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Second Order PDEs

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Linear PDEs Of The Second Order With Constant Coefficients

A second order lnear PDE with constant coefficients is given by:

auxx+buxy+cuyy+dux+euy+fu=g(x,y)

where at least one of a, b and c is non-zero.

  • If b24ac>0, then the equation is called hyperbolic.

The wave equation a2uxx=utt is an example.

  • If b24ac=0, then the equation is called parabolic.

The heat equation α2uxx=ut is an example.

  • b24ac<0, then the equation is called elliptic.

The Laplace equation uxx+uyy=0 is an example.

Canonical Forms

A second-order linear PDE can be reduced to its canonical form by an appropriate change of variables ξ=ξ(x,y) and η=η(x,y).

If the Jacobian J=|ξxξyηxηy| is non-zero, we can solve for x and y as functions of ξ and η.

Using these transformations, the second order PDE a(x,y)uxx+2b(x,y)uxy+c(x,y)uyy+d(x,y)ux+e(x,y)uy+f(x,y)u=g(x,y), can be reduced to the below canonical form:

Awξξ+2Bwξη+Cwηη+Dwξ+Ewη+Fw=G(ξ,η)

where

A=aξx2+2bξxξy+cξy2
B=aξxηx+b(ξxηy+ξyηx)+cξyηy
C=aηx2+2bηxηy+cηy2
D=aξxx+2bξxy+cξyy+dξx+eξy
E=aηxx+2bηxy+cηyy+dηx+eηy
F=f(x(ξ,η),y(ξ,η))
G=g(x(ξ,η),y(ξ,η))

PYQs

Linear PDEs Of The Second Order With Constant Coefficients

1) Solve the partial differential equation:

(2D25DD+2D2)z=5sin(2x+y)+24(yx)+e3x+4y

where Dx, Dy.

[2018, 15M]


2) Solve (D22DDD2)z=ex+2y+x3+sin2x where Dx, Dy, D22x2, D22y2.

[2017, 10M]


3) Let Γ be a closed curve in xyplane and let S denote the region bounded by the curve Γ. Let 2wx2+2wy2=f(x,y)(x,y)S. If f is prescribed at each point (x,y) of S and w is prescribed on the boundary Γ of S then prove that any solution w=w(x,y), satisfying these conditions, is unique.

[2017, 10M]


4) Solve the partial differential equation 3zx323zx2y3zxy2+23zy3=ex+y.

[2016, 15M]


5) Solve: (D2+DD2D)u=ex+y, where D=x and D=y.

[2015, 15M]


6) Solve the partial differential equation (2D25DD+2D2)z=24(yx).

[2014, 10M]


7) Solve (D2+DD6D2)z=x2sin(x+y) where D and D denote x and y.

[2013, 15M]


8) Solve the PDE (D2D2+D+3D2)z=e(xy)x2y.

[2011, 12M]


9) Solve the PDE (D2D)(D2D)Z=e2x+y+xy.

[2010, 12M]


10) Solve: (D2DD2D2)z=(2x2+xyy2)sinxycosxy where D and D represent x and y and y.

[2009, 15M]


11) Find the general solution of the partial differential equation: (D2+DD6D2)z=ycosx, where Dx, Dy.

[2008, 12M]


12) Solve: 3zx343zx2y+43zxy2=2sin(3x+2y).

[2006, 12M]


13) Solve the partial differential equation: 2zx22zxy22zy2=(y1)ex.

[2004, 15M]


14) Find the general solution of 2zx2+32zxy+22zy2=x+y+cos(2x+3y).

[2003, 12M]


Canonical Forms

1) Reduce the following second order partial differential equation to canonical from and find the general solution:

2ux22x2uxy+x22uy2=uy+12x

[2019, 20M]


2) Reduce the equation y22zx22xy2zxy+x22zy2=y2xzx+x2yzy to canonical form and hence solve it.

[2017, 15M]


3) Find the solution of the initial-boundary value problem:
utuxx+u=0, 0<x<l, t>0 u(0,t)=u(l,t)=0, t0
u(x,0)=x(lx), 0<x<l

[2015, 15M]


4) Reduce the second-order partial differential equation x22ux22xy2uxy+y22uy2+xux+yuy=0 into canonical form. Hence, find its general solution.

[2015, 15M]


5) Reduce the equation 2zx2=x22zy2 to canonical form.

[2014, 15M]


6) Reduce the equation y2zx2+(x+y)2zxy+x2zy2=0 to its canonical from when xy.

[2013, 10M]


7) Reduce the following 2nd order partial differential equation into canonical form and find find its general solution.
xuxx+2x2uxyux=0.

[2010, 20M]


8) Reduce 2zx2=x22zy2 canonical form.

[2008, 15M]


9) Solve the equation x22zx2y22zy2+xzxyzy=x2y4 by reducing it to the equation with constant coefficients.

[2001, 20M]


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