Solving the simples Partial Differential Equations

In summary: The general solution is F(x+ y) = constant. The solution may be in the form of F(x+y) = f(y) + g(x). This method only works for constant coefficients and no x or y terms in the coefficients. This method is more difficult to carry out.
  • #1
theneedtoknow
176
0
Hello,

I joined a class on Partial Differential Equations 3 or 4 lectures late, so I have missed the classes outlining how to solve the simplest forms of them. I am trying to understand the textbook, but so far it is not going so well for me, so I just need to ask some questions about the whole process.

First, let's start with something of the form a*ux + b*uy = 0 (where ux and uy are the partial derivatives of u)
We have a geometric method (find equations of all lines parallel to the vector (a, b) and assume the function u is constant on those lines (given that the directional derivative in the direction of (a, b) is 0
so, solving for the constant (in the expression for all lines parallel to the vector), we find u = f (bx - ay)

This method I understand somewhat alright - I managed to apply it to some other cases, such as x*ux + y*uy = 0 , by finding the equations of all curves with the vector (x,y) tangent to them, and assuming the function u is constant on those curves.

However, in the same chapter, we have some examples where the right side is not 0
such as ux + uy = 1
Here, I can no longer assume that the function u is a constant on the curves with V = (1,1) as a tangent (since the derivative is not 0). How do I extend this geometric method to cases when the right hand side is not zero?
Also, we have a 2nd method outlined- Coordinate Method.
For example, in the a*ux + b*uy = 0 case, we change coordinates such that the x-axis is alligned with the direction of the vector of the directional derivative (a, b)

so x' = ax + by and y' = bx - ay

(Aside... to change to prime coordinates that align the x-axis to any vector (c, d), do we just have to carry out A*x = x' where

A = | c d |
| d -c |
?
Can someone confirm if this is how we rotate coordinate axes to align with a vector?
End of Aside)

Then, using these primed coordinates, we rewrite the original equation, solve for ux and uy in terms of ux' and uy', plug them in, and solve.
this gives us ux' = 0 ------> u = f(y') = f (bx-ay)

Now, this method is a bit harder for me to carry out. I assume this method only works when the coefficient before ux and uy are constants (contain no x or y terms), so that the vector to which we need to rotate is constant.

I tried using this method to solve the above mentioned question ux + uy = 1, which i couldnt' solve using the geometric method (or at least I don't know how to)

giving me x' = x + y , y' = x - y
then ux = du/dx' * dx'/dx + du/dy'*dy'/dx
and likewise for uy
solving for these i get ux = ux' + uy' and uy = ux' - uy'
now ux + uy = 2ux' = 1

or ux' = 1/2

or u = (1/2) x' + f(y')
or u = (1/2) (x+y) + f(x-y) + C

with f an arbitrary function

then ux = 1/2 + f'(x-y) and uy = 1/2 -f' (x-y)

ux + uy = 1, which is what i needed, but I still don't know if the method I arrived at it via is right ... I thought my solution was supposed to be some kind of arbitrary function f, but in this case, my solution has both a specific function in it (1/2x + 1/2 y) and an arbitrary function f(x-y) ... is this normal for PDEs?

sorry for the long post...I hope someone can help clear up all the small confusions regarding this topic that I have mentioned. Thank you in advance :)
 
Last edited:
Physics news on Phys.org
  • #2
theneedtoknow said:
However, in the same chapter, we have some examples where the right side is not 0
such as ux + uy = 1
Here, I can no longer assume that the function u is a constant on the curves with V = (1,1) as a tangent (since the derivative is not 0). How do I extend this geometric method to cases when the right hand side is not zero?

One can find a particular solution and then use the general solution to ux + uy = 0.
 

FAQ: Solving the simples Partial Differential Equations

What is a partial differential equation (PDE)?

A partial differential equation is a mathematical equation that involves multiple variables and their partial derivatives. It describes the relationship between a function and its derivatives, and is commonly used to model physical systems in various fields such as physics, engineering, and economics.

How do you solve a PDE?

The process of solving a PDE involves finding a function that satisfies the equation. This can be done analytically, using mathematical techniques such as separation of variables and Fourier transforms, or numerically, using computer algorithms. The choice of method depends on the complexity of the PDE and the desired level of accuracy.

What are boundary conditions in PDEs?

Boundary conditions are additional information that is needed to uniquely determine a solution to a PDE. They specify the behavior of the function at the boundaries of the domain, and can be either Dirichlet conditions (prescribing the value of the function) or Neumann conditions (prescribing the value of the derivative).

What are some real-world applications of PDEs?

PDEs are used to model a wide range of physical phenomena, such as heat transfer, fluid dynamics, and electromagnetism. They are also commonly used in financial mathematics to model the behavior of stock prices and interest rates, and in image processing to enhance and analyze images.

Are there any limitations to solving PDEs?

Yes, there are several limitations to solving PDEs. Some PDEs have no closed-form solutions and can only be solved numerically. Additionally, the accuracy of the solution depends on the choice of numerical method and the size of the grid used for the approximation. Moreover, PDEs can become computationally expensive to solve for complex systems with high-dimensional domains.

Similar threads

Back
Top