# Does this PDE has solutions

Mathematics Asked on December 17, 2020

Consider the cauchy problem, $$5u_x+2u_y=7$$ with $$u=1$$ in line $$2x-5y=0$$.

Does it has no solution, unique solution or infinitely many solution?

Characteristic is parallel to the initial line. So I try to take different initial line and solve it but it seems something is not compatible and I couldn’t find any solution. please give me a hint

$$5u_x+2u_y=7$$ The general solution (from method of characteristics or other method) is : $$u(x,y)=frac72 y+F(2x-5y)$$ $$F$$ is an arbitrary function (to be determined according to some boundary condition).

You can check it directly in putting $$u=frac72 y+F(2x-5y)$$ into $$5u_x+2u_y=7$$.

CONDITION : $$u=1$$ on $$2x-5y=0$$ $$1=frac72 y+F(0)$$ This is impossible because $$frac72 y+F(0)$$ which is not constant cannot be equal to a constant $$=1$$ , what ever the function $$F$$ is.

Thus there is no solution which satisfies both the PDE and the specified condition.

WITH a DIFFERENT CONDITION : $$u(x,0)=h(x)$$

$$h(x)=0+F(2x-0)$$ With $$X=2xquad impliesquad h(X/2)=F(X).quad$$The function $$F$$ is determined. We put it into the above general solution where $$X=2x-5y$$. $$u(x,y)=frac72 y+h(frac{2x-5y}{2})$$ This is valid what ever $$h(0)$$ is.

Moreover, if $$h(0)=1$$ then $$u(0,0)=h(0)=1.quad$$ And $$u(x,y)$$ is not constant on the line $$2x-5y=0$$ but is : $$u=frac72 y+1=frac75 x+1$$.

Correct answer by JJacquelin on December 17, 2020