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My Professor's notes had this problem:

Solve the PDE $u_x + u_y = 0$ in the domain $y > \phi(x)$, $x \in \mathbb{R}$ given that $u = g(x)$ on the curve $y = \phi(x)$, where $\phi(x) = \frac{x}{1 + |x|}$.

Her notes say that the general solution is $u(x, y) = A(x - y)$, where $A()$ is an arbitrary function. However, if I try to solve this myself, then I am unable to get this.

My class only started learning this, so please forgive me for any errors.

So based on what we have learned I think that the first two characteristic equations are $\dfrac{dx}{dt} = 1$ and $\dfrac{dy}{dt} = 1$. If we use separation of variables we get $x(t) = t + C_1$ and $y(t) = t + C_2$. One key point we were told here is that $C_1$ and $C_2$ are constants that are the same for any single characteristic curve but change between different characteristic curves.

I then used change of variables to get $x(0) = s = C_1$ and $y(0) = \dfrac{s}{1 + |s|} = C_2$ which must mean that the change of variables for the map $(s, t) \to (x, y)$ is $x(s, t) = t + s$ and $y(s, t) = t + \dfrac{s}{1 + |s|}$.

Ok, I think I have done everything up to now correctly.

I now solve the last characteristic equation $\dfrac{du}{dt} = 0$. If we again use separation of variables then we get $u(x,y) = C_3$.

But now I wonder how she got $u(x, y) = A(x - y)$ as the general solution? And where did the function $A$ come from?

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    $\begingroup$ First, try differentiating the result $u = A(x-y)$ to check that it is indeed a solution. Secondly, $x = t + C_{1}$ and $y = t + C_{2} \implies x = y + C_{3}$ and hence $C_{3} = x - y$. Solving the ODE $u' = 0$ yields $u = f(C_{3}) = f(x-y)$ which gives the result. Now apply the initial condition. $\endgroup$ – mattos Aug 7 '18 at 0:08
  • $\begingroup$ @Mattos this is how I thought about it in my mind. Please tell me if this is correct. $\dfrac{du}{dt} = 0$ so $u = A(C_3)$. Why is $A$ dependent of $C_3$? The values of $C_1$ and $C_2$ in the characteristic equations change from one characteristic to another but are constant on any single characteristic. In other words, the values of $C_1$ and $C_2$ distinguish the different characteristics. $A$ is also a constant along any single characteristic but will be different for different characteristics. Therefore, $A$ will change depending on $C_1$ and $C_2$ in the characteristic equations. $\endgroup$ – user575678 Aug 7 '18 at 8:06
  • $\begingroup$ @Mattos Therefore, let $C_3 = C_1 - C_2$. We then have $x - y = C_1 - C_2 = C_3$. Therefore, we have our result that $u = A(C_3) = A(x - y)$ $\endgroup$ – user575678 Aug 7 '18 at 8:07
  • $\begingroup$ @Mattos If it is not too much trouble can you please comment on whether I am correct above in my reasoning? If you would like, I will create a new question for this purpose. Thanks $\endgroup$ – user575678 Aug 8 '18 at 6:41
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Let $p:=x-y$ and $q:=x+y$. Then, $x=\dfrac{1}{2}q+\dfrac{1}{2}p$ and $y=\dfrac{1}{2}q-\dfrac{1}{2}p$. The two new variables $p$ and $q$ are independent: $$\frac{\partial p}{\partial q}=\frac{\partial{x}}{\partial{q}}\,\left(\frac{\partial p}{\partial x}\right)+\frac{\partial{y}}{\partial{q}}\,\left(\frac{\partial p}{\partial y}\right)=\left(\frac12\right)\cdot 1+\left(\frac12\right)\cdot(-1)=0$$ and $$\frac{\partial q}{\partial p}=\frac{\partial{x}}{\partial{p}}\,\left(\frac{\partial q}{\partial x}\right)+\frac{\partial{y}}{\partial{p}}\,\left(\frac{\partial p}{\partial y}\right)=\left(\frac12\right)\cdot 1+\left(-\frac12\right)\cdot1=0\,.$$ That is, if a function $U(p,q)$ satisfies $\dfrac{\partial U}{\partial q}=0$, then $U(p,q)=A(p)$ for some function $A$.

Now, let $U(p,q)$ denote $u\left(\frac{1}{2}q+\frac{1}{2}p,\frac{1}{2}q-\frac{1}{2}p\right)=u(x,y)$ in your question. Since $$\frac{\partial U}{\partial q}=\frac{\partial x}{\partial q}\,\left(\frac{\partial U}{\partial x}\right)+\frac{\partial y}{\partial q}\,\left(\frac{\partial U}{\partial y}\right)=\frac{1}{2}\,\left(\frac{\partial u}{\partial x}\right)+\frac{1}{2}\,\left(\frac{\partial u}{\partial y}\right)=\frac12\,\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\right)=0\,,$$ we have that $U(p,q)=A(p)$ for some function $A$. Thus, $$u(x,y)=A(x-y)\,.$$

In particular, if $u(x,y)=g(x)$ when $y=\phi(x)=\dfrac{x}{1+|x|}$, then $$g(x)=A\big(x-\phi(x)\big)=A\left(\frac{x\,|x|}{1+|x|}\right)\,.$$ If $t:=\dfrac{x|x|}{1+|x|}$, then $x=\gamma(t)$, where $$\gamma(t)=\left\{\begin{array}{ll}\frac{t}{2}\,\left(1+\sqrt{1+\frac{4}{|t|}}\right)&\text{if }t\neq 0\,,\\ 0&\text{if }t=0\,.\end{array}\right.\,.$$ Hence, $$A(t)=g\big(\gamma(t)\big)\,.$$ That is, $$u(x,y)=A(x-y)=g\big(\gamma(x-y)\big)=\left\{\begin{array}{ll}g\Biggl(\frac{x-y}{2}\,\left(1+\sqrt{1+\frac{4}{|x-y|}}\right)\Biggr)&\text{if }x\neq y\,,\\ g(0)&\text{if }x=y\,.\end{array}\right.\,.$$

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  • $\begingroup$ Hello Batominoski. I wonder how $\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\right)=0$? $\endgroup$ – user575678 Aug 7 '18 at 3:38
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    $\begingroup$ That is your given condition: $u_x+u_y=0$. $\endgroup$ – Batominovski Aug 7 '18 at 3:44
  • $\begingroup$ Oh of course yes. I am sorry. I also have the concern I told mattos: Is it true that $p = x - y$ and $q =x + y$ are independent? They both have constants that depend on the characteristic curve, yes? See my response to him in the comments. I think this is my main point of confusion. $\endgroup$ – user575678 Aug 7 '18 at 3:57
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    $\begingroup$ I proved that they are independent in the first paragraph of my answer. $\endgroup$ – Batominovski Aug 7 '18 at 3:58
  • $\begingroup$ Yes I am sorry. Your work makes sense. I am thankful for your help! $\endgroup$ – user575678 Aug 7 '18 at 4:01

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