Solve PDE: Find F to Satisfy \lambda F + (\frac{\partial F}{\partial y})^{2}=0

[Karlisbad]

Solving this PDE :(

Hello i have a question about this..let be a function $$F(x(t),y(t),z(t),t)$$ then if we use the "total derivative" respect to t and partial derivatives..could we find an F so it satisfies:

$$\frac{d (\frac{\partial F}{\partial x})}{dt}+\lambda F + (\frac{\partial F}{\partial y})^{2}=0$$ ??

how could you solve that ??.. my big problem is that this involves "total" and partial derivatives respect to x and y all mixed up.:yuck: :yuck:

 

[Amr Morsi]

Karlis,

Since the equation includes total derivative with respect to t, then x, y and z are functions of t as already included in the argument of F; X(t) ....

However, you still can solve it, but in terms of dx/dt, dy/dt, dz/dt ... etc. But, here, only function in dx/dt, dy/dt and dz/dt.

Try this:
d[T(r>,t)]/dt=dx/dt*p[T(r>,t)]/px+dy/dt*p[T(r>,t)]/py+dz/dt*p[T(r>,t)]/pz.
where p/px is the partial derivative.
This can be done by the rule of differentials.
Here, T(r>,t)=pF/px

You will get a normal partial differential function in x, y and z with 3 time-dependent functions (considered to be constants in the equation). Solve it, if this form has an analytical solution (or any other sort) in PDEs.


Engineer\ Amr Morsi.

 

[tacman]

Hello, solving a PDE involves finding a function that satisfies the given equation. In this case, the PDE is:

\lambda F + \left(\frac{\partial F}{\partial y}\right)^2 = 0

To solve this, we can use the method of separation of variables. We assume that the function F can be expressed as a product of two functions, one depending only on x and the other depending only on y:

F(x,y) = X(x)Y(y)

We then substitute this into the PDE and rearrange:

\lambda X(x)Y(y) + \left(\frac{dY(y)}{dy}\right)^2 = 0

We can now separate the variables, one on each side of the equation:

\lambda X(x) + \left(\frac{dY(y)}{dy}\right)^2 = 0

\frac{dY(y)}{dy} = \pm \sqrt{-\lambda X(x)}

Integrating both sides with respect to y, we get:

Y(y) = C_1 \pm \sqrt{-\lambda X(x)}y

where C_1 is an arbitrary constant. Now, we can substitute this back into our original equation and solve for X(x):

\lambda X(x) + \left(\frac{dY(y)}{dy}\right)^2 = 0

\lambda X(x) + \lambda C_1^2 y^2 = 0

X(x) = -\frac{C_1^2 y^2}{\lambda}

Combining our solutions for X(x) and Y(y), we get:

F(x,y) = -\frac{C_1^2 y^2}{\lambda} \pm C_1 \sqrt{-\lambda X(x)} y

which is the general solution to our PDE. Depending on the boundary conditions of the problem, we can determine the value of C_1 and thus obtain a specific solution for F(x,y). I hope this helps!