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Consider the following PDE: $$-3U_x+4U_y=0$$ where $U=3x$ on curve $y=x+1$.

First we invoke the method of characteristics: $$\frac{dx}{-3}=\frac{dy}{4}=\frac{dU}{0}.$$ From this we get $c_1=\frac{x}{3}+\frac{y}{4}$ and $U(x,y)=c_2$. So $F(c_1)=c_2$ or $F(\frac{x}{3}+\frac{y}{4})=u(x,y)$.

Applying the condition, we know that $3x=u(x,x+1)=F(\frac{x}{3}+\frac{x+1}{4})=F(\frac{x+3}{12}).$ So $F(\frac{x+3}{12})=3x.$ This is now where I'm stuck.

emka
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The general solution of $-3U_x+4U_y=0$

obviously is $U(x,y)=F(\frac{x}{3}+\frac{y}{4}) \space$ where $F$ is any derivable function.

I don't understand what you mean "incorporate the curve". If you are talking of a boundary condition on the curve $y=x+1$, please, make clear what is the boundary condition.

With the condition $U(x,y=x+1)=3x\space$ one have to find a function $F\space$ so that $$F(\frac{x}{3}+\frac{x+1}{4})=3x$$ $$F(\frac{7x}{12}+\frac{1}{4})=3x$$ It is easy to see that the equality is obtained witth a function $F(X)=aX+b$ where $a=\frac{36}{7}$ and $b=-\frac{36}{7} \frac{1}{4}$ because : $$\frac{36}{7}(\frac{7x}{12}+\frac{1}{4})-\frac{9}{7}=3x$$ As a consequence, the final result is : $$U(x,y)=\frac{36}{7}(\frac{x}{3}+\frac{y}{4})-\frac{9}{7}$$

JJacquelin
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  • The question posed simply said subject to the conditiln u(x,0)=3x and the curve y=x+1. – emka Sep 22 '14 at 17:43
  • If the condition is $U(x,o)=3x$, the condition is set on the curve $y=0$, not on $y=x+1$. And this implies $U(x,y)=3x+\frac{9}{4}y$. Sorry, I continue to not understand your wording "and the curve $y=x+1$ " – JJacquelin Sep 22 '14 at 18:00
  • Sorry, I kept haphazardly reading the question. I have made the appropriate correction. – emka Sep 22 '14 at 18:50
  • Well, I suppose that I understand the bounding condition now. The corresponding result is added to my previous answer. – JJacquelin Sep 22 '14 at 21:07