PDE and the separation of variables

[Magnetons]

Homework Statement

Apply the method of separation of variables ##u(x,y)=f(x)g(y)## to solve the equation .

Relevant Equations

##u_{t}=c^{2}(u_{xx}+u_{yy})##

using the equation ##u(x,y)=f(x)g(y)##, first, I substitute the value of ##u_{xx}## and ##u_{yy}## in the given PDE. after that solve the ODEs but I can't understand about the ##u_{t}##.In my solution, I put ##u_{t}=0## because u is only the function of x and y. Is it the right approach, to me it seems wrong

 

[BvU]

Aren't you supposed to separate in a different way ? In u(t) and u(x,y) for example

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[BvU]

Google 'heat equation'

 

[Magnetons]

Aren't you supposed to separate in a different way ? In u(t) and u(x,y) for example

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don't know it is how the question is given in the book

 

[BvU]

Well, you run into trouble with ##u_t=0##, so I suggest to try something different.

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[Magnetons]

Well, you run into trouble with ##u_t=0##, so I suggest to try something different.

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something different ..

 

[pasmith]

Homework Statement: Apply the method of separation of variables ##u(x,y)=f(x)g(y)## to solve the equation .
Relevant Equations: ##u_{t}=c^{2}(u_{xx}+u_{yy})##

using the equation ##u(x,y)=f(x)g(y)##, first, I substitute the value of ##u_{xx}## and ##u_{yy}## in the given PDE. after that solve the ODEs but I can't understand about the ##u_{t}##.In my solution, I put ##u_{t}=0## because u is only the function of x and y. Is it the right approach, to me it seems wrong

If they wanted you to assume [itex]u_t = 0[/itex], would they not have just asked for [itex]u_{xx} + u_{yy} = 0[/itex]?

Perhaps you are expected to make the leap to [itex]u = h(t)f(x,y)[/itex].

 

[Magnetons]

If they wanted you to assume [itex]u_t = 0[/itex], would they not have just asked for [itex]u_{xx} + u_{yy} = 0[/itex]?

Perhaps you are expected to make the leap to [itex]u = h(t)f(x,y)[/itex].

No ## u_t = 0 ## doesn't mention in question i assume it .

 

[BvU]

don't know it is how the question is given in the book

Yeah, they managed to confuse you (on purpose?) writing ##u(x,y)=f(x)g(y)## instead of ##u(p,q)=f(p)g(q)## or something less suggestive...

Your post #8 explains why. (and post#7 IS exercise 25 (h) ! )

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[Magnetons]

Yeah, they managed to confuse you (on purpose?) writing ##u(x,y)=f(x)g(y)## instead of ##u(p,q)=f(p)g(q)## or something less suggestive...

Your post #8 explains why. (and post#7 IS exercise 25 (h) ! )

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how should I solve this equation

 

[BvU]

make the leap to [itex]u = h(t)f(x,y)[/itex].

 

[BvU]

Found a satisfactory solution ?

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