Nonhomogeneous Semi-Linear PDE with the Characteristic Method











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I need to solve this semi-linear PDE:



$u_x - 3u_y = sin (y) + cos (x)$



The initial condition provided is:



$ u (t,t)= t^2$



I need to use the Characteristic Method. I learned the method from this video.



I have reached an answer. However, I am not sure if it is wright. I am insecure in two different moments of my solution.



My intermediate steps are:



First constant: $c_1= y + 3x $



Second constant: $c_2= xcos y -cos x-u$



The second constant above is my first source of insecurity.



I needed to solve this equality as an ODE:



$frac{dx}{1}= frac{du}{sin x + cos y}$



I did so with this approach:



$int (sin x + cos y) dx = int 1 du $



Using integral properties:
$(int sin x dx) + (int cos y dx) = int 1 du $



Which results in:
$c_2= xcos y -cos x-u$



I am not sure if this is right. I don't know why, but I am affraid I can't treat cos y as a constant on the second integral.



Supposing this was correct, I tried to go on. So I used an arbitrary function G to make the relation between both constants. Hence, $c_2 =G(c_1) $ and we have that:



$ xcos y -cos x-u = G(y + 3x) $



With the initial condition we have:



$G(4x) = xcos y -cos x- x^2$



This result above is my second source of insecurity.



Before, I was dealing with clear arbitraty functions like $G(x)$ or $G(y)$.
Now, the result is different. I did some manipulation to put it on the more casual form. So:



$G(x) = frac{x}{4}cos y -cosfrac{x}{4}- frac{x^2}{16}$



After the definition of $G(x)$ above , I inputed the value of $c_1$ , having:



$G(y + 3x) = frac{y+3x}{4}cos y -cosfrac{y+3x}{4} - frac{(y+3x)^2}{16} $.



Finally, solving for $u$:



$u(x,y) = -frac{y+3x}{4}cos y +cosfrac{y+3x}{4} + frac{(y+3x)^2}{16} + xcos y -cos x $



Is this right?



If I did something wrong, what was it?



Thanks in advance!






pde characteristics







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    up vote
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    I need to solve this semi-linear PDE:



    $u_x - 3u_y = sin (y) + cos (x)$



    The initial condition provided is:



    $ u (t,t)= t^2$



    I need to use the Characteristic Method. I learned the method from this video.



    I have reached an answer. However, I am not sure if it is wright. I am insecure in two different moments of my solution.



    My intermediate steps are:



    First constant: $c_1= y + 3x $



    Second constant: $c_2= xcos y -cos x-u$



    The second constant above is my first source of insecurity.



    I needed to solve this equality as an ODE:



    $frac{dx}{1}= frac{du}{sin x + cos y}$



    I did so with this approach:



    $int (sin x + cos y) dx = int 1 du $



    Using integral properties:
    $(int sin x dx) + (int cos y dx) = int 1 du $



    Which results in:
    $c_2= xcos y -cos x-u$



    I am not sure if this is right. I don't know why, but I am affraid I can't treat cos y as a constant on the second integral.



    Supposing this was correct, I tried to go on. So I used an arbitrary function G to make the relation between both constants. Hence, $c_2 =G(c_1) $ and we have that:



    $ xcos y -cos x-u = G(y + 3x) $



    With the initial condition we have:



    $G(4x) = xcos y -cos x- x^2$



    This result above is my second source of insecurity.



    Before, I was dealing with clear arbitraty functions like $G(x)$ or $G(y)$.
    Now, the result is different. I did some manipulation to put it on the more casual form. So:



    $G(x) = frac{x}{4}cos y -cosfrac{x}{4}- frac{x^2}{16}$



    After the definition of $G(x)$ above , I inputed the value of $c_1$ , having:



    $G(y + 3x) = frac{y+3x}{4}cos y -cosfrac{y+3x}{4} - frac{(y+3x)^2}{16} $.



    Finally, solving for $u$:



    $u(x,y) = -frac{y+3x}{4}cos y +cosfrac{y+3x}{4} + frac{(y+3x)^2}{16} + xcos y -cos x $



    Is this right?



    If I did something wrong, what was it?



    Thanks in advance!






    pde characteristics







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      I need to solve this semi-linear PDE:



      $u_x - 3u_y = sin (y) + cos (x)$



      The initial condition provided is:



      $ u (t,t)= t^2$



      I need to use the Characteristic Method. I learned the method from this video.



      I have reached an answer. However, I am not sure if it is wright. I am insecure in two different moments of my solution.



      My intermediate steps are:



      First constant: $c_1= y + 3x $



      Second constant: $c_2= xcos y -cos x-u$



      The second constant above is my first source of insecurity.



      I needed to solve this equality as an ODE:



      $frac{dx}{1}= frac{du}{sin x + cos y}$



      I did so with this approach:



      $int (sin x + cos y) dx = int 1 du $



      Using integral properties:
      $(int sin x dx) + (int cos y dx) = int 1 du $



      Which results in:
      $c_2= xcos y -cos x-u$



      I am not sure if this is right. I don't know why, but I am affraid I can't treat cos y as a constant on the second integral.



      Supposing this was correct, I tried to go on. So I used an arbitrary function G to make the relation between both constants. Hence, $c_2 =G(c_1) $ and we have that:



      $ xcos y -cos x-u = G(y + 3x) $



      With the initial condition we have:



      $G(4x) = xcos y -cos x- x^2$



      This result above is my second source of insecurity.



      Before, I was dealing with clear arbitraty functions like $G(x)$ or $G(y)$.
      Now, the result is different. I did some manipulation to put it on the more casual form. So:



      $G(x) = frac{x}{4}cos y -cosfrac{x}{4}- frac{x^2}{16}$



      After the definition of $G(x)$ above , I inputed the value of $c_1$ , having:



      $G(y + 3x) = frac{y+3x}{4}cos y -cosfrac{y+3x}{4} - frac{(y+3x)^2}{16} $.



      Finally, solving for $u$:



      $u(x,y) = -frac{y+3x}{4}cos y +cosfrac{y+3x}{4} + frac{(y+3x)^2}{16} + xcos y -cos x $



      Is this right?



      If I did something wrong, what was it?



      Thanks in advance!






      pde characteristics







      share|cite|improve this question















      I need to solve this semi-linear PDE:



      $u_x - 3u_y = sin (y) + cos (x)$



      The initial condition provided is:



      $ u (t,t)= t^2$



      I need to use the Characteristic Method. I learned the method from this video.



      I have reached an answer. However, I am not sure if it is wright. I am insecure in two different moments of my solution.



      My intermediate steps are:



      First constant: $c_1= y + 3x $



      Second constant: $c_2= xcos y -cos x-u$



      The second constant above is my first source of insecurity.



      I needed to solve this equality as an ODE:



      $frac{dx}{1}= frac{du}{sin x + cos y}$



      I did so with this approach:



      $int (sin x + cos y) dx = int 1 du $



      Using integral properties:
      $(int sin x dx) + (int cos y dx) = int 1 du $



      Which results in:
      $c_2= xcos y -cos x-u$



      I am not sure if this is right. I don't know why, but I am affraid I can't treat cos y as a constant on the second integral.



      Supposing this was correct, I tried to go on. So I used an arbitrary function G to make the relation between both constants. Hence, $c_2 =G(c_1) $ and we have that:



      $ xcos y -cos x-u = G(y + 3x) $



      With the initial condition we have:



      $G(4x) = xcos y -cos x- x^2$



      This result above is my second source of insecurity.



      Before, I was dealing with clear arbitraty functions like $G(x)$ or $G(y)$.
      Now, the result is different. I did some manipulation to put it on the more casual form. So:



      $G(x) = frac{x}{4}cos y -cosfrac{x}{4}- frac{x^2}{16}$



      After the definition of $G(x)$ above , I inputed the value of $c_1$ , having:



      $G(y + 3x) = frac{y+3x}{4}cos y -cosfrac{y+3x}{4} - frac{(y+3x)^2}{16} $.



      Finally, solving for $u$:



      $u(x,y) = -frac{y+3x}{4}cos y +cosfrac{y+3x}{4} + frac{(y+3x)^2}{16} + xcos y -cos x $



      Is this right?



      If I did something wrong, what was it?



      Thanks in advance!






      pde characteristics




      pde characteristics






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          $$u_x - 3u_y = sin (y) + cos (x),qquad (1)\
          u(t,t)=t^2 qquadqquadqquadqquadquad (2)$$
          Solution of $u_x - 3u_y =0$ is
          $$u_h=sin{(x)}+frac{cos{(y)}}{3}.$$
          Particular solution of $(1)$ is
          $$u_p=G(y+3x).$$
          Then general solution of $(1)$ is
          $$u=u_p+u_h=G(y+3x)+sin{(x)}+frac{cos{(y)}}{3}.$$
          From $(2)$ we get
          $$G(4t) +sin{(t)}+frac{cos{(t)}}{3}={{t}^{2}}.$$
          Then
          $$G(t)=frac{t^2}{16}-sinleft(frac{t}{4}right)-frac{cos{(frac{t}{4})}}{3}.$$



          Solution of problem $(1), (2)$ is
          $$u=frac{{{left( y+3 xright) }^{2}}}{16}-sin{left( frac{y+3 x}{4}right) }-frac{cos{left( frac{y+3 x}{4}right) }}{3}+sin{(x)}+frac{cos{(y)}}{3}$$






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            $$u_x - 3u_y = sin (y) + cos (x),qquad (1)\
            u(t,t)=t^2 qquadqquadqquadqquadquad (2)$$
            Solution of $u_x - 3u_y =0$ is
            $$u_h=sin{(x)}+frac{cos{(y)}}{3}.$$
            Particular solution of $(1)$ is
            $$u_p=G(y+3x).$$
            Then general solution of $(1)$ is
            $$u=u_p+u_h=G(y+3x)+sin{(x)}+frac{cos{(y)}}{3}.$$
            From $(2)$ we get
            $$G(4t) +sin{(t)}+frac{cos{(t)}}{3}={{t}^{2}}.$$
            Then
            $$G(t)=frac{t^2}{16}-sinleft(frac{t}{4}right)-frac{cos{(frac{t}{4})}}{3}.$$



            Solution of problem $(1), (2)$ is
            $$u=frac{{{left( y+3 xright) }^{2}}}{16}-sin{left( frac{y+3 x}{4}right) }-frac{cos{left( frac{y+3 x}{4}right) }}{3}+sin{(x)}+frac{cos{(y)}}{3}$$






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              $$u_x - 3u_y = sin (y) + cos (x),qquad (1)\
              u(t,t)=t^2 qquadqquadqquadqquadquad (2)$$
              Solution of $u_x - 3u_y =0$ is
              $$u_h=sin{(x)}+frac{cos{(y)}}{3}.$$
              Particular solution of $(1)$ is
              $$u_p=G(y+3x).$$
              Then general solution of $(1)$ is
              $$u=u_p+u_h=G(y+3x)+sin{(x)}+frac{cos{(y)}}{3}.$$
              From $(2)$ we get
              $$G(4t) +sin{(t)}+frac{cos{(t)}}{3}={{t}^{2}}.$$
              Then
              $$G(t)=frac{t^2}{16}-sinleft(frac{t}{4}right)-frac{cos{(frac{t}{4})}}{3}.$$



              Solution of problem $(1), (2)$ is
              $$u=frac{{{left( y+3 xright) }^{2}}}{16}-sin{left( frac{y+3 x}{4}right) }-frac{cos{left( frac{y+3 x}{4}right) }}{3}+sin{(x)}+frac{cos{(y)}}{3}$$






              share|cite|improve this answer























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                $$u_x - 3u_y = sin (y) + cos (x),qquad (1)\
                u(t,t)=t^2 qquadqquadqquadqquadquad (2)$$
                Solution of $u_x - 3u_y =0$ is
                $$u_h=sin{(x)}+frac{cos{(y)}}{3}.$$
                Particular solution of $(1)$ is
                $$u_p=G(y+3x).$$
                Then general solution of $(1)$ is
                $$u=u_p+u_h=G(y+3x)+sin{(x)}+frac{cos{(y)}}{3}.$$
                From $(2)$ we get
                $$G(4t) +sin{(t)}+frac{cos{(t)}}{3}={{t}^{2}}.$$
                Then
                $$G(t)=frac{t^2}{16}-sinleft(frac{t}{4}right)-frac{cos{(frac{t}{4})}}{3}.$$



                Solution of problem $(1), (2)$ is
                $$u=frac{{{left( y+3 xright) }^{2}}}{16}-sin{left( frac{y+3 x}{4}right) }-frac{cos{left( frac{y+3 x}{4}right) }}{3}+sin{(x)}+frac{cos{(y)}}{3}$$






                share|cite|improve this answer












                $$u_x - 3u_y = sin (y) + cos (x),qquad (1)\
                u(t,t)=t^2 qquadqquadqquadqquadquad (2)$$
                Solution of $u_x - 3u_y =0$ is
                $$u_h=sin{(x)}+frac{cos{(y)}}{3}.$$
                Particular solution of $(1)$ is
                $$u_p=G(y+3x).$$
                Then general solution of $(1)$ is
                $$u=u_p+u_h=G(y+3x)+sin{(x)}+frac{cos{(y)}}{3}.$$
                From $(2)$ we get
                $$G(4t) +sin{(t)}+frac{cos{(t)}}{3}={{t}^{2}}.$$
                Then
                $$G(t)=frac{t^2}{16}-sinleft(frac{t}{4}right)-frac{cos{(frac{t}{4})}}{3}.$$



                Solution of problem $(1), (2)$ is
                $$u=frac{{{left( y+3 xright) }^{2}}}{16}-sin{left( frac{y+3 x}{4}right) }-frac{cos{left( frac{y+3 x}{4}right) }}{3}+sin{(x)}+frac{cos{(y)}}{3}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 5 hours ago









                Aleksas Domarkas

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