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I am given the following

$$ \begin{cases} u_x+u_y=1-u, x>0\\ u(x,x+x^2)=sinx\\ \end{cases} $$

I am trying to solve it using method of characteristics.

So I got the following equations:

$$ \begin{cases} x_t=1\Rightarrow x(t,s)=t+f_1(s)\\ y_t=1\Rightarrow x(t,s)=t+f_2(s)\\ u_t=1-u \end{cases} $$

But the third equation is again a pde

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    $\begingroup$ Yes but $u=u(t,s)$ and you can see it as $u=\tilde{u}(t)$ since there is no explicit appearence of $s$, hence $u_t=\tilde{u}'$ so you can solve the ODE in the classical way. Then as in the first two equations you've solved the dependency on $s$ appears with the initial data $f_3(s)=\sin{s}$ $\endgroup$
    – Dadeslam
    Apr 22, 2020 at 10:51
  • $\begingroup$ @Dadeslam So I get $u(t,s)=t-ut+f_3(s)$? $\endgroup$
    – newhere
    Apr 22, 2020 at 10:57
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    $\begingroup$ I did not do the full computation but I think rewriting the equation as $(e^tu)_t = e^t$ you get $u(t,s)=1+e^{-t}(\sin{(s)}-1)$ $\endgroup$
    – Dadeslam
    Apr 22, 2020 at 10:59
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    $\begingroup$ By the way I've just seen in the PDE there was $u+1$ and not $1-u$ so we need to rewrite it as $(ue^{-t})=e^{-t}$ hence the result is different $\endgroup$
    – Dadeslam
    Apr 22, 2020 at 11:02
  • $\begingroup$ @Dadeslam Yes, sorry it is $1-u$ $\endgroup$
    – newhere
    Apr 22, 2020 at 11:04

1 Answer 1

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The problem reduces to solve the following system: $$ \dfrac{dx}{1}=\dfrac{dy}{1}=\dfrac{du}{1-u}. $$ Thus, solving the first equality we obtain $$ x=y+c_1 \quad \implies \quad x-y=c_1. $$ On the other hand, solving the equation for $u$ we have $$ dx=\dfrac{du}{1-u} \quad \implies \quad -\log(1-u)=x+c_2 \quad \implies \quad u=1-e^{-x+\tilde{c}_2}. $$ Making $\tilde{c}_2=-c_2=\phi(c_1)$, where $\phi$ is a function to be determined according to the boundary conditions, we obtain $$ u(x,y)=1-e^{-x+\phi(x-y)}. $$ I let you to find $\phi$ by replacing $u(x,y)$ into the boundary conditions. Notice that this is correct since $u$ solves the PDE. In fact, the left-hand side is $$ u_x=e^{-x+\phi}-\phi'e^{-x+\phi},\quad u_y=\phi'e^{-x+\phi} \quad \implies u_x+u_y=e^{-x+\phi} $$ On the other hand, the right-hand side is $$ 1-u=1-\big(1-e^{-x+\phi}\big)=e^{-x+\phi}. $$

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  • $\begingroup$ Why can we set $-c_2=\phi(c_1)$? $\endgroup$
    – newhere
    Apr 22, 2020 at 11:43
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    $\begingroup$ Kind of general procedure. Once you solve the equation for $x$ and $y$ you will get an integration constant. Then, after solving the equation for the unknown $u$ (and hence obtaining a second integration constant), you replace the latter constant for an arbitrary function of the first one $\widetilde{c}_2=\phi(c_1)$. $\endgroup$
    – Sharik
    Apr 22, 2020 at 11:49
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    $\begingroup$ This is related to the fact that when you integrate (in one variable) a multi-dimensional function, you obtain an integration constant depending on the remaining variables (i.e. an arbitrary function depending on the remaining variables): for instance $$ \int \partial_y f(x,y)dy=f(x,y)+c(x).$$ I recommend you to take a look in en.wikipedia.org/wiki/Method_of_characteristics $\endgroup$
    – Sharik
    Apr 22, 2020 at 11:52
  • $\begingroup$ I see, So we got $u(x,y)=1-\phi(x-y)e^{-x}$ using the boundary condition we get $sinx=1-\phi(-x^2)e^{-x}\Rightarrow e^x(-sinx+1)=\phi(-x^2)$ $\endgroup$
    – newhere
    Apr 22, 2020 at 12:02
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    $\begingroup$ Looks okay for me $\endgroup$
    – Sharik
    Apr 22, 2020 at 12:30

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