First Order PDEs
We will cover following topics
Types of first order PDEs
First order PDEs are classified into following four types:
(i) Linear: Linear in p, q and z, and coefficients of p and q are independent functions of x and y.
It is of the form P(x,y)p+Q(x,y)q=R(x,y)z+S(x,y). For example, p+(x+y)q−z=ex.
(ii) Semilinear: Linear in p and q, and and coefficients of p and q are independent functions of x and y. It is of the form P(x,y)p+Q(x,y)q=R(x,y,z).
For example, p+(x+y)q−z3xy.
(iii) Quasilinear: Linear in p and q. It is of the form P(x,y,z)p+Q(x,y,z)q=R(x,y,z).
For example, pz+(x+y)q−z3xy=ex.
(iv) Nonlinear: PDEs which are not Linear. For example, p2+q2=1.
Types of solutions of PDEs
(i) Complete Solution or Complete Integral: An expression of the type f(x,y,z,a,b)=0 which is also a solution of the PDE.
For example, the expression f(x,y,z,a,b)=z−(ax+by+a2+b2) is a solution to the PDE z−px−qy−p2−q2=0.
(ii) General Solution or General Integral: An expression of the form f(ϕ,ψ)=0 which is also a solution to the PDE.
For example, the expression f(ϕ,ψ)=0, where ϕ=yx and ψ=zx, is a solution to the PDE z−xp−yq.
(iii) Singular Solution or Singular Integral: It is the envelope of the complete integral, if the envelope exists.
Quasilinear first order PDEs
The general solution of a quasilinear first order PDE a(x,y,u)ux+b(x,y,u)uy=c(x,y,u) is given by f(ϕ,ψ)=0, where f is an arbitrary function of ϕ(x,y,u) and ψ(x,y,u), and ϕ=c1, ψ=c2 are the solution curves of the characteristic equations
dxa=dyb=dzcThe solution curves defined by ϕ(x,y,u)=c1 and ψ(x,y,u)=c2 are called the families of characteristic curves of the given quasilinear PDE.
Non-linear first order PDEs
Charpit’s method for nonlinear first order PDEs
Let f(x,y,z,p,q)=0 be the given PDE.
We first write down the Charpit’s auxiliary equations, i.e.,
dp∂f∂x+p∂f∂z=dp∂f∂y+q∂f∂z=dz−p(∂f∂p)−q(∂f∂q)=dx−∂f∂p=dy−∂f∂q=dF0,
and select any two partial fractions from these equations to find p and q. We then write dz=pdx+qdy and integrate this to find the complete integral of the PDE.
Cauchy’s Method Of Characteristics
According to Cauchy’s method of characteristics, the characteristic curves of the quasilinear PDE a(x,y,z)∂z∂x+b(x,y,z)∂z∂y=c(x,y,z) are given by the equations:
dxa(x,y,z)=dyb(x,y,z)=dzc(x,y,z)PYQs
Quasilinear first order PDEs
1) Find a complete integral of the partial differential equation 2(pq+yp+qx)+x2+y2=0.
[2017, 15M]
2) Final the general integral of the partial differential equation (y+zx)p−(x+yz)q=x2−y2.
[2016, 10M]
3) Solve the partial differential equation: (y2+z2−x2)p−2xyq+2xz=0, where p=∂z∂x and q=∂z∂y and q=∂z∂y.
[2015, 10M]
4) Solve for the general solution pcos(x+y)+qsin(x+y)=z, where p=∂z∂x and q=∂z∂y.
[2015, 15M]
5) Solve the PDE (x+2z)∂z∂x+(4zx−y)∂z∂y=2x2+y.
[2011, 12M]
6) Find the surface satisfying the PDE (D2−2DD′+D′2)Z=0 and the conditions that bZ=y2 when x=0 and aZ=x2 when y=0.
[2010, 12M]
7) Find complete and singular integrals of 2xz−px2−2qxy+pq=0 using Charpit’s method.
[2008, 15M]
8) Solve 2zx−px2−2qxy+pq=0.
[2007, 6M]
9) Solve: px(z−2y2)=(z−qy)(z−y2−2x3).
[2006, 12M]
10) Solve the equation p2x+q2y=z, p=∂z∂x, q=∂z∂y.
[2006, 15M]
11) Obtain the general solution of (D−3D′−2)2z=2e2xsin(y+3x) where D=∂∂x and D′=∂∂y.
[2005, 30M]
12) Using Charpit’s method, find the complete solution of the partial differential equation p2x+q2y=z.
[2004, 15M]
13) Find two complete integrals of the partial differential equation x2p2+y2q2−4=0.
[2002, 12M]
14) Find the solution of the equation z=12(p2+q2)+(p−x)(q−y).
[2002, 12M]
15) Find the complete integral of the partial differential equation 2p2q2+3x2y2=8x2q2(x2+y2).
[2001, 12M]
16) Find the general integral of the equation {my(x+y)−nz2}∂z∂x−{lx(x+y)−nz2}∂z∂y=(lx−my)z.
[2001, 12M]
Cauchy’s Method of Characteristics
1) Solve the first order quasilinear partial differential equation by the method of characteristics:
x∂u∂x+(u−x−y)∂u∂y=x+2y in x>0,−∞<u<∞ with u=1+y on x=1.
[2019, 15M]
2) Find the general solution of the partial differential equation
(y3x−2x4)p+(2y4−x3y)q=9z(x3−y3),
where p=∂z∂x, q=∂z∂y, and its integral surface that passes through the curve: x=t, y=t2, z=1.
[2018, 15M]
3) Determine the characteristics of the equation z=p2−q2 and find the integral surface which passes though the parabola 4z+x2=0.
[2016, 15M]
4) Solve the following partial differential equation, zp+yq=x, x0(s)=s, y0(s)=1, z0(s)=2s, by the method of characteristics.
[2010, 20M]
5) Find the characteristics of: y2r−x2t=0 where r and t have their usual meanings.
[2009, 15M]
6) Find the general solution of the partial differential equation (2xy−1)p+(z−2x2)q=2(x−yz) and also find the particular solution which passes through the lines x=1, y=0.
[2008, 12M]
7) Find the particular integral of x(y−z)p+y(z−x)q=z(x−y) which represents a surface passing through x=y=z.
[2005, 12M]
8) Find the complete integral of the partial differential equation (p2+q2)x=pz and deduce the solution which passes through the curve x=0, z2=4y.
[2004, 12M]
9) Find the characteristic strips of the equation xp+yq−pq=0 and then find the equation of the integral surface through the curve z=x2, y=0.
[2002, 20M]
10) Prove that for the equation z+px+qy−1−pqx2y2=0 the characteristic strips are given by x(t)=1B+Ce−t, y(t)=1A+De−t, z(t)=E−(AC+BD)e−t, p(t)=A(B+Ce−t)2, q(t)=B(A+De−t)2 where A, B, C, D and E are arbitrary constants. Hence find the values of these arbitrary constants if the integral surface passes the line z=0, x=y.
[2001, 30M]