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

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