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Prove the following

Let $a,b,c,d \in R$ such that $c^2+d^2 \neq0$. Then the cauchy problem

$a u_x + b u_y= e^{x+y}, \ \ x,y \in \mathbb R,$

$u(x,y)=0$ on $cx+dy=0$

has a no solution if $ac+bd = 0$

Edit : [and unique solution if $ac+bd \neq 0$]

My attempt :

I am trying the find equation of $u(x,y)$ from Lagrange's method

$\frac{dx}{a}=\frac{dy}{b}=\frac{dz}{e^{x+y}}$

$\frac{dx}{a}=\frac{dy}{b} \implies x=\frac{ay}{b}+c_1$

Now $\frac{dz}{e^{x+y}}=\frac{dy}{b} \implies e^{\frac{ay}{b}+y+c_1}dy=bdz$

I stuck here how to proceed further or is there any short method to proceed?

prashant dattatrey
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1 Answers1

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The problem seems to be incorrectly stated. Take the line parametrized as $(x,y) = (as,bs)$, where $s$ is a parameter. Suppose $u$ is a solution. The derivative $$ \frac{d}{ds}u(as,bs) = au_x+bu_y = e^{as+bs} \ne 0. $$ But on this line you also have $$ cx+dy = cas+dbx = (ca+db)s = 0,$$ where $$ u = 0. $$ This is a contradiction. Therefore there is no solution.

Bob Terrell
  • 3,812