Change of variables in differential equation?












1












$begingroup$


I have the following formula:



$$f(x) = frac{d^2w(x)}{dx^2}$$



Now I would like to normalize $x$ by dividing it by $L$? This would be the substitution: $$hat{x}=x/L$$



How would my formula change? (step by step please)






substitution







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I have the following formula:



    $$f(x) = frac{d^2w(x)}{dx^2}$$



    Now I would like to normalize $x$ by dividing it by $L$? This would be the substitution: $$hat{x}=x/L$$



    How would my formula change? (step by step please)






    substitution







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I have the following formula:



      $$f(x) = frac{d^2w(x)}{dx^2}$$



      Now I would like to normalize $x$ by dividing it by $L$? This would be the substitution: $$hat{x}=x/L$$



      How would my formula change? (step by step please)






      substitution







      share|cite|improve this question











      $endgroup$




      I have the following formula:



      $$f(x) = frac{d^2w(x)}{dx^2}$$



      Now I would like to normalize $x$ by dividing it by $L$? This would be the substitution: $$hat{x}=x/L$$



      How would my formula change? (step by step please)






      substitution




      substitution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 14:53









      rafa11111

      1,1251417




      1,1251417










      asked Dec 1 '18 at 13:45









      jamesjames

      1519




      1519






















          1 Answer
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          1












          $begingroup$

          Using $x = L hat{x}$ we have $dx = L dhat{x}$ and $dx^2 = L^2 dhat{x}^2$. Therefore,
          $$
          frac{d}{dx^2} = frac{1}{L^2} frac{d}{dhat{x}^2}.
          $$
          This simple "substitution" is not mathematically rigorous, but you could use the chain rule twice to obtain the same thing; since the change of variable consists only in the multiplication by a constant, this "heuristic" works.



          The final expression is
          $$
          f(L hat{x}) = frac{1}{L^2} frac{d}{dhat{x}^2} w(Lhat{x}).
          $$



          If, for example, $w(x) = A exp(Bx)$, we have $f(x) = AB^2 exp(Bx)$. With the new variable,
          $$
          A (BL)^2 exp(BL hat{x}) = frac{d}{dhat{x}^2} A exp(BL hat{x}).
          $$
          If you define $hat{B}=BL$, $hat{f} = hat{B}^2 exp(hat{B} hat{x})$ and $hat{w}=exp(hat{B} hat{x})$, the equation is
          $$
          hat{f} = frac{d hat{w}}{dhat{x}^2} .
          $$
          Therefore, after you normalize $x$ with $hat{x}$, you should also normalize the functions. If $w$ has units of Kelvin, for example, $f$ has units of $K/m^2$. After the normalization, both $hat{w}$ and $hat{f}$ are nondimensional.



          Appendix: rigorous change of variable using chain rule



          Let $x=L hat{x}$. Therefore, $hat{x}=x/L$. From the chain rule,
          $$
          frac{dw}{dx} = frac{dhat{x}}{dx} frac{dw}{dhat{x}} = frac{1}{L} frac{dw}{dhat{x}}
          $$
          and
          $$
          frac{d^2w}{dx^2}=frac{d}{dx} left(frac{dw}{dx}right) = frac{dhat{x}}{dx} frac{d}{dhat{x}}left(frac{1}{L} frac{dw}{dhat{x}} right) = frac{1}{L^2} frac{d^2 w}{d hat{x}^2}.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much for your help ! Your answer is great !
            $endgroup$
            – james
            Dec 1 '18 at 15:36










          • $begingroup$
            After thinking more about it, I am still a little confused of how to use the chain rule rigorously in this case. I therefore opened a new question: math.stackexchange.com/questions/3040296/…
            $endgroup$
            – james
            Dec 15 '18 at 9:04










          • $begingroup$
            In your example to normalize the function, where did "A" go ?
            $endgroup$
            – james
            Dec 15 '18 at 9:19











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          $begingroup$

          Using $x = L hat{x}$ we have $dx = L dhat{x}$ and $dx^2 = L^2 dhat{x}^2$. Therefore,
          $$
          frac{d}{dx^2} = frac{1}{L^2} frac{d}{dhat{x}^2}.
          $$
          This simple "substitution" is not mathematically rigorous, but you could use the chain rule twice to obtain the same thing; since the change of variable consists only in the multiplication by a constant, this "heuristic" works.



          The final expression is
          $$
          f(L hat{x}) = frac{1}{L^2} frac{d}{dhat{x}^2} w(Lhat{x}).
          $$



          If, for example, $w(x) = A exp(Bx)$, we have $f(x) = AB^2 exp(Bx)$. With the new variable,
          $$
          A (BL)^2 exp(BL hat{x}) = frac{d}{dhat{x}^2} A exp(BL hat{x}).
          $$
          If you define $hat{B}=BL$, $hat{f} = hat{B}^2 exp(hat{B} hat{x})$ and $hat{w}=exp(hat{B} hat{x})$, the equation is
          $$
          hat{f} = frac{d hat{w}}{dhat{x}^2} .
          $$
          Therefore, after you normalize $x$ with $hat{x}$, you should also normalize the functions. If $w$ has units of Kelvin, for example, $f$ has units of $K/m^2$. After the normalization, both $hat{w}$ and $hat{f}$ are nondimensional.



          Appendix: rigorous change of variable using chain rule



          Let $x=L hat{x}$. Therefore, $hat{x}=x/L$. From the chain rule,
          $$
          frac{dw}{dx} = frac{dhat{x}}{dx} frac{dw}{dhat{x}} = frac{1}{L} frac{dw}{dhat{x}}
          $$
          and
          $$
          frac{d^2w}{dx^2}=frac{d}{dx} left(frac{dw}{dx}right) = frac{dhat{x}}{dx} frac{d}{dhat{x}}left(frac{1}{L} frac{dw}{dhat{x}} right) = frac{1}{L^2} frac{d^2 w}{d hat{x}^2}.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much for your help ! Your answer is great !
            $endgroup$
            – james
            Dec 1 '18 at 15:36










          • $begingroup$
            After thinking more about it, I am still a little confused of how to use the chain rule rigorously in this case. I therefore opened a new question: math.stackexchange.com/questions/3040296/…
            $endgroup$
            – james
            Dec 15 '18 at 9:04










          • $begingroup$
            In your example to normalize the function, where did "A" go ?
            $endgroup$
            – james
            Dec 15 '18 at 9:19
















          1












          $begingroup$

          Using $x = L hat{x}$ we have $dx = L dhat{x}$ and $dx^2 = L^2 dhat{x}^2$. Therefore,
          $$
          frac{d}{dx^2} = frac{1}{L^2} frac{d}{dhat{x}^2}.
          $$
          This simple "substitution" is not mathematically rigorous, but you could use the chain rule twice to obtain the same thing; since the change of variable consists only in the multiplication by a constant, this "heuristic" works.



          The final expression is
          $$
          f(L hat{x}) = frac{1}{L^2} frac{d}{dhat{x}^2} w(Lhat{x}).
          $$



          If, for example, $w(x) = A exp(Bx)$, we have $f(x) = AB^2 exp(Bx)$. With the new variable,
          $$
          A (BL)^2 exp(BL hat{x}) = frac{d}{dhat{x}^2} A exp(BL hat{x}).
          $$
          If you define $hat{B}=BL$, $hat{f} = hat{B}^2 exp(hat{B} hat{x})$ and $hat{w}=exp(hat{B} hat{x})$, the equation is
          $$
          hat{f} = frac{d hat{w}}{dhat{x}^2} .
          $$
          Therefore, after you normalize $x$ with $hat{x}$, you should also normalize the functions. If $w$ has units of Kelvin, for example, $f$ has units of $K/m^2$. After the normalization, both $hat{w}$ and $hat{f}$ are nondimensional.



          Appendix: rigorous change of variable using chain rule



          Let $x=L hat{x}$. Therefore, $hat{x}=x/L$. From the chain rule,
          $$
          frac{dw}{dx} = frac{dhat{x}}{dx} frac{dw}{dhat{x}} = frac{1}{L} frac{dw}{dhat{x}}
          $$
          and
          $$
          frac{d^2w}{dx^2}=frac{d}{dx} left(frac{dw}{dx}right) = frac{dhat{x}}{dx} frac{d}{dhat{x}}left(frac{1}{L} frac{dw}{dhat{x}} right) = frac{1}{L^2} frac{d^2 w}{d hat{x}^2}.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much for your help ! Your answer is great !
            $endgroup$
            – james
            Dec 1 '18 at 15:36










          • $begingroup$
            After thinking more about it, I am still a little confused of how to use the chain rule rigorously in this case. I therefore opened a new question: math.stackexchange.com/questions/3040296/…
            $endgroup$
            – james
            Dec 15 '18 at 9:04










          • $begingroup$
            In your example to normalize the function, where did "A" go ?
            $endgroup$
            – james
            Dec 15 '18 at 9:19














          1












          1








          1





          $begingroup$

          Using $x = L hat{x}$ we have $dx = L dhat{x}$ and $dx^2 = L^2 dhat{x}^2$. Therefore,
          $$
          frac{d}{dx^2} = frac{1}{L^2} frac{d}{dhat{x}^2}.
          $$
          This simple "substitution" is not mathematically rigorous, but you could use the chain rule twice to obtain the same thing; since the change of variable consists only in the multiplication by a constant, this "heuristic" works.



          The final expression is
          $$
          f(L hat{x}) = frac{1}{L^2} frac{d}{dhat{x}^2} w(Lhat{x}).
          $$



          If, for example, $w(x) = A exp(Bx)$, we have $f(x) = AB^2 exp(Bx)$. With the new variable,
          $$
          A (BL)^2 exp(BL hat{x}) = frac{d}{dhat{x}^2} A exp(BL hat{x}).
          $$
          If you define $hat{B}=BL$, $hat{f} = hat{B}^2 exp(hat{B} hat{x})$ and $hat{w}=exp(hat{B} hat{x})$, the equation is
          $$
          hat{f} = frac{d hat{w}}{dhat{x}^2} .
          $$
          Therefore, after you normalize $x$ with $hat{x}$, you should also normalize the functions. If $w$ has units of Kelvin, for example, $f$ has units of $K/m^2$. After the normalization, both $hat{w}$ and $hat{f}$ are nondimensional.



          Appendix: rigorous change of variable using chain rule



          Let $x=L hat{x}$. Therefore, $hat{x}=x/L$. From the chain rule,
          $$
          frac{dw}{dx} = frac{dhat{x}}{dx} frac{dw}{dhat{x}} = frac{1}{L} frac{dw}{dhat{x}}
          $$
          and
          $$
          frac{d^2w}{dx^2}=frac{d}{dx} left(frac{dw}{dx}right) = frac{dhat{x}}{dx} frac{d}{dhat{x}}left(frac{1}{L} frac{dw}{dhat{x}} right) = frac{1}{L^2} frac{d^2 w}{d hat{x}^2}.
          $$






          share|cite|improve this answer











          $endgroup$



          Using $x = L hat{x}$ we have $dx = L dhat{x}$ and $dx^2 = L^2 dhat{x}^2$. Therefore,
          $$
          frac{d}{dx^2} = frac{1}{L^2} frac{d}{dhat{x}^2}.
          $$
          This simple "substitution" is not mathematically rigorous, but you could use the chain rule twice to obtain the same thing; since the change of variable consists only in the multiplication by a constant, this "heuristic" works.



          The final expression is
          $$
          f(L hat{x}) = frac{1}{L^2} frac{d}{dhat{x}^2} w(Lhat{x}).
          $$



          If, for example, $w(x) = A exp(Bx)$, we have $f(x) = AB^2 exp(Bx)$. With the new variable,
          $$
          A (BL)^2 exp(BL hat{x}) = frac{d}{dhat{x}^2} A exp(BL hat{x}).
          $$
          If you define $hat{B}=BL$, $hat{f} = hat{B}^2 exp(hat{B} hat{x})$ and $hat{w}=exp(hat{B} hat{x})$, the equation is
          $$
          hat{f} = frac{d hat{w}}{dhat{x}^2} .
          $$
          Therefore, after you normalize $x$ with $hat{x}$, you should also normalize the functions. If $w$ has units of Kelvin, for example, $f$ has units of $K/m^2$. After the normalization, both $hat{w}$ and $hat{f}$ are nondimensional.



          Appendix: rigorous change of variable using chain rule



          Let $x=L hat{x}$. Therefore, $hat{x}=x/L$. From the chain rule,
          $$
          frac{dw}{dx} = frac{dhat{x}}{dx} frac{dw}{dhat{x}} = frac{1}{L} frac{dw}{dhat{x}}
          $$
          and
          $$
          frac{d^2w}{dx^2}=frac{d}{dx} left(frac{dw}{dx}right) = frac{dhat{x}}{dx} frac{d}{dhat{x}}left(frac{1}{L} frac{dw}{dhat{x}} right) = frac{1}{L^2} frac{d^2 w}{d hat{x}^2}.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 14:43

























          answered Dec 1 '18 at 14:27









          rafa11111rafa11111

          1,1251417




          1,1251417












          • $begingroup$
            Thank you very much for your help ! Your answer is great !
            $endgroup$
            – james
            Dec 1 '18 at 15:36










          • $begingroup$
            After thinking more about it, I am still a little confused of how to use the chain rule rigorously in this case. I therefore opened a new question: math.stackexchange.com/questions/3040296/…
            $endgroup$
            – james
            Dec 15 '18 at 9:04










          • $begingroup$
            In your example to normalize the function, where did "A" go ?
            $endgroup$
            – james
            Dec 15 '18 at 9:19


















          • $begingroup$
            Thank you very much for your help ! Your answer is great !
            $endgroup$
            – james
            Dec 1 '18 at 15:36










          • $begingroup$
            After thinking more about it, I am still a little confused of how to use the chain rule rigorously in this case. I therefore opened a new question: math.stackexchange.com/questions/3040296/…
            $endgroup$
            – james
            Dec 15 '18 at 9:04










          • $begingroup$
            In your example to normalize the function, where did "A" go ?
            $endgroup$
            – james
            Dec 15 '18 at 9:19
















          $begingroup$
          Thank you very much for your help ! Your answer is great !
          $endgroup$
          – james
          Dec 1 '18 at 15:36




          $begingroup$
          Thank you very much for your help ! Your answer is great !
          $endgroup$
          – james
          Dec 1 '18 at 15:36












          $begingroup$
          After thinking more about it, I am still a little confused of how to use the chain rule rigorously in this case. I therefore opened a new question: math.stackexchange.com/questions/3040296/…
          $endgroup$
          – james
          Dec 15 '18 at 9:04




          $begingroup$
          After thinking more about it, I am still a little confused of how to use the chain rule rigorously in this case. I therefore opened a new question: math.stackexchange.com/questions/3040296/…
          $endgroup$
          – james
          Dec 15 '18 at 9:04












          $begingroup$
          In your example to normalize the function, where did "A" go ?
          $endgroup$
          – james
          Dec 15 '18 at 9:19




          $begingroup$
          In your example to normalize the function, where did "A" go ?
          $endgroup$
          – james
          Dec 15 '18 at 9:19


















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