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

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.

• See here on page 16. – Mhenni Benghorbal Dec 10 '12 at 20:50
• Thanks.It has been very useful. – learner Dec 11 '12 at 3:55

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

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.

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

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.

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