1. ## Partial differntial equation

In this problem , I dont understand why at t = 0 , the temp at x = 1m , 2m and 3m are = 10 degree ?

Second question , it's stated earlier in the question that the end B is maintained at 10 degree , so shouldn't the temp at , t = 0.2 and t = 0.4 , the temperature at end B = 10 degree ? While at t = 0 , the temp at end B = 0 degree as stated in the question .

2. ## Re: Partial differntial equation

Are you reading the problem correctly? "One end, A, of an insulated metal bar. AB, of length 4 m, is kept at 0C while the other end, B is maintained at 10C until a steady state temperature is achieved. At t= 0, the end, B, is suddenly reduced to 0C and and kept at that temperature."

That is, the two ends are at 0 and 10 C for t less than 0 at which time there is a "steady state temperature is achieved." That means that your initial temperature is the steady state u(x, 0)= (10/4)x= (5/2)x. At that time, B is dropped to 0C. Your problem is to determine the temperature, u(x, t), for t greater than 0.

The "initial condition" for this problem u(x, 0)= (5/2)x (for all x except x= 4) while the "boundary conditions" are u(0, t)= 0 and u(4, t)= 0. The initial condition in not continuous but since it is discontinuous only at one end point that does not cause any difficulty.

3. ## Re: Partial differntial equation

Originally Posted by HallsofIvy
Are you reading the problem correctly? "One end, A, of an insulated metal bar. AB, of length 4 m, is kept at 0C while the other end, B is maintained at 10C until a steady state temperature is achieved. At t= 0, the end, B, is suddenly reduced to 0C and and kept at that temperature."

That is, the two ends are at 0 and 10 C for t less than 0 at which time there is a "steady state temperature is achieved." That means that your initial temperature is the steady state u(x, 0)= (10/4)x= (5/2)x. At that time, B is dropped to 0C. Your problem is to determine the temperature, u(x, t), for t greater than 0.

The "initial condition" for this problem u(x, 0)= (5/2)x (for all x except x= 4) while the "boundary conditions" are u(0, t)= 0 and u(4, t)= 0. The initial condition in not continuous but since it is discontinuous only at one end point that does not cause any difficulty.
How could the u(x, 0)= (10/4)x= (5/2)x ?? We can see that at u(1,0) , u(2,0 ) and u(3,0) , the t = 10 degree

What do you mean by the colour part ? I still didnt get it .

4. ## Re: Partial differntial equation

No, we don't "see that at u(1,0) , u(2,0 ) and u(3,0) , the t = 10 degree". The two ends are held at 0 and 10 degrees [b]until a steady state temperature is achieved." "Steady state" means that as long as conditions are maintained (the bar is insulated and the end A, at x= 0 is 0, the end at B, x= 4, is at 10) the temperature will not change- the given d.e. is satisfied by a function in x that does not depend on t.

That means that $\displaystyle \frac{\partial^2 u}{\partial t^2}$ and so, for all x, between 0 and 4, $\displaystyle \frac{\partial^2 u}{\partial t^2}$ must also be equal to 0 in order that the differential equation is satisfied. Since the second derivative is 0, the first derivative is a constant and the function itself is linear: u(x, 0)= ax+ b. So u(0, 0)= b= 0 and u(4, 0)= 4a= 10. a= 10/4= 5/2. u(x, 0)= (5/2)x.

5. ## Re: Partial differntial equation

Originally Posted by HallsofIvy
No, we don't "see that at u(1,0) , u(2,0 ) and u(3,0) , the t = 10 degree". The two ends are held at 0 and 10 degrees [b]until a steady state temperature is achieved." "Steady state" means that as long as conditions are maintained (the bar is insulated and the end A, at x= 0 is 0, the end at B, x= 4, is at 10) the temperature will not change- the given d.e. is satisfied by a function in x that does not depend on t.

That means that $\displaystyle \frac{\partial^2 u}{\partial t^2}$ and so, for all x, between 0 and 4, $\displaystyle \frac{\partial^2 u}{\partial t^2}$ must also be equal to 0 in order that the differential equation is satisfied. Since the second derivative is 0, the first derivative is a constant and the function itself is linear: u(x, 0)= ax+ b. So u(0, 0)= b= 0 and u(4, 0)= 4a= 10. a= 10/4= 5/2. u(x, 0)= (5/2)x.
By saying that u(x, 0)= (5/2)x , do you mean the ans uploaded u(1,0) , u(2,0) , u(3,0) is wrong ?

but , it's given in the question that at t = 0 , x = 4 m , the temp is 0 , which is not 0

6. ## Re: Partial differntial equation

Originally Posted by HallsofIvy
No, we don't "see that at u(1,0) , u(2,0 ) and u(3,0) , the t = 10 degree". The two ends are held at 0 and 10 degrees [b]until a steady state temperature is achieved." "Steady state" means that as long as conditions are maintained (the bar is insulated and the end A, at x= 0 is 0, the end at B, x= 4, is at 10) the temperature will not change- the given d.e. is satisfied by a function in x that does not depend on t.

That means that $\displaystyle \frac{\partial^2 u}{\partial t^2}$ and so, for all x, between 0 and 4, $\displaystyle \frac{\partial^2 u}{\partial t^2}$ must also be equal to 0 in order that the differential equation is satisfied. Since the second derivative is 0, the first derivative is a constant and the function itself is linear: u(x, 0)= ax+ b. So u(0, 0)= b= 0 and u(4, 0)= 4a= 10. a= 10/4= 5/2. u(x, 0)= (5/2)x.
you say that "Steady state" means that as long as conditions are maintained" ,
But , why at t = 0 , the temp at x = 4 is T = 0 ?
It's stated in the problem earlier that the temp is maintained at 10 degree , why it will suddenly decraese to 0 ?

7. ## Re: Partial differntial equation

Originally Posted by HallsofIvy
Are you reading the problem correctly? "One end, A, of an insulated metal bar. AB, of length 4 m, is kept at 0C while the other end, B is maintained at 10C until a steady state temperature is achieved. At t= 0, the end, B, is suddenly reduced to 0C and and kept at that temperature."

That is, the two ends are at 0 and 10 C for t less than 0 at which time there is a "steady state temperature is achieved." That means that your initial temperature is the steady state u(x, 0)= (10/4)x= (5/2)x. At that time, B is dropped to 0C. Your problem is to determine the temperature, u(x, t), for t greater than 0.

The "initial condition" for this problem u(x, 0)= (5/2)x (for all x except x= 4) while the "boundary conditions" are u(0, t)= 0 and u(4, t)= 0. The initial condition in not continuous but since it is discontinuous only at one end point that does not cause any difficulty.
Do you mean for the time where at x = 4 , T = 4 degree , is t less than 0 ( time taken for the end span to become 0 degree ) , it 's actually t < 0 .. ...So , now we are only interested when after the steady state achieved ( the end span drop 0 degree from 10 degree ) ? So , t > 0 is considered ?

8. ## Re: Partial differntial equation

Originally Posted by HallsofIvy
Are you reading the problem correctly? "One end, A, of an insulated metal bar. AB, of length 4 m, is kept at 0C while the other end, B is maintained at 10C until a steady state temperature is achieved. At t= 0, the end, B, is suddenly reduced to 0C and and kept at that temperature."

That is, the two ends are at 0 and 10 C for t less than 0 at which time there is a "steady state temperature is achieved." That means that your initial temperature is the steady state u(x, 0)= (10/4)x= (5/2)x. At that time, B is dropped to 0C. Your problem is to determine the temperature, u(x, t), for t greater than 0.

The "initial condition" for this problem u(x, 0)= (5/2)x (for all x except x= 4) while the "boundary conditions" are u(0, t)= 0 and u(4, t)= 0. The initial condition in not continuous but since it is discontinuous only at one end point that does not cause any difficulty.
why at time t = 0 , the temperature at x = 1 , x = 2 and x = 3 is T = 10 degree ?

Bump

10. ## Re: Partial differntial equation

I explained to you, four weeks ago, that you had the wrong initial condition. Your problem is to solve the equation $\displaystyle \frac{\partial T}{\partial t}= \frac{\partial^2 T}{\partial x^2}$ with boundary conditions T(0, t)= T(4, t)= 0 and initial condition T(x, 0)= (5/2)x. Have you tried solving the problem with that initial condition?

11. ## Re: Partial differntial equation

Originally Posted by HallsofIvy
I explained to you, four weeks ago, that you had the wrong initial condition. Your problem is to solve the equation $\displaystyle \frac{\partial T}{\partial t}= \frac{\partial^2 T}{\partial x^2}$ with boundary conditions T(0, t)= T(4, t)= 0 and initial condition T(x, 0)= (5/2)x. Have you tried solving the problem with that initial condition?
do you mean that at u(1,0) . T = (5/2)(1) = 2.5 degree , at u(2,0) , T = (5/2)(2) = 5 ? at u(3,0) , T = (5/2)(3) = 7.5 ?

If so , then the ans provided by the book of at u(1,0) , u(2,0) ,and u(3,0) , T = 10 degree is wrong ???

12. ## Re: Partial differntial equation

Originally Posted by HallsofIvy
I explained to you, four weeks ago, that you had the wrong initial condition. Your problem is to solve the equation $\displaystyle \frac{\partial T}{\partial t}= \frac{\partial^2 T}{\partial x^2}$ with boundary conditions T(0, t)= T(4, t)= 0 and initial condition T(x, 0)= (5/2)x. Have you tried solving the problem with that initial condition?
I did try , but my ans is . T = (5/2)(1) = 2.5 degree , at u(2,0) , T = (5/2)(2) = 5 ? at u(3,0) , T = (5/2)(3) = 7.5 ? Is that correct ?