Solution of the good PDE ?

[pde]

Solution of the good PDE ?

Find the solution of u of the equation

u_x + x u_y = u + x if u(x,0)=x^2,x>0.

 

[kosovtsov]

The general solution to your PDE is

$u(x,y) = -1-x+e^x F(x^2-2y),$

where F is an arbitrary function.

Your boundary condition leads to

$F(t) = (1+t^{1/2}+t)exp(-t^{1/2}),$

where $t=x^2-2y.$

 

[fluidistic]

The general solution to your PDE is

$u(x,y) = -1-x+e^x F(x^2-2y),$

where F is an arbitrary function.

Your boundary condition leads to

$F(t) = (1+t^{1/2}+t)exp(-t^{1/2}),$

where $t=x^2-2y.$

I'm not the original poster, but nice. May I know which method you used to obtain the general solution? Thank you!

 

[kosovtsov]

The main idea: If the homogeneous DE

$\alpha(x,y)\frac{\partial s(x,y)}{\partial x}+\beta(x,y)\frac{\partial s(x,y)}{\partial x}=0$

is solvable (it is sufficient to find any particular solution), then DE of the following type

$\alpha(x,y)\frac{\partial p(x,y)}{\partial x}+\beta(x,y)\frac{\partial p(x,y)}{\partial x}=\xi(x,y)p(x,y)+f(x,y)$

is solvable (at least formally) too. And general solution for p(x,y) can be found in principle from s(x,y) by means of only algebraic manipulations.

For given DE first of all we have to reduce the homogeneous part of initial DE

$\frac{\partial u(x,y)}{\partial x}+x\frac{\partial u(x,y)}{\partial x}-u(x,y)=0$

to ($u=\exp(v)$)

$\frac{\partial v(x,y)}{\partial x}+x\frac{\partial v(x,y)}{\partial x}-1=0.$

The (particular) polynomial solution of homogeneous part of the last DE

$\frac{\partial v(x,y)}{\partial x}+x\frac{\partial v(x,y)}{\partial x}=0.$

is $v(x,y)=x^2-2y$, so its general solution is $v(x,y)=F(x^2-2y).$

To find general solutions for DEs on the way back it is quite enough here to seek particular solutions of nonhomogeneous DEs in forms $v(x,y)=a(x),u(x,y)=b(x)$.

 

[blue_raver22]

The solution of this partial differential equation (PDE) can be found by using the method of characteristics. This method involves finding a set of curves, known as characteristics, along which the PDE reduces to an ordinary differential equation (ODE). The solution of the ODE can then be used to find the solution of the PDE.

In this case, the characteristics are given by the equations dx/dt = 1 and dy/dt = x. Solving these equations, we get x = t and y = t^2/2 + c, where c is an arbitrary constant. This means that the PDE reduces to an ODE along the curves x = t and y = t^2/2 + c.

Solving the ODE, we get u(x,y) = e^y (x^2 + 2c). Using the initial condition, u(x,0) = x^2, we can determine the value of c as 0. Therefore, the solution of the PDE is u(x,y) = e^y (x^2).

In summary, the solution of the good PDE is u(x,y) = e^y (x^2), where x > 0 and y = t^2/2. This solution satisfies the given PDE and the initial condition.