Solve PDE using method of characteristics with non-local boundary conditions.Given the population model by the following linear first order PDE in u(a,t) with constants b and μ: u a + u t = − μ t u a , t &gt; 0 u ( a , 0 ) = u 0 ( a ) a ≥ 0 u ( 0 , t ) = F ( t ) = b ∫ 0 ∞ u ( a , t ) d aWe can split the integral in two with our non-local boundary data: F ( t ) = b ∫ 0 t u ( a , t ) d a + b ∫ t ∞ u ( a , t ) d aChoosing the characteristic coordinates ( ξ , τ ) and re-arranging the expression to form the normal to the solution surface we have the following equation with initial conditions: ( u a , u t , − 1 ) ∙ ( 1 , 1 , − μ t u ) = 0 x ( 0 ) = ξ , t ( 0 ) = 0 , u ( 0 ) = u 0 ( ξ )Characteristic equations: d a d τ = 1 , d t d τ = 1 , d u d τ = − μ t uSolving each of these ODE's in τ gives the following: ( 1 ) ∫ d a = ∫ d τ ( 2 ) ∫ d t = ∫ d τ ( 3 ) ∫ d u = − ∫ μ t u d τ a = τ + F ( ξ ) t = τ + F ( ξ ) ∴ a = τ + ξ ∴ t = τ ∫ d u = − ∫ μ τ u d τ ∫ 1 u d u = − ∫ μ τ d τ ln ⁡ u = − 1 2 μ τ 2 + F ( ξ ) u = G ( ξ ) e − 1 2 μ τ 2 ∴ u = u 0 ( ξ ) e − 1 2 μ τ 2 Substituting back the original coordinates we can re-write this expression with a coordinate change: ξ = a − t τ = t ∴ u ( a , t ) = u 0 ( a − t ) e − 1 2 t 2 Now this is where I get stuck, how do I use the boundary data to come up with a well-posed solution? u ( 0 , t ) = u 0 ( − t ) e − 1 2 μ t 2 = b ∫ 0 t u ( a , t ) d a + b ∫ t ∞ u ( a , t ) d a

Question

Solve PDE using method of characteristics with non-local boundary conditions.Given the population model by the following linear first order PDE in u(a,t) with constants b and   μ:      u    a    +      u    t    =  −  μ  t  u            a  ,  t  &amp;gt;  0  u  (  a  ,  0  )  =      u    0    (  a  )        a  ≥  0  u  (  0  ,  t  )  =  F  (  t  )  =  b      ∫    0    ∞    u  (  a  ,  t  )        d    aWe can split the integral in two with our non-local boundary data:  F  (  t  )  =  b      ∫    0    t    u  (  a  ,  t  )        d    a  +  b      ∫    t    ∞    u  (  a  ,  t  )        d    aChoosing the characteristic coordinates   (  ξ  ,  τ  ) and re-arranging the expression to form the normal to the solution surface we have the following equation with initial conditions:            (            u    a    ,      u    t    ,  −  1            )        ∙            (        1  ,  1  ,  −  μ  t  u            )        =  0  x  (  0  )  =  ξ  ,        t  (  0  )  =  0  ,        u  (  0  )  =      u    0    (  ξ  )Characteristic equations:                    d            a                      d            τ        =  1  ,                          d            t                      d            τ        =  1  ,                          d            u                      d            τ        =  −  μ  t  uSolving each of these ODE&#039;s in   τ gives the following:  (  1  )  ∫      d    a  =  ∫      d    τ                      (  2  )  ∫      d    t  =  ∫      d    τ                      (  3  )  ∫      d    u  =  −  ∫  μ  t  u        d    τ  a  =  τ  +  F  (  ξ  )                      t  =  τ  +  F  (  ξ  )  ∴  a  =  τ  +  ξ                      ∴  t  =  τ  ∫      d    u  =  −  ∫  μ  τ  u        d    τ  ∫      1    u          d    u  =  −  ∫  μ  τ        d    τ  ln  ⁡  u  =  −      1    2    μ      τ    2    +  F  (  ξ  )  u  =  G  (  ξ  )        e          −              1        2            μ              τ        2              ∴  u  =      u    0    (  ξ  )        e          −              1        2            μ              τ        2            Substituting back the original coordinates we can re-write this expression with a coordinate change:  ξ  =  a  −  t                      τ  =  t  ∴  u  (  a  ,  t  )  =      u    0    (  a  −  t  )        e          −              1        2                    t        2            Now this is where I get stuck, how do I use the boundary data to come up with a well-posed solution?  u  (  0  ,  t  )  =      u    0    (  −  t  )        e          −              1        2            μ              t        2              =  b      ∫    0    t    u  (  a  ,  t  )        d    a  +  b      ∫    t    ∞    u  (  a  ,  t  )        d    a

Julius Haley · Accepted Answer

Step 1      u    a    +      u    t    =  −  μ  t  uCharpit-Lagrange system of characteristic ODEs:            d      a        1    =            d      t        1    =            d      u              −      μ      t      u      First characteristic equation, from             d      a        1    =            d      t        1  :  a  −  t  =      c    1  Second characteristic equation, from             d      t        1    =            d      u              −      μ      t      u      :  u      e          μ              t        2                    /            2        =      c    2  General solution of the PDE on the form of implicit equation       c    2    =  Φ  (      c    1    ):  u      e          μ              t        2                    /            2        =  Φ  (  a  −  t  )where   Φ is an arbitrary function, to be determined according to a boundary condition.  u  (  a  ,  t  )  =      e          −      μ              t        2                    /            2        Φ  (  a  −  t  )CONDITION:   u  (  a  ,  0  )  =      u    0    (  a  )  =      e          −      μ              0        2                    /            2        Φ  (  a  −  0  )  =  Φ  (  a  )Step 2Now the function   u  (  a  ,  0  )  =      u    0    (  a  )  =      e          −      μ              0        2                    /            2        Φ  (  a  −  0  )  =  Φ  (  a  ) is determined. We put it into the above general solution where   x  =  a  −  t:                              u          (          a          ,          t          )          =                      e                          −              μ                              t                2                                            /                            2                                            u            0                    (          a          −          t          )                    The solution of the PDE fitting to the condition   u  (  a  ,  0  )  =      u    0    (  a  ) is fully determined if the function       u    0    (  a  ) is known. In this case the initial condition   u  (  0  ,  t  )  =  F  (  t  ) is superfluous. Moreover, this condition might introduce a contradiction with the condition   u  (  a  ,  0  )  =      u    0    (  a  ).So, if the two conditions are specified without relationship between them, the problem has no solution in general.The problem is likely to have a solution if the next relationship was satisfied :  u  (  0  ,  t  )  =  b      ∫    0    ∞    u  (  a  ,  t  )        d    a  =      e          −      μ              t        2                    /            2            u    0    (  0  −  t  )  =      e          −      μ              t        2                    /            2            u    0    (  −  t  )  b      ∫    0    ∞        e          −      μ              t        2                    /            2            u    0    (  a  −  t  )        d    a  =      e          −      μ              t        2                    /            2            u    0    (  −  t  )  b      ∫    0    ∞        u    0    (  a  −  t  )        d    a  =      u    0    (  −  t  )Conclusion:If       u    0    (  a  ) is not a function of general form, but is a particular function which satisfies the equation   b      ∫    0    ∞        u    0    (  a  −  t  )        d    a  =      u    0    (  −  t  ) the problem has a solution which is:   u  (  a  ,  t  )  =      e          −      μ              t        2                    /            2            u    0    (  a  −  t  ).

Solve PDE using method of characteristics with non-local boundary conditions.

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