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 $u_t = 0$, would they not have just asked for $u_{xx} + u_{yy} = 0$?

Perhaps you are expected to make the leap to $u = h(t)f(x,y)$.

 

[Magnetons]

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

Perhaps you are expected to make the leap to $u = h(t)f(x,y)$.

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 $u = h(t)f(x,y)$.

 

[BvU]

Found a satisfactory solution ?

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