Partial Differential Equation (partial-y/partialt = k * partial^2-y/partial-x^2)

mario99

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[math]\frac{\partial y}{\partial t} = k\frac{\partial^2 y}{\partial x^2}[/math]


Any help would be appreciated.
 
[math]\frac{\partial y}{\partial t} = k\frac{\partial^2 y}{\partial x^2}[/math]


Any help would be appreciated.
Please at least try to make an attempt. Whatever source you are looking at probably went over the method.

Hint: Try separation of variables.

Give it a try and post your results.

-Dan
 
Thank you topsquark, khansaheb, and nasi112 for helping me.


Please at least try to make an attempt. Whatever source you are looking at probably went over the method.

Hint: Try separation of variables.

Give it a try and post your results.

-Dan
It's not about the method. I have already solved the differential equation before I posted it here by Fourier transform. My solution is different than the given one. I want you to solve it by your own way to see where I went wrong.



This is a heat equation in one dimension. Do you want any solution that satisfies the differential equation? Or Do you want a unique solution?
What do you mean by one dimension? There are already four variables in the equation, [imath]x[/imath], [imath]y[/imath], [imath]t[/imath], [imath]k[/imath]. Why not four dimensions? What is the different between a unique solution and not a unique solution? Please, solve the equation and let me see your solution, so that I can compare it with my solution and know where I went wrong.



What are the boundary conditions?

What is the initial condition?
[imath]y(0,t) = 0[/imath]
[imath]y(x,0) = f(x)[/imath]

Do you want analytical or numerical method of solution?
Please, show both, analytical and numerical solution.
 
It's not about the method. I have already solved the differential equation before I posted it here by Fourier transform. My solution is different than the given one. I want you to solve it by your own way to see where I went wrong.
We know that's what you want. You've made that very clear.

You just don't listen, do you? We are going to ask you leading questions or give you hints. How many times does the "new Mario" need to be told this? Post your solution and we will tell you where you might have gone wrong.

-Dan
 
[imath]y(0,t) = 0[/imath]
[imath]y(x,0) = f(x)[/imath]


Please, show both, analytical and numerical solution.
If you had used these (and only these conditions) to derive the solution (of the given PDE) - then your derived solution/s is INCORRECT.
 
I want you to solve it by your own way to see where I went wrong....

Please, solve the equation and let me see your solution, so that I can compare it with my solution and know where I went wrong....

Please, show both, analytical and numerical solution.
I'm sorry, but that's not how it works, as has (repeatedly) been explained to you. Kindly please begin to cooperate with the helpers, in accordance with the rules of this forum. Thank you.
 
What do you mean by one dimension?
I mean that the heat equation can come in one dimension, two dimensions, or three dimensions.

To show you that, I will use the variable \(\displaystyle u(x,t)\) because I need \(\displaystyle y\) to be an independent variable.

One dimension:
\(\displaystyle \frac{1}{k}\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}\)

Two dimensions:
\(\displaystyle \frac{1}{k}\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\)

Three dimensions:
\(\displaystyle \frac{1}{k}\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\)

Although it is possible to call this differential equation \(\displaystyle \frac{\partial y}{\partial t} = k\frac{\partial^2 y}{\partial x^2}\) two dimensional, we usually count the dimension with respect to the space variable. The space variable \(\displaystyle x\) here is one dimensional, so we say it is a one dimensional heat equation.

There are already four variables in the equation, [imath]x[/imath], [imath]y[/imath], [imath]t[/imath], [imath]k[/imath]. Why not four dimensions?
This is not the right way to tell the dimensions of an equation. I showed you above how to find the dimension. It will be an interesting partial differential equation if \(\displaystyle k\) is a variable which I don't think so.


What is the different between a unique solution and not a unique solution? Please, solve the
Why did khansaheb ask you for boundary and initial conditions? Because without them, there can be a million solution to this differential equation. To get one and only one solution (unique solution), you have to write a complete boundary and initial conditions with a domain. What I can see in your first post is like a chicken without feathers!


Please, solve the equation and let me see your solution, so that I can compare it with my solution and know where I went wrong.
I don't mind to solve the equation, but I will not. Not because the rules of the forum forbids that, it is because you are just throwing naked questions, you don't even understand the basic idea behind them.


To prove to you that, you understand nothing, you cannot even answer this simple question!
There are already four variables in the equation, [imath]x[/imath], [imath]y[/imath], [imath]t[/imath], [imath]k[/imath].
Classify which ones of these are the dependent variables, the independent variables, and the constants.
 
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