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I was thinking about the problem that says:

The initial value problem $u_{x}+u_{y}=1,u(s,s)=1,0\leq s\leq1,$ has

(a) two solutions,

(b) a unique solution,

(c) no solution,

(d) infinitely many solutions.

My attempts: By using Lagrange's method, we see $\frac{dx}{1}=\frac{dy}{1}=\frac{du}{1}.$ Hence we get, $x=y+c_1$ and $u=x+c_2.$ Now since $u(x,y)=x+c_2 ,$ we get by the given condition that $u(s,s)=1=s+c_2$ and so $c_2=1-s.$ So, finally we get, $u(x,y)=x+(1-s).$ Since $0\leq s\leq1,$we get infinitely many solutions for various choice of s. Am i going in the right direction? Please help.Thanks in advance for your time.

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    $\begingroup$ See here on page 16. $\endgroup$ – Mhenni Benghorbal Dec 10 '12 at 20:50
  • $\begingroup$ Thanks.It has been very useful. $\endgroup$ – learner Dec 11 '12 at 3:55
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In fact this PDE belongs to the PDE of the form http://eqworld.ipmnet.ru/en/solutions/fpde/fpde1201.pdf

The general solution is $u(x,y)=x+F(x-y)$ or $u(x,y)=y+F(x-y)$

$u(s,s)=1$ :

$s+F(0)=1$

$F(0)=1-s$

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

$\therefore$ The initial value problem has no solution

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The solution of the differential equation is

$$ u \left( x,y \right) =x+{\it F} \left( y-x \right),$$

where $F$ is an arbitrary function. Now, apply the initial condition you have been given and see the reference to finish the problem.

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  • $\begingroup$ Thank you,sir. I have got it. $\endgroup$ – learner Dec 11 '12 at 3:54
  • $\begingroup$ @learner: You are welcome. $\endgroup$ – Mhenni Benghorbal Dec 11 '12 at 11:12
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Geometric approach

$u_x+u_y$ is the derivative of $u$ in the direction $(1,1)$; that is, $$ u_x+u_y=\nabla u\cdot(1,1)\tag{1} $$ Therefore, $u_x+u_y=1$ specifies that the rate of change in the direction of $(1,1)$ is $1$.

Since the initial conditions specify $u(s,s)=1$, the rate of change in the direction $(1,1)$ is $0$. Thus, there can be no solutions.


Another approach

If we rotate coordinates so that $t=\frac{x+y}{2}$ and $s=\frac{x-y}{2}$, the equation $u_x+u_y=1$ becomes $$ u_t=1\tag{2} $$ Thus, $u(t,s)=t+f(s)$ for any function $f$. Derotating yields $$ u(x,y)=\frac{x+y}{2}+f\left(\frac{x-y}{2}\right)\tag{3} $$ To match up the initial conditions, we need that for all $s\in[0,1]$, $$ 1=u(s,s)=s+f(0)\tag{4} $$ Again, this says there can be no solutions.

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  • $\begingroup$ thank your sir for the explanation. $\endgroup$ – learner Dec 11 '12 at 9:16

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