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Consider the partial differential equation $$u_x+2xu_y=0$$

Describe the existence and uniqueness properties for each auxiliary condition below . Include solutions possible or explain why none exists.

$(a). u(x,x^2)=2\\ (b). u(x,x^2)=e^{-x}\\ (c). u(0,y)=e^{-y}$

My attempt: Given PDE $u_x+2xu_y=0$

then $$\frac{dx}{1}=\frac{dy}{2x}=\frac{du}{0}$$

then $$2x-y-c=1, u=c_2$$

From option (a) $u(x,x^2)=2$ then $$x=t,y=t^2, u=2$$ also $$2t-t^2=c_1,2=c_2$$

Since we are not eliminate $c_1, c_2$ pde have no slotuion

is my approach is right?

Ѕᴀᴀᴅ
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Inverse Problem
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2 Answers2

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$$\frac{dx}{1}=\frac{dy}{2x}=\frac{du}{0}$$ First characteristics from $\quad\frac{dx}{1}=\frac{dy}{2x}\quad\implies\quad y-x^2=c_1$.

Second characteristics from $\quad\frac{du}{0}=$finite $\quad\implies\quad u=c_2$.

General solution : $$u(x,y)=F(y-x^2)$$ with any differentiable function $F$.

Case $(a)$ :

$u(x,x^2)=2=F(x^2-x^2)=F(0)\quad$

All functions $F(X)$ with condition $F(0)=2$ are convenient. There is an infinity of solutions. For example :

$u(x,y)=2$

$u(x,y)=2+(y-x^2)$

$u(x,y)=2+\sin(y-x^2)$

and so on...

Case $(b)$ :

$u(x,x^2)=e^{-x}=F(x^2-x^2)=F(0)$

It is impossible that $e^{-x}=$constant. There is no solution.

Case $(c)$ :

$u(0,y)=e^{-y}=F(y-0)=F(y)$

The function is determined : $\quad F(X)=e^{-X}\quad$ Puting it into the general solution where $X=y-x^2$ leads to a unique solution :

$u(x,y)=e^{-(y-x^2)}$

JJacquelin
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  • @JJacquelin.....Thank yo sir........so much – Inverse Problem Feb 06 '18 at 07:50
  • perhaps this is a naive question, but it seems that you set the initial data for the characteristics $x(t)$ and $y(t)$ as $x_0=y_0=0$. Why is this justified? – operatorerror Feb 07 '18 at 16:11
  • @qbert : Where do you see "initial data" in my answer ? There is no $t$ , no $x_0$, no $y_0$ anywhere. – JJacquelin Feb 07 '18 at 17:07
  • @JJacquelin right, because you are using this shortcut $\frac{dx}{1}=\frac{dy}{2x}$ which I assume reduces to the usual method of characteristics. I was trying to understand, since I am unfamiliar with the shortcut, not criticize your answer (I upvoted it!). For me $t$ is the usual parameter to be solved for later, as $x_0$ and $y_0$ are the data for the curve $x(t),y(t)$ also ideally to be solved for later – operatorerror Feb 07 '18 at 17:16
  • Nevermind, I have reconciled the two and will provide my own answer. – operatorerror Feb 07 '18 at 17:39
  • @qbert: I used this method in my answer because this was the method used in the suresh's question, in order to show him how to go further on the same way. Of course, the other method should be more convenient in more complicated cases. But it is sufficient and straightforward in many usual cases. – JJacquelin Feb 07 '18 at 18:22
  • @JJacquelin I understand, I just had not seen it before and made an error in calculating – operatorerror Feb 07 '18 at 18:24
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Caveat: this is a longer answer, but I find it easier to understand just what's going on here.

Using the method of characteristics, you examine the function $u$ along a trajectory $(x(t),y(t))$ through some point $(x_0,y_0)$ (usually in a boundary value problem you choose a convenient pair for the initial conditions, i.e. where you have boundary data) converting the partial differential equation into an ordinary differential equation. Namely, you want $$ x'(t)=1\\ y'(t)=2x(t) $$ since then $\frac{d}{dt}u(x(t),y(t))=u_x+2x(t)u_y=0$ for $u$ which solve the pde. So along trajectories of the above sort, as $u(x(t),y(t))=u(x_0,y_0)$.

Solving the coupled system, we find $$ x(t)=t+x_0\\ y(t)=t^2+2x_0t+y_0 $$ and eliminating the parameter $t$, we have that $u$ is constant as long as $$ y=(x-x_0)^2+2x_0(x-x_0)+y_0=x^2-x_0^2+y_0 $$

We know that $$ u(x,y)=u(x,x^2+y_0-x_0^2)=u(x_0,y_0) $$

From here, it is obvious that the first condition is easy to satisfy. We know that the desired function is constant along the parabola $y=x^2$, this condition just fixes such a constant to be $2$.

For the second, it is clearly impossible, since we know that $u$ is constant on the parabola $y=x^2$, but $e^{-x}$ is not constant.

For the third, we have data on the line $x=0$, so it is convenient to set $x_0=0$. Then your condition yields $$ u(x,y)=u(0,y_0)=e^{-y_0} $$ but our choice of characteristic is valid for any $y_0$, so what is $y_0$ when $x_0$ is zero in terms of $x$ and $y$? It is $y-x^2$. Giving us the answer $$ u(x,y)=e^{-(y-x^2)} $$

operatorerror
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