PDE Example: Where Did $u(x, t) - t^2 = 0$ Come From?

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My lecture notes give the following:

Consider the PDE $dfrac partialu(x, t) partialt = 2t$.

Writing this as $dfrac partial partialt (u(x, t) - t^2) = 0$,

we see that $u(x, t) - t^2$ must not vary in time and hence solutions of this PDE must be of the form $u(x, t) = t^2 + A(x)$, where $A(x)$ is an arbitrary function.

To make the solution unique, we need more information, in this a boundary condition of the form $u(x, 0) = g(x)$,

so that the unique solution to the PDE with the associate boundary (initial) condition is $u(x, t) = t^2 + g(x)$.

I'm not sure where $dfrac partial partialt (u(x, t) - t^2) = 0$ came from?

I tried to derive these calculations as follows:

$dfrac partialu(x, t) partialt = 2t$

$therefore int dfrac partialu(x, t) partialt dt = int 2t dt$

$rightarrow u(x, t) = t^2 + A(x, t)$ (ERROR: this should be $A(x)$, as per Botond's comment!!!)

$rightarrow u(x, t) - t^2 = A(x, t)$

$therefore dfrac partial partialt (u(x, t) - t^2) = dfracpartialpartialt A(x, t)$

So I got $dfracpartialpartialt A(x, t)$, whereas the lecture notes have $0$.

I would greatly appreciate it if people could please take the time to review this. Am I misunderstanding something? Or is there an error in the lecture slides?

pde

share|cite|improve this question

edited Jul 22 at 13:06

asked Jul 22 at 12:52

The Pointer

2,4442829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you have $A(x,t)$? I think it should just be $A(x)$.
  – Botond
  Jul 22 at 12:55
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Isn't it clear that the partial derivative w.r.t. of $u(x,t)-t^2$ is $dfracpartial(u(x,t)partial t-dfracpartial(t^2)partial t=dfracpartial(u(x,t)partial t-2t$?
  – Bernard
  Jul 22 at 12:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Botond Hmm, but $u$ is a function of both $x$ and $t$, so wouldn't the arbitrary function have to be $A(x, t)$?
  – The Pointer
  Jul 22 at 12:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Bernard Oh wow, I completely misread it. My apologies. Just to confirm, shouldn't the arbitrary function be $A(x, t)$ in this situation?
  – The Pointer
  Jul 22 at 13:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  No. Try to differentiate your equation and see if you get the initial equation.
  – Botond
  Jul 22 at 13:03
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

favorite

My lecture notes give the following:

Consider the PDE $dfrac partialu(x, t) partialt = 2t$.

Writing this as $dfrac partial partialt (u(x, t) - t^2) = 0$,

we see that $u(x, t) - t^2$ must not vary in time and hence solutions of this PDE must be of the form $u(x, t) = t^2 + A(x)$, where $A(x)$ is an arbitrary function.

To make the solution unique, we need more information, in this a boundary condition of the form $u(x, 0) = g(x)$,

so that the unique solution to the PDE with the associate boundary (initial) condition is $u(x, t) = t^2 + g(x)$.

I'm not sure where $dfrac partial partialt (u(x, t) - t^2) = 0$ came from?

I tried to derive these calculations as follows:

$dfrac partialu(x, t) partialt = 2t$

$therefore int dfrac partialu(x, t) partialt dt = int 2t dt$

$rightarrow u(x, t) = t^2 + A(x, t)$ (ERROR: this should be $A(x)$, as per Botond's comment!!!)

$rightarrow u(x, t) - t^2 = A(x, t)$

$therefore dfrac partial partialt (u(x, t) - t^2) = dfracpartialpartialt A(x, t)$

So I got $dfracpartialpartialt A(x, t)$, whereas the lecture notes have $0$.

I would greatly appreciate it if people could please take the time to review this. Am I misunderstanding something? Or is there an error in the lecture slides?

pde

share|cite|improve this question

edited Jul 22 at 13:06

asked Jul 22 at 12:52

The Pointer

2,4442829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you have $A(x,t)$? I think it should just be $A(x)$.
  – Botond
  Jul 22 at 12:55
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Isn't it clear that the partial derivative w.r.t. of $u(x,t)-t^2$ is $dfracpartial(u(x,t)partial t-dfracpartial(t^2)partial t=dfracpartial(u(x,t)partial t-2t$?
  – Bernard
  Jul 22 at 12:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Botond Hmm, but $u$ is a function of both $x$ and $t$, so wouldn't the arbitrary function have to be $A(x, t)$?
  – The Pointer
  Jul 22 at 12:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Bernard Oh wow, I completely misread it. My apologies. Just to confirm, shouldn't the arbitrary function be $A(x, t)$ in this situation?
  – The Pointer
  Jul 22 at 13:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  No. Try to differentiate your equation and see if you get the initial equation.
  – Botond
  Jul 22 at 13:03
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

favorite

up vote
1
down vote

favorite

My lecture notes give the following:

Consider the PDE $dfrac partialu(x, t) partialt = 2t$.

Writing this as $dfrac partial partialt (u(x, t) - t^2) = 0$,

we see that $u(x, t) - t^2$ must not vary in time and hence solutions of this PDE must be of the form $u(x, t) = t^2 + A(x)$, where $A(x)$ is an arbitrary function.

To make the solution unique, we need more information, in this a boundary condition of the form $u(x, 0) = g(x)$,

so that the unique solution to the PDE with the associate boundary (initial) condition is $u(x, t) = t^2 + g(x)$.

I'm not sure where $dfrac partial partialt (u(x, t) - t^2) = 0$ came from?

I tried to derive these calculations as follows:

$dfrac partialu(x, t) partialt = 2t$

$therefore int dfrac partialu(x, t) partialt dt = int 2t dt$

$rightarrow u(x, t) = t^2 + A(x, t)$ (ERROR: this should be $A(x)$, as per Botond's comment!!!)

$rightarrow u(x, t) - t^2 = A(x, t)$

$therefore dfrac partial partialt (u(x, t) - t^2) = dfracpartialpartialt A(x, t)$

So I got $dfracpartialpartialt A(x, t)$, whereas the lecture notes have $0$.

I would greatly appreciate it if people could please take the time to review this. Am I misunderstanding something? Or is there an error in the lecture slides?

pde

share|cite|improve this question

edited Jul 22 at 13:06

asked Jul 22 at 12:52

The Pointer

2,4442829

My lecture notes give the following:

Consider the PDE $dfrac partialu(x, t) partialt = 2t$.

Writing this as $dfrac partial partialt (u(x, t) - t^2) = 0$,

we see that $u(x, t) - t^2$ must not vary in time and hence solutions of this PDE must be of the form $u(x, t) = t^2 + A(x)$, where $A(x)$ is an arbitrary function.

To make the solution unique, we need more information, in this a boundary condition of the form $u(x, 0) = g(x)$,

so that the unique solution to the PDE with the associate boundary (initial) condition is $u(x, t) = t^2 + g(x)$.

I'm not sure where $dfrac partial partialt (u(x, t) - t^2) = 0$ came from?

I tried to derive these calculations as follows:

$dfrac partialu(x, t) partialt = 2t$

$therefore int dfrac partialu(x, t) partialt dt = int 2t dt$

$rightarrow u(x, t) = t^2 + A(x, t)$ (ERROR: this should be $A(x)$, as per Botond's comment!!!)

$rightarrow u(x, t) - t^2 = A(x, t)$

$therefore dfrac partial partialt (u(x, t) - t^2) = dfracpartialpartialt A(x, t)$

So I got $dfracpartialpartialt A(x, t)$, whereas the lecture notes have $0$.

I would greatly appreciate it if people could please take the time to review this. Am I misunderstanding something? Or is there an error in the lecture slides?

pde

share|cite|improve this question

edited Jul 22 at 13:06

asked Jul 22 at 12:52

The Pointer

2,4442829

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 13:06

edited Jul 22 at 13:06

edited Jul 22 at 13:06

asked Jul 22 at 12:52

The Pointer

2,4442829

asked Jul 22 at 12:52

The Pointer

2,4442829

asked Jul 22 at 12:52

The Pointer

2,4442829

2,4442829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you have $A(x,t)$? I think it should just be $A(x)$.
  – Botond
  Jul 22 at 12:55
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Isn't it clear that the partial derivative w.r.t. of $u(x,t)-t^2$ is $dfracpartial(u(x,t)partial t-dfracpartial(t^2)partial t=dfracpartial(u(x,t)partial t-2t$?
  – Bernard
  Jul 22 at 12:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Botond Hmm, but $u$ is a function of both $x$ and $t$, so wouldn't the arbitrary function have to be $A(x, t)$?
  – The Pointer
  Jul 22 at 12:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Bernard Oh wow, I completely misread it. My apologies. Just to confirm, shouldn't the arbitrary function be $A(x, t)$ in this situation?
  – The Pointer
  Jul 22 at 13:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  No. Try to differentiate your equation and see if you get the initial equation.
  – Botond
  Jul 22 at 13:03
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you have $A(x,t)$? I think it should just be $A(x)$.
  – Botond
  Jul 22 at 12:55
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Isn't it clear that the partial derivative w.r.t. of $u(x,t)-t^2$ is $dfracpartial(u(x,t)partial t-dfracpartial(t^2)partial t=dfracpartial(u(x,t)partial t-2t$?
  – Bernard
  Jul 22 at 12:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Botond Hmm, but $u$ is a function of both $x$ and $t$, so wouldn't the arbitrary function have to be $A(x, t)$?
  – The Pointer
  Jul 22 at 12:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Bernard Oh wow, I completely misread it. My apologies. Just to confirm, shouldn't the arbitrary function be $A(x, t)$ in this situation?
  – The Pointer
  Jul 22 at 13:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  No. Try to differentiate your equation and see if you get the initial equation.
  – Botond
  Jul 22 at 13:03
  

  
  
  
  

Why do you have $A(x,t)$? I think it should just be $A(x)$.
– Botond
Jul 22 at 12:55

Why do you have $A(x,t)$? I think it should just be $A(x)$.
– Botond
Jul 22 at 12:55

4

4

Isn't it clear that the partial derivative w.r.t. of $u(x,t)-t^2$ is $dfracpartial(u(x,t)partial t-dfracpartial(t^2)partial t=dfracpartial(u(x,t)partial t-2t$?
– Bernard
Jul 22 at 12:56

Isn't it clear that the partial derivative w.r.t. of $u(x,t)-t^2$ is $dfracpartial(u(x,t)partial t-dfracpartial(t^2)partial t=dfracpartial(u(x,t)partial t-2t$?
– Bernard
Jul 22 at 12:56

@Botond Hmm, but $u$ is a function of both $x$ and $t$, so wouldn't the arbitrary function have to be $A(x, t)$?
– The Pointer
Jul 22 at 12:59

@Botond Hmm, but $u$ is a function of both $x$ and $t$, so wouldn't the arbitrary function have to be $A(x, t)$?
– The Pointer
Jul 22 at 12:59

@Bernard Oh wow, I completely misread it. My apologies. Just to confirm, shouldn't the arbitrary function be $A(x, t)$ in this situation?
– The Pointer
Jul 22 at 13:01

@Bernard Oh wow, I completely misread it. My apologies. Just to confirm, shouldn't the arbitrary function be $A(x, t)$ in this situation?
– The Pointer
Jul 22 at 13:01

2

2

No. Try to differentiate your equation and see if you get the initial equation.
– Botond
Jul 22 at 13:03

No. Try to differentiate your equation and see if you get the initial equation.
– Botond
Jul 22 at 13:03

 | 
show 2 more comments

1 Answer
1

active

oldest

votes

up vote
3
down vote



accepted

It is as Bernard wrote in the comments:
$fracpartialpartial tt^2 = 2t$. Hence, you get
$$
fracpartialpartial tu(x,t)=2tLeftrightarrow fracpartialpartial tu(x,t)=fracpartialpartial tt^2Leftrightarrow fracpartialpartial tu(x,t)-fracpartialpartial tt^2=0Leftrightarrow fracpartialpartial tleft(u(x,t)-t^2right)=0.
$$
At the last step, you use that the derivative is a linear operator.

In your argument, there is a mistake at
$$
int 2tdt = t^2+A(x,t).
$$
The function $A$ can't depend on $t$. You might think of the additive constant at the integration. But it is a constant wrt the integration variable (in this case $t$). On the other hand, the constant can depend on the remaining variables, like $x$ in your example. So, you should write
$$
int 2t~dt = t^2+A(x)
$$
such that $A$ just depends on $x$. Naturally, so get $fracpartialpartial tA(x)=0$ as you need.

share|cite|improve this answer

edited Jul 22 at 13:05

answered Jul 22 at 13:00

Mundron Schmidt

7,1162727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I completely misread it. My apologies everyone.
  – The Pointer
  Jul 22 at 13:02
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

It is as Bernard wrote in the comments:
$fracpartialpartial tt^2 = 2t$. Hence, you get
$$
fracpartialpartial tu(x,t)=2tLeftrightarrow fracpartialpartial tu(x,t)=fracpartialpartial tt^2Leftrightarrow fracpartialpartial tu(x,t)-fracpartialpartial tt^2=0Leftrightarrow fracpartialpartial tleft(u(x,t)-t^2right)=0.
$$
At the last step, you use that the derivative is a linear operator.

In your argument, there is a mistake at
$$
int 2tdt = t^2+A(x,t).
$$
The function $A$ can't depend on $t$. You might think of the additive constant at the integration. But it is a constant wrt the integration variable (in this case $t$). On the other hand, the constant can depend on the remaining variables, like $x$ in your example. So, you should write
$$
int 2t~dt = t^2+A(x)
$$
such that $A$ just depends on $x$. Naturally, so get $fracpartialpartial tA(x)=0$ as you need.

share|cite|improve this answer

edited Jul 22 at 13:05

answered Jul 22 at 13:00

Mundron Schmidt

7,1162727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I completely misread it. My apologies everyone.
  – The Pointer
  Jul 22 at 13:02
  
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

It is as Bernard wrote in the comments:
$fracpartialpartial tt^2 = 2t$. Hence, you get
$$
fracpartialpartial tu(x,t)=2tLeftrightarrow fracpartialpartial tu(x,t)=fracpartialpartial tt^2Leftrightarrow fracpartialpartial tu(x,t)-fracpartialpartial tt^2=0Leftrightarrow fracpartialpartial tleft(u(x,t)-t^2right)=0.
$$
At the last step, you use that the derivative is a linear operator.

In your argument, there is a mistake at
$$
int 2tdt = t^2+A(x,t).
$$
The function $A$ can't depend on $t$. You might think of the additive constant at the integration. But it is a constant wrt the integration variable (in this case $t$). On the other hand, the constant can depend on the remaining variables, like $x$ in your example. So, you should write
$$
int 2t~dt = t^2+A(x)
$$
such that $A$ just depends on $x$. Naturally, so get $fracpartialpartial tA(x)=0$ as you need.

share|cite|improve this answer

edited Jul 22 at 13:05

answered Jul 22 at 13:00

Mundron Schmidt

7,1162727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I completely misread it. My apologies everyone.
  – The Pointer
  Jul 22 at 13:02
  
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

It is as Bernard wrote in the comments:
$fracpartialpartial tt^2 = 2t$. Hence, you get
$$
fracpartialpartial tu(x,t)=2tLeftrightarrow fracpartialpartial tu(x,t)=fracpartialpartial tt^2Leftrightarrow fracpartialpartial tu(x,t)-fracpartialpartial tt^2=0Leftrightarrow fracpartialpartial tleft(u(x,t)-t^2right)=0.
$$
At the last step, you use that the derivative is a linear operator.

In your argument, there is a mistake at
$$
int 2tdt = t^2+A(x,t).
$$
The function $A$ can't depend on $t$. You might think of the additive constant at the integration. But it is a constant wrt the integration variable (in this case $t$). On the other hand, the constant can depend on the remaining variables, like $x$ in your example. So, you should write
$$
int 2t~dt = t^2+A(x)
$$
such that $A$ just depends on $x$. Naturally, so get $fracpartialpartial tA(x)=0$ as you need.

share|cite|improve this answer

edited Jul 22 at 13:05

answered Jul 22 at 13:00

Mundron Schmidt

7,1162727

It is as Bernard wrote in the comments:
$fracpartialpartial tt^2 = 2t$. Hence, you get
$$
fracpartialpartial tu(x,t)=2tLeftrightarrow fracpartialpartial tu(x,t)=fracpartialpartial tt^2Leftrightarrow fracpartialpartial tu(x,t)-fracpartialpartial tt^2=0Leftrightarrow fracpartialpartial tleft(u(x,t)-t^2right)=0.
$$
At the last step, you use that the derivative is a linear operator.

In your argument, there is a mistake at
$$
int 2tdt = t^2+A(x,t).
$$
The function $A$ can't depend on $t$. You might think of the additive constant at the integration. But it is a constant wrt the integration variable (in this case $t$). On the other hand, the constant can depend on the remaining variables, like $x$ in your example. So, you should write
$$
int 2t~dt = t^2+A(x)
$$
such that $A$ just depends on $x$. Naturally, so get $fracpartialpartial tA(x)=0$ as you need.

share|cite|improve this answer

edited Jul 22 at 13:05

answered Jul 22 at 13:00

Mundron Schmidt

7,1162727

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 13:05

edited Jul 22 at 13:05

edited Jul 22 at 13:05

answered Jul 22 at 13:00

Mundron Schmidt

7,1162727

answered Jul 22 at 13:00

Mundron Schmidt

7,1162727

answered Jul 22 at 13:00

Mundron Schmidt

7,1162727

7,1162727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I completely misread it. My apologies everyone.
  – The Pointer
  Jul 22 at 13:02
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I completely misread it. My apologies everyone.
  – The Pointer
  Jul 22 at 13:02
  
  

  
  
  
  

Yes, I completely misread it. My apologies everyone.
– The Pointer
Jul 22 at 13:02

Yes, I completely misread it. My apologies everyone.
– The Pointer
Jul 22 at 13:02

add a comment | 

 

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