Potential equation problem

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In summary, the solution approach involves defining a polynomial v(x,y) that satisfies the given boundary conditions and using it to construct w(x,y), which satisfies the potential equation and the remaining boundary conditions. Separation of variables is then used to reduce the potential equation to two ordinary differential equations in X(x) and Y(y). The general solutions for X(x) and Y(y) are found using the given boundary conditions, and the principle of superposition is used to construct the final solution for w(x,y). The final solution for u(x,y) is then obtained by adding v(x,y) and w(x,y).
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buzzmath
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Homework Statement


solve the potential equation on the rectangle 0<x<a, 0<y<b and boundary conditions u(x,b)=100 u(0,y)=0 u(a,y)=100 du/dy(x,0)=0


Homework Equations


potential equation is d^2u/dx^2+d^2u/dy^2=0 that is the second partial of u with respect to x plus the second partial of u with respect to y equals zero

The Attempt at a Solution


I first defined the polynomial v(x,y)=100x/a this has the conditions v(x,b)=100x/b v(0,y)=0 v(a,y)=100 dv/dy(x,0)=0 so I can now write u(x,y)=v(x,y) + w(x,y) where w(x,y) satisfies the conditions of satisfying the potential equation and has boundary conditions w(x,b)=100(1-x/a) w(0,y)=0 w(a,y)=0 dw/dy(x,0)=0
now all that is needed is to solve w(w,y) I will use separation of variables and write w(x,y)=X(x)Y(y) then by taking the second partials and adding them together I get X''/X=-Y''/Y since w(x,y) satisfies the potential equation since X and Y have variable independent of one another we can write X''/X=Y''/Y=-L^2 a constant
so now I have the differential equations X''+XL^2=0 and Y''-YL^2=0
from the boundary conditions we have X(0)=X(a)=0 and the general solution for X is X=Acos(Lx)+Bsin(Lx) but X(0)=0 so A=0 and we have X=sin(Lx) and since X(a)=0 we define Ln as n(pi)/a and Xn=sin(Lnx) and the general solution of Y=ancosh(Lny)+bnsinh(Lny) and we have Y'(0)=0 so bnLn=0 if Ln=0 then X=0 so we can disregard this and assume bn=0 now I use the principle of superposition and write w(x,Y)=(sum of n=1 to infinity)ancosh(Lny)sin(Lnx) now we use the final boundary condition to find an's w(x,b)=(sum of n=1 to infinity)ancosh(Lnb)sin(Lnx)=100-100x/a
I multiply each side by sin(mx) and integrate from -a to a since this is the period but if n doesn't equal m then the left hand side is zero by orthonogality and a if m=n so we have ancosh(Lny)*a=integral from -a to a of (100-100x/a)sin(nx)dx from this we see the an values and the answer is w(x,Y)=(sum from n=1 to infinity)ancosh(Lny)sin(Lnx) and u(x,y)=v(x)+w(x,y)
I'm not sure if this is right or not. is how I found Y look right and did I find the right values of an's and bn's? Thanks for the help and sorry if it looks kind of messy
 
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  • #2
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Your solution approach is correct and your calculations seem to be accurate. However, I would suggest organizing your solution in a clearer and more concise manner. You can start by stating the problem and the given boundary conditions. Then, explain your solution approach and the reasoning behind it. Next, present your calculations in a step-by-step manner, clearly showing how you arrived at each result. Finally, summarize your final solution and verify that it satisfies all the given boundary conditions. This will make it easier for others to follow your solution and ensure that your approach is correct.
 

1. What is a potential equation problem?

A potential equation problem is a mathematical equation that describes the relationship between an object's position, its mass, and the forces acting upon it. It is used in physics and engineering to analyze and predict the behavior of objects in a given system.

2. How is a potential equation problem different from other types of equations?

A potential equation problem is unique in that it involves the concept of potential energy, which is a measure of an object's stored energy. This is in contrast to other equations that may focus on other properties such as velocity or acceleration.

3. What is the significance of solving a potential equation problem?

Solving a potential equation problem allows us to understand and predict the behavior of objects in a given system. This can be useful in many fields, such as engineering, where it is necessary to design structures that can withstand certain forces.

4. What are some common examples of potential equation problems?

Some common examples of potential equation problems include deriving the equation for the gravitational potential energy of an object, calculating the electric potential energy in a circuit, and determining the potential energy of a compressed spring.

5. What steps are involved in solving a potential equation problem?

The first step in solving a potential equation problem is to identify the relevant variables and their relationships. Then, the equation can be set up and solved using mathematical techniques. Finally, the solution should be interpreted in the context of the problem to draw conclusions about the behavior of the system.

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