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finding maximum value of BVP

2

$begingroup$

Find the maximum value of $u$ in the following BVP

beginalign*
u_t &= u_xx, ; ; t,x in (0,1) \
u(0,t) &= 2t^2-t \
u(1,t) &= sin pi t \
u(x,0) &= x(1-x) \ endalign*

Attempt

We know maximum can only occur at the boundaries. So if we call $A = max u(0,t) $, $B = max u(1,t)$, $C= max u(x,0)$, then

$$ max u = max (A,B,C) $$

Now, clearly, $max u(1,t) = 1$ for any $t$ and $max u(x,0) = 1/4$ since it is parabola opens down with vertex 1/2.

Now, for $max u(0,t) = 2t^2 - t$, notice that it is increasing so its maximum must occur at $t=1$ thus $max u(0,t) = 2-1 = 1$

So, we have $max u = max(1,1/4,1) = boxed1$

Is this solution correct?

proof-verification pde heat-equation maximum-principle

share|cite|improve this question

edited Apr 1 at 6:02

Alex Ravsky

43k32583

asked Mar 24 at 6:08

JamesJames

2,636425

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Squares have 4 sides, not 3.
  $endgroup$
  – Paul Sinclair
  Mar 24 at 15:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The point of the exercise is not to solve the equation. The point is to use the maximum principle
  $endgroup$
  – ILoveMath
  Mar 30 at 4:49
  

  
  
  
  

add a comment | 

2

$begingroup$

Find the maximum value of $u$ in the following BVP

beginalign*
u_t &= u_xx, ; ; t,x in (0,1) \
u(0,t) &= 2t^2-t \
u(1,t) &= sin pi t \
u(x,0) &= x(1-x) \ endalign*

Attempt

We know maximum can only occur at the boundaries. So if we call $A = max u(0,t) $, $B = max u(1,t)$, $C= max u(x,0)$, then

$$ max u = max (A,B,C) $$

Now, clearly, $max u(1,t) = 1$ for any $t$ and $max u(x,0) = 1/4$ since it is parabola opens down with vertex 1/2.

Now, for $max u(0,t) = 2t^2 - t$, notice that it is increasing so its maximum must occur at $t=1$ thus $max u(0,t) = 2-1 = 1$

So, we have $max u = max(1,1/4,1) = boxed1$

Is this solution correct?

proof-verification pde heat-equation maximum-principle

share|cite|improve this question

edited Apr 1 at 6:02

Alex Ravsky

43k32583

asked Mar 24 at 6:08

JamesJames

2,636425

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Squares have 4 sides, not 3.
  $endgroup$
  – Paul Sinclair
  Mar 24 at 15:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The point of the exercise is not to solve the equation. The point is to use the maximum principle
  $endgroup$
  – ILoveMath
  Mar 30 at 4:49
  

  
  
  
  

add a comment | 

$begingroup$

Find the maximum value of $u$ in the following BVP

beginalign*
u_t &= u_xx, ; ; t,x in (0,1) \
u(0,t) &= 2t^2-t \
u(1,t) &= sin pi t \
u(x,0) &= x(1-x) \ endalign*

Attempt

We know maximum can only occur at the boundaries. So if we call $A = max u(0,t) $, $B = max u(1,t)$, $C= max u(x,0)$, then

$$ max u = max (A,B,C) $$

Now, clearly, $max u(1,t) = 1$ for any $t$ and $max u(x,0) = 1/4$ since it is parabola opens down with vertex 1/2.

Now, for $max u(0,t) = 2t^2 - t$, notice that it is increasing so its maximum must occur at $t=1$ thus $max u(0,t) = 2-1 = 1$

So, we have $max u = max(1,1/4,1) = boxed1$

Is this solution correct?

proof-verification pde heat-equation maximum-principle

share|cite|improve this question

edited Apr 1 at 6:02

Alex Ravsky

43k32583

asked Mar 24 at 6:08

JamesJames

2,636425

$endgroup$

share|cite|improve this question

edited Apr 1 at 6:02

Alex Ravsky

43k32583

asked Mar 24 at 6:08

JamesJames

2,636425

share|cite|improve this question

edited Apr 1 at 6:02

Alex Ravsky

43k32583

asked Mar 24 at 6:08

JamesJames

2,636425

edited Apr 1 at 6:02

Alex Ravsky

43k32583

edited Apr 1 at 6:02

Alex Ravsky

43k32583

asked Mar 24 at 6:08

JamesJames

2,636425

asked Mar 24 at 6:08

JamesJames

2,636425