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Mathemathical model equation PDE

2

2

$begingroup$

I am studying models in DPE an the professor give us this problem:

$begincasesu_t+au_x=f(x,t)\ u(x,0)=g(x)endcases$

I've studied the transport equation and the Burger's equation. About heat equation, a little. Hoewever, this one is not included in those kinds of problems.

I will be very mercy if someone could help me with indications to this solution.

Thanks.

pde mathematical-modeling hyperbolic-equations

share|cite|improve this question

edited Apr 2 at 17:05

Na'omi

asked Apr 2 at 16:59

Na'omiNa'omi

27813

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: what happens along the line $x = at + c$?
  $endgroup$
  – Robert Israel
  Apr 2 at 17:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $u'(x(t),t)=u_t(x(t),t)+u_x(x(t),t)x'(t)=u_t(x(t),t)+au_x(x(t),t)=f(x,t)$...? I could not continue yet.
  $endgroup$
  – Na'omi
  Apr 2 at 17:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you done a Google search for a similar PDE? See web.stanford.edu/class/math220a/handouts/pracfinal1sols.pdf?
  $endgroup$
  – Axion004
  Apr 2 at 17:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Further hint: if $v(t) = u(at+c, t)$, then $$dfracdvdt = a u_x(at+c,t) + u_t(at+c,t) = ldots$$
  $endgroup$
  – Robert Israel
  Apr 2 at 17:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, @Axion004.
  $endgroup$
  – Na'omi
  Apr 2 at 22:52
  

  
  
  
  

 | 
show 1 more comment

2

2

$begingroup$

I am studying models in DPE an the professor give us this problem:

$begincasesu_t+au_x=f(x,t)\ u(x,0)=g(x)endcases$

I've studied the transport equation and the Burger's equation. About heat equation, a little. Hoewever, this one is not included in those kinds of problems.

I will be very mercy if someone could help me with indications to this solution.

Thanks.

pde mathematical-modeling hyperbolic-equations

share|cite|improve this question

edited Apr 2 at 17:05

Na'omi

asked Apr 2 at 16:59

Na'omiNa'omi

27813

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: what happens along the line $x = at + c$?
  $endgroup$
  – Robert Israel
  Apr 2 at 17:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $u'(x(t),t)=u_t(x(t),t)+u_x(x(t),t)x'(t)=u_t(x(t),t)+au_x(x(t),t)=f(x,t)$...? I could not continue yet.
  $endgroup$
  – Na'omi
  Apr 2 at 17:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you done a Google search for a similar PDE? See web.stanford.edu/class/math220a/handouts/pracfinal1sols.pdf?
  $endgroup$
  – Axion004
  Apr 2 at 17:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Further hint: if $v(t) = u(at+c, t)$, then $$dfracdvdt = a u_x(at+c,t) + u_t(at+c,t) = ldots$$
  $endgroup$
  – Robert Israel
  Apr 2 at 17:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, @Axion004.
  $endgroup$
  – Na'omi
  Apr 2 at 22:52
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I am studying models in DPE an the professor give us this problem:

$begincasesu_t+au_x=f(x,t)\ u(x,0)=g(x)endcases$

I've studied the transport equation and the Burger's equation. About heat equation, a little. Hoewever, this one is not included in those kinds of problems.

I will be very mercy if someone could help me with indications to this solution.

Thanks.

pde mathematical-modeling hyperbolic-equations

share|cite|improve this question

edited Apr 2 at 17:05

Na'omi

asked Apr 2 at 16:59

Na'omiNa'omi

27813

$endgroup$

share|cite|improve this question

edited Apr 2 at 17:05

Na'omi

asked Apr 2 at 16:59

Na'omiNa'omi

27813

share|cite|improve this question

edited Apr 2 at 17:05

Na'omi

asked Apr 2 at 16:59

Na'omiNa'omi

27813

asked Apr 2 at 16:59

Na'omiNa'omi

27813

asked Apr 2 at 16:59

Na'omiNa'omi

27813

1 Answer
1

active

oldest

votes

2

$begingroup$

We can find the characteristics with this system of ODE's:

$dfracdt1=dfracdxa=dfracduf(x,t)$

From the first proportion, $x=c_1+at$. Now:

$dfracdt1=dfracduf(at+c_1,t)$ or

$f(at+c_1,t)dt=du$ Integrating,

$$u(x,t)=int_0^tf(ar+c_1,r)dr+c_2=int_0^tf(ar+x-at,r)dr+c_2$$

For the general solution we have to consider $c_1$ and $c_2$ are somehow related: $c_2=h(c_1)$, with $h$ some single variable differentiable function to determine with th initial conditions: $c_2=h(x-at)$:

$$u(x,t)=int_0^tf(ar+x-at,r)dr+h(x-at)$$

Finally we can impose the i.c. $u(x,0)=g(x)$

$u(x,0)=g(x)=h(x)$ leading to

$$u(x,t)=int_0^tf(ar+x-at,r)dr+g(x-at)$$

Added
I just finished and see the link in the comments to your post. There is the answer, but it is proposed and checked, not deduced, so I post mine as it shows a way to get the solution using the method of characteristics.

$$u(x,t)=g(x-at)+int_0^tf(x-a(t-r),r)dr$$

share|cite|improve this answer

answered Apr 2 at 19:12

Rafa BudríaRafa Budría

6,0521825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I will study this.
  $endgroup$
  – Na'omi
  Apr 2 at 22:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, this kind of proportions can always be used? This will help a lot... but I did not study this yet. I can get $dfracpartial uf (x+y)=dfrac1frac1partial t+fracapartial x$. But this I could not get those proportions. Many thanks.
  $endgroup$
  – Na'omi
  Apr 2 at 23:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can find how to pose and why can be used to solve the PDE if you google with "method of characteristics".
  $endgroup$
  – Rafa Budría
  Apr 3 at 4:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, many thanks.
  $endgroup$
  – Na'omi
  Apr 3 at 17:51
  

  
  
  
  

add a comment | 

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2

$begingroup$

We can find the characteristics with this system of ODE's:

$dfracdt1=dfracdxa=dfracduf(x,t)$

From the first proportion, $x=c_1+at$. Now:

$dfracdt1=dfracduf(at+c_1,t)$ or

$f(at+c_1,t)dt=du$ Integrating,

$$u(x,t)=int_0^tf(ar+c_1,r)dr+c_2=int_0^tf(ar+x-at,r)dr+c_2$$

For the general solution we have to consider $c_1$ and $c_2$ are somehow related: $c_2=h(c_1)$, with $h$ some single variable differentiable function to determine with th initial conditions: $c_2=h(x-at)$:

$$u(x,t)=int_0^tf(ar+x-at,r)dr+h(x-at)$$

Finally we can impose the i.c. $u(x,0)=g(x)$

$u(x,0)=g(x)=h(x)$ leading to

$$u(x,t)=int_0^tf(ar+x-at,r)dr+g(x-at)$$

Added
I just finished and see the link in the comments to your post. There is the answer, but it is proposed and checked, not deduced, so I post mine as it shows a way to get the solution using the method of characteristics.

$$u(x,t)=g(x-at)+int_0^tf(x-a(t-r),r)dr$$

share|cite|improve this answer

answered Apr 2 at 19:12

Rafa BudríaRafa Budría

6,0521825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I will study this.
  $endgroup$
  – Na'omi
  Apr 2 at 22:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, this kind of proportions can always be used? This will help a lot... but I did not study this yet. I can get $dfracpartial uf (x+y)=dfrac1frac1partial t+fracapartial x$. But this I could not get those proportions. Many thanks.
  $endgroup$
  – Na'omi
  Apr 2 at 23:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can find how to pose and why can be used to solve the PDE if you google with "method of characteristics".
  $endgroup$
  – Rafa Budría
  Apr 3 at 4:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, many thanks.
  $endgroup$
  – Na'omi
  Apr 3 at 17:51
  

  
  
  
  

add a comment | 

2

$begingroup$

We can find the characteristics with this system of ODE's:

$dfracdt1=dfracdxa=dfracduf(x,t)$

From the first proportion, $x=c_1+at$. Now:

$dfracdt1=dfracduf(at+c_1,t)$ or

$f(at+c_1,t)dt=du$ Integrating,

$$u(x,t)=int_0^tf(ar+c_1,r)dr+c_2=int_0^tf(ar+x-at,r)dr+c_2$$

For the general solution we have to consider $c_1$ and $c_2$ are somehow related: $c_2=h(c_1)$, with $h$ some single variable differentiable function to determine with th initial conditions: $c_2=h(x-at)$:

$$u(x,t)=int_0^tf(ar+x-at,r)dr+h(x-at)$$

Finally we can impose the i.c. $u(x,0)=g(x)$

$u(x,0)=g(x)=h(x)$ leading to

$$u(x,t)=int_0^tf(ar+x-at,r)dr+g(x-at)$$

Added
I just finished and see the link in the comments to your post. There is the answer, but it is proposed and checked, not deduced, so I post mine as it shows a way to get the solution using the method of characteristics.

$$u(x,t)=g(x-at)+int_0^tf(x-a(t-r),r)dr$$

share|cite|improve this answer

answered Apr 2 at 19:12

Rafa BudríaRafa Budría

6,0521825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I will study this.
  $endgroup$
  – Na'omi
  Apr 2 at 22:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, this kind of proportions can always be used? This will help a lot... but I did not study this yet. I can get $dfracpartial uf (x+y)=dfrac1frac1partial t+fracapartial x$. But this I could not get those proportions. Many thanks.
  $endgroup$
  – Na'omi
  Apr 2 at 23:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can find how to pose and why can be used to solve the PDE if you google with "method of characteristics".
  $endgroup$
  – Rafa Budría
  Apr 3 at 4:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, many thanks.
  $endgroup$
  – Na'omi
  Apr 3 at 17:51
  

  
  
  
  

add a comment | 

$begingroup$

We can find the characteristics with this system of ODE's:

$dfracdt1=dfracdxa=dfracduf(x,t)$

From the first proportion, $x=c_1+at$. Now:

$dfracdt1=dfracduf(at+c_1,t)$ or

$f(at+c_1,t)dt=du$ Integrating,

$$u(x,t)=int_0^tf(ar+c_1,r)dr+c_2=int_0^tf(ar+x-at,r)dr+c_2$$

For the general solution we have to consider $c_1$ and $c_2$ are somehow related: $c_2=h(c_1)$, with $h$ some single variable differentiable function to determine with th initial conditions: $c_2=h(x-at)$:

$$u(x,t)=int_0^tf(ar+x-at,r)dr+h(x-at)$$

Finally we can impose the i.c. $u(x,0)=g(x)$

$u(x,0)=g(x)=h(x)$ leading to

$$u(x,t)=int_0^tf(ar+x-at,r)dr+g(x-at)$$

Added
I just finished and see the link in the comments to your post. There is the answer, but it is proposed and checked, not deduced, so I post mine as it shows a way to get the solution using the method of characteristics.

$$u(x,t)=g(x-at)+int_0^tf(x-a(t-r),r)dr$$

share|cite|improve this answer

answered Apr 2 at 19:12

Rafa BudríaRafa Budría

6,0521825

$endgroup$

share|cite|improve this answer

answered Apr 2 at 19:12

Rafa BudríaRafa Budría

6,0521825

answered Apr 2 at 19:12

Rafa BudríaRafa Budría

6,0521825

answered Apr 2 at 19:12

Rafa BudríaRafa Budría

6,0521825