎜ ⎜⎜⎜ = := e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x x 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k _C2 s k k + + s k k ⎞ ⎠ ⎟⎟ ⎟⎟⎟ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ per1:=subs_C1=0,pers2; per1 u , x s = := 1 2 − + + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x x 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k _C2 s k k s k k invlaplaceper1,x,s; invlaplace , , u , x s x s 1 2 ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ invlaplace , , e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x x x s − ⎛ ⎝ ⎜⎜ ⎜⎜⎜ = ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ invlaplace , , e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x x x s 2 _C2 s k k ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Dirac − s s k k + + ⎞ ⎠ ⎟⎟ ⎟⎟⎟ s k k ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Interval terbatas pada kasus parabolik restart;withPDEtools: D2yx+yx=fx; = + D 2 y x y x f x p:=Axcoshx+Bxsinhx; := p + A x cosh x B x sinh x p1:=DAxcoshx+DBxsinhx=0; := p1 = + D A x cosh x D B x sinh x p2:=DAxsinhx+DBxcoshx=fx; := p2 = + D A x sinh x D B x cosh x f x p3:=solve{p1,p2},{DAx,DBx}; := p3 ⎧ ⎩ ⎪⎪ ⎪⎪⎨ ⎫ ⎭ ⎪⎪ ⎪⎪⎬ , = D A x − 1 2 f x − e x 2 1 e x = D B x 1 2 + e x 2 1 f x e x s1:=p3[1]; := s1 = D A x − 1 2 f x − e x 2 1 e x s2:=p3[2]; := s2 = D B x 1 2 + e x 2 1 f x e x sol1:=dsolve{s1},Ax; := sol1 = A x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ 1 2 f x − + e x e −x x _C1 sol2:=dsolve{s2},Bx; := sol2 = B x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ 1 2 f x + e x e −x x _C1 sol3:=subs_C1=_C2,sol2; := sol3 = B x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ 1 2 f x + e x e −x x _C2 pers1:=subssol1,sol3,p; pers1 := + ⎛ ⎝ ⎜⎜ ⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟ + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ 1 2 f x − + e x e −x x _C1 cosh x ⎛ ⎝ ⎜⎜ ⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟ + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ 1 2 f x + e x e −x x _C2 sinh x pers2:=simplifypers1; pers2 1 2 cosh x d ⌠ ⌡ ⎮ ⎮ f x − + e x e −x x cosh x _C1 1 2 sinh x d ⌠ ⌡ ⎮ ⎮ f x + e x e −x x + + := sinh x _C2 + restart;withPDEtools:withinttrans: p:=diffux,s,x2-skux,s=-fxk; := p = − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ ∂ ∂ 2 x 2 u , x s s u , x s k − f x k p1:=dsolvep,ux,s; p1 u , x s d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − 1 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k f x s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ 1 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k f x s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k + = := _F1 s e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k _F2 s e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k + + pp1:=subs_F1=Ax,_F2=Bx,p1; pp1 u , x s d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − 1 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k f x s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ 1 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k f x s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k + = := A x s e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k B x s e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k + + p2:=DAxcoshsqrtskx+DBxsinhsqrt skx=0; := p2 = + D A x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ cosh s k x D B x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ sinh s k x p3:=sqrtskDAxsinhsqrtskx+sqrtsk DBxcoshsqrtskx=-fxk; := p3 = + s k D A x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ sinh s k x s k D B x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ cosh s k x − f x k s1:=solve{p2,p3},{DAx,DBx}; := s1 ⎧ ⎩ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎨ ⎫ ⎭ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎬ , = D A x 1 2 f x ⎛ ⎝ ⎜⎜ ⎜ ⎞ ⎠ ⎟⎟ ⎟ − ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x 2 1 s k k e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x = D B x − 1 2 f x ⎛ ⎝ ⎜⎜ ⎜ ⎞ ⎠ ⎟⎟ ⎟ + ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x 2 1 s k k e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x s2:=s1[1]; := s2 = D A x 1 2 f x ⎛ ⎝ ⎜⎜ ⎜ ⎞ ⎠ ⎟⎟ ⎟ − ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x 2 1 s k k e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x s3:=s1[2]; := s3 = D B x − 1 2 f x ⎛ ⎝ ⎜⎜ ⎜ ⎞ ⎠ ⎟⎟ ⎟ + ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x 2 1 s k k e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x sol1:=dsolves2,Ax; := sol1 = A x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − 1 2 f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x s k k x _C1 sol2:=dsolves3,Bx; := sol2 = B x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − 1 2 f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x s k k x _C1 sol3:=subs_C1=_C2,sol2; := sol3 = B x + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − 1 2 f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x s k k x _C2 solu1:=subssol1,sol3,pp1; solu1 u , x s d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − 1 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k f x s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ 1 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k f x s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k + = := ⎛ ⎝ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − 1 2 f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x s k k x _C1 s e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k + ⎛ ⎝ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜ ⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎟⎟ + d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ − 1 2 f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x s k k x _C2 s e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k + solu2:=simplifysolu1; solu2 u , x s 1 2 d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k f x x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k s s k s k s − ⎛ ⎝ ⎜⎜ ⎜⎜⎜ = := d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k f x x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k s s k s k s + s k e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x x s − 2 s k e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k _C1 s s s k s k s + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k s k ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x x s − 2 e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k s k _C2 s s s k s k s + s k s s k s k s ⎞ ⎠ ⎟⎟ ⎟⎟⎟ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ solu3:=invlaplacesolu2,x,s; solu3 invlaplace , , u , x s x s 1 2 ⎛ ⎝ ⎜⎜ ⎜⎜⎜ = := s s k s k s ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ invlaplace , , d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k f x x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k x s − s s k s k s ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ invlaplace , , d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k f x x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k x s + s k ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x x s ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Dirac + s s k k − 2 s k _C1 s s s k s k s ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Dirac + s s k k + s k ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x x s ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Dirac − + s s k k − 2 s k _C2 s s s k s k s ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Dirac − + s s k k + s k s s k s k s ⎞ ⎠ ⎟⎟ ⎟⎟⎟ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ simplifysolu3; invlaplace , , u , x s x s 1 2 s s k s k s ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ invlaplace , , d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k f x x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k x s ⎛ ⎝ ⎜⎜ ⎜⎜⎜ − = s s k s k s ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ invlaplace , , d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x k f x x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x k x s − s k ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x x s ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Dirac + s k s k k + 2 s k _C1 s s s k s k s ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Dirac + s k s k k − s k ⎛ ⎝ ⎜⎜ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟⎟⎟ d ⌠ ⌡ ⎮ ⎮ ⎮ ⎮ f x ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ + e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ s k x e ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − s k x x s ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Dirac − + s k s k k + 2 s k _C2 s s s k s k s ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Dirac − + s k s k k − s k s s k s k s ⎞ ⎠ ⎟⎟ ⎟⎟⎟ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ Contoh 2. Gambarlah u pada contoh 1 sebagai sebuah permukaan dalam ruang yang mempuyai temperatur dan persamaan konduksi panasnya C t o o 27 = 2 2 2 x u t u ∂ ∂ = ∂ ∂ untuk 2 ≤ ≤ t dengan panjang L x ≤ ≤ , 2 , = L . Penyelesaian: withplots: Warning, the name changecoords has been redefined u1:=ux,t=gammaerfx2sqrtkt; := u1 = u , x t γ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ erf 1 2 x k t u2:=subsgamma=27,k=2,u1; := u2 = u , x t 27 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ erf 1 4 x 2 t plot3drhsu2,x=0..0.4,t=0..2,style=hidden,orienta tion=[25,30],title=Gambar 2; Contoh 3. Gambarlah u pada contoh 1 sebagai sebuah permukaan dalam ruang yang mempuyai temperatur dan persamaan konduksi panasnya C t o o 35 = 2 2 8 x u t u ∂ ∂ = ∂ ∂ untuk 8 , ≤ ≤ t dengan panjang L x ≤ ≤ , 1 = L . Penyelesaian: withplots: Warning, the name changecoords has been redefined p1:=ux,t=gammaerfx2sqrtkt; := p1 = u , x t γ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ erf 1 2 x k t p2:=subsgamma=35,k=8,p1; := p2 = u , x t 35 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ erf 1 16 x 8 t plot3drhsp2,x=0..1,t=0..0.8,style=hidden,orienta tion=[35,40],title=Gambar 3; Contoh 4. Gambarlah u pada contoh 1 sebagai sebuah permukaan dalam ruang yang mempuyai temperatur dan pada persamaan konduksi panas C t o o 37 = C t o o 38 = 2 2 8 x u t u ∂ ∂ = ∂ ∂ untuk 8 , ≤ ≤ t dengan panjang L x ≤ ≤ , 1 = L . Penyelesaian: withplots: Warning, the name changecoords has been redefined

p:=ux,t=gammaerfx2sqrtkt;

:= p = u , x t γ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ erf 1 2 x k t p1:=subsgamma=37,k=8,p; := p1 = u , x t 37 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ erf 1 16 x 8 t p1:=subsgamma=38,k=8,p; := p1 = u , x t 38 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ erf 1 16 x 8 t plot3d{rhsp1,rhsp2},x=0..1,t=0..0.8,style=hidd en,orientation=[35,40],axes=FRAMED,title=Gambar 4;

BAB V PENUTUP

A. Simpulan

Pemodelan persamaan konduksi panas dimensi satu adalah , dimana k adalah konstan. Bentuk transformasi Laplace dari masalah nilai batas pada persamaan konduksi panas dimensi satu adalah. tt t ku u = 1. Interval setengah tak terbatas pada kasus parabolik ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ∫ ∫ − − − dx e x f e k s k dx e x f e k s k e c s x u x k s x k s x k s x k s x k s 2 1 2 1 , ~ 2 dengan syarat batas: ~ , ~ , ~ s g dx s u d s u = + β α . 2. Interval terbatas pada kasus parabolik x k s dx e e x f k s k c s x u x k s x k s cosh 2 1 , ~ 1 ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − − = ∫ − x k s dx e e x f k s k c x k s x k s sinh 2 1 2 ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − + ∫ − , dengan syarat batas: ~ , ~ , ~ 1 s g dx s u d s u = + β α , dan 53 ~ , ~ , ~ 2 s g dx s l u d s l u = + β α . Sedangkan penyelesaian bentuk transformasi Laplace dari masalah nilai batas pada persamaan konduksi panas dimensi satu adalah. 1. Interval setengah tak terbatas pada kasus parabolik = , t x u L -1 ⎪ ⎪ ⎩ ⎪⎪ ⎨ ⎧ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ∫ − − dx e x f e k s k e c x k s x k s x k s 2 1 2 ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∫ − dx e x f e k s k x k s x k s 2 1 2. Interval terbatas pada kasus parabolik = , t x u L -1 ⎪ ⎪ ⎩ ⎪⎪ ⎨ ⎧ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − − ∫ − x k s dx e e x f k s k c x k s x k s cosh 2 1 1 ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − + ∫ − x k s dx e e x f k s k c x k s x k s sinh 2 1 2

B. Saran

Pada penulisan skripsi ini, permasalahan hanya dibatasi penyelesaian masalah nilai batas pada persamaan diferensial parsial linear orde dua dengan kasus parabolik pada persamaan konduksi panas dimensi satu