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Consider the cauchy problem:

$U_x+U_y=1$

with $U(s,s)=\sin s$

Since $\dfrac{dx}{1}=\dfrac{dy}{1}=\dfrac{dz}{1}$

Thus $U-x=f(U-y)$

use initial condition we get $\sin s-s=f(\sin s-s)$

I think $f(x)=x$

Please tell me I am right or not.

doraemonpaul
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user120386
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  • As stated, the problem has no solution because the initial data is inconsistent with the PDE. If $U(s,s)=\sin s$ then along the line $y=x$ we have $U_x+U_y=2\cos s$, not $1$. The problem is that the values of $U$ are prescribed along a characteristic line; they should be prescribed on a line that is transverse to characteristics. – you can call me Al Jan 31 '14 at 04:43

1 Answers1

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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{dy}{dt}=1$ , letting $y(0)=y_0$ , we have $y=t+y_0=x+y_0$

$\dfrac{dU}{dt}=1$ , letting $U(0)=f(y_0)$ , we have $U=t+f(y_0)=x+f(y-x)$

$U(s,s)=\sin s$ :

$s+f(0)=\sin s$

$f(0)=\sin s-s$

which does not make sense to have constant$=$function

$\therefore$ The initial value problem has no solution

doraemonpaul
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