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I am studying PDEs using the book "PDEs An Introduction 2nd edition" by Walter A. Strauss. In Chapter 2, a "geometric method" is described in order to solve linear PDEs of the type:

$$ (x,y)\mapsto u_x + yu_y = 0 $$

This is said to be equivalent to the directional derivative of $u$ in the direction of the vector $(1,y)$ being set to 0. Then characteristic curves having as tangent vectors $(1,y)$ are found

$$\frac{dy}{dx} = \frac{y}{1} \implies y = Ce^x $$

Since $u(x,y)$ is constant on these curves:

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

is the general solutions of the PDE, where $f$ is an arbitrary function. Now the same method is applied to solve more general equations, such as:

$$u_x + u_y + u = f(x,y)$$

I have tried to use the same method to solve the following differential equation:

$$yu_x -xu_y = 3x$$ where $ u(x,0) = x^2$

Then applying the method:

$$ \frac{dy}{dx} = \frac{-x}{y} \implies C = x^2 + y^2$$

The PDE reduces to an ODE:

$$\frac{du}{dx} = 3x/y \implies u (x,y) = 3x^2/2y + f(C)$$

However with the boundary imposed ($u(x,0) = x^2$) this seem impossible to solve since to find the particular solution I would have to divide the a term by zero. Does the method described in the book have limited scope? How can I solve this differential equation? Also, why is $u(x,y)$ constant on the "characteristic curves"?

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    $\begingroup$ If you introduce polar coordinates $x=r\cos \theta, y=r\sin \theta$, then $y\partial_x-x\partial_y=-\partial_\theta$. This is not the method you asked but it leads to a quick integration. $\endgroup$ – Giuseppe Negro Apr 2 '18 at 21:41
  • $\begingroup$ Yeah that is a good trick, however I was more interested in the method and not the solution itself. $\endgroup$ – daljit97 Apr 2 '18 at 21:54
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I cannot see where you found an impossibility or an exception in the rules.

I the case of $\quad yu_x-xu_y=3x \tag 1$

$$\frac{dx}{y}=\frac{dy}{-x}=\frac{du}{3x}$$ A first family of characteristic curves comes from $\quad\frac{dx}{y}=\frac{dy}{-u}\quad$ which leads to $$x^2+y^2=c_1$$ A second family of characteristic curves comes from $\quad\frac{dy}{-x}=\frac{du}{3x}\quad$ which leads to $$u+3y=c_2$$ The general solution of the PDE $(1)$ expressed on the form of implicit equation is : $$\Phi(x^2+y^2\:,\:u+3y)=0$$ where $\Phi$ is an arbitrary fonction of two variables.

Or equivalently, the general solution of PDE $(1)$ on explicit form is $\quad u+3y=F(x^2+y^2)$ $$u(x,y)=-3y+F(x^2+y^2)$$

where $F(X)$ is an arbitrary function of the variable $X$ with $X=x^2+y^2$.

Boundary condition : $\quad u(x,0)=x^2=F(x^2+0^2)=F(x^2)$.

So, the function $F$ is now determined : $\quad F(X)=X.\quad$ Putting it into the above general solution where $X=x^2+y^2$ leads to the particular solution which fits the boundary condition : $$u(x,y)=-3y+(x^2+y^2)$$

OTHER EXAMPLE : $\quad u_x+u_y+u=e^{x+2y} \tag 2$ $$u_x+u_y=e^{x+2y}-u$$ $$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u}$$ A first family of characteristic curves comes from $\quad\frac{dx}{1}=\frac{dy}{1}\quad$ which leads to $$x-y=c_1$$ A second family of characteristic curves comes from $\frac{dy}{1}=\frac{du}{e^{x+2y}-u}=\frac{du}{e^{(c_1+y)+2y}-u}=\frac{du}{e^{c_1+3y}-u}$ $$\frac{du}{dy}+u=e^{c_1+3y}$$ This is a first order linear ODE easy to solve : $\quad u=\frac14 e^{c_1+3y}+c_2e^{-y}=\frac14 e^{(x-y)+3y}+c_2e^{-y}$ $$ue^y-\frac14 e^{x+3y}=c_2$$ The general solution of the PDE $(2)$ expressed on the form of implicit equation is : $$\Phi\left((x-y)\:,\:(ue^y-\frac14 e^{x+3y})\right)=0$$ where $\Phi$ is an arbitrary fonction of two variables.

Or equivalently, the general solution of PDE $(2)$ on explicit form is from $\quad ue^y-\frac14 e^{x+3y}=F(x-y)$ $$u(x,y)=\frac14 e^{x+2y}+e^{-y}F(x-y)$$ where $F(X)$ is an arbitrary function of the variable $X=x-y$.

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  • $\begingroup$ I believe that the author of my book seems to use a different method to solve the PDE, have a look here at an example. If you look at the last few lines of my question, there seems to be a problem (using the method used in the link) $\endgroup$ – daljit97 Apr 2 '18 at 21:29
  • $\begingroup$ It is fundamentally the same method, with some variant of symbols and presentation. See :en.wikipedia.org/w/… and see en.wikipedia.org/wiki/Method_of_characteristics. You will have to do a bit of "gymnastic of the brain" in order to make the link. I wish you bon courage. $\endgroup$ – JJacquelin Apr 2 '18 at 21:43
  • $\begingroup$ You can see that the general solution in the book is : $$u(x,y)=\frac14 e^{x+2y}+e^{-y}f(x-y)$$ which is exactly the same than in my answer. I will not repeat what is already in the book to determine the function $f$ according to the boundary condition. The method is the same. $\endgroup$ – JJacquelin Apr 2 '18 at 21:52
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    $\begingroup$ $\frac{du}{dx} = 3x/y \implies u (x,y) = 3x^2/2y + f(C)$ is false because you integrate $3x/y$ without taking account that $y$ is not constant. You cannot solve it that way. This not a correct application of the method of characteristic. One have to find two independent families of characteristic curves. You got one :$x^2+y^2=C_1$ but not a second one. $\endgroup$ – JJacquelin Apr 2 '18 at 22:11
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    $\begingroup$ @Isham yes you are right, I was too invested into thinking about why the method led me to that point that it didn't occur to me that a simple substitution was needed. $\endgroup$ – daljit97 Apr 2 '18 at 22:40
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You can't have this: $$\frac{du}{dx} = 3x/y ==> u (x,y) = 3x^2/y + f(C)$$ Because you have a differential equation ( du,dx) but three variables namely x,y,u. Y shouldnt be there...Apart for that mistake you did a very good job.

Usisng the method of characteristics:

$$\frac {dx}{y}=\frac {dy}{-x}=\frac {dz}{3x}$$ $$-xdx=ydy \implies x^2+y^2=K_1$$ $$3xdy=-xdz \implies y+z/3 =K_2$$ $$f(x^2+y^2)=y+u/3$$ $$u(x,y)=3f(x^2+y^2)-3y$$ we have $u(x,0)=x^2$ $$f(x^2)=x^2/3 \implies f(x)=x/3$$ Therefore $$\boxed{u(x,y)=x^2+y^2-3y}$$

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  • $\begingroup$ Hmm in the book in order to solve $u_x + u_y + u = e^{x+2y}$ the author does setup the following equation: $\frac{\partial u}{\partial x} + u = e^{x+2y}$ Would that be an exception to the rule? $\endgroup$ – daljit97 Apr 2 '18 at 19:39
  • $\begingroup$ @daljit97 I don't know I dont have the book right now. Thats weird that Strauss was so complicated about first order linear pde.. when it's easy to solve directly with characteristics method.. $\endgroup$ – Isham Apr 2 '18 at 19:42
  • $\begingroup$ Here is the link with the exercise and the solution provided stemjock.com/STEM%20Books/Strauss%20PDEs%202e/Chapter%201/… $\endgroup$ – daljit97 Apr 2 '18 at 19:48
  • $\begingroup$ Oh thats another method..I dont know that method...I deleted my comments on that method you used since I was completely wrong about it... $\endgroup$ – Isham Apr 2 '18 at 19:53
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    $\begingroup$ Oh ok, could you provide a reference that explains and show why the method of characteristic works? Looking around I couldn't find anything satisfying. $\endgroup$ – daljit97 Apr 2 '18 at 20:03

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