Matematika 4 4. vaja B. Jurˇciˇc Zlobec

Matematika 4

4. vaja

B. Jurˇciˇc Zlobec¹

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

Matematika FE, Ljubljana, Slovenija 17. april 2013

Reˇsi parcialno diferencialno enaˇ cbo

u(x , y ) = 0

I ∂²

∂x²u(x,y) = 0.

I ∂

∂xu(x,y) =f(y).

I u(x,y) =f(y)x+g(y).

Reˇsi parcialno diferencialno enaˇ cbo

= 0

I ∂²

∂x∂yu(x,y) = 0.

I ∂

∂xu(x,y) =f⁰(x).

I u(x,y) =f(x) +g(y).

Reˇsi parcialno diferencialno enaˇ cbo

+

= 0

I ∂²

∂x∂yu(x,y) + ∂

∂xu(x,y) = 0.

I Uvedemo novo spremenljivkou(x,y)_x =v(x,y).

I v(x,y)_y+v(x,y) = 0→v(x,y) =f⁰(x)e^−y.

I ux(x,y) =f⁰(x)e^y →u(x,y) =f(x)e^−y+g(y).

I u(x,y) =f(x)e^−y +g(y).

Reˇsi parcialno diferencialno enaˇ cbo

+ u(x, y ) = 0

I ∂²

∂x²u(x,y) +u(x,y) = 0.

I u(x,y) =f(y) cosx+g(y) sinx.

Reˇsi parcialno diferencialno enaˇ cbo

= x

I ∂

∂yu(x,y) =x

I u(x,y) =x y+f(x).

Reˇsi parcialno diferencialno enaˇ cbo

∂²u(x,y)

∂x∂y

+

+ x + y = 0

I ∂²u

∂x∂yu(x,y) +∂u

∂xu(x,y) +x+y = 0.

I Uvedemo novo spremenljivko

u(x,y)_x =v(x,y)→v(x,y)_y +v =−x−y.

I Reˇsitev homogene enaˇcbe v_hy +v_h= 0 jev_h=f⁰(x)e^−y.

I Partikularna reˇsitevv(x,y) =C(y)e^y.

I v(x,y)y+v(x,y) =−x−y →C⁰(y)e^−y−C(y)e^−y + C(y)e^−y =−x−y →C⁰(y) = (−x−y)e^y →

I C(y) =−xe^y −ye^y+e^y →v(x,y) =−x−y+ 1.

I u(x,y)x =v(x,y) =f⁰(x)e^−y−x−y+ 1→

I u(x,y) =f(x)e^−y −x²

2 −x y −x+g(y).

Reˇsi parcialno diferencialno enaˇ cbo

= 2x y u(x , y )

I Vzamemo, da jey konstanta.

I Navadna diferencialna enaˇcba ima loˇcljive spremenljivke.

I Piˇsemo v(x) =u(x,y)→

I v⁰= 2x y v → ^dv_v = 2x y dx.

I ln|v|=x²y+ ln|C| →v(x) =C e^x2y.

I u(x,y) =C(y)e^x2y.

Reˇsi parcialno diferencialno enaˇ cbo

+ x

= y

I Uvedemo novo spremenljivko in vzamemo, da je y konstanta.

I v(x) =u(x,y)x →v+xv⁰ =y →x^dv_dx =y−v.

I Loˇcimo spremenljivke _y−v^dv = ^dx_x .

I ln|y−v|= ln|x|+ ln|C| →y−v =C x→v =y−Cx.

I u(x,y)x =y+f(y)x →u(x,y) =x y +f(y)^x₂² +g(y).

I u(x,y) =x y+f(y)x²

2 +g(y).

Reˇsi parcialno diferencialno enaˇ cbo

+

= xy

I Uvedemo novo spremenljivko in vzamemo, da je x konstanta.

I v(y) =u(x,y)y →v⁰+v =x y.

I Reˇsitev homogene enaˇcbe jev(y) =Ce^−y.

I Nastavek za partikularno reˇsitevv(y) =C(y)e^−y.

I C⁰(y)e−y =x y →C⁰(y) =x y e^y →C(y) =x(ye^y−e^y)

I v(y) =x y −x→v(y) =x y −x+Ce^−y.

I u(x,y)x =x y −x+f(x)e^−y →u(x,y) =

1

2x y²−x y +f(x)e^−y +g(x).

I u(x,y) = 1

2x y²−x y+f(x)e^−y +g(x).

Vpelji nove spremenljivke in reˇsi enaˇ cbo

∂²u(x,y)

∂x²

− 2

+

= 0, t = x , z = x + y .

I ∂u

∂x = ∂u

∂t

∂t

∂x +∂u

∂z

∂z

∂x.

I ∂u

∂y = ^∂u_∂t∂y^∂t +^∂u_∂z^∂z_∂y.

I u_x =u_t+u_z u_y =u_z→u_xx = (u_t+u_z)_t+ (u_t+u_z)_z= utt+ 2utz +uzz, uyy =uzz →uxy = (ut+uz)z =utz+uzz.

I uxx−2u_xy+uyy =utt+2utz+uzz−2u_tz−2u_zz+uzz =utt = 0.

I u =f(z)t+g(z)→u(x,y) =f(x+y)x+g(x+y).

I u(x,y) =f(x+y)x+g(x+y).

Vpelji nove spremenljivke in reˇsi enaˇ cbo

x

− y

= 2u(x , y ), t = x

, z = x y

I ∂u

∂x = 2x^∂u_∂x +y^∂u_z → ^∂u_∂y =x^∂u_z

I ∂u

∂x = 2√

t^∂u_∂x +^√z_t^∂u_z → ^∂u_∂y =√ t^∂u_∂z.

I √ t2√

tu_t+√ t^√z

tu_z−^√z

t

√tu_z = 2u

I 2tut = 2u → ^du_u = ^dt_t →u=tf(z)

I u(x,y) =x²f(xy).

Vpelji nove spremenljivke in reˇsi enaˇ cbo

∂²u(x,y)

∂x²

+

− 2

= 0, t = x + y , z = 2x − y

.

I u_x =u_t+ 2u_z u_y =u_t−u_z →

I u_xx = (u_t+ 2u_z)_t+ 2(u_t+ 2u_z)_z =u_tt+ 4u_tz + 4u_zz, u_yy = (u_t−u_z)_t−(u_t−u_z)_z =u_tt −2u_tz+u_zz, uxy = (ut+ 2uz)t−(ut+ 2uz)z.

I uxx−2uxy+uyy =utt + 2utz+uzz−2utz −2uzz+uzz → utt = 0.

I u =f(z)t+g(z)→u(x,y) =f(2x−y)(x+y) +g(2x−y).

I u(x,y) =f(2x−y)(x+y) +g(2x−y).

Poiˇsˇ ci reˇsitev diferencialne enaˇ cbe

+

= 0 v obliki u(x , y ) = X (x)Y (y )

I Vstavimou(x,y) =X(x)Y(y), X⁰(x)Y(y) +X(x)Y⁰(y) = 0→

I X⁰(x)

X(x) +Y⁰(y) Y(y) = 0.

I Velja X⁰(x)

X(x) =−Y⁰(y) Y(y) =k,

I kjer je k parameter nodvisen od x in y.

I dX

X =k dx in ^dY_Y =−k dy →

I X(x) =Ae^kx inY(y) =Be^−ky →

I u(x,y) =Ce^k(x−y)

Poiˇsˇ ci reˇsitev diferencialne enaˇ cbe x

u

+ 3y

u = 0 v obliki u = XY

I x²X⁰Y⁰+ 3y²XY = 0→

I x²X⁰ X

Y⁰

Y + 3y² = 0→

I x^2X_X⁰ =−3y²_Y^Y0 =k,

I kjer je k parameter nodvisen od x in y.

I dX

X =k^dx_x2 in ^dY_Y =−3y^2dy_k dy →

I X(x) =Ae^−kx in Y(y) =Be^−y

3

k →

I u(x,y) =Ce^−kx−y

3 k

Poiˇsˇ ci reˇsitev diferencialne enaˇ cbe u

+ yu

= 0 v obliki u = XY , ki ustreza pogojema u(1, 0) = 1 in u(0, 1) = 2.

I X⁰Y +yXY⁰ = 0→ X⁰

X +yY⁰

Y = 0→

I X⁰

X =−y^Y_Y⁰ =k, kjer je k parameter nodvisen od x in y.

I dX

X =k dx in ^dY_Y =−y dy →

I X(x) =Ae^kx inY(y) =Be^−ky

2

2 → u(x,y) =Ce^kx−ky

2 2 .

I u(1,0) =Ce^k = 1 in u(0,1) =Ce^−k2 = 2

I k =−¹₃log 4,C = 4^1/3.