Question

solve the following by monges method , xy( t-r)+(x^2-y^2)(s-2)=py-qx

Comments

Are you assuming p=∂z/∂x, q=∂z/∂y, r=∂²z/∂x², s=∂²z/∂x∂y, t=∂²z/∂y²?

Answer 1

Using the equivalences in the comments, we can also write the following partial derivatives with some manipulation:

r=∂p/∂x, s=∂p/∂y and s=∂q/∂x, t=∂q/∂y from the definitions of p=∂z/∂x and q=∂z/∂y.

p and q are functions of x and y only, so only partial derivatives wrt x and y are required.

And we can relate partial to full derivatives (for p(x,y) and q(x,y)):

dz=(∂z/∂x)dx+(∂z/∂y)dy=pdx+qdy; similarly:

dp=rdx+sdy, dq=sdx+tdy. From these we get: r=(dp-sdy)/dx and t=(dq-sdx)/dy.

The given PDE needs to be written in the form: Rr+Ss+Tt=V:

-xyr+(x²-y²)s+xyt=2x²-2y²+py-qx, so:

R=-xy, S=x²-y², T=xy, V=2x²-2y²+py-qx.

Rr+Ss+Tt=R(dp-sdy)/dx+Ss+T(dq-sdx)/dy=V,

Rdy(dp-sdy)+Ssdxdy+Tdx(dq-sdx)=Vdxdy,

Rdydp-sRdy²+sSdxdy+Tdxdq-sTdx²-Vdxdy=0,

Rdydp+Tdxdq-Vdxdy=s(Rdy²-Sdxdy+Tdx²).

One possible solution to these equations is:

Rdydp+Tdxdq-Vdxdy=0 and Rdy²-Sdxdy+Tdx²=0 (Monge's subsidiary equations).

That is:

-xydydp+xydxdq-(2x²-2y²+py-qx)dxdy=0 and:

-xy(dy)²-(x²-y²)dxdy+xy(dx)²=0,

(dy)²+(x/y-y/x)dxdy-(dx)²=0.

Consider (dy-m₁dx)(dy-m₂dx)=(dy)²-(m₁+m₂)dxdy+m₁m₂(dx)²=0.

So m₁+m₂=y/x-x/y and m₁m₂=-1; m₁=y/x, m₂=-x/y, making m₁m₂=-1.

dy-(ydx/x)=0, xdy=ydx; or, dy+xdx/y=0, ydy=-xdx. We have two auxiliary integrals:

(1) dy/y-dx/x=0; integrating: ln|y|-ln|x|=k, ln|y/x|=k, or y/x=e^k, which we can write y/x=a, a constant, so y=ax and dy=adx.

(2) ydy+xdx=0; integrating: ½y²+½x²=c, a positive constant (because the sum of the squares of two real numbers is always positive), which can be written x²+y²=b², where c=b².

Therefore, ydy=-xdx, and y=√(b²-x²), so dy=-xdx/y=-xdx/√(b²-x²).

We can use these to make substitutions in the other subsidiary equation.

(1)⇒(1i) -x(ax)(adx)dp+x(ax)dxdq-(2x²-2(ax)²+p(ax)-qx)dx(adx)=0, (substituting y=ax and dy=adx)

-a²x²dxdp+ax²dxdq-a(2x²-2a²x²+apx-qx)(dx)²=0,

-a²x²dxdp+ax²dxdq-2ax²(dx)²+2a³x²(dx)²-a²px(dx)²+aqx(dx)²=0,

axdx(-axdp+xdq-2xdx+2a²xdx-apdx+qdx)=0

-axdp+xdq-2xdx+2a²xdx-apdx+qdx=0,

-a(xdp+pdx)+(xdq+qdx)+2xdx(a²-1)=0.

This is integrable: -apx+qx+(a²-1)x²=c, a constant.

We have a=y/x and c=-apx+qx+(a²-1)x², so, if c=f₁(a), where f₁ is an arbitrary function:

-apx+qx+(a²-1)x²=f₁(y/x).

(2)⇒(2i) -xydydp+xydxdq-(2x²-2y²+py-qx)dxdy=0 is the subsidiary equation. 

Substitute y=√(b²-x²) and dy=-xdx/√(b²-x²) (also ydy=-xdx).

-x(-xdx)dp+x√(b²-x²)dxdq-(2x²-2(b²-x²)+p√(b²-x²)-qx)dx(-xdx/√(b²-x²))=0.

We can divide through by xdx:

xdp+√(b²-x²)dq+(4x²-2b²+p√(b²-x²)-qx)dx/√(b²-x²)=0,

xdp+√(b²-x²)dq+(4x²-2b²)dx/√(b²-x²)+pdx-qxdx/√(b²-x²)=0,

d(px)+d(q√(b²-x²))+(4x²-2b²)dx/√(b²-x²)=0.

Integrating: px+q√(b²-x²)+∫[(4x²-2b²)/√(b²-x²)]dx=c, a constant.

Let F(x)=(4x²-2b²)dx/√(b²-x²), then c=px+q√(b²-x²)+F(x). (F(x) is integrable, but to save space the integration is not shown here.)

We have b²=x²+y² and c=px+q√(b²-x²)+F(x). If c=f₂(b²) then:

px+q√(b²-x²)+F(x)=f₂(x²+y²). But -apx+qx+(a²-1)x²=f₁(y/x), q=(apx+f₁(y/x)-(a²-1)x²)/x from (1i).

px+(apx+f₁(y/x)-(a²-1)x²)/x)√(b²-x²)+F(x)=f₂(x²+y²),

(px²+apx+f₁(y/x)-(a²-1)x²)√(b²-x²)+xF(x)=xf₂(x²+y²),

p(x²+ax)√(b²-x²)=xf₂(x²+y²)-xF(x)-(f₁(y/x)-(a²-1)x²)√(b²-x²),

p=(xf₂(x²+y²)-xF(x)-(f₁(y/x)-(a²-1)x²)√(b²-x²))/((x²+ax)√(b²-x²)).

dz=pdx+qdy, so p and q can be substituted (!) to get dz in terms of x and y.

The above calculations need to be verified and corrected if necessary...