# Solving this 1st Order PDE [closed]

I am trying to solve the following PDE with an initial condition:

$$u_x + u_y = x + y$$ with $$u(x, 0) = 0$$

I am not sure which method to use to solve this.

Thanks

## closed as off-topic by vonbrand, colormegone, Michael Galuza, Strants, user91500Aug 26 '15 at 5:25

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• Try this method – Giovanni Aug 25 '15 at 21:48

The idea behind the method of characteristics is to reduce the problem to solving ODEs instead of PDEs, which is something that is usually easier. All the work that follows is done in that direction: we want to define a function of a single variable and use the PDE to write a Cauchy problem for this new function. The idea is to then recover the solution to the original problem from this auxiliary function.

Let $z(t) := u(x(t), y(t))$, assuming a solution exists, where $x(t)$ and $y(t)$ are yet to be defined. Let's compute $z'$ to get an idea of what $x(t)$ and $y(t)$ should be: $$z'(t) = u_x(x(t), y(t))x'(t) + u_y(x(t), y(t))y'(t).$$

Notice that this is exactly the LHS of our equation, provided $x'(t) = y'(t) = 1$!

Hence we are tempted to look at the values of $u$ along the lines $(t + x_0, t)$. To find these lines I have solved the Cauchy problems $$x'(t) = 1, x(0) = x_0$$ $$y'(t) = 1, y(0) = 0;$$ the choices for the initial values will be clear in a second.

Let's go back to $z'$. Using what we have done so far we observe that $$z'(t) = u_x(x(t), y(t))\cdot 1 + u_y(x(t), y(t))\cdot 1 = x(t) + y(t) = 2t + x_0$$ Moreover $$z(0) = u(x(0),y(0)) = u(x_0,0) = 0.$$ This provides everything that is needed in order to write down a Cauchy problem for $z$ and also explains the choices made for the initial values $(x_0,0)$: we used the PDE to write the ODE for $z$ and we used the boundary condition to write an initial condition for $z$!

Solving the IVP we get $z(t) = t(t + x_0)$.

Now let $(x,y) \in \mathbb{R^2}$, and let $x_0$ be such that for some $t$, $(x,y) = (t + x_0,t)$. (consider the line parallel to $y = x$ passing through $(x,y)$ and let $x_0$ be the intersection between this line and the $x$-axis.) Solve for $t$: $$y = t = x - x_0.$$

Finally plug into the formula for $z$: $$u(x,y) = z(y) = xy.$$

(I would have just commented, but I don't yet have enough reputation)

The solution is $u = xy$. Uniqueness tells us that this is the only solution. The "quickest" way (if you have been doing this for a little bit) is by inspection, which is a fancy name for an educated guess. You get better at doing this with simple PDE's as you work more with pdes. Like a lot of things, some pople are naturally better at finding them, but it is something that gets easier as you work more.

You could in theory use a more "advanced" method that fits the situation, but it should get you the same answer.

• Just for reputations... +1 – johannesvalks Aug 25 '15 at 22:54

disclaimer: I use the notation $K'$ to denote the derivative of the function $K$ if it only depends on a single variable.

Well, there's a plus sign, so why not assume that

$$u(x,y) = X(x)+Y(y)$$

and give it a try?

From $u(x,0)=0$ one could tell now that $$X(x) +Y(0) = 0$$ which leads to $$X(x) = -Y(0)$$ The problem is that the derivative thereof is 0:

$$X'=0$$

That makes satisfying the following equations problematic

$$u_x+u_y = x+y$$ $$X'+Y' = x+y$$ $$Y' = x+y$$ Because $Y(y)$ (by definition) only depends on $y$, not $x$ as implied by the above formula for $Y'$, the above doesn't work out.

Another common pattern is to see it multiplication works: $$u(x,y) = X(x)\cdot Y(y)$$

From

$$X(x)Y(0) = 0$$

it makes sense to conclude that

$$Y(0) = 0$$

Now what's with the $u_x+u_y = x+y$?

$$u_x = X'(x)Y(y)$$ $$u_y = X(x)Y'(y)$$

thus

$$X'(x)Y(y) + X(x)Y'(y) = x+y$$

If trying to find some correspondence as the equaition stands now however

$$\color{purple}{X'(x)Y(y)} + \color{green}{X(x)Y'(y)} = \color{purple}{x}+\color{green}{y}$$

things don't work out, because in either color the nonderivative term must be 1, because it doesn't (by definition) depend on the other variable), take purple for example

$$\color{purple}{X'(x)}\color{red}{Y(y)} = \color{purple}{x}\color{red}{1}$$

That $x$ on the right side cannot come from $Y(y)$.

But then what if the correspondence was different? Let's try some different colors on different things to see if it makes any difference.

$$\color{blue}{X'(x)Y(y)} + \color{orange}{X(x)Y'(y)} = \color{orange}{x}+\color{blue}{y}$$

Take it apart $$\color{red}{X'(x)}\color{blue}{Y(y)} = \color{red}{1} \color{blue}{y}$$ $$\color{orange}{X(x)}\color{red}{Y'(y)} = \color{orange}{x}\color{red}{1}$$

Works for me:

$$X(x) = x$$ $$X'(x) = 1$$ $$Y(y) = y$$ $$Y'(y) = 1$$ $$Y(0) = 0$$ $$u=X(x)\cdot Y(y)$$

The general idea of the splitting-that-darn-thing-apart business is to be able to reason about which parts can go where. When a function only depends on one variable but not the other, it's easier to say if some equation can work or not.

If in doubt, try different colors.

An alternative to the method of characteristics. I like to call it common sense.

Note that the PDE is linear and with constant coefficients. Hence, the structure of the solution is given by $u(x,y) = u_h + u_p$, where $u_h$ is a solution of $u_x + u_y = 0$ and $u_p$ is a particular solution.

Trying solutions of the form $u_h = A \exp{(a x+ by)}$ leads to the condition $a + b = 0$ leading to $u_h = A \exp{[a(x-y)]}$, which can be extended up to $u_h = f(x-y)$, for any function $f$ (why?).

A particular solution is a polynomial in $x$ and $y$. For instance, $u_p = C x + D y + E \, xy$ satisfies the PDE provided $C = D = 0$ and $E = 1$. Therefore,

$$u = f(x-y) + xy$$

is the solution of your PDE. How do we choose $f$ to satisfy the boundary condition?