Question

solve ( r-s)y + (s-t)x +  q-p =0, by monges method

Comments

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

Answer 1

You can find some details of derived equations using the link:

https://www.mathhomeworkanswers.org/289987/solve-the-following-by-monges-method-xy-t-r-x-2-y-2-s-2-py-qx

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For any z(x,y):

dz=(∂z/∂x)dx+(∂z/∂y)dy=pdx+qdy, from the definitions of p and q.

In the form Rr+Ss+Tt=V the given PDE becomes:

ry-sy+sx-tx+q-p=0⇒R=y, S=-y+x, T=-x, V=p-q.

r=(dp-sdy)/dx and t=(dq-sdx)/dy (see other question).

Therefore we can replace r and t:

y(dp-sdy)/dx+s(x-y)-x(dq-sdx)/dy=p-q, multiplying through by dxdy:

ydy(dp-sdy)+s(x-y)dxdy-xdx(dq-sdx)=(p-q)dxdy.

Monge's subsidiary equations are (see other question):

Rdydp+Tdxdq-Vdxdy=0 and Rdy²-Sdxdy+Tdx²=0, that is:

ydydp-xdxdq-(p-q)dxdy=0 and y(dy)²-(x-y)dxdy-x(dx)²=0.

The second of these equations can be written:

(dy)²-(x/y-1)dxdy-(x/y)(dx)²=0. We would like to factorise it:

(dy-m₁dx)(dy-m₂dx)=(dy)²-(m₁+m₂)dxdy+m₁m₂(dx)²≡(dy)²-(x/y-1)dxdy-(x/y)(dx)²=0.

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

This gives us two possible equations:

dy-xdx/y=0⇒ydy-xdx=0; and dy+dx=0, both of which are integrable:

(1) y²/2-x²/2=k, y²-x²=a, a constant, y=√(x²+a), dy=xdx/√(x²+a);

(2) y+x=b, another constant, y=b-x; dy=-dx.

We can now make some substitutions in ydydp-xdxdq-(p-q)dxdy=0 to eliminate y:

(1)⇒(1.1) xdxdp-xdxdq-(p-q)(dx)²x/√(x²+a)=0, which reduces to:

dp-dq-(p-q)dx/√(x²+a)=0,

(dp-dq)/(p-q)=dx/(√(x²+a)) which is integrable:

Let a=c², then ln|p-q|=(1/c)tan^-1(x/c), or p-q=e^{(1/c)tan⁻¹(x/c)}. We can write this as p-q=f(x).

(2)⇒(2.1) -(b-x)dxdp-xdxdq+(p-q)(dx)²=0,

-(b-x)dp-xdq+(p-q)dx=0,

-bdp+xdp-xdq+pdx-qdx=0,

-bdq+d(px)-d(qx)=0 which is integrable:

-bq+px-qx=d, a constant.

d=g(b), where g is an arbitrary function.

-bq+px-qx=g(x+y),

We have -bq+px-qx=g(x+y) and p-q=f(x), i.e., p=q+f(x).

-bq+qx+xf(x)-qx=g(x+y), q=(xf(x)-g(x+y))/b, p=(bf(x)+xf(x)-g(x+y))/b.

dz=pdx+qdy=[bf(x)+xf(x)-g(x+y))/b]dx+[(xf(x)-g(x+y))/b]dy, the integral of which can be written:

z=F(x,y)+G(x,y)+H, where H is a constant and F=∫[bf(x)+xf(x)-g(x+y))/b]dx, G=∫[(xf(x)-g(x+y))/b]dy.

More specifically:

(p-q)x=bq, p-q=bq/x, p=q+qb/x=q(1+b/x).

So bq/x=e^{(1/c)tan⁻¹(x/c)}, q=(x/b)e^{(1/c)tan⁻¹(x/c)},

so p=(1+b/x)(x/b)e^{(1/c)tan⁻¹(x/c)},

p=((x+b)/b)e^{(1/c)tan⁻¹(x/c)}.

So p and q are each defined as functions of x.

dz=pdx+qdy=((x+b)/b)e^{(1/c)tan⁻¹(x/c)}dx+(x/b)e^{(1/c)tan⁻¹(x/c)}dy,

dz=e^{(1/c)tan⁻¹(x/c)}((x+b)dx+xdy)/b, from which z can be derived.