I've seen several problems with $u_x^2 + u_y^2$, such as here Characteristics method applied to the PDE $u_x^2 + u_y^2=u$.

But what about this equation: $$u_x + \frac{1}{2}u_y^2+\frac{1}{2}x^2=0$$ with initial conditions $u(0,y)=f(y)$.

I am told that that fully eliminating parameters is not possible with $f$ unspecified. Apparently the best you can do it is with two equations with one extra variable.

Anyone able to help me out?

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    $\begingroup$ This problem doesn't seem to be well-defined. $x,y$ are generally spatial variables. Even if we assume that an initial condition implies that $x > 0$, there is no information on what $y$ does. $\endgroup$ – Gregory Feb 14 '18 at 0:29
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    $\begingroup$ That being said as a start, you can maybe use the substitution $v = u + \frac{1}{6} x^3$, which will remove the $x^2$ term. $\endgroup$ – Gregory Feb 14 '18 at 0:29
  • $\begingroup$ You can use Charpit mathod for solving this type of PDE. $\endgroup$ – SAHEB PAL Feb 14 '18 at 0:31

First, we reduce the equation as Gregory pointed out:


Now we have:


Edit. The general solution.

It seems to me this equation can be reduced to the Burgers' equation. Let us take another partial derivative w.r.t. $y$:

$$v_{xy}+v_y v_{yy}=0$$

Introducing a new function:


We can write:

$$w_x+w w_y=0$$

Which is precisely the Burgers' equation. Using the method of characteristics as stated here, we have the implicit general solution as:

$$w(x,y+w(0,y) x)=w(0,y)$$

Where $w(0,y)$ is the initial condition (which can probably be obtained from $v(0,y)$ as $w(0,y)=\frac{d}{dy} v(0,y)$).

If the initial condition is nice enough, we can find the explicit solution, by setting:

$$z=y+w(0,y) x$$



Then we find $v(x,z)$ as:

$$v(x,y)=\int w(x,y) dy+C$$

Where $C$ is a constant. You can try this method with various initial conditions.

Some easier ways to get particular solutions in case of simple initial condition:

$1)$ Let's assume the solution to have the form:



$$g'+\frac{1}{2} h'^2=0$$

But the functions depend on different variables, so for the equation to hold, they have to be linear:

$$g(x)=ax+b \\ h(y)=cy+d$$

The equation gives us:

$$a+\frac{1}{2} c^2=0$$


$$v(x,y)=-\frac{1}{2} c^2x+cy+B$$

Where $B=b+d$.

I do not see how any kind of initial conditions except linear will work here.

$2)$ But this is not the only form we can assume, let's try another:


Now we have:

$$hg'+\frac{1}{2}g^2 h'^2=0$$

Dividing by $hg^2$:


Both parts of the equation have to be constant:



Solving, we obtain:

$$g(x)=\frac{1}{\mu x +a}\\ h(y)= \left(\sqrt{\frac{\mu}{2}} y+b \right)^2$$

So we have a solution:

$$v(x,y)=\frac{1}{\mu x +a} \left(\sqrt{\frac{\mu}{2}} y+b \right)^2$$

We can get rid of one of the constants, for example:

$$v(x,y)=\frac{1}{x +A} \left(\sqrt{\frac{1}{2}} y+B \right)^2$$

This is a different solution than $1)$ and can also satisfy the initial conditions for a particular $f(y)$.


Assume $p=\frac{\partial v}{\partial x}$, $q=\frac{\partial v}{\partial y}$, and applying the transformation $v=u+x^3/6$ as mentioned Gregory Sir, the above PDE reduce to $2p+q^2 =0$. Then by using Charpit auxiliary equation we get $p=a$ and $q=b$ where $a$ and $b$ are arbitrary constant satisfy the relation $2a+b^2=0$. Now using $dv=pdx+qdy$, we get $v=ax+by+c$ and so $u=-\frac {x^3} 6+ax+by+c$ is the general solution. Using the relation $2a+b^2 =0$ and given initial condition you got the particular solution.





Let $v=u_y$ ,

Then $v_x+vv_y=0$ with $v(0,y)=f_y(y)$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{dv}{dt}=0$ , letting $v(0)=v_0$ , we have $v=v_0$

$\dfrac{dy}{dt}=v=v_0$ , letting $y(0)=f(v_0)$ , we have $y=v_0t+f(v_0)=xv+f(v)$ i.e. $v=F(y-xv)$


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