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I solved/analyzed the below PDE

$$\left\{\begin{matrix} u_x+u_y+u=e^{x+2y}\\ u(x,0)=0 \end{matrix}\right.$$

and have a question to the one of the steps involving the integration, see below


Using method of characteristic $$ \frac{dx}{dt}=1, \ \ \frac{dy}{dt}=1, \ \ \frac{du}{dt}=-u+e^{x+2y} $$

from the first and second ODE $$ dx=dy, \quad y=x+c. \quad c=y-x$$

and from the first and third ODE $$\frac{du}{-u+e^{x+2y}}=dx$$ which gives $$u_x+u=e^{x+2y}$$ multiplying both sides by $e^x$ and using product rule gives $$(e^x u)'=e^{2y+x}$$ substituting for y=x+c $$(e^x u)'=e^{3x+2c}$$ integrating both sides $$e^x u = \frac{1}{4} e^{4x+2c}+ \color{red}{f(c)}$$

Question:
The only way to get the correct general solution is to add f(c) as the constant part of the above integration. My question is how do one know it should be a function of c f(c) and not just some constant c here? Using c only gives a different equation.

Furhter substituting for c gives the general solution to the PDE $$u(x,y)=\frac{1}{4} e^{x+2y}+e^{-x}f(y-x)$$

Using the initial condition one can get the specific solution $$u(x,y)=\frac{1}{4} e^{x+2y}+\frac{1}{4} e^{x-2y}$$

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  • $\begingroup$ Also in your differential system under the header "Using the method of characteristic" the third equation should be du / dt not dy /dt. $\endgroup$ – Tucker Jul 20 '16 at 1:33
  • $\begingroup$ seriously? you just need to use another C $\endgroup$ – Anonymous Jul 20 '16 at 3:38
  • $\begingroup$ @ Tucker, typo, corrected to du/dt. Thanks $\endgroup$ – Michal Jul 20 '16 at 9:51
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@ Michal : your solution is correct. But it isn't the only way. It might be less disturbing if one can find a particular solution in order to change of function:

$$u_x+u_y+u=e^{x+2y}$$ The form of the right term suggests to look for a particular solution on the form $Ce^{x+2y}$ which leads to $C=\frac{1}{4} \quad\to\quad U=\frac{1}{4}e^{x+2y}$

Change of function : $\quad u(x,y)=v+U=v(x,y)+ \frac{1}{4}e^{x+2y}$ $$v_x+v_y=-v$$ The set of characteristic equations is : $$\frac{dx}{1}=\frac{dy}{1}=\frac{dv}{-v}$$ The equation of a first characteristic curve comes from $\frac{dx}{1}=\frac{dy}{1}\quad\to\quad x-y=c_1$

The equation of a second characteristic curve comes from $\frac{dx}{1}=\frac{dv}{-v}\quad\to\quad ve^x=c_2$

The general solution expressed on the form of implicit equation is : $$\Phi(x-y,ve^x)=0$$ where $\Phi$ is any differentiable function of two variables. Solving for the second variable leads to the explicit equation : $$ve^x=f(x-y)$$ where $f$ is any differentiable function. $$v=e^{-x}f(x-y)$$ $$ u(x,y)=e^{-x}f(x-y)+\frac{1}{4}e^{x+2y}$$ This is equivalent to what you already found (with $f(y-x)$, which is equivalent since $f$ is any function).

NOTE, corrected mistake in my first answer :

(Method without change of function) $$u_x+u_y=-u+e^{x+2y}$$ $$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{-u+e^{x+2y}}$$ First characteristic curve : $\frac{dx}{1}=\frac{dy}{1} \quad\to\quad x-y=c_1$

Second characteristic curve, from $\frac{dx}{1}=\frac{du}{-u+e^{x+2y}}$ $$\frac{du}{dx}=-u+e^{x+2y}=-u+e^{x+2(x-c_1)}$$ Solution of this ODE : $$ue^x-\frac{1}{4}e^{4x-2c_1}=c_2$$ $$ue^x-\frac{1}{4}e^{2x+2y}=c_2$$

The general solution expressed on the form of implicit equation is : $$\Phi(x-y,ue^x-\frac{1}{4}e^{2x+2y})=0$$ where $\Phi$ is any differentiable function of two variables. Solving for the second variable leads to the explicit equation : $$ue^x-\frac{1}{4}e^{2x+2y}=f(x-y)$$ where $f$ is any differentiable function. $$u=\frac{1}{4}e^{x+2y}+e^{-x}f(x-y)$$

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  • $\begingroup$ @ JJacquelin, thanks for the great explanations. I understand all the steps you presented but the first one. "The form of the right term suggests to look for a particular solution on the form $Ce^{x+2y}$ which leads $C=\frac{1}{4} \quad\to\quad y_p=\frac{1}{4}e^{x+2y}$" How do we know we are looking for this solution? How do we know $C=\frac{1}{4}$ and what is $y_p$? $\endgroup$ – Michal Jul 20 '16 at 10:15
  • $\begingroup$ "Suggests" doesn't mean that we are sure. It only suggests to try. If it doesn't works, then we have to try something else. In the present case, we try a particular function $U=Ce^{x+2y}$ with symbol $U$ instead of $u$ to remind that it would be a particular solution (The previous suffix $p$ was confusing : it didn't mean a differentiation). We put this function into $u_x+u_y+u=e^{x+2y}=(C+2C+C)e^{x+2y}$ which gives $4C=1$. Also, sorry for the typo $y_p$ instead of $u_p$ now corrected and replaced by $U$. $\endgroup$ – JJacquelin Jul 20 '16 at 11:37
  • $\begingroup$ sorry I don't get the intuition behind the $(C+2C+C)e^{x+2y}$ $\endgroup$ – Michal Jul 20 '16 at 11:59
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    $\begingroup$ There is no intuition behind the $(C+2C+4C)e^{x+2y}$ because it is pur calculus as a consequence of the initial intuition which is the form of function in trial : $$u=Ce^{x+2y}\: ; \: u_x=Ce^{x+2y}\: ;\: u_y=2Ce^{x+2y}$$ $$u_x+u_y+u=Ce^{x+2y}+2Ce^{x+2y}+Ce^{x+2y}=4Ce^{x+2y}$$ $$u_x+u_y+u=e^{x+2y}=4Ce^{x+2y}$$ Hense $1=4C$ hence $C=\frac{1}{4}$ hense $$u=\frac{1}{4}e^{x+2y}\:\text{ is a particular solution}$$ $\endgroup$ – JJacquelin Jul 20 '16 at 14:05
  • $\begingroup$ is the purpose of the change of function to "conceal" the U function and create a simple homogeneous equation, correct? is it a special case or rather a common method? I mean if you look at a new problem what makes you think that this could be a way to solve? is it the function of e to something on the RHS? $\endgroup$ – Michal Jul 20 '16 at 14:53
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You can use some constant $k$, but then one realizes that along the characteristics there are two constants or invariants $k,c$. If $k$ is an arbitrary number and $c$ is an arbitrary number then couldn't we express $k$ as some function of $c$?

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